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12 Static Equilibrium and Elasticity

12.2 Examples of Static Equilibrium

Learning objectives.

By the end of this section, you will be able to:

  • Identify and analyze static equilibrium situations
  • Set up a free-body diagram for an extended object in static equilibrium
  • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations

All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Figure to Figure . We introduced a problem-solving strategy in Figure to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.

Problem-Solving Strategy: Static Equilibrium

  • Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
  • Set up a free-body diagram for the object. (a) Choose the xy -reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x – and y -directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign [latex](+)[/latex] means that the working direction is the actual direction. A minus sign [latex](-)[/latex] means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis.
  • Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition Figure for force components in the x -direction. (b) Use the free-body diagram to write a correct equilibrium condition Figure for force components in the y -direction. (c) Use the free-body diagram to write a correct equilibrium condition Figure for torques along the axis of rotation. Use Figure to evaluate torque magnitudes and senses.
  • Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved.
  • Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in Figure .

Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

The Torque Balance

Three masses are attached to a uniform meter stick, as shown in Figure . The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are [latex]{m}_{1}=50.0\,\text{g}[/latex] and [latex]{m}_{2}=75.0\,\text{g}.[/latex] Find the mass [latex]{m}_{3}[/latex] that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

Figure is a schematic drawing of a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Mass 3 is 30 cm to the right of S. Mass 2 is 40 cm to the left of S. Mass 1 is 30 cm to the left of Mass 2.

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:

[latex]{w}_{1}={m}_{1}g[/latex] is the weight of mass [latex]{m}_{1};[/latex] [latex]{w}_{2}={m}_{2}g[/latex] is the weight of mass [latex]{m}_{2};[/latex]

[latex]w=mg[/latex] is the weight of the entire meter stick; [latex]{w}_{3}={m}_{3}g[/latex] is the weight of unknown mass [latex]{m}_{3};[/latex]

[latex]{F}_{S}[/latex] is the normal reaction force at the support point S .

We choose a frame of reference where the direction of the y -axis is the direction of gravity, the direction of the x -axis is along the meter stick, and the axis of rotation (the z -axis) is perpendicular to the x -axis and passes through the support point S . In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot Figure . At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w , at the 50-cm mark.

Figure is a schematic drawing of a force distribution for a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Force Fs at the point S is pointing upward. Force w3, to the right of point S and separated by distance r3 is pointing downward. Forces w, w2, and w1 are to the left of point S and are pointing downward. They are separated by distance r, r2, and r1, respectively.

With Figure and Figure for reference, we begin by finding the lever arms of the five forces acting on the stick:

Now we can find the five torques with respect to the chosen pivot:

The second equilibrium condition (equation for the torques) for the meter stick is

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

Selecting the [latex]+y[/latex]-direction to be parallel to [latex]{\mathbf{\overset{\to }{F}}}_{S},[/latex] the first equilibrium condition for the stick is

Substituting the forces, the first equilibrium condition becomes

We solve these equations simultaneously for the unknown values [latex]{m}_{3}[/latex] and [latex]{F}_{S}.[/latex] In Figure , we cancel the g factor and rearrange the terms to obtain

To obtain [latex]{m}_{3}[/latex] we divide both sides by [latex]{r}_{3},[/latex] so we have

To find the normal reaction force, we rearrange the terms in Figure , converting grams to kilograms:

Significance

Notice that Figure is independent of the value of g . The torque balance may therefore be used to measure mass, since variations in g -values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.

Check Your Understanding

Repeat Figure using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.

316.7 g; 5.8 N

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Figure and Figure . We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.

Forces in the Forearm

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in Figure . His forearm is positioned at [latex]\beta =60^\circ[/latex] with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

Figure is a schematic drawing of a forearm rotated around the elbow. A 50 pound ball is held in the palm. The distance between the elbow and the ball is 13 inches. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 1.5 inches. Forearm forms a 60 degree angle with the upper arm.

We identify three forces acting on the forearm: the unknown force [latex]\mathbf{\overset{\to }{F}}[/latex] at the elbow; the unknown tension [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}}[/latex] in the muscle; and the weight [latex]\mathbf{\overset{\to }{w}}[/latex] with magnitude [latex]w=50\,\text{lb}.[/latex] We adopt the frame of reference with the x -axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x -axis makes an angle [latex]\beta =60^\circ[/latex] with the vertical. The y -axis is perpendicular to the x -axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle [latex]\beta[/latex] and represent each force by its x – and y -components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in Figure . At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x – and y -components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force T is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Forces F and T form angle beta with the x axis. Force W forms an angle beta with line connecting it with its projection to the y axis.

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y -components of the forces because the x -components of the forces are all parallel to their lever arms, so that for any of them we have [latex]\text{sin}\,\theta =0[/latex] in Figure . For the y -components we have [latex]\theta =\pm90^\circ[/latex] in Figure . Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of [latex]{T}_{y}[/latex] and of [latex]{w}_{y}.[/latex]

We see from the free-body diagram that the x -component of the net force satisfies the equation

and the y -component of the net force satisfies

Figure and Figure are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is

Figure is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are [latex]{r}_{T}=1.5\,\text{in}\text{.}[/latex] and [latex]{r}_{w}=13.0\,\text{in}\text{.}[/latex] At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Figure , they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into Figure , Figure , and Figure to obtain, respectively,

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T , because Figure for the x -component is equivalent to Figure for the y -component. In this way, we obtain the first equilibrium condition for forces

and the second equilibrium condition for torques

The magnitude of tension in the muscle is obtained by solving Figure :

The force at the elbow is obtained by solving Figure :

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction , and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.

Suppose we adopt a reference frame with the direction of the y -axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y -components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in Figure , indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles [latex]{\theta }_{T}[/latex] and [latex]{\theta }_{w}[/latex] that the forces [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}}[/latex] and [latex]\mathbf{\overset{\to }{w}}[/latex] (respectively) make with their lever arms. In the definition of torque given by Figure , the angle [latex]{\theta }_{T}[/latex] is the direction angle of the vector [latex]{\mathbf{\overset{\to }{T}}}_{\text{M}},[/latex] counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle [latex]{\theta }_{w}[/latex] is measured counterclockwise from the radial direction of the lever arm to the vector [latex]\mathbf{\overset{\to }{w}}.[/latex] Done this way, the non-zero torques are most easily computed by directly substituting into Figure as follows:

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force Tm is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Force Tm forms and angle theta tau that is equal to beta with the direction of the lever arm. Force W forms an angle theta w that is equal to the sum of beta and Pi with the direction of the lever arm.

The second equilibrium condition, [latex]{\tau }_{T}+{\tau }_{w}=0,[/latex] can be now written as

From the free-body diagram, the first equilibrium condition (for forces) is

Figure is identical to Figure and gives the result [latex]T=433.3\,\text{lb}.[/latex] Figure gives

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Repeat Figure assuming that the forearm is an object of uniform density that weighs 8.896 N.

[latex]T=\text{1963 N};\,\text{F}=1732\,\text{N}[/latex]

A Ladder Resting Against a Wall

A uniform ladder is [latex]L=5.0\,\text{m}[/latex] long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in Figure . The inclination angle between the ladder and the rough floor is [latex]\beta =53^\circ.[/latex] Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction [latex]{\mu }_{\text{s}}[/latex] at the interface of the ladder with the floor that prevents the ladder from slipping.

Figure is a schematic drawing of a 5.0-m-long ladder resting against a wall. Ladder forms a 53 degree angle with the floor.

We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force [latex]f={\mu }_{\text{s}}N[/latex] directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y -axis in the vertical direction (parallel to the wall) and the x -axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in Figure . With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.

Figure is a free-body diagram for a ladder that forms an angle beta with the floor and rests against a wall. Force N is applied at the point at the floor and is perpendicular to the floor. Force W is applied at the mid-point of the ladder. Force F is applied at the point resting at the wall and is perpendicular to the wall. Force W forms an angle theta w with the direction of the lever arm. Theta w is equal to the sum of Pi and half Pi with the beta subtracted. Force F forms an angle theta F with the direction of the lever arm. Theta F is equal to the Pi minus beta.

From the free-body diagram, the net force in the x -direction is

the net force in the y -direction is

and the net torque along the rotation axis at the pivot point is

where [latex]{\tau }_{w}[/latex] is the torque of the weight w and [latex]{\tau }_{F}[/latex] is the torque of the reaction F . From the free-body diagram, we identify that the lever arm of the reaction at the wall is [latex]{r}_{F}=L=5.0\,\text{m}[/latex] and the lever arm of the weight is [latex]{r}_{w}=L\,\text{/}\,2=2.5\,\text{m}.[/latex] With the help of the free-body diagram, we identify the angles to be used in Figure for torques: [latex]{\theta }_{F}=180^\circ-\beta[/latex] for the torque from the reaction force with the wall, and [latex]{\theta }_{w}=180^\circ+(90^\circ-\beta )[/latex] for the torque due to the weight. Now we are ready to use Figure to compute torques:

We substitute the torques into Figure and solve for [latex]F:[/latex]

We obtain the normal reaction force with the floor by solving Figure : [latex]N=w=400.0\,\text{N}.[/latex] The magnitude of friction is obtained by solving Figure : [latex]f=F=150.7\,\text{N}.[/latex] The coefficient of static friction is [latex]{\mu }_{\text{s}}=f\,\text{/}\,N=150.7\,\text{/}\,400.0=0.377.[/latex]

The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:

Its magnitude is

and its direction is

We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use Figure for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in Figure is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, Figure gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and Figure expresses the rectangular component of this vector product along the axis of rotation.

This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, Figure . No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is [latex]53^\circ,[/latex] our results hold. But the ladder will slip if the net torque becomes negative in Figure . This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

For the situation described in Figure , determine the values of the coefficient [latex]{\mu }_{\text{s}}[/latex] of static friction for which the ladder starts slipping, given that [latex]\beta[/latex] is the angle that the ladder makes with the floor.

[latex]{\mu }_{s} \lt 0.5\,\text{cot}\,\beta[/latex]

Forces on Door Hinges

A swinging door that weighs [latex]w=400.0\,\text{N}[/latex] is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges Figure . The door has a width of [latex]b=1.00\,\text{m},[/latex] and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance [latex]a=2.00\,\text{m}.[/latex] Find the forces on the hinges when the door rests half-open.

Figure is a schematic drawing of a swinging vertical door supported by two hinges attached at points A and B. The distance between points A and B is 2 meters. Door is one meter wide.

The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force [latex]\mathbf{\overset{\to }{A}}[/latex] from hinge [latex]A,[/latex] an unknown force [latex]\mathbf{\overset{\to }{B}}[/latex] from hinge [latex]B,[/latex] and the known weight [latex]\mathbf{\overset{\to }{w}}[/latex] attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y -axis along the direction of gravity and the x -axis in the plane of the slab, as shown in panel (a) of Figure , and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force [latex]\mathbf{\overset{\to }{A}}[/latex] [latex]({A}_{x}[/latex] and [latex]{A}_{y}),[/latex] and two components of force [latex]\mathbf{\overset{\to }{B}}[/latex] [latex]({B}_{x}[/latex] and [latex]{B}_{y}).[/latex] In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns [latex]({A}_{x},[/latex] [latex]{B}_{x},[/latex] [latex]{A}_{y},[/latex] and [latex]{B}_{y}),[/latex] we must set up four independent equations. One equation is the equilibrium condition for forces in the x -direction. The second equation is the equilibrium condition for forces in the y -direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, [latex]{A}_{y}={B}_{y}.[/latex] To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of Figure . Finally, we solve the equations for the unknown force components and find the forces.

Figure A is a geometrical representation for a swinging vertical door supported by two hinges attached at points A and B. Forces A and B are applied at the points A and B. Projections of these forces to the x and y axes are shown. Force w is applied at the point CM. Point CM is lower than point A by half-a and to the right of point A by half-b. Line from point A to CM forms an angle beta with the edge of the wall. Figure B is a free-body diagram for a swinging vertical door is supported by two hinges attached at points A and B. Force Ay forms an angle beta with the line connecting points P and CM. Force By forms an angle beta with the line connecting points B and CM. Force W forms an angle beta with the line that is the continuation of the line connecting points P and CM. Distance between points P and CM is d.

From the free-body diagram for the door we have the first equilibrium condition for forces:

We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P :

We use the free-body diagram to find all the terms in this equation:

In evaluating [latex]\text{sin}\,\beta ,[/latex] we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Figure and compute [latex]{B}_{x}:[/latex]

Therefore the magnitudes of the horizontal component forces are [latex]{A}_{x}={B}_{x}=100.0\,\text{N}.[/latex] The forces on the door are

The forces on the hinges are found from Newton’s third law as

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.

Solve the problem in Figure by taking the pivot position at the center of mass.

[latex]{\mathbf{\overset{\to }{F}}}_{\text{door on}\,A}=100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}\,\text{;}\,{\mathbf{\overset{\to }{F}}}_{\text{door on}\,B}=-100.0\,\text{N}\mathbf{\hat{i}}-200.0\,\text{N}\mathbf{\hat{j}}[/latex]

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

Figure is a schematic drawing of a woman standing 1.5 m away from one end and 4.5 m away from another end of a scaffold.

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut is 4.0 m long and is supported by a 5.0 m long cable tied to the wall at a point 3.0 m above the left end of the strut.

  • A variety of engineering problems can be solved by applying equilibrium conditions for rigid bodies.
  • In applications, identify all forces that act on a rigid body and note their lever arms in rotation about a chosen rotation axis. Construct a free-body diagram for the body. Net external forces and torques can be clearly identified from a correctly constructed free-body diagram. In this way, you can set up the first equilibrium condition for forces and the second equilibrium condition for torques.
  • In setting up equilibrium conditions, we are free to adopt any inertial frame of reference and any position of the pivot point. All choices lead to one answer. However, some choices can make the process of finding the solution unduly complicated. We reach the same answer no matter what choices we make. The only way to master this skill is to practice.

Conceptual Questions

Is it possible to rest a ladder against a rough wall when the floor is frictionless?

Show how a spring scale and a simple fulcrum can be used to weigh an object whose weight is larger than the maximum reading on the scale.

A painter climbs a ladder. Is the ladder more likely to slip when the painter is near the bottom or near the top?

A uniform plank rests on a level surface as shown below. The plank has a mass of 30 kg and is 6.0 m long. How much mass can be placed at its right end before it tips? ( Hint: When the board is about to tip over, it makes contact with the surface only along the edge that becomes a momentary axis of rotation.)

Figure schematic drawing of uniform plank rests on a level surface. Part of the plank that is 4.2 meters long is supported by the plank. Part of the plank that is 1.8 meters long is hanging over it.

The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board?

Figure is a schematic drawing of two boys on the seesaw. One boy sits on the edge of the seesaw three meters from the center. Another boys sits at the opposite edge of the seesaw, five meters from the center.

In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tree 15 m away, as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car?

Figure is a schematic drawing that shows a rope tied to the front bumper and the other end to a tree 15 m away. A force of 400 N is applied to the center of the rope and causes it to get displaced 0.30 m.

A uniform 40.0-kg scaffold of length 6.0 m is supported by two light cables, as shown below. An 80.0-kg painter stands 1.0 m from the left end of the scaffold, and his painting equipment is 1.5 m from the right end. If the tension in the left cable is twice that in the right cable, find the tensions in the cables and the mass of the equipment.

Figure is a schematic drawing of a man standing at the left side and the bucket placed at the right side of a scaffold.

When the structure shown below is supported at point P , it is in equilibrium. Find the magnitude of force F and the force applied at P . The weight of the structure is negligible.

Figure shows the distribution of forces applied to point P. Force of 2000 N, two meters to the left of the point P, moves it downwards. Force F, two meters to the left and two meters above of the point P, moves it to the right. Force of 1000 N, two meters to the right and three meters below of the point P, moves it to the left.

To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base.

784 N, 376 N

A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs 200.0 N. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut forms a 30 degree angle with the cable tied to the wall above the left end of the strut.

The forearm shown below is positioned at an angle [latex]\theta[/latex] with respect to the upper arm, and a 5.0-kg mass is held in the hand. The total mass of the forearm and hand is 3.0 kg, and their center of mass is 15.0 cm from the elbow. (a) What is the magnitude of the force that the biceps muscle exerts on the forearm for [latex]\theta =60^\circ\text{?}[/latex] (b) What is the magnitude of the force on the elbow joint for the same angle? (c) How do these forces depend on the angle [latex]\theta ?[/latex]

Figure is a schematic drawing of a forearm rotated around the elbow. A 5 kilogram ball is held in the palm. The distance between the elbow and the ball is 35 centimeters. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 4 centimeters. Forearm forms a theta angle with the upper arm.

The uniform boom shown below weighs 3000 N. It is supported by the horizontal guy wire and by the hinged support at point A . What are the forces on the boom due to the wire and due to the support at A ? Does the force at A act along the boom?

Figure is a schematic drawing of a 2000 N weight that is supported by the horizontal guy wire and by the hinged support at point A. Hinged support forms a 45 degree angle with the ground.

The uniform boom shown below weighs 700 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?

Figure is a schematic drawing of a 400 N weight that is by a cable and by a hinge at the wall. Hinge forms a 20 degree angle with the line perpendicular to the wall. Cable forms a 45 degree angle with the line perpendicular to the wall.

A 12.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 1000 kg. For the position shown, calculate tension T in the cable and the force at the axle A .

Figure is a schematic drawing of a crane lifting a 3000-kg load. Arm of a crane forms a 30 degree angle with the line parallel to the ground. Cable supporting load forms a 10 degree angle with the arm.

A uniform trapdoor shown below is 1.0 m by 1.5 m and weighs 300 N. It is supported by a single hinge (H), and by a light rope tied between the middle of the door and the floor. The door is held at the position shown, where its slab makes a [latex]30^\circ[/latex] angle with the horizontal floor and the rope makes a [latex]20^\circ[/latex] angle with the floor. Find the tension in the rope and the force at the hinge.

Figure is a schematic drawing of a trapdoor that is 1.0 m by 1.5 m. Door is supported by a single hinge labeled H, and by a light rope tied between the middle of the door and the floor. The door makes a 30 degree angle with the floor and the rope makes a 20 degree angle with the floor.

A 90-kg man walks on a sawhorse, as shown below. The sawhorse is 2.0 m long and 1.0 m high, and its mass is 25.0 kg. Calculate the normal reaction force on each leg at the contact point with the floor when the man is 0.5 m from the far end of the sawhorse. ( Hint: At each end, find the total reaction force first. This reaction force is the vector sum of two reaction forces, each acting along one leg. The normal reaction force at the contact point with the floor is the normal (with respect to the floor) component of this force.)

Figure is a schematic drawing of a man walks on a sawhorse. Each side of the sawhorse is supported by two connected legs. There are 60 degree angles between the legs.

12.2 Examples of Static Equilibrium Copyright © 2016 by OpenStax. All Rights Reserved.

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  • 9.4 Applications of Statics, Including Problem-Solving Strategies
  • Introduction to Science and the Realm of Physics, Physical Quantities, and Units
  • 1.1 Physics: An Introduction
  • 1.2 Physical Quantities and Units
  • 1.3 Accuracy, Precision, and Significant Figures
  • 1.4 Approximation
  • Section Summary
  • Conceptual Questions
  • Problems & Exercises
  • Introduction to One-Dimensional Kinematics
  • 2.1 Displacement
  • 2.2 Vectors, Scalars, and Coordinate Systems
  • 2.3 Time, Velocity, and Speed
  • 2.4 Acceleration
  • 2.5 Motion Equations for Constant Acceleration in One Dimension
  • 2.6 Problem-Solving Basics for One-Dimensional Kinematics
  • 2.7 Falling Objects
  • 2.8 Graphical Analysis of One-Dimensional Motion
  • Introduction to Two-Dimensional Kinematics
  • 3.1 Kinematics in Two Dimensions: An Introduction
  • 3.2 Vector Addition and Subtraction: Graphical Methods
  • 3.3 Vector Addition and Subtraction: Analytical Methods
  • 3.4 Projectile Motion
  • 3.5 Addition of Velocities
  • Introduction to Dynamics: Newton’s Laws of Motion
  • 4.1 Development of Force Concept
  • 4.2 Newton’s First Law of Motion: Inertia
  • 4.3 Newton’s Second Law of Motion: Concept of a System
  • 4.4 Newton’s Third Law of Motion: Symmetry in Forces
  • 4.5 Normal, Tension, and Other Examples of Forces
  • 4.6 Problem-Solving Strategies
  • 4.7 Further Applications of Newton’s Laws of Motion
  • 4.8 Extended Topic: The Four Basic Forces—An Introduction
  • Introduction: Further Applications of Newton’s Laws
  • 5.1 Friction
  • 5.2 Drag Forces
  • 5.3 Elasticity: Stress and Strain
  • Introduction to Uniform Circular Motion and Gravitation
  • 6.1 Rotation Angle and Angular Velocity
  • 6.2 Centripetal Acceleration
  • 6.3 Centripetal Force
  • 6.4 Fictitious Forces and Non-inertial Frames: The Coriolis Force
  • 6.5 Newton’s Universal Law of Gravitation
  • 6.6 Satellites and Kepler’s Laws: An Argument for Simplicity
  • Introduction to Work, Energy, and Energy Resources
  • 7.1 Work: The Scientific Definition
  • 7.2 Kinetic Energy and the Work-Energy Theorem
  • 7.3 Gravitational Potential Energy
  • 7.4 Conservative Forces and Potential Energy
  • 7.5 Nonconservative Forces
  • 7.6 Conservation of Energy
  • 7.8 Work, Energy, and Power in Humans
  • 7.9 World Energy Use
  • Introduction to Linear Momentum and Collisions
  • 8.1 Linear Momentum and Force
  • 8.2 Impulse
  • 8.3 Conservation of Momentum
  • 8.4 Elastic Collisions in One Dimension
  • 8.5 Inelastic Collisions in One Dimension
  • 8.6 Collisions of Point Masses in Two Dimensions
  • 8.7 Introduction to Rocket Propulsion
  • Introduction to Statics and Torque
  • 9.1 The First Condition for Equilibrium
  • 9.2 The Second Condition for Equilibrium
  • 9.3 Stability
  • 9.5 Simple Machines
  • 9.6 Forces and Torques in Muscles and Joints
  • Introduction to Rotational Motion and Angular Momentum
  • 10.1 Angular Acceleration
  • 10.2 Kinematics of Rotational Motion
  • 10.3 Dynamics of Rotational Motion: Rotational Inertia
  • 10.4 Rotational Kinetic Energy: Work and Energy Revisited
  • 10.5 Angular Momentum and Its Conservation
  • 10.6 Collisions of Extended Bodies in Two Dimensions
  • 10.7 Gyroscopic Effects: Vector Aspects of Angular Momentum
  • Introduction to Fluid Statics
  • 11.1 What Is a Fluid?
  • 11.2 Density
  • 11.3 Pressure
  • 11.4 Variation of Pressure with Depth in a Fluid
  • 11.5 Pascal’s Principle
  • 11.6 Gauge Pressure, Absolute Pressure, and Pressure Measurement
  • 11.7 Archimedes’ Principle
  • 11.8 Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action
  • 11.9 Pressures in the Body
  • Introduction to Fluid Dynamics and Its Biological and Medical Applications
  • 12.1 Flow Rate and Its Relation to Velocity
  • 12.2 Bernoulli’s Equation
  • 12.3 The Most General Applications of Bernoulli’s Equation
  • 12.4 Viscosity and Laminar Flow; Poiseuille’s Law
  • 12.5 The Onset of Turbulence
  • 12.6 Motion of an Object in a Viscous Fluid
  • 12.7 Molecular Transport Phenomena: Diffusion, Osmosis, and Related Processes
  • Introduction to Temperature, Kinetic Theory, and the Gas Laws
  • 13.1 Temperature
  • 13.2 Thermal Expansion of Solids and Liquids
  • 13.3 The Ideal Gas Law
  • 13.4 Kinetic Theory: Atomic and Molecular Explanation of Pressure and Temperature
  • 13.5 Phase Changes
  • 13.6 Humidity, Evaporation, and Boiling
  • Introduction to Heat and Heat Transfer Methods
  • 14.2 Temperature Change and Heat Capacity
  • 14.3 Phase Change and Latent Heat
  • 14.4 Heat Transfer Methods
  • 14.5 Conduction
  • 14.6 Convection
  • 14.7 Radiation
  • Introduction to Thermodynamics
  • 15.1 The First Law of Thermodynamics
  • 15.2 The First Law of Thermodynamics and Some Simple Processes
  • 15.3 Introduction to the Second Law of Thermodynamics: Heat Engines and Their Efficiency
  • 15.4 Carnot’s Perfect Heat Engine: The Second Law of Thermodynamics Restated
  • 15.5 Applications of Thermodynamics: Heat Pumps and Refrigerators
  • 15.6 Entropy and the Second Law of Thermodynamics: Disorder and the Unavailability of Energy
  • 15.7 Statistical Interpretation of Entropy and the Second Law of Thermodynamics: The Underlying Explanation
  • Introduction to Oscillatory Motion and Waves
  • 16.1 Hooke’s Law: Stress and Strain Revisited
  • 16.2 Period and Frequency in Oscillations
  • 16.3 Simple Harmonic Motion: A Special Periodic Motion
  • 16.4 The Simple Pendulum
  • 16.5 Energy and the Simple Harmonic Oscillator
  • 16.6 Uniform Circular Motion and Simple Harmonic Motion
  • 16.7 Damped Harmonic Motion
  • 16.8 Forced Oscillations and Resonance
  • 16.10 Superposition and Interference
  • 16.11 Energy in Waves: Intensity
  • Introduction to the Physics of Hearing
  • 17.2 Speed of Sound, Frequency, and Wavelength
  • 17.3 Sound Intensity and Sound Level
  • 17.4 Doppler Effect and Sonic Booms
  • 17.5 Sound Interference and Resonance: Standing Waves in Air Columns
  • 17.6 Hearing
  • 17.7 Ultrasound
  • Introduction to Electric Charge and Electric Field
  • 18.1 Static Electricity and Charge: Conservation of Charge
  • 18.2 Conductors and Insulators
  • 18.3 Coulomb’s Law
  • 18.4 Electric Field: Concept of a Field Revisited
  • 18.5 Electric Field Lines: Multiple Charges
  • 18.6 Electric Forces in Biology
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  • Introduction to Electric Potential and Electric Energy
  • 19.1 Electric Potential Energy: Potential Difference
  • 19.2 Electric Potential in a Uniform Electric Field
  • 19.3 Electrical Potential Due to a Point Charge
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  • 19.6 Capacitors in Series and Parallel
  • 19.7 Energy Stored in Capacitors
  • Introduction to Electric Current, Resistance, and Ohm's Law
  • 20.1 Current
  • 20.2 Ohm’s Law: Resistance and Simple Circuits
  • 20.3 Resistance and Resistivity
  • 20.4 Electric Power and Energy
  • 20.5 Alternating Current versus Direct Current
  • 20.6 Electric Hazards and the Human Body
  • 20.7 Nerve Conduction–Electrocardiograms
  • Introduction to Circuits and DC Instruments
  • 21.1 Resistors in Series and Parallel
  • 21.2 Electromotive Force: Terminal Voltage
  • 21.3 Kirchhoff’s Rules
  • 21.4 DC Voltmeters and Ammeters
  • 21.5 Null Measurements
  • 21.6 DC Circuits Containing Resistors and Capacitors
  • Introduction to Magnetism
  • 22.1 Magnets
  • 22.2 Ferromagnets and Electromagnets
  • 22.3 Magnetic Fields and Magnetic Field Lines
  • 22.4 Magnetic Field Strength: Force on a Moving Charge in a Magnetic Field
  • 22.5 Force on a Moving Charge in a Magnetic Field: Examples and Applications
  • 22.6 The Hall Effect
  • 22.7 Magnetic Force on a Current-Carrying Conductor
  • 22.8 Torque on a Current Loop: Motors and Meters
  • 22.9 Magnetic Fields Produced by Currents: Ampere’s Law
  • 22.10 Magnetic Force between Two Parallel Conductors
  • 22.11 More Applications of Magnetism
  • Introduction to Electromagnetic Induction, AC Circuits and Electrical Technologies
  • 23.1 Induced Emf and Magnetic Flux
  • 23.2 Faraday’s Law of Induction: Lenz’s Law
  • 23.3 Motional Emf
  • 23.4 Eddy Currents and Magnetic Damping
  • 23.5 Electric Generators
  • 23.6 Back Emf
  • 23.7 Transformers
  • 23.8 Electrical Safety: Systems and Devices
  • 23.9 Inductance
  • 23.10 RL Circuits
  • 23.11 Reactance, Inductive and Capacitive
  • 23.12 RLC Series AC Circuits
  • Introduction to Electromagnetic Waves
  • 24.1 Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
  • 24.2 Production of Electromagnetic Waves
  • 24.3 The Electromagnetic Spectrum
  • 24.4 Energy in Electromagnetic Waves
  • Introduction to Geometric Optics
  • 25.1 The Ray Aspect of Light
  • 25.2 The Law of Reflection
  • 25.3 The Law of Refraction
  • 25.4 Total Internal Reflection
  • 25.5 Dispersion: The Rainbow and Prisms
  • 25.6 Image Formation by Lenses
  • 25.7 Image Formation by Mirrors
  • Introduction to Vision and Optical Instruments
  • 26.1 Physics of the Eye
  • 26.2 Vision Correction
  • 26.3 Color and Color Vision
  • 26.4 Microscopes
  • 26.5 Telescopes
  • 26.6 Aberrations
  • Introduction to Wave Optics
  • 27.1 The Wave Aspect of Light: Interference
  • 27.2 Huygens's Principle: Diffraction
  • 27.3 Young’s Double Slit Experiment
  • 27.4 Multiple Slit Diffraction
  • 27.5 Single Slit Diffraction
  • 27.6 Limits of Resolution: The Rayleigh Criterion
  • 27.7 Thin Film Interference
  • 27.8 Polarization
  • 27.9 *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
  • Introduction to Special Relativity
  • 28.1 Einstein’s Postulates
  • 28.2 Simultaneity And Time Dilation
  • 28.3 Length Contraction
  • 28.4 Relativistic Addition of Velocities
  • 28.5 Relativistic Momentum
  • 28.6 Relativistic Energy
  • Introduction to Quantum Physics
  • 29.1 Quantization of Energy
  • 29.2 The Photoelectric Effect
  • 29.3 Photon Energies and the Electromagnetic Spectrum
  • 29.4 Photon Momentum
  • 29.5 The Particle-Wave Duality
  • 29.6 The Wave Nature of Matter
  • 29.7 Probability: The Heisenberg Uncertainty Principle
  • 29.8 The Particle-Wave Duality Reviewed
  • Introduction to Atomic Physics
  • 30.1 Discovery of the Atom
  • 30.2 Discovery of the Parts of the Atom: Electrons and Nuclei
  • 30.3 Bohr’s Theory of the Hydrogen Atom
  • 30.4 X Rays: Atomic Origins and Applications
  • 30.5 Applications of Atomic Excitations and De-Excitations
  • 30.6 The Wave Nature of Matter Causes Quantization
  • 30.7 Patterns in Spectra Reveal More Quantization
  • 30.8 Quantum Numbers and Rules
  • 30.9 The Pauli Exclusion Principle
  • Introduction to Radioactivity and Nuclear Physics
  • 31.1 Nuclear Radioactivity
  • 31.2 Radiation Detection and Detectors
  • 31.3 Substructure of the Nucleus
  • 31.4 Nuclear Decay and Conservation Laws
  • 31.5 Half-Life and Activity
  • 31.6 Binding Energy
  • 31.7 Tunneling
  • Introduction to Applications of Nuclear Physics
  • 32.1 Diagnostics and Medical Imaging
  • 32.2 Biological Effects of Ionizing Radiation
  • 32.3 Therapeutic Uses of Ionizing Radiation
  • 32.4 Food Irradiation
  • 32.5 Fusion
  • 32.6 Fission
  • 32.7 Nuclear Weapons
  • Introduction to Particle Physics
  • 33.1 The Yukawa Particle and the Heisenberg Uncertainty Principle Revisited
  • 33.2 The Four Basic Forces
  • 33.3 Accelerators Create Matter from Energy
  • 33.4 Particles, Patterns, and Conservation Laws
  • 33.5 Quarks: Is That All There Is?
  • 33.6 GUTs: The Unification of Forces
  • Introduction to Frontiers of Physics
  • 34.1 Cosmology and Particle Physics
  • 34.2 General Relativity and Quantum Gravity
  • 34.3 Superstrings
  • 34.4 Dark Matter and Closure
  • 34.5 Complexity and Chaos
  • 34.6 High-temperature Superconductors
  • 34.7 Some Questions We Know to Ask
  • A | Atomic Masses
  • B | Selected Radioactive Isotopes
  • C | Useful Information
  • D | Glossary of Key Symbols and Notation

Learning Objectives

By the end of this section, you will be able to:

  • Discuss the applications of Statics in real life.
  • State and discuss various problem-solving strategies in Statics.

Statics can be applied to a variety of situations, ranging from raising a drawbridge to bad posture and back strain. We begin with a discussion of problem-solving strategies specifically used for statics. Since statics is a special case of Newton’s laws, both the general problem-solving strategies and the special strategies for Newton’s laws, discussed in Problem-Solving Strategies , still apply.

Problem-Solving Strategy: Static Equilibrium Situations

  • The first step is to determine whether or not the system is in static equilibrium . This condition is always the case when the acceleration of the system is zero and accelerated rotation does not occur .
  • It is particularly important to draw a free body diagram for the system of interest . Carefully label all forces, and note their relative magnitudes, directions, and points of application whenever these are known.
  • Solve the problem by applying either or both of the conditions for equilibrium (represented by the equations net F = 0 net F = 0 and net τ = 0 net τ = 0 , depending on the list of known and unknown factors. If the second condition is involved, choose the pivot point to simplify the solution . Any pivot point can be chosen, but the most useful ones cause torques by unknown forces to be zero. (Torque is zero if the force is applied at the pivot (then r = 0 r = 0 ), or along a line through the pivot point (then θ = 0 θ = 0 )). Always choose a convenient coordinate system for projecting forces.
  • Check the solution to see if it is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step never diminishes, although in unfamiliar applications, it is usually more difficult to judge reasonableness. These judgments become progressively easier with experience.

Now let us apply this problem-solving strategy for the pole vaulter shown in the three figures below. The pole is uniform and has a mass of 5.00 kg. In Figure 9.18 , the pole’s cg lies halfway between the vaulter’s hands. It seems reasonable that the force exerted by each hand is equal to half the weight of the pole, or 24.5 N. This obviously satisfies the first condition for equilibrium (net F = 0) (net F = 0) . The second condition (net τ = 0) (net τ = 0) is also satisfied, as we can see by choosing the cg to be the pivot point. The weight exerts no torque about a pivot point located at the cg, since it is applied at that point and its lever arm is zero. The equal forces exerted by the hands are equidistant from the chosen pivot, and so they exert equal and opposite torques. Similar arguments hold for other systems where supporting forces are exerted symmetrically about the cg. For example, the four legs of a uniform table each support one-fourth of its weight.

In Figure 9.18 , a pole vaulter holding a pole with its cg halfway between his hands is shown. Each hand exerts a force equal to half the weight of the pole, F R = F L = w / 2 F R = F L = w / 2 . (b) The pole vaulter moves the pole to his left, and the forces that the hands exert are no longer equal. See Figure 9.18 . If the pole is held with its cg to the left of the person, then he must push down with his right hand and up with his left. The forces he exerts are larger here because they are in opposite directions and the cg is at a long distance from either hand.

Similar observations can be made using a meter stick held at different locations along its length.

If the pole vaulter holds the pole as shown in Figure 9.19 , the situation is not as simple. The total force he exerts is still equal to the weight of the pole, but it is not evenly divided between his hands. (If F L = F R F L = F R , then the torques about the cg would not be equal since the lever arms are different.) Logically, the right hand should support more weight, since it is closer to the cg. In fact, if the right hand is moved directly under the cg, it will support all the weight. This situation is exactly analogous to two people carrying a load; the one closer to the cg carries more of its weight. Finding the forces F L F L and F R F R is straightforward, as the next example shows.

If the pole vaulter holds the pole from near the end of the pole ( Figure 9.20 ), the direction of the force applied by the right hand of the vaulter reverses its direction.

Example 9.2

What force is needed to support a weight held near its cg.

For the situation shown in Figure 9.19 , calculate: (a) F R F R , the force exerted by the right hand, and (b) F L F L , the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand.

Figure 9.19 includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (net F = 0 (net F = 0 ), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (net τ = 0 ) (net τ = 0 ) if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand.

Solution for (a)

There are now only two nonzero torques, those from the gravitational force ( τ w τ w ) and from the push or pull of the right hand ( τ R τ R ). Stating the second condition in terms of clockwise and counterclockwise torques,

or the algebraic sum of the torques is zero.

Here this is

since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise torque. Using the definition of torque, τ = rF sin θ τ = rF sin θ , noting that θ = 90º θ = 90º , and substituting known values, we obtain

Solution for (b)

The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law:

From this we can conclude:

Solving for F L F L , we obtain

F L F L is seen to be exactly half of F R F R , as we might have guessed, since F L F L is applied twice as far from the cg as F R F R .

If the pole vaulter holds the pole as he might at the start of a run, shown in Figure 9.20 , the forces change again. Both are considerably greater, and one force reverses direction.

Take-Home Experiment

This is an experiment to perform while standing in a bus or a train. Stand facing sideways. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Now stand facing forward. How do you move your body to readjust the distribution of your mass as the bus accelerates and decelerates? Why is it easier and safer to stand facing sideways rather than forward? Note: For your safety (and those around you), make sure you are holding onto something while you carry out this activity!

PhET Explorations

Balancing act.

Play with objects on a teeter totter to learn about balance. Test what you've learned by trying the Balance Challenge game.

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Access for free at https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
  • Authors: Paul Peter Urone, Roger Hinrichs
  • Publisher/website: OpenStax
  • Book title: College Physics 2e
  • Publication date: Jul 13, 2022
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/college-physics-2e/pages/1-introduction-to-science-and-the-realm-of-physics-physical-quantities-and-units
  • Section URL: https://openstax.org/books/college-physics-2e/pages/9-4-applications-of-statics-including-problem-solving-strategies

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solving static equilibrium problems

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solving static equilibrium problems

If an object is at equilibrium, then the forces are balanced. Balanced is the key word that is used to describe equilibrium situations. Thus, the net force is zero and the acceleration is 0 m/s/s. Objects at equilibrium must have an acceleration of 0 m/s/s. This extends from Newton's first law of motion . But having an acceleration of 0 m/s/s does not mean the object is at rest. An object at equilibrium is either ...

  • at rest and staying at rest, or
  • in motion and continuing in motion with the same speed and direction.

This too extends from Newton's first law of motion .

Analyzing a Static Equilibrium Situation

If an object is at rest and is in a state of equilibrium, then we would say that the object is at "static equilibrium." "Static" means stationary or at rest . A common physics lab is to hang an object by two or more strings and to measure the forces that are exerted at angles upon the object to support its weight. The state of the object is analyzed in terms of the forces acting upon the object. The object is a point on a string upon which three forces were acting. See diagram at right. If the object is at equilibrium, then the net force acting upon the object should be 0 Newton. Thus, if all the forces are added together as vectors, then the resultant force (the vector sum) should be 0 Newton. (Recall that the net force is "the vector sum of all the forces" or the resultant of adding all the individual forces head-to-tail.) Thus, an accurately drawn vector addition diagram can be constructed to determine the resultant. Sample data for such a lab are shown below.

For most students, the resultant was 0 Newton (or at least very close to 0 N). This is what we expected - since the object was at equilibrium , the net force (vector sum of all the forces) should be 0 N.

Another way of determining the net force (vector sum of all the forces) involves using the trigonometric functions to resolve each force into its horizontal and vertical components. Once the components are known, they can be compared to see if the vertical forces are balanced and if the horizontal forces are balanced. The diagram below shows vectors A, B, and C and their respective components. For vectors A and B, the vertical components can be determined using the sine of the angle and the horizontal components can be analyzed using the cosine of the angle. The magnitude and direction of each component for the sample data are shown in the table below the diagram.

The data in the table above show that the forces nearly balance. An analysis of the horizontal components shows that the leftward component of A nearly balances the rightward component of B. An analysis of the vertical components show that the sum of the upward components of A + B nearly balance the downward component of C. The vector sum of all the forces is ( nearly ) equal to 0 Newton. But what about the 0.1 N difference between rightward and leftward forces and the 0.2 N difference between the upward and downward forces? Why do the components of force only nearly balance? The sample data used in this analysis are the result of measured data from an actual experimental setup. The difference between the actual results and the expected results is due to the error incurred when measuring force A and force B. We would have to conclude that this low margin of experimental error reflects an experiment with excellent results. We could say it's "close enough for government work."

Analyzing a Hanging Sign

The above analysis of the forces acting upon an object in equilibrium is commonly used to analyze situations involving objects at static equilibrium . The most common application involves the analysis of the forces acting upon a sign that is at rest. For example, consider the picture at the right that hangs on a wall. The picture is in a state of equilibrium, and thus all the forces acting upon the picture must be balanced. That is, all horizontal components must add to 0 Newton and all vertical components must add to 0 Newton. The leftward pull of cable A must balance the rightward pull of cable B and the sum of the upward pull of cable A and cable B must balance the weight of the sign.

Suppose the tension in both of the cables is measured to be 50 N and that the angle that each cable makes with the horizontal is known to be 30 degrees. What is the weight of the sign? This question can be answered by conducting a force analysis using trigonometric functions . The weight of the sign is equal to the sum of the upward components of the tension in the two cables. Thus, a trigonometric function can be used to determine this vertical component. A diagram and accompanying work is shown below.

Since each cable pulls upwards with a force of 25 N, the total upward pull of the sign is 50 N. Therefore, the force of gravity (also known as weight) is 50 N, down. The sign weighs 50 N.

In the above problem, the tension in the cable and the angle that the cable makes with the horizontal are used to determine the weight of the sign. The idea is that the tension, the angle, and the weight are related. If the any two of these three are known, then the third quantity can be determined using trigonometric functions.

Thinking Conceptually

There is an important principle that emanates from some of the trigonometric calculations performed above. The principle is that as the angle with the horizontal increases, the amount of tensional force required to hold the sign at equilibrium decreases. To illustrate this, consider a 10-Newton picture held by three different wire orientations as shown in the diagrams below. In each case, two wires are used to support the picture; each wire must support one-half of the sign's weight (5 N). The angle that the wires make with the horizontal is varied from 60 degrees to 15 degrees. Use this information and the diagram below to determine the tension in the wire for each orientation. When finished, click the button to view the answers.

At 60 degrees, the tension is 5.8 N. (5 N / sin 60 degrees).

At 45 degrees, the tension is 7.1 N. (5 N / sin 45 degrees).

At 15 degrees, the tension is 19.3 N (5 N / sin 15 degrees).

In conclusion, equilibrium is the state of an object in which all the forces acting upon it are balanced. In such cases, the net force is 0 Newton. Knowing the forces acting upon an object, trigonometric functions can be utilized to determine the horizontal and vertical components of each force. If at equilibrium, then all the vertical components must balance and all the horizontal components must balance.

   

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solving static equilibrium problems

Check Your Understanding

The following questions are meant to test your understanding of equilibrium situations. Click the button to view the answers to these questions.

1. The following picture is hanging on a wall. Use trigonometric functions to determine the weight of the picture.

The weight of the sign is 42.4 N .

The tension is 30.0 N and the angle is 45 degrees. Thus,

sine (45 degrees) = (F vert ) / (30.0 N).

The proper use of algebra leads to the equation:

F vert = (30.0 N) • sine (45 degrees) = 21.2 N

Each cable pulls upward with 21.2 N of force. Thus, the sign must weigh twice this - 42.4 N.

2. The sign below hangs outside the physics classroom, advertising the most important truth to be found inside. The sign is supported by a diagonal cable and a rigid horizontal bar. If the sign has a mass of 50 kg, then determine the tension in the diagonal cable that supports its weight.

The tension is 980 Newtons .

Since the mass is 50 kg, the weight is 490 N. Since there is only one "upward-pulling" cable, it must supply all the upward force. This cable pulls upwards with approximately 490 N of force. Thus,

sine (30 degrees) = (490 N ) / (F tens ).

Proper use of algebra leads to the equation

F tens = (490 N) / [ sine 30 (degrees) ] = 980 N.

3. The following sign can be found in Glenview. The sign has a mass of 50 kg. Determine the tension in the cables.

The tension is 346 Newtons .

Since the mass is 50.0 kg, the weight is 490 N. Each cable must pull upwards with 245 N of force.

Thus, sine (45 degrees) = (245 N ) / (F tens ).

F tens = (245 N) / [sine (45 degrees)] = 346 N.

4. After its most recent delivery, the infamous stork announces the good news. If the sign has a mass of 10 kg, then what is the tensional force in each cable? Use trigonometric functions and a sketch to assist in the solution.

The tension 56.6 Newtons .

Since the mass is 10.0 kg, the weight is 98.0 N. Each cable must pull upwards with 49.0 N of force. Thus,

sine 60 (degrees) = (49.0 N) / (F tens ).

F tens = (49.0 N) / [ sine 60 (degrees) ] = 56.6 N.

5. Suppose that a student pulls with two large forces (F 1 and F 2 ) in order to lift a 1-kg book by two cables. If the cables make a 1-degree angle with the horizontal, then what is the tension in the cable?

The tension 281 Newtons!

Since the mass is 1 kg, the weight is 9.8 N. Each cable must pull upwards with 4.9 N of force. Thus,

sine (1 degree) = (4.9 N) / (F tens ).

F tens = (4.9 N) / [ sine (1 degree) ] = 281 N.

practice problem 1

For all solutions, let T 1 be the cable on the left and T 2 be the cable on the right. The sign always has weight ( W ), which points down. The sign isn't going anywhere (it's not accelerating), therefore the three forces are in equilibrium. Describe this state using the language of physics — equations; in particular, component analysis equations. As always, make a nice drawing to show what's going on. Use a ruler and a protractor if you wish.

The two upward components should equal one another. Together they should equal the weight, which means each one is carrying half the load.

Weight points down (270°) and T 1 points to the left (180°). These are both good vectors — good in the sense that they are easy to deal with. T 1 is the troublemaker. Break it up into components and state the conditions for equilibrium in the vertical and horizontal directions. I like to put negative vectors on the left side of the equals sign and positive vectors on the right side. I also suggest working through the vertical equation first.

Weight is the only force with a convenient direction. Resolve the tensions into their components. State the equilibrium condition along both axes. I suggest working with the horizontal equation first.

That's the end of the physics. The rest of the work is math. Solve the horizontal equation for T 1 .

Substitute the result into the vertical equation.

Solve that for T 2 , substitute values, and compute T 2 .

Substitute back into the horizontal equation and compute T 1 .

My, that last one wasn't very much fun. Let's see if there isn't a simpler solution. We used component analysis since it's the default approach. Whenever you're given a pile of vectors and you need to combine them, components is the way to go — especially if you have no expectation of any special relationships among the vectors. We use this brainless, brute force approach to problems all the time. Understand the rules, describe them using commands a computer understands, put numbers in, get answers out.

Sometimes, however, there are clever solutions available. They don't work all the time, but when they do we should use them. In this practice problem, the vectors are rigged so that the alternate solution is easier than the default solution. The graphical method for addition of vectors requires placing them head to tail. The sum would be the resultant vector connecting the tail of the first vector to the head of the last. When forces are in equilibrium, their sum is zero and their will be no resultant. This means, it should be possible to arrange the three vectors in this practice problem into a closed figure — a triangle. Let's try it.

This is what we call a degenerate triangle. Sure it has three sides, but it covers no area. The two short sides lie on top of the long side. Symmetry tells us the two short sides should have equal length. Thus each tension equals half the weight. We already said this, so there is no advantage to this method over the previous one.

The horizontal tension and the vertical weight are the legs of a 45–45–90 triangle whose hypotenuse is the diagonal tension. These forces should form the ratio 1:1:√2.

The two tensions are the legs of a 30–60–90 triangle and weight is the hypotenuse. This means the sides should form the ratio 1:√3:2. Just be sure to get the tensions to correspond to the correct parts of this ratio.

And there are probably other ways to solve this problem.

practice problem 2

  • What is the net force?
  • What fourth force will put the point in equilibrium?

Compute the x and y components of each vector. Arrange the results in a table like this one.

Add the components…

Use pythagorean theorem to get the magnitude of the resultant force…

Use tangent to get the direction…

Add these numbers to the table…

The fourth force that would put this arrangement in equilibrium (the equilibrant) is equal and opposite the resultant. The components work this way too. To get the opposite direction angle, add on 180°.

practice problem 3

  • Draw a free body diagram of the crate.
  • the component of the crate's weight that is perpendicular to the ramp
  • the component of the crate's weight that is parallel to the ramp
  • the normal force between the crate and the ramp
  • the static friction force between the crate and the ramp
  • At what angle will the crate just begin to slip?

This is an example of a classic physics problem that students have been solving since the 17th century. It starts as an equilibrium problem, since the crate isn't going anywhere.

The component of the crate's weight perpendicular to the ramp is found using the cosine function. An object's weight is entirely pushing into a surface when the surface is level (a 0° angle of inclination). None of that weight is pushing into the surface when the surface is vertical, like a wall (a 90° angle of inclination). Cosine is a maximum when the angle is zero and zero when the angle is 90°. This is how the perpendicular component works.

The component of the crate's weight parallel to the ramp is found using the sine function. An object's weight has no sideways component on a level floor (a floor with no inclination). An object's weight is entirely parallel to a wall (a floor with a 90° inclination, in a sense). Sine is zero when the angle is zero and a maximum when the angle is 90°. This is how the parallel component works.

Normal forces are normal — that is, perpendicular to a tangent drawn to a curve or surface. This crate isn't currently going anywhere, so all the forces perpendicular to the incline must cancel. For a static crate on an incline, the force normal to the incline equals the perpendicular component of its weight.

Friction is a sideways, lateral, or tangential force — that is, parallel to a tangent drawn to a curve or surface. I'll say it again, this crate isn't going anywhere, so all the forces parallel to the incline should cancel. For a static crate on an incline, the static friction force equals the parallel component of the crate's weight.

The component of the crate's weight parallel to the incline pulls the crate down the incline while the frictional force tries to keep it in place. Since nothing is going anywhere, these two forces must balance each other.

As the angle of inclination increases, so to does the static friction, but it can't keep doing this forever. At some angle, the parallel component of the weight will equal the maximum static friction. Friction won't be strong enough and the crate will slip.

Cancel the weight.

μ s  cos θ = sin θ

Do some trig.

tan θ = μ s

Enter numbers.

tan θ =  0.28

Compute. The angle at which the crate just begins to slip is…

θ =  16°

This number is known as the critical angle (because it marks a critical value separating two types of behavior — sticking vs. sliding), angle of friction (because you gotta call it something), angle of repose (because granular materials will settle, or repose, in conical piles with this angle), or critical angle of repose (because adding grains to a pile with this angle will make it slump ).

practice problem 4

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PHYS 8 3 1a

An 80.0 kg painter climbs 85% of the way up a uniform 25.0 kg ladder before it starts to slip backwards along the ground. What is the coefficient of static friction between the ladder and ground? Assume the wall is frictionless.

PHYS 8 3 2

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Topic Notes

In this lesson, we will learn:

  • Solving statics problems using both translational and rotational equilibrium
  • An object or group of objects that are not moving are in static equilibrium .
  • In static equilibrium, the conditions for both translational and rotational equilibrium must be met.

Σ F = 0 N \Sigma F = 0 N Σ F = 0 N

or equivalently:

Σ F x = 0 N \Sigma F_{x} = 0 N Σ F x ​ = 0 N and Σ F y = 0 N \Sigma F_{y} = 0 N Σ F y ​ = 0 N

Σ F : \Sigma F: Σ F : sum of all forces, in newtons (N)

Σ F x : \Sigma F_{x}: Σ F x ​ : sum of all force components in x direction, in newtons (N)

Σ F y : \Sigma F_{y}: Σ F y ​ : sum of all force components in y direction, in newtons (N)

τ = F ⊥ d \tau = F_{\perp}d τ = F ⊥ ​ d

τ \tau τ : torque, in newton meters (N·m)

F ⊥ : F_{\perp}: F ⊥ ​ : component of force perpendicular to d d d , in newtons (N)

d : d: d : distance from point of rotation, in meters (m)

Σ τ = 0 \Sigma \tau = 0 Σ τ = 0 N·m

or simpler equation:

total C W CW C W τ \tau τ = total C C W CCW CC W τ \tau τ

Σ τ : \Sigma \tau : Σ τ : sum of all torques, in newton meters (N·m)

total C W CW C W τ \tau τ : magnitude of all torques in the clockwise direction, in newton meters (N·m)

total C C W CCW CC W τ \tau τ : magnitude of all torques in the counterclockwise direction, in newton meters (N·m)

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Library Home

Engineering Statics: Open and Interactive

(5 reviews)

solving static equilibrium problems

Daniel W. Baker, Colorado State University

William Haynes, Massachusetts Maritime Academy

Copyright Year: 2020

Publisher: Daniel Baker and William Haynes

Language: English

Formats Available

Conditions of use.

Attribution-NonCommercial-ShareAlike

Learn more about reviews.

Reviewed by Mohammad Valipour, Assistant Professor, Metropolitan State University of Denver on 8/28/23

I found this book very interesting and thorough. Fortunately, the book covers most of the syllabus for Statics courses that I teach. The order of the chapters is also logical and in a standard form including Introduction to Statics, Forces and... read more

Comprehensiveness rating: 4 see less

I found this book very interesting and thorough. Fortunately, the book covers most of the syllabus for Statics courses that I teach. The order of the chapters is also logical and in a standard form including Introduction to Statics, Forces and Other Vectors, Equilibrium of Particles, Moments and Static Equivalence, Rigid Body Equilibrium, Equilibrium of Structures, Centroids and Centers of Gravity, Internal Forces, Friction, and Moments of Inertia. There are interactive problems at the end of each chapter of the book which can be considered as a quick review of the book. Also, I found the trigonometric tables very useful for students who have forgotten basic geometry and trigonometry. One of the interesting parts of this book is using QR codes for which you can have access to dynamic pictures of the book. The only negative point that I can mention is the lack of enough sample problems for each content. Students who take Engineering Statics need to practice many problems/examples to get ready for their exams. Only one sample problem for most of the topics is not enough particularly for those who need time to understand the concepts through solving sample problems. I think students can use this book as a study guide very well but they also need an auxiliary textbook/pamphlet to fill the gap of not solving multiple problems for each topic. Adding an index/glossary is also helpful.

Content Accuracy rating: 5

In the current format, the book looks accurate. I could not find any problem with a wrong answer neither in the solution nor in the final answers.

Relevance/Longevity rating: 5

I found the relevancy aspect of this book pretty high. Since the topics rely on the fundamentals of physics and follow Euclidean geometry, we cannot expect any change in the near future. However, the authors try to stay up to date by using matrix solutions which is highly appreciated. In addition, the users can take advantage of the graphical solutions if they use technology such as GeoGebra or a CAD program to make the diagram, then their answers will be precise.

Clarity rating: 4

Using subsections facilitates the readability aspect of the book. Supporting graphs are also useful. Thinking Deeper sections are also interesting for those students who want to know more about the concepts behind the problems. I like the friendly tone that has been used by the authors and the vocabulary and terminology are understandable for all readers even international students. Using more colorful figures may be attractive for students who are less interested in engineering problems.

Consistency rating: 4

This is my favorite part of this book. It is very consistent and readable like a story from chapter 1 to chapter 10. The overall structure of the book holds together quite well. I like that figures in each chapter have their own numbering system with respect to the chapter number. However, figure captions look too short, and adding some more information may be useful.

Modularity rating: 5

There are 10 chapters in this book. The author has done a nice job by putting sections and subsections in the right place. Different parts of the text are useable to be presented to students in different topics which might not be even the same in the book. The book includes occasional references to other subsections for further information, but such self-references do not look disruptive.

Organization/Structure/Flow rating: 5

I think the material is provided and put together realistically. The organization of this book is logical. The sections and even subsections flow easily together with the previous and following sections/subsections. I like the idea of using key questions at the beginning of each chapter. This book is organized and follows a clear structure. Each chapter starts with a chapter description and a list of sections. I found the online version of the book flow better due to the interactive design. The PDF version is also well formatted. Also, the topics are presented in a short concise fashion.

Interface rating: 5

An advantage of this book is having a user-friendly interface which makes it much easier for students to follow up with the materials discussed in the book. Unlike many other OER books, the PDF version does look like a regular book. In my opinion, the navigation of the book is easy for both the PDF version and the online version. All figures and graphs are clear and readable. The sections and subsections are loaded quickly, and the figures and diagrams can be loaded without any issues.

Grammatical Errors rating: 5

I could not find any grammatical errors in the book.

Cultural Relevance rating: 4

I did not find the book offensive/insensitive in any way. It would be better to cover more examples from different parts of the world to be fairer for international readers/students.

Reviewed by James Book, Assistant Instructional Professor, Pittsburg State University on 12/15/22

This text covers most all topics under the statics umbrella with the notable exception of virtual work. The end of chapter interactive problem sets are excellent and make for a good review of the topics covered. The option to "randomize" each... read more

This text covers most all topics under the statics umbrella with the notable exception of virtual work. The end of chapter interactive problem sets are excellent and make for a good review of the topics covered. The option to "randomize" each problem is a great help in exam preparation. The back matter covers useful trigonometric functions and provides handy steel sections tables.

I found no obvious errors or biases in the text. The back matter equations and reference tables are accurate.

The basic principles of statics covered by this text will remain relevant for a very long time. The end of chapter problem "randomization" feature will provide fresh challenges to the student and teacher. The prose used in writing the text is modern and up to date. Since it is provided on-line, the text should be relatively easy to keep current.

Clarity rating: 5

The text is written in an easy-to-read style with a fairly basic vocabulary. The technical terms used are well explained. The illustrations are adequate but could be expanded to include some simple video content. A feature to vary the font type and size would make the text more accessible. The use of color in many of the illustrations helps with clarity. The method of concealing the answer and solution guides to the example problems is clever.

Consistency rating: 5

The text follows a consistent, repeated pattern throughout. The presentation of each topic adequately builds on ideas and concepts from previous chapters. The overall structure of the text holds together well.

The text is divided into very manageable segments that can stand on their own. It does not appear to be overly self-referential. There are no enormous blocks of text or sections that run-on unnecessarily.

The text is well organized with deliberate and logical progression through statics concepts. The on-line version of the text seems to flow better due to the interactive design, but both on-line and pdf versions are well organized.

Both the on-line and pdf versions of the text provide easy and clear navigation. A check of several of the text hyperlinks showed accurate navigation. Charts, graphs, and other images were clear and readable.

No obvious grammatical errors were encountered.

Cultural Relevance rating: 5

The topic of statics does not lend itself to much discussion of cultural topics. The examples used in the text were generic and did not seem to be culturally biased, offensive, or insensitive.

Reviewed by Michael Pastor, Assistant Professor, Tidewater Community College on 11/27/22

A fully hyperlinked and intuitive table of contents is available for this text. I could find no index, however, the PDF version is searchable, and a search bar exists in the browser edition. For the most part, this text covers all topics normally... read more

A fully hyperlinked and intuitive table of contents is available for this text. I could find no index, however, the PDF version is searchable, and a search bar exists in the browser edition. For the most part, this text covers all topics normally associated with a typical engineering undergraduate class in mechanics dealing with the state of bodies at rest. The text begins with Newton's Laws, Forces, and Vectors, then moves on to the analysis of particles and rigid bodies. The text also includes chapters on centroids and moments of inertia, as well as chapters on internal loading and friction. A section on virtual work and energy methods is however, missing.

I have not noticed in inaccuracies and have used this text as an optional online resource in my sophomore level statics class for a number of semesters now. Students have never reported any issues.

The relevance for this text is quite high. The material presented here has not changed in at least the last 4 decades. The online mode of delivery, however, is quite refreshing and a relative new achievement. This text is licensed under CC Attribution-Non Commercial-Share Alike.

Chapters are divided into sections and the information is brief and to the point. Occasionally supporting graphics are presented and these can be easily magnified. Very little time is spent on derivation or formulation of relationships presented. However, occasionally extra "thinking deeper" information is presented at a mouse click. Here optional background information is presented on certain subjects. There are some example problems associated with topics and these are somewhat interactive... usually involving showing an answer then showing a more detailed solution on mouse click . Occasionally, there is an interactive diagram demonstrating some concept visually. However, I did not always find these intuitive, and in some cases did not understand how to effectively manipulate them.

I did not notice any inconsistencies in terminology or framework. The work is authored in PreTeXt and powered by MathJax. I have always found it quite easy to navigate chapters and sections consistently in this text. However, I have never tried using it on a cell phone or pad. In the appendix, there is a notation chapter outlining many if not all symbols used in the text.

Modularity rating: 4

The text is divided into chapters (common to most texts of this type). Each chapter is then broken down into sections. An appendix with math formulas and steel section properties is also included. This helps comply with ABET standards. I, however, could not locate simple shape properties in any tables or diagrams either in the chapters or appendices.

Organization/Structure/Flow rating: 4

The topics are present is a short concise fashion. This in my opinion is appropriate for an online resource. However, I would like to see the availability of more details concerning the derivation and/or development of some concepts and equations. Perhaps this could be added in optional user interactive sections.

Navigation of chapters, sections, and pages is quite easy and intuitive for this text. Some of the interactive diagrams were confusing and not well explained. A search bar is available to help locate specific ideas. However, this material is so consistently organized that navigation with the interactive Contents menu is all that I have ever needed.

I saw no issues here.

I noticed no insensitive or offensive areas in this text.

I have used this text for a number of semesters as a secondary resource for students in my engineering static classes. I think it would also work well as an instructor resource. The license allows for it to be upgraded and specialized to a users needs in a non-commercial and open way.

Reviewed by Anahita Khodadadi, Assistant Professor, Portland State University on 6/22/22

The textbook covers the fundamental concepts of statics including, force and vector analysis, equilibrium, internal reactions, and geometrical properties. The textbook also includes required steel section tables and a review of trigonometry... read more

The textbook covers the fundamental concepts of statics including, force and vector analysis, equilibrium, internal reactions, and geometrical properties. The textbook also includes required steel section tables and a review of trigonometry principles. I would suggest this textbook in combination with a textbook on basics of mechanics and material properties to students who seek learning about basic concepts of engineering design.

To the best of my understanding the content of the textbook and the interactive exercises look accurate, unbiased, and thorough.

The textbook contains basic principles of statics which are not expected to be changed unless a groundbreaking theory in physics emerges in the future! The text itself is arranged in a way that can easily be edited and extended. I appreciate the efforts that the author has made to create and embed the interactive diagrams and exercises within the textbook instead of inserting the link of available items across different references. This will maintain the configuration of the textbook in the long term.

The text is written in a friendly tone and even students’ presumptions and concerns are early discussed in first chapters. Technical terms are well explained for those who may not have any background in engineering. In the future, the author may include relevant videos to the textbook as well to enhance the clarity of the materials and better engage audio-visual learners.

The text and even visuals are consistent in terms of terminology, format, and graphics. All figures are numbered and mentioned within the text.

The textbook is appropriately organized in 10 chapters. Each chapter is explained in multiple subsections that allow readers focusing on small chunks of learning materials. The text includes occasional references to other subsections for further information but such self-references do not look disruptive.

Both online and PDF version of the book are presented in a fine, clear and logical fashion. The online version allows easy navigation between different sections and subsections. The PDF version is also clearly formatted. It is helpful that each chapter begins with a series of key questions and ends with a number of exercises.

I reviewed the online version both on a computer and smartphone. The interface looks fine on a computer but on a smartphone some of the interactive diagrams cannot be displayed. However, I think students may rarely use a smartphone as a primary means of accessing the textbook.

The textbook looks well proofed.

The textbook is focused on math and physics and doesn’t discuss culturally sensitive topics. Examples intrinsically do not have the capacity to demonstrate diversity and inclusion matters.

Reviewed by Peter Kazarinoff, Professor, Portland Community College on 12/16/21

The topics covered in a typical college Engineering Statics course are present. The chapters follow a common Statics textbook pattern of concepts, starting with forces and particles and ending with friction and moments of inertia. Chapter 6... read more

Comprehensiveness rating: 5 see less

The topics covered in a typical college Engineering Statics course are present. The chapters follow a common Statics textbook pattern of concepts, starting with forces and particles and ending with friction and moments of inertia. Chapter 6 includes the method of cuts and the method of joints. The only thing that many commercial Statics textbooks have compared to this book is an extensive number of problems at the end of each chapter (the fiction chapter, in particular, had few practice problems) and more reference material at the end of the book such as the centroid of common shapes. What this book has that those commercial books lack are interactive problems.

To the best of my knowledge, the content in the book is accurate. The interactive problems I attempted showed the same auto-generated answer as I recorded using pencil and paper. The equations seem accurate throughout. The reference material at the end of the book which contains things like trig identities and properties of steel sections seems accurate.

The fundamentals of Engineering Statics, like introductory Physics and Chemistry, have not changed in a decade. So the content in the book is relevant to a current Statics course and will be relevant to future Statics classes. The only reason the book could become dated is that the interactive animations and interactive problems are no longer supported by new web browsers or new web browsing tools that I can’t even imagine will be in place in 10 years. The book has a pdf version that can be printed.

The clarity of the writing is high, the font and spacing are easy to read. The book is written in a formal academic style which is clear but can seem terse. The diagrams in the book are easy to read and use a common style to show forces, angles, and geometry.

The book is consistent from chapter to chapter and the formatting is consistent from chapter to chapter. The book has a clear numbering system for chapters, sections, and subsections. Each of the diagrams and pictures in the book follows the same captioning format. Equations in the book are formatted consistently and labeled in the same way. Each section within a chapter in the book contains a set of “key questions” that section addresses.

The book is broken up into chapters and each chapter is broken down into sections. A typical quarter or semester-long Statics course would cover almost all of the book. It would be possible to only cover a few chapters. These chapters would need to start at the beginning of the book. It wouldn’t make sense to try and pull out just the middle or end chapters as the material in the book builds up chapter to chapter. One way the book could be used is to just assign the interactive problems for practice.

This book is organized and follows a clear structure. Each chapter starts with a chapter description and a list of sections. Each section starts with “Key Questions” and then proceeds with the section content. There are interactive problems at the end of each chapter.

Interface rating: 4

The online book interface is easy to navigate. Each chapter and section is clickable and it is easy to determine which part of the book you are reading. The sections load quickly and the images, diagrams, and interactive problems load without issue. In particular, the interactive problems are pretty slick. The only reason I don’t rate the interface as a 5 is that there is no search function. I don’t know how hard it would be to add a search bar to the online version of the book, but I do think a search function would be helpful. On my device, the book only took up the left half of my screen. This may be related to the browser/device I use, but in my reading, it seems like half of the screen real estate is wasted and a lot of scrolling is needed.

No grammatical or structural errors were found. The book seems to be free of typos and seems well-proofed. There also don’t seem to be any formatting inconsistencies chapter to chapter or section to section.

Cultural Relevance rating: 3

From what I read, I didn't notice any insensitive or offensive passages in the book. However, the lens of diversity and cultural relevance is not addressed in this book. Some of the pictures in the book depicting statics topics cover common “male-dominated” examples such as motorcycles, and football training sleds.

This is a well-written, high-quality, and organized book. It is a great resource for both instructors and students in undergraduate courses in Engineering Statics. For our needs, at a community college with a 2-year program in Mechanical Engineering and Civil Engineering, this book is a good alternative to commercial offerings from Pearson or McGraw-Hill. It’s a high-quality and interesting book with fantastic interactive problems. The only knock against it is that there could be more worked examples and problems at the end of each chapter for student practice.

Table of Contents

  • 1 Introduction to Statics
  • 2 Forces and Other Vectors
  • 3 Equilibrium of Particles
  • 4 Moments and Static Equivalence
  • 5 Rigid Body Equilibrium
  • 6 Equilibrium of Structures
  • 7 Centroids and Centers of Gravity
  • 8 Internal Loadings
  • 10 Moments of Inertia

Ancillary Material

  • Daniel Baker and William Haynes

About the Book

Engineering Statics  is a free, open-source textbook appropriate for anyone who wishes to learn more about vectors, forces, moments, static equilibrium, and the properties of shapes. Specifically, it has been written to be the textbook for Engineering Mechanics: Statics, the first course in the Engineering Mechanics series offered in most university-level engineering programs.

This book’s content should prepare you for subsequent classes covering Engineering Mechanics: Dynamics and Mechanics of Materials. At its core,  Engineering Statics  provides the tools to solve static equilibrium problems for rigid bodies. The additional topics of resolving internal loads in rigid bodies and computing area moments of inertia are also included as stepping stones for later courses. We have endeavored to write in an approachable style and provide many questions, examples, and interactives for you to engage with and learn from.

About the Contributors

Daniel W. Baker , Colorado State University

William Haynes , Massachusetts Maritime Academy

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12 Static Equilibrium and Elasticity

12.2 examples of static equilibrium, learning objectives.

By the end of this section, you will be able to:

  • Identify and analyze static equilibrium situations
  • Set up a free-body diagram for an extended object in static equilibrium
  • Set up and solve static equilibrium conditions for objects in equilibrium in various physical situations

All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of (Figure) to (Figure) . We introduced a problem-solving strategy in (Figure) to illustrate the physical meaning of the equilibrium conditions. Now we generalize this strategy in a list of steps to follow when solving static equilibrium problems for extended rigid bodies. We proceed in five practical steps.

Problem-Solving Strategy: Static Equilibrium

  • Identify the object to be analyzed. For some systems in equilibrium, it may be necessary to consider more than one object. Identify all forces acting on the object. Identify the questions you need to answer. Identify the information given in the problem. In realistic problems, some key information may be implicit in the situation rather than provided explicitly.
  • Set up a free-body diagram for the object. (a) Choose the xy -reference frame for the problem. Draw a free-body diagram for the object, including only the forces that act on it. When suitable, represent the forces in terms of their components in the chosen reference frame. As you do this for each force, cross out the original force so that you do not erroneously include the same force twice in equations. Label all forces—you will need this for correct computations of net forces in the x – and y -directions. For an unknown force, the direction must be assigned arbitrarily; think of it as a ‘working direction’ or ‘suspected direction.’ The correct direction is determined by the sign that you obtain in the final solution. A plus sign [latex] (+) [/latex] means that the working direction is the actual direction. A minus sign [latex] (-) [/latex] means that the actual direction is opposite to the assumed working direction. (b) Choose the location of the rotation axis; in other words, choose the pivot point with respect to which you will compute torques of acting forces. On the free-body diagram, indicate the location of the pivot and the lever arms of acting forces—you will need this for correct computations of torques. In the selection of the pivot, keep in mind that the pivot can be placed anywhere you wish, but the guiding principle is that the best choice will simplify as much as possible the calculation of the net torque along the rotation axis.
  • Set up the equations of equilibrium for the object. (a) Use the free-body diagram to write a correct equilibrium condition (Figure) for force components in the x -direction. (b) Use the free-body diagram to write a correct equilibrium condition (Figure) for force components in the y -direction. (c) Use the free-body diagram to write a correct equilibrium condition (Figure) for torques along the axis of rotation. Use (Figure) to evaluate torque magnitudes and senses.
  • Simplify and solve the system of equations for equilibrium to obtain unknown quantities. At this point, your work involves algebra only. Keep in mind that the number of equations must be the same as the number of unknowns. If the number of unknowns is larger than the number of equations, the problem cannot be solved.
  • Evaluate the expressions for the unknown quantities that you obtained in your solution. Your final answers should have correct numerical values and correct physical units. If they do not, then use the previous steps to track back a mistake to its origin and correct it. Also, you may independently check for your numerical answers by shifting the pivot to a different location and solving the problem again, which is what we did in (Figure) .

Note that setting up a free-body diagram for a rigid-body equilibrium problem is the most important component in the solution process. Without the correct setup and a correct diagram, you will not be able to write down correct conditions for equilibrium. Also note that a free-body diagram for an extended rigid body that may undergo rotational motion is different from a free-body diagram for a body that experiences only translational motion (as you saw in the chapters on Newton’s laws of motion). In translational dynamics, a body is represented as its CM, where all forces on the body are attached and no torques appear. This does not hold true in rotational dynamics, where an extended rigid body cannot be represented by one point alone. The reason for this is that in analyzing rotation, we must identify torques acting on the body, and torque depends both on the acting force and on its lever arm. Here, the free-body diagram for an extended rigid body helps us identify external torques.

The Torque Balance

Three masses are attached to a uniform meter stick, as shown in (Figure) . The mass of the meter stick is 150.0 g and the masses to the left of the fulcrum are [latex] {m}_{1}=50.0\,\text{g} [/latex] and [latex] {m}_{2}=75.0\,\text{g}. [/latex] Find the mass [latex] {m}_{3} [/latex] that balances the system when it is attached at the right end of the stick, and the normal reaction force at the fulcrum when the system is balanced.

Figure is a schematic drawing of a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Mass 3 is 30 cm to the right of S. Mass 2 is 40 cm to the left of S. Mass 1 is 30 cm to the left of Mass 2.

Figure 12.9 In a torque balance, a horizontal beam is supported at a fulcrum (indicated by S) and masses are attached to both sides of the fulcrum. The system is in static equilibrium when the beam does not rotate. It is balanced when the beam remains level.

For the arrangement shown in the figure, we identify the following five forces acting on the meter stick:

[latex] {w}_{1}={m}_{1}g [/latex] is the weight of mass [latex] {m}_{1}; [/latex] [latex] {w}_{2}={m}_{2}g [/latex] is the weight of mass [latex] {m}_{2}; [/latex]

[latex] w=mg [/latex] is the weight of the entire meter stick; [latex] {w}_{3}={m}_{3}g [/latex] is the weight of unknown mass [latex] {m}_{3}; [/latex]

[latex] {F}_{S} [/latex] is the normal reaction force at the support point S .

We choose a frame of reference where the direction of the y -axis is the direction of gravity, the direction of the x -axis is along the meter stick, and the axis of rotation (the z -axis) is perpendicular to the x -axis and passes through the support point S . In other words, we choose the pivot at the point where the meter stick touches the support. This is a natural choice for the pivot because this point does not move as the stick rotates. Now we are ready to set up the free-body diagram for the meter stick. We indicate the pivot and attach five vectors representing the five forces along the line representing the meter stick, locating the forces with respect to the pivot (Figure) . At this stage, we can identify the lever arms of the five forces given the information provided in the problem. For the three hanging masses, the problem is explicit about their locations along the stick, but the information about the location of the weight w is given implicitly. The key word here is “uniform.” We know from our previous studies that the CM of a uniform stick is located at its midpoint, so this is where we attach the weight w , at the 50-cm mark.

Figure is a schematic drawing of a force distribution for a torque balance, a horizontal beam supported at a fulcrum (indicated by S) and three masses are attached to both sides of the fulcrum. Force Fs at the point S is pointing upward. Force w3, to the right of point S and separated by distance r3 is pointing downward. Forces w, w2, and w1 are to the left of point S and are pointing downward. They are separated by distance r, r2, and r1, respectively.

Figure 12.10 Free-body diagram for the meter stick. The pivot is chosen at the support point S.

With (Figure) and (Figure) for reference, we begin by finding the lever arms of the five forces acting on the stick:

Now we can find the five torques with respect to the chosen pivot:

The second equilibrium condition (equation for the torques) for the meter stick is

When substituting torque values into this equation, we can omit the torques giving zero contributions. In this way the second equilibrium condition is

Selecting the [latex] +y [/latex]-direction to be parallel to [latex] {\overset{\to }{F}}_{S}, [/latex] the first equilibrium condition for the stick is

Substituting the forces, the first equilibrium condition becomes

We solve these equations simultaneously for the unknown values [latex] {m}_{3} [/latex] and [latex] {F}_{S}. [/latex] In (Figure) , we cancel the g factor and rearrange the terms to obtain

To obtain [latex] {m}_{3} [/latex] we divide both sides by [latex] {r}_{3}, [/latex] so we have

To find the normal reaction force, we rearrange the terms in (Figure) , converting grams to kilograms:

Significance

Notice that (Figure) is independent of the value of g . The torque balance may therefore be used to measure mass, since variations in g -values on Earth’s surface do not affect these measurements. This is not the case for a spring balance because it measures the force.

Check Your Understanding

Repeat (Figure) using the left end of the meter stick to calculate the torques; that is, by placing the pivot at the left end of the meter stick.

316.7 g; 5.8 N

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by (Figure) and (Figure) . We present this solution to illustrate the importance of a suitable choice of reference frame. Although all inertial reference frames are equivalent and numerical solutions obtained in one frame are the same as in any other, an unsuitable choice of reference frame can make the solution quite lengthy and convoluted, whereas a wise choice of reference frame makes the solution straightforward. We show this in the equivalent solution to the same problem. This particular example illustrates an application of static equilibrium to biomechanics.

Forces in the Forearm

A weightlifter is holding a 50.0-lb weight (equivalent to 222.4 N) with his forearm, as shown in (Figure) . His forearm is positioned at [latex] \beta =60\text{°} [/latex] with respect to his upper arm. The forearm is supported by a contraction of the biceps muscle, which causes a torque around the elbow. Assuming that the tension in the biceps acts along the vertical direction given by gravity, what tension must the muscle exert to hold the forearm at the position shown? What is the force on the elbow joint? Assume that the forearm’s weight is negligible. Give your final answers in SI units.

Figure is a schematic drawing of a forearm rotated around the elbow. A 50 pound ball is held in the palm. The distance between the elbow and the ball is 13 inches. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 1.5 inches. Forearm forms a 60 degree angle with the upper arm.

Figure 12.11 The forearm is rotated around the elbow (E) by a contraction of the biceps muscle, which causes tension [latex] {\overset{\to }{T}}_{\text{M}}. [/latex]

We identify three forces acting on the forearm: the unknown force [latex] \overset{\to }{F} [/latex] at the elbow; the unknown tension [latex] {\overset{\to }{T}}_{\text{M}} [/latex] in the muscle; and the weight [latex] \overset{\to }{w} [/latex] with magnitude [latex] w=50\,\text{lb}. [/latex] We adopt the frame of reference with the x -axis along the forearm and the pivot at the elbow. The vertical direction is the direction of the weight, which is the same as the direction of the upper arm. The x -axis makes an angle [latex] \beta =60\text{°} [/latex] with the vertical. The y -axis is perpendicular to the x -axis. Now we set up the free-body diagram for the forearm. First, we draw the axes, the pivot, and the three vectors representing the three identified forces. Then we locate the angle [latex] \beta [/latex] and represent each force by its x – and y -components, remembering to cross out the original force vector to avoid double counting. Finally, we label the forces and their lever arms. The free-body diagram for the forearm is shown in (Figure) . At this point, we are ready to set up equilibrium conditions for the forearm. Each force has x – and y -components; therefore, we have two equations for the first equilibrium condition, one equation for each component of the net force acting on the forearm.

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force T is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Forces F and T form angle beta with the x axis. Force W forms an angle beta with line connecting it with its projection to the y axis.

Figure 12.12 Free-body diagram for the forearm: The pivot is located at point E (elbow).

Notice that in our frame of reference, contributions to the second equilibrium condition (for torques) come only from the y -components of the forces because the x -components of the forces are all parallel to their lever arms, so that for any of them we have [latex] \text{sin}\,\theta =0 [/latex] in (Figure) . For the y -components we have [latex] \theta =±90\text{°} [/latex] in (Figure) . Also notice that the torque of the force at the elbow is zero because this force is attached at the pivot. So the contribution to the net torque comes only from the torques of [latex] {T}_{y} [/latex] and of [latex] {w}_{y}. [/latex]

We see from the free-body diagram that the x -component of the net force satisfies the equation

and the y -component of the net force satisfies

(Figure) and (Figure) are two equations of the first equilibrium condition (for forces). Next, we read from the free-body diagram that the net torque along the axis of rotation is

(Figure) is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are [latex] {r}_{T}=1.5\,\text{in}\text{.} [/latex] and [latex] {r}_{w}=13.0\,\text{in}\text{.} [/latex] At this point, we do not need to convert inches into SI units, because as long as these units are consistent in (Figure) , they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into (Figure) , (Figure) , and (Figure) to obtain, respectively,

When we simplify these equations, we see that we are left with only two independent equations for the two unknown force magnitudes, F and T , because (Figure) for the x -component is equivalent to (Figure) for the y -component. In this way, we obtain the first equilibrium condition for forces

and the second equilibrium condition for torques

The magnitude of tension in the muscle is obtained by solving (Figure) :

The force at the elbow is obtained by solving (Figure) :

The negative sign in the equation tells us that the actual force at the elbow is antiparallel to the working direction adopted for drawing the free-body diagram. In the final answer, we convert the forces into SI units of force. The answer is

Two important issues here are worth noting. The first concerns conversion into SI units, which can be done at the very end of the solution as long as we keep consistency in units. The second important issue concerns the hinge joints such as the elbow. In the initial analysis of a problem, hinge joints should always be assumed to exert a force in an arbitrary direction , and then you must solve for all components of a hinge force independently. In this example, the elbow force happens to be vertical because the problem assumes the tension by the biceps to be vertical as well. Such a simplification, however, is not a general rule.

Suppose we adopt a reference frame with the direction of the y -axis along the 50-lb weight and the pivot placed at the elbow. In this frame, all three forces have only y -components, so we have only one equation for the first equilibrium condition (for forces). We draw the free-body diagram for the forearm as shown in (Figure) , indicating the pivot, the acting forces and their lever arms with respect to the pivot, and the angles [latex] {\theta }_{T} [/latex] and [latex] {\theta }_{w} [/latex] that the forces [latex] {\overset{\to }{T}}_{\text{M}} [/latex] and [latex] \overset{\to }{w} [/latex] (respectively) make with their lever arms. In the definition of torque given by (Figure) , the angle [latex] {\theta }_{T} [/latex] is the direction angle of the vector [latex] {\overset{\to }{T}}_{\text{M}}, [/latex] counted counterclockwise from the radial direction of the lever arm that always points away from the pivot. By the same convention, the angle [latex] {\theta }_{w} [/latex] is measured counterclockwise from the radial direction of the lever arm to the vector [latex] \overset{\to }{w}. [/latex] Done this way, the non-zero torques are most easily computed by directly substituting into (Figure) as follows:

Figure is a free-body diagram for the forearm. Force F is applied at the point E. Force Tm is applied at the distance r tau from the point E. Force W is applied at the opposite side separated by r w from the point E. Projections of the forces at the x and y axes are shown. Force Tm forms and angle theta tau that is equal to beta with the direction of the lever arm. Force W forms an angle theta w that is equal to the sum of beta and Pi with the direction of the lever arm.

Figure 12.13 Free-body diagram for the forearm for the equivalent solution. The pivot is located at point E (elbow).

The second equilibrium condition, [latex] {\tau }_{T}+{\tau }_{w}=0, [/latex] can be now written as

From the free-body diagram, the first equilibrium condition (for forces) is

(Figure) is identical to (Figure) and gives the result [latex] T=433.3\,\text{lb}. [/latex] (Figure) gives

We see that these answers are identical to our previous answers, but the second choice for the frame of reference leads to an equivalent solution that is simpler and quicker because it does not require that the forces be resolved into their rectangular components.

Repeat (Figure) assuming that the forearm is an object of uniform density that weighs 8.896 N.

[latex] T=\text{1963 N};\,\text{F}=1732\,\text{N} [/latex]

A Ladder Resting Against a Wall

A uniform ladder is [latex] L=5.0\,\text{m} [/latex] long and weighs 400.0 N. The ladder rests against a slippery vertical wall, as shown in (Figure) . The inclination angle between the ladder and the rough floor is [latex] \beta =53\text{°}. [/latex] Find the reaction forces from the floor and from the wall on the ladder and the coefficient of static friction [latex] {\mu }_{\text{s}} [/latex] at the interface of the ladder with the floor that prevents the ladder from slipping.

Figure is a schematic drawing of a 5.0-m-long ladder resting against a wall. Ladder forms a 53 degree angle with the floor.

Figure 12.14 A 5.0-m-long ladder rests against a frictionless wall.

We can identify four forces acting on the ladder. The first force is the normal reaction force N from the floor in the upward vertical direction. The second force is the static friction force [latex] f={\mu }_{\text{s}}N [/latex] directed horizontally along the floor toward the wall—this force prevents the ladder from slipping. These two forces act on the ladder at its contact point with the floor. The third force is the weight w of the ladder, attached at its CM located midway between its ends. The fourth force is the normal reaction force F from the wall in the horizontal direction away from the wall, attached at the contact point with the wall. There are no other forces because the wall is slippery, which means there is no friction between the wall and the ladder. Based on this analysis, we adopt the frame of reference with the y -axis in the vertical direction (parallel to the wall) and the x -axis in the horizontal direction (parallel to the floor). In this frame, each force has either a horizontal component or a vertical component but not both, which simplifies the solution. We select the pivot at the contact point with the floor. In the free-body diagram for the ladder, we indicate the pivot, all four forces and their lever arms, and the angles between lever arms and the forces, as shown in (Figure) . With our choice of the pivot location, there is no torque either from the normal reaction force N or from the static friction f because they both act at the pivot.

Figure is a free-body diagram for a ladder that forms an angle beta with the floor and rests against a wall. Force N is applied at the point at the floor and is perpendicular to the floor. Force W is applied at the mid-point of the ladder. Force F is applied at the point resting at the wall and is perpendicular to the wall. Force W forms an angle theta w with the direction of the lever arm. Theta w is equal to the sum of Pi and half Pi with the beta subtracted. Force F forms an angle theta F with the direction of the lever arm. Theta F is equal to the Pi minus beta.

Figure 12.15 Free-body diagram for a ladder resting against a frictionless wall.

From the free-body diagram, the net force in the x -direction is

the net force in the y -direction is

and the net torque along the rotation axis at the pivot point is

where [latex] {\tau }_{w} [/latex] is the torque of the weight w and [latex] {\tau }_{F} [/latex] is the torque of the reaction F . From the free-body diagram, we identify that the lever arm of the reaction at the wall is [latex] {r}_{F}=L=5.0\,\text{m} [/latex] and the lever arm of the weight is [latex] {r}_{w}=L\,\text{/}\,2=2.5\,\text{m}. [/latex] With the help of the free-body diagram, we identify the angles to be used in (Figure) for torques: [latex] {\theta }_{F}=180\text{°}-\beta [/latex] for the torque from the reaction force with the wall, and [latex] {\theta }_{w}=180\text{°}+(90\text{°}-\beta ) [/latex] for the torque due to the weight. Now we are ready to use (Figure) to compute torques:

We substitute the torques into (Figure) and solve for [latex] F: [/latex]

We obtain the normal reaction force with the floor by solving (Figure) : [latex] N=w=400.0\,\text{N}. [/latex] The magnitude of friction is obtained by solving (Figure) : [latex] f=F=150.7\,\text{N}. [/latex] The coefficient of static friction is [latex] {\mu }_{\text{s}}=f\,\text{/}\,N=150.7\,\text{/}\,400.0=0.377. [/latex]

The net force on the ladder at the contact point with the floor is the vector sum of the normal reaction from the floor and the static friction forces:

Its magnitude is

and its direction is

We should emphasize here two general observations of practical use. First, notice that when we choose a pivot point, there is no expectation that the system will actually pivot around the chosen point. The ladder in this example is not rotating at all but firmly stands on the floor; nonetheless, its contact point with the floor is a good choice for the pivot. Second, notice when we use (Figure) for the computation of individual torques, we do not need to resolve the forces into their normal and parallel components with respect to the direction of the lever arm, and we do not need to consider a sense of the torque. As long as the angle in (Figure) is correctly identified—with the help of a free-body diagram—as the angle measured counterclockwise from the direction of the lever arm to the direction of the force vector, (Figure) gives both the magnitude and the sense of the torque. This is because torque is the vector product of the lever-arm vector crossed with the force vector, and (Figure) expresses the rectangular component of this vector product along the axis of rotation.

This result is independent of the length of the ladder because L is cancelled in the second equilibrium condition, (Figure) . No matter how long or short the ladder is, as long as its weight is 400 N and the angle with the floor is [latex] 53\text{°}, [/latex] our results hold. But the ladder will slip if the net torque becomes negative in (Figure) . This happens for some angles when the coefficient of static friction is not great enough to prevent the ladder from slipping.

For the situation described in (Figure) , determine the values of the coefficient [latex] {\mu }_{\text{s}} [/latex] of static friction for which the ladder starts slipping, given that [latex] \beta [/latex] is the angle that the ladder makes with the floor.

[latex] {\mu }_{s}<0.5\,\text{cot}\,\beta [/latex]

Forces on Door Hinges

A swinging door that weighs [latex] w=400.0\,\text{N} [/latex] is supported by hinges A and B so that the door can swing about a vertical axis passing through the hinges (Figure) . The door has a width of [latex] b=1.00\,\text{m}, [/latex] and the door slab has a uniform mass density. The hinges are placed symmetrically at the door’s edge in such a way that the door’s weight is evenly distributed between them. The hinges are separated by distance [latex] a=2.00\,\text{m}. [/latex] Find the forces on the hinges when the door rests half-open.

Figure is a schematic drawing of a swinging vertical door supported by two hinges attached at points A and B. The distance between points A and B is 2 meters. Door is one meter wide.

Figure 12.16 A 400-N swinging vertical door is supported by two hinges attached at points A and B.

The forces that the door exerts on its hinges can be found by simply reversing the directions of the forces that the hinges exert on the door. Hence, our task is to find the forces from the hinges on the door. Three forces act on the door slab: an unknown force [latex] \overset{\to }{A} [/latex] from hinge [latex] A, [/latex] an unknown force [latex] \overset{\to }{B} [/latex] from hinge [latex] B, [/latex] and the known weight [latex] \overset{\to }{w} [/latex] attached at the center of mass of the door slab. The CM is located at the geometrical center of the door because the slab has a uniform mass density. We adopt a rectangular frame of reference with the y -axis along the direction of gravity and the x -axis in the plane of the slab, as shown in panel (a) of (Figure) , and resolve all forces into their rectangular components. In this way, we have four unknown component forces: two components of force [latex] \overset{\to }{A} [/latex] [latex] ({A}_{x} [/latex] and [latex] {A}_{y}), [/latex] and two components of force [latex] \overset{\to }{B} [/latex] [latex] ({B}_{x} [/latex] and [latex] {B}_{y}). [/latex] In the free-body diagram, we represent the two forces at the hinges by their vector components, whose assumed orientations are arbitrary. Because there are four unknowns [latex] ({A}_{x}, [/latex] [latex] {B}_{x}, [/latex] [latex] {A}_{y}, [/latex] and [latex] {B}_{y}), [/latex] we must set up four independent equations. One equation is the equilibrium condition for forces in the x -direction. The second equation is the equilibrium condition for forces in the y -direction. The third equation is the equilibrium condition for torques in rotation about a hinge. Because the weight is evenly distributed between the hinges, we have the fourth equation, [latex] {A}_{y}={B}_{y}. [/latex] To set up the equilibrium conditions, we draw a free-body diagram and choose the pivot point at the upper hinge, as shown in panel (b) of (Figure) . Finally, we solve the equations for the unknown force components and find the forces.

Figure A is a geometrical representation for a swinging vertical door supported by two hinges attached at points A and B. Forces A and B are applied at the points A and B. Projections of these forces to the x and y axes are shown. Force w is applied at the point CM. Point CM is lower than point A by half-a and to the right of point A by half-b. Line from point A to CM forms an angle beta with the edge of the wall. Figure B is a free-body diagram for a swinging vertical door is supported by two hinges attached at points A and B. Force Ay forms an angle beta with the line connecting points P and CM. Force By forms an angle beta with the line connecting points B and CM. Force W forms an angle beta with the line that is the continuation of the line connecting points P and CM. Distance between points P and CM is d.

Figure 12.17 (a) Geometry and (b) free-body diagram for the door.

From the free-body diagram for the door we have the first equilibrium condition for forces:

We select the pivot at point P (upper hinge, per the free-body diagram) and write the second equilibrium condition for torques in rotation about point P :

We use the free-body diagram to find all the terms in this equation:

In evaluating [latex] \text{sin}\,\beta , [/latex] we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into (Figure) and compute [latex] {B}_{x}: [/latex]

Therefore the magnitudes of the horizontal component forces are [latex] {A}_{x}={B}_{x}=100.0\,\text{N}. [/latex] The forces on the door are

The forces on the hinges are found from Newton’s third law as

Note that if the problem were formulated without the assumption of the weight being equally distributed between the two hinges, we wouldn’t be able to solve it because the number of the unknowns would be greater than the number of equations expressing equilibrium conditions.

Solve the problem in (Figure) by taking the pivot position at the center of mass.

[latex] {\overset{\to }{F}}_{\text{door on}\,A}=100.0\,\text{N}\hat{i}-200.0\,\text{N}\hat{j}\,\text{;}\,{\overset{\to }{F}}_{\text{door on}\,B}=-100.0\,\text{N}\hat{i}-200.0\,\text{N}\hat{j} [/latex]

A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg. Find the tensions in the two vertical ropes supporting the scaffold.

Figure is a schematic drawing of a woman standing 1.5 m away from one end and 4.5 m away from another end of a scaffold.

A 400.0-N sign hangs from the end of a uniform strut. The strut is 4.0 m long and weighs 600.0 N. The strut is supported by a hinge at the wall and by a cable whose other end is tied to the wall at a point 3.0 m above the left end of the strut. Find the tension in the supporting cable and the force of the hinge on the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut is 4.0 m long and is supported by a 5.0 m long cable tied to the wall at a point 3.0 m above the left end of the strut.

  • A variety of engineering problems can be solved by applying equilibrium conditions for rigid bodies.
  • In applications, identify all forces that act on a rigid body and note their lever arms in rotation about a chosen rotation axis. Construct a free-body diagram for the body. Net external forces and torques can be clearly identified from a correctly constructed free-body diagram. In this way, you can set up the first equilibrium condition for forces and the second equilibrium condition for torques.
  • In setting up equilibrium conditions, we are free to adopt any inertial frame of reference and any position of the pivot point. All choices lead to one answer. However, some choices can make the process of finding the solution unduly complicated. We reach the same answer no matter what choices we make. The only way to master this skill is to practice.

Conceptual Questions

Is it possible to rest a ladder against a rough wall when the floor is frictionless?

Show how a spring scale and a simple fulcrum can be used to weigh an object whose weight is larger than the maximum reading on the scale.

A painter climbs a ladder. Is the ladder more likely to slip when the painter is near the bottom or near the top?

A uniform plank rests on a level surface as shown below. The plank has a mass of 30 kg and is 6.0 m long. How much mass can be placed at its right end before it tips? ( Hint: When the board is about to tip over, it makes contact with the surface only along the edge that becomes a momentary axis of rotation.)

Figure schematic drawing of uniform plank rests on a level surface. Part of the plank that is 4.2 meters long is supported by the plank. Part of the plank that is 1.8 meters long is hanging over it.

The uniform seesaw shown below is balanced on a fulcrum located 3.0 m from the left end. The smaller boy on the right has a mass of 40 kg and the bigger boy on the left has a mass 80 kg. What is the mass of the board?

Figure is a schematic drawing of two boys on the seesaw. One boy sits on the edge of the seesaw three meters from the center. Another boys sits at the opposite edge of the seesaw, five meters from the center.

In order to get his car out of the mud, a man ties one end of a rope to the front bumper and the other end to a tree 15 m away, as shown below. He then pulls on the center of the rope with a force of 400 N, which causes its center to be displaced 0.30 m, as shown. What is the force of the rope on the car?

Figure is a schematic drawing that shows a rope tied to the front bumper and the other end to a tree 15 m away. A force of 400 N is applied to the center of the rope and causes it to get displaced 0.30 m.

A uniform 40.0-kg scaffold of length 6.0 m is supported by two light cables, as shown below. An 80.0-kg painter stands 1.0 m from the left end of the scaffold, and his painting equipment is 1.5 m from the right end. If the tension in the left cable is twice that in the right cable, find the tensions in the cables and the mass of the equipment.

Figure is a schematic drawing of a man standing at the left side and the bucket placed at the right side of a scaffold.

When the structure shown below is supported at point P , it is in equilibrium. Find the magnitude of force F and the force applied at P . The weight of the structure is negligible.

Figure shows the distribution of forces applied to point P. Force of 2000 N, two meters to the left of the point P, moves it downwards. Force F, two meters to the left and two meters above of the point P, moves it to the right. Force of 1000 N, two meters to the right and three meters below of the point P, moves it to the left.

To get up on the roof, a person (mass 70.0 kg) places a 6.00-m aluminum ladder (mass 10.0 kg) against the house on a concrete pad with the base of the ladder 2.00 m from the house. The ladder rests against a plastic rain gutter, which we can assume to be frictionless. The center of mass of the ladder is 2.00 m from the bottom. The person is standing 3.00 m from the bottom. Find the normal reaction and friction forces on the ladder at its base.

784 N, 376 N

A uniform horizontal strut weighs 400.0 N. One end of the strut is attached to a hinged support at the wall, and the other end of the strut is attached to a sign that weighs 200.0 N. The strut is also supported by a cable attached between the end of the strut and the wall. Assuming that the entire weight of the sign is attached at the very end of the strut, find the tension in the cable and the force at the hinge of the strut.

Figure is a schematic drawing of a sign which hangs from the end of a uniform strut. The strut forms a 30 degree angle with the cable tied to the wall above the left end of the strut.

The forearm shown below is positioned at an angle [latex] \theta [/latex] with respect to the upper arm, and a 5.0-kg mass is held in the hand. The total mass of the forearm and hand is 3.0 kg, and their center of mass is 15.0 cm from the elbow. (a) What is the magnitude of the force that the biceps muscle exerts on the forearm for [latex] \theta =60\text{°}\text{?} [/latex] (b) What is the magnitude of the force on the elbow joint for the same angle? (c) How do these forces depend on the angle [latex] \theta ? [/latex]

Figure is a schematic drawing of a forearm rotated around the elbow. A 5 kilogram ball is held in the palm. The distance between the elbow and the ball is 35 centimeters. The distance between the elbow and the biceps muscle, which causes a torque around the elbow, is 4 centimeters. Forearm forms a theta angle with the upper arm.

The uniform boom shown below weighs 3000 N. It is supported by the horizontal guy wire and by the hinged support at point A . What are the forces on the boom due to the wire and due to the support at A ? Does the force at A act along the boom?

Figure is a schematic drawing of a 2000 N weight that is supported by the horizontal guy wire and by the hinged support at point A. Hinged support forms a 45 degree angle with the ground.

The uniform boom shown below weighs 700 N, and the object hanging from its right end weighs 400 N. The boom is supported by a light cable and by a hinge at the wall. Calculate the tension in the cable and the force on the hinge on the boom. Does the force on the hinge act along the boom?

Figure is a schematic drawing of a 400 N weight that is by a cable and by a hinge at the wall. Hinge forms a 20 degree angle with the line perpendicular to the wall. Cable forms a 45 degree angle with the line perpendicular to the wall.

A 12.0-m boom, AB , of a crane lifting a 3000-kg load is shown below. The center of mass of the boom is at its geometric center, and the mass of the boom is 1000 kg. For the position shown, calculate tension T in the cable and the force at the axle A .

Figure is a schematic drawing of a crane lifting a 3000-kg load. Arm of a crane forms a 30 degree angle with the line parallel to the ground. Cable supporting load forms a 10 degree angle with the arm.

A uniform trapdoor shown below is 1.0 m by 1.5 m and weighs 300 N. It is supported by a single hinge (H), and by a light rope tied between the middle of the door and the floor. The door is held at the position shown, where its slab makes a [latex] 30\text{°} [/latex] angle with the horizontal floor and the rope makes a [latex] 20\text{°} [/latex] angle with the floor. Find the tension in the rope and the force at the hinge.

Figure is a schematic drawing of a trapdoor that is 1.0 m by 1.5 m. Door is supported by a single hinge labeled H, and by a light rope tied between the middle of the door and the floor. The door makes a 30 degree angle with the floor and the rope makes a 20 degree angle with the floor.

A 90-kg man walks on a sawhorse, as shown below. The sawhorse is 2.0 m long and 1.0 m high, and its mass is 25.0 kg. Calculate the normal reaction force on each leg at the contact point with the floor when the man is 0.5 m from the far end of the sawhorse. ( Hint: At each end, find the total reaction force first. This reaction force is the vector sum of two reaction forces, each acting along one leg. The normal reaction force at the contact point with the floor is the normal (with respect to the floor) component of this force.)

Figure is a schematic drawing of a man walks on a sawhorse. Each side of the sawhorse is supported by two connected legs. There are 60 degree angles between the legs.

  • OpenStax University Physics. Authored by : OpenStax CNX. Located at : https://cnx.org/contents/[email protected]:Gofkr9Oy@15 . License : CC BY: Attribution . License Terms : Download for free at http://cnx.org/contents/[email protected]

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8.4: Solving Statics Problems

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learning objectives

  • Formulate and apply six steps to solve static problems

Statics is the study of forces in equilibrium. Recall that Newton’s second law states:

\[\mathrm{∑F=ma}\]

Therefore, for all objects moving at constant velocity (including a velocity of 0 — stationary objects), the net external force is zero. There are forces acting, but they are balanced — that is to say, they are “in equilibrium. ”

When solving equilibrium problems, it might help to use the following steps:

  • First, ensure that the problem you’re solving is in fact a static problem—i.e., that no acceleration (including angular acceleration) is involved Remember:\(\mathrm{∑F=ma=0}\) for these situations. If rotational motion is involved, the condition \(\mathrm{∑τ=Iα=0}\) must also be satisfied, where is torque, is the moment of inertia, and is the angular acceleration.
  • Choose a pivot point. Often this is obvious because the problem involves a hinge or a fixed point. If the choice is not obvious, pick the pivot point as the location at which you have the most unknowns. This simplifies things because forces at the pivot point create no torque because of the cross product:\(\mathrm{τ=rF}\)
  • Write an equation for the sum of torques, and then write equations for the sums of forces in the x and y directions. Set these sums equal to 0. Be careful with your signs.
  • Solve for your unknowns.
  • Insert numbers to find the final answer.
  • Check if the solution is reasonable by examining the magnitude, direction, and units of the answer. The importance of this last step cannot be overstated, although in unfamiliar applications, it can be more difficult to judge reasonableness. However, these judgments become progressively easier with experience.
  • First, ensure that the problem you’re solving is in fact a static problem—i.e., that no acceleration (including angular acceleration ) is involved.
  • Choose a pivot point — use the location at which you have the most unknowns.
  • Write equations for the sums of torques and forces in the x and y directions.
  • Solve the equations for your unknowns algebraically, and insert numbers to find final answers.
  • torque : A rotational or twisting effect of a force; (SI unit newton-meter or Nm; imperial unit foot-pound or ft-lb)
  • moment of inertia : A measure of a body’s resistance to a change in its angular rotation velocity

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  • OpenStax College, College Physics. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42173/latest/?collection=col11406/1.7 . License : CC BY: Attribution
  • OpenStax College, College Physics. September 17, 2013. Provided by : OpenStax CNX. Located at : http://cnx.org/content/m42167/latest/?collection=col11406/1.7 . License : CC BY: Attribution
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Chemistry LibreTexts

6.7: Solving Equilibrium Problems

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  • Page ID 127253

  • David Harvey
  • DePauw University

Ladder diagrams are a useful tool for evaluating chemical reactivity and for providing a reasonable estimate of a chemical system’s composition at equilibrium. If we need a more exact quantitative description of the equilibrium condition, then a ladder diagram is insufficient; instead, we need to find an algebraic solution. In this section we will learn how to set‐up and solve equilibrium problems. We will start with a simple problem and work toward more complex problems.

A Simple Problem: The Solubility of Pb(IO 3 ) 2

If we place an insoluble compound such as Pb(IO 3 ) 2 in deionized water, the solid dissolves until the concentrations of Pb 2 + and \(\text{IO}_3^-\) satisfy the solubility product for Pb(IO 3 ) 2 . At equilibrium the solution is saturated with Pb(IO 3 ) 2 , which means simply that no more solid can dissolve. How do we determine the equilibrium concentrations of Pb 2 + and \(\text{IO}_3^-\), and what is the molar solubility of Pb(IO 3 ) 2 in this saturated solution?

When we first add solid Pb(IO 3 ) 2 to water, the concentrations of Pb 2 + and \(\text{IO}_3^-\) are zero and the reaction quotient, Q r , is

\[Q_r = \left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_{3}^{-}\right]^{2}=0 \nonumber\]

As the solid dissolves, the concentrations of these ions increase, but Q r remains smaller than K sp . We reach equilibrium and “satisfy the solubility product” when Q r = K sp .

We begin by writing the equilibrium reaction and the solubility product expression for Pb(IO 3 ) 2 .

\[\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{IO}_{3}^{-}(a q) \nonumber\]

e \[K_{\mathrm{sp}}=\left[\mathrm{Pb}^{2+}\right]\left[\mathrm{IO}_{3}^{-}\right]^{2}=2.5 \times 10^{-13} \label{6.1}\]

As Pb(IO 3 ) 2 dissolves, two \(\text{IO}_3^-\) ions form for each ion of Pb 2 + . If we assume that the change in the molar concentration of Pb 2 + at equilibrium is x , then the change in the molar concentration of \(\text{IO}_3^-\) is 2 x . The following table helps us keep track of the initial concentrations, the change in con‐ centrations, and the equilibrium concentrations of Pb2+ and \(\text{IO}_3^-\).

Because a solid, such as Pb(IO 3 ) 2 , does not appear in the solubility product expression, we do not need to keep track of its concentration. Remember, however, that the K sp value applies only if there is some solid Pb(IO 3 ) 2 present at equilibrium.

Substituting the equilibrium concentrations into Equation \ref{6.1} and solving gives

\[(x)(2 x)^{2}=4 x^{3}=2.5 \times 10^{-13} \nonumber\]

\[x=3.97 \times 10^{-5} \nonumber\]

Substituting this value of x back into the equilibrium concentration expressions for Pb 2 + and \(\text{IO}_3^-\) gives their concentrations as

\[\left[\mathrm{Pb}^{2+}\right]=x=4.0 \times 10^{-5} \mathrm{M} \text { and }\left[\mathrm{IO}_{3}^{-}\right]=2 x=7.9 \times 10^{-5} \nonumber\]

Because one mole of Pb(IO 3 ) 2 contains one mole of Pb 2 + , the molar solubility of Pb(IO 3 ) 2 is equal to the concentration of Pb 2 + , or \(4.0 \times 10^{-5}\) M.

We can express a compound’s solubility in two ways: as its molar solubility (mol/L) or as its mass solubility (g/L). Be sure to express your answer clearly.

Exercise 6.7.1

Calculate the molar solubility and the mass solubility for Hg 2 Cl 2 , given the following solubility reaction and K sp value.

\[\mathrm{Hg}_{2} \mathrm{Cl}_{2}(s)\rightleftharpoons \mathrm{Hg}_{2}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad K_{\mathrm{sp}}=1.2 \times 10^{-8} \nonumber\]

When Hg 2 Cl 2 dissolves, two Cl – are produced for each ion of \(\text{Hg}_2^{2+}\). If we assume x is the change in the molar concentration of \(\text{Hg}_2^{2+}\), then the change in the molar concentration of Cl – is 2 x . The following table helps us keep track of our solution to this problem.

Substituting the equilibrium concentrations into the K sp expression forHg 2 Cl 2 gives

\[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}=(x)(2 \mathrm{x})^{2}=4 x^{3}=1.2 \times 10^{-18} \nonumber\]

\[x=6.69 \times 10^{-7} \nonumber\]

Substituting x back into the equilibrium expressions for \(\text{Hg}_2^{2+}\) and Cl – gives their concentrations as

\[\left[\mathrm{Hg}_{2}^{2+}\right]=x=6.7 \times 10^{-7} \ \mathrm{M} \quad\left[\mathrm{Cl}^{-}\right]=2 x=1.3 \times 10^{-6} \ \mathrm{M} \nonumber\]

The molar solubility is equal to [\(\text{Hg}_2^{2+}\)], or \(6.7 \times 10^{-7}\) mol/L.

A More Complex Problem: The Common Ion Effect

Calculating the solubility of Pb(IO 3 ) 2 in deionized water is a straightforward problem because the solid’s dissolution is the only source of Pb 2 + and \(\text{IO}_3^-\). But what if we add Pb(IO 3 ) 2 to a solution of 0.10 M Pb(NO 3 ) 2 ? Before we set‐up and solve this problem algebraically, think about the system’s chemistry and decide whether the solubility of Pb(IO 3 ) 2 will increase, decrease, or remain the same. Beginning a problem by thinking about the likely answer is a good habit to develop. Knowing what answers are reasonable will help you spot errors in your calculations and give you more confidence that your solution to a problem is correct. Because the solution already contains a source of Pb 2 + , we can use Le Châtelier’s principle to predict that the solubility of Pb(IO 3 ) 2 is smaller than that in our previous problem.

We begin by setting up a table to help us keep track of the concentrations of Pb 2 + and \(\text{IO}_3^-\) as this system moves toward and reaches equilibrium.

Substituting the equilibrium concentrations into Equation \ref{6.1}

\[(0.10+x)(2 x)^{2}=2.5 \times 10^{-13} \nonumber\]

and multiplying out the terms on the equation’s left side leaves us with

\[4 x^{3}+0.40 x^{2}=2.5 \times 10^{-13} \label{6.2}\]

This is a more difficult equation to solve than that for the solubility of Pb(IO 3 ) 2 in deionized water, and its solution is not immediately obvious. We can find a rigorous solution to Equation \ref{6.2} using computational software packages and spreadsheets, some of which are described in Chapter 6.10 .

There are several approaches to solving cubic equations, but none are computationally easy using just paper and pencil.

How might we solve Equation \ref{6.2} if we do not have access to a computer? One approach is to use our understanding of chemistry to simplify the problem. From Le Châtelier’s principle we know that a large initial concentration of Pb 2 + will decrease significantly the solubility of Pb(IO 3 ) 2 . One reasonable assumption is that the initial concentration of Pb 2 + is very close to its equilibrium concentration. If this assumption is correct, then the following approximation is reasonable

\[\left[\mathrm{Pb}^{2+}\right]=0.10+x \approx 0.10 \nonumber\]

Substituting this approximation into Equation \ref{6.1} and solving for x gives

\[(0.10)(2 x)^{2}=0.4 x^{2}=2.5 \times 10^{-13} \nonumber\]

\[x=7.91 \times 10^{-7} \nonumber\]

Before we accept this answer, we must verify that our approximation is reasonable. The difference between the actual concentration of Pb 2 + , which is 0.10 + x M, and our assumption that the concentration of Pb 2 + is 0.10 M is \(7.9 \times 10^{-7}\), or \(7.9 \times 10^{-4}\) % of the assumed concentration. This is a negligible error. If we accept the result of our calculation, we find that the equilibrium concentrations of Pb 2 + and \(\text{IO}_3^-\) are

\[\left[\mathrm{Pb}^{2+}\right]=0.10+x \approx 0.10 \ \mathrm{M} \text { and }\left[\mathrm{IO}_{3}^{-}\right]=2 x=1.6 \times 10^{-6} \ \mathrm{M} \nonumber\]

\[\begin{aligned} \% \text { error } &=\frac{\text { actual }-\text { assumed }}{\text { assumed }} \times 100 \\ &=\frac{(0.10+x)-0.10}{0.10} \times 100 \\ &=\frac{7.91 \times 10^{-7}}{0.10} \times 100 \\ &=7.91 \times 10^{-4} \% \end{aligned} \nonumber\]

The molar solubility of Pb(IO 3 ) 2 is equal to the additional concentration of Pb 2 + in solution, or \(7.9 \times 10^{-4}\) mol/L. As expected, we find that Pb(IO 3 ) 2 is less soluble in the presence of a solution that already contains one of its ions. This is known as the common ion effect .

As outlined in the following example, if an approximation leads to an error that is unacceptably large, then we can extend the process of making and evaluating approximations.

One “rule of thumb” when making an approximation is that it should not introduce an error of more than ±5%. Although this is not an unreasonable choice, what matters is that the error makes sense within the context of the problem you are solving.

Example 6.7.1

Calculate the solubility of Pb(IO 3 ) 2 in \(1.0 \times 10^{-4}\) M Pb(NO 3 ) 2 .

If we let x equal the change in the concentration of Pb 2 + , then the equilibrium concentrations of Pb 2 + and \(\text{IO}_3^-\) are

\[\left[\mathrm{P} \mathrm{b}^{2+}\right]=1.0 \times 10^{-4}+ \ x \text { and }\left[\mathrm{IO}_{3}^-\right]=2 x \nonumber\]

Substituting these concentrations into Equation \ref{6.1} leaves us with

\[\left(1.0 \times 10^{-4}+ \ x\right)(2 x)^{2}=2.5 \times 10^{-13} \nonumber\]

To solve this equation for x , let’s make the following assumption

\[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ x \approx 1.0 \times 10^{-4} \ \mathrm{M} \nonumber\]

Solving for x gives its value as \(2.50 \times 10^{-5}\); however, when we substitute this value for x back, we find that the calculated concentration of Pb 2 + at equilibrium

\[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ x=1.0 \times 10^{-4}+ \ 2.50 \times 10^{-5}=1.25 \times 10^{-4} \ \mathrm{M} \nonumber\]

is 25% greater than our assumption of \(1.0 \times 10^{-4}\) M. This error is unreasonably large.

Rather than shouting in frustration, let’s make a new assumption. Our first assumption—that the concentration of Pb 2 + is \(1.0 \times 10^{-4}\) M—was too small. The calculated concentration of \(1.25 \times 10^{-4}\) M, therefore, probably is a too large, but closer to the correct concentration than was our first assumption. For our second approximation, let’s assume that

\[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ x=1.25 \times 10^{-4} \mathrm{M} \nonumber\]

Substituting into Equation \ref{6.1} and solving for x gives its value as \(2.24 \times 10^{-5}\). The resulting concentration of Pb 2 + is

\[\left[\mathrm{Pb}^{2+}\right]=1.0 \times 10^{-4}+ \ 2.24 \times 10^{-5}=1.22 \times 10^{-4} \ \mathrm{M} \nonumber\]

which differs from our assumption of \(1.25 \times 10^{-4}\) M by 2.4%. Because the original concentration of Pb 2 + is given to two significant figure, this is a more reasonable error. Our final solution, to two significant figures, is

\[\left[\mathrm{Pb}^{2+}\right]=1.2 \times 10^{-4} \ \mathrm{M} \text { and }\left[\mathrm{IO}_{3}\right]=4.5 \times 10^{-5} \ \mathrm{M} \nonumber\]

and the molar solubility of Pb(IO 3 ) 2 is \(2.2 \times 10^{-5}\) mol/L. This iterative approach to solving the problems is known as the method of successive approximations .

Exercise 6.7.2

Calculate the molar solubility for Hg 2 Cl 2 in 0.10 M NaCl and compare your answer to its molar solubility in deionized water (see Exercise 6.7.1 ).

We begin by setting up a table to help us keep track of the concentrations \(\text{Hg}_2^{2+}\) and Cl – as this system moves toward and reaches equilibrium.

Substituting the equilibrium concentrations into the K sp expression for Hg 2 Cl 2 leaves us with a difficult to solve cubic equation.

\[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}=(x)(0.10+2 x)^{2}=4 x^{3}+0.40 x^{2}+0.010 x \nonumber\]

Let’s make an assumption to simplify this problem. Because we expect the value of x to be small, let’s assume that

\[\left[\mathrm{Cl}^{-}\right]=0.10+2 x \approx 0.10 \nonumber\]

This simplifies our problem to

\[K_{\mathrm{sp}}=\left[\mathrm{Hg}_{2}^{2+}\right]\left[\mathrm{Cl}^{-}\right]^{2}=(x)(0.10)^{2}=0.010 x=1.2 \times 10^{-18} \nonumber\]

which gives the value of x as \(1.2 \times 10^{-16}\) M. The difference between the actual concentration of Cl – , which is (0.10 + 2 x ) M, and our assumption that it is 0.10 M introduces an error of \(2.4 \times 10^{-13}\) %. This is a negligible error. The molar solubility of Hg 2 Cl 2 is the same as the concentration of \(\text{Hg}_2^{2+}\), or \(1.2 \times 10^{-16}\) M. As expected, the molar solubility in 0.10 M NaCl is less than \(6.7 \times 10^{-7}\) mol/L, which is its solubility in water (see solution to Exercise 6.7.1 ).

A Systematic Approach to Solving Equilibrium Problems

Calculating the solubility of Pb(IO 3 ) 2 in a solution of Pb(NO 3 ) 2 is more complicated than calculating its solubility in deionized water. The calculation, however, is still relatively easy to organize and the simplifying assumptions are fairly obvious. This problem is reasonably straightforward because it involves only one equilibrium reaction and one equilibrium constant.

Determining the equilibrium composition of a system with multiple equilibrium reactions is more complicated. In this section we introduce a systematic approach to setting‐up and solving equilibrium problems. As shown in Table 6.7.1 , this approach involves four steps.

In addition to equilibrium constant expressions, two other equations are important to this systematic approach to solving an equilibrium problem. The first of these equations is a mass balance equation , which simply is a statement that matter is conserved during a chemical reaction. In a solution of acetic acid, for example, the combined concentrations of the conjugate weak acid, CH 3 COOH, and the conjugate weak base, CH 3 COO – , must equal acetic acid’s initial concentration, \(C_{\text{CH}_3\text{COOH}}\).

\[C_{\mathrm{CH}_{\mathrm{3}} \mathrm{COOH}}=\left[\mathrm{CH}_{3} \mathrm{COOH}\right]+\left[\mathrm{CH}_{3} \mathrm{COO}^{-}\right] \nonumber\]

You may recall from Chapter 2 that this is the difference between a formal concentration and a molar concentration. The variable C represents a formal concentration.

The second equation is a charge balance equation , which requires that the total positive charge from the cations equal the total negative charge from the anions. Mathematically, the charge balance equation is

\[\sum_{i=1}^{n}\left(z^{+}\right)_{i}\left[{C^{z}}^+\right]_{i} = -\sum_{j=1}^{m}(z^-)_{j}\left[{A^{z}}^-\right]_{j} \nonumber\]

where [ C z + ] i and [ A z - ] j are, respectively, the concentrations of the i th cation and the j th anion, and ( z + ) i and ( z – ) j are the charges for the i th cation and the j th anion. Every ion in solution, even if it does not appear in an equilibrium reaction, must appear in the charge balance equation. For example, the charge balance equation for an aqueous solution of Ca(NO 3 ) 2 is

\[2 \times\left[\mathrm{Ca}^{2+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{+}\right]+\left[\mathrm{NO}_{3}^-\right] \nonumber\]

Note that we multiply the concentration of Ca 2 + by two and that we include the concentrations of H 3 O + and OH – .

A charge balance is a conservation of a charge. The minus sign in front of the summation term on the right side of the charge balance equation ensures that both summations are positive. There are situations where it is impossible to write a charge balance equation because we do not have enough information about the solution’s composition. For example, suppose we fix a solution’s pH using a buffer. If the buffer’s composition is not specified, then we cannot write a charge balance equation.

Example 6.7.2

Write mass balance equations and a charge balance equation for a 0.10 M solution of NaHCO 3 .

It is easier to keep track of the species in solution if we write down the reactions that define the solution’s composition. These reactions are the dissolution of a soluble salt

\[\mathrm{NaHCO}_{3}(s) \rightarrow \mathrm{Na}^{+}(a q)+\mathrm{HCO}_{3}^{-}(a q) \nonumber\]

and the acid–base dissociation reactions of \(\text{HCO}_3^-\) and H 2 O

\[\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{CO}_{3}^{2-}(a q) \nonumber\]

\[\mathrm{HCO}_{3}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{CO}_{3}(a q) \nonumber\]

\[2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\]

The mass balance equations are

\[0.10 \mathrm{M}=\left[\mathrm{H}_{2} \mathrm{CO}_{3}\right]+\left[\mathrm{HCO}_{3}^{-}\right]+\left[\mathrm{CO}_{3}^{2-}\right] \nonumber\]

\[0.10 \ \mathrm{M}=\left[\mathrm{Na}^{+}\right] \nonumber\]

and the charge balance equation is

\[\left[\mathrm{Na}^{+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]+\left[\mathrm{HCO}_{3}^-\right]+2 \times\left[\mathrm{CO}_{3}^{2-}\right] \nonumber\]

Exercise 6.7.3

Write appropriate mass balance and charge balance equations for a solution containing 0.10 M KH 2 PO 4 and 0.050 M Na 2 HPO 4 .

To help us determine what ions are in solution, let’s write down all the reaction needed to prepare the solutions and the equilibrium reactions that take place within these solutions. These reactions are the dissolution of two soluble salts

\[\mathrm{KH}_{2} \mathrm{PO}_{4}(s) \longrightarrow \mathrm{K}^{+}(a q)+\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \nonumber\]

\[\mathrm{NaHPO}_{4}(s) \longrightarrow \mathrm{Na}^{+}(a q)+\mathrm{HPO}_{4}^{2-}(a q) \nonumber\]

and the acid–base dissociation reactions for \(\text{H}_2\text{PO}_4^-\), \(\text{HPO}_4^{2-}\). and H 2 O.

\[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons\mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{HPO}_{4}^{2-}(a q) \nonumber\]

\[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \nonumber\]

\[\mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{PO}_{4}^{3-}(a q) \nonumber\]

Note that we did not include the base dissociation reaction for \(\text{HPO}_4^{2-}\) because we already accounted for its product, \(\text{H}_2\text{PO}_4^-\), in another reaction. The mass balance equations for K + and Na + are straightforward

\[\left[\mathrm{K}^{+}\right]=0.10 \ \mathrm{M} \text { and }\left[\mathrm{Na}^{+}\right]=0.10 \ \mathrm{M} \nonumber\]

but the mass balance equation for phosphate takes a bit more thought. Both \(\text{H}_2\text{PO}_4^-\) and \(\text{HPO}_4^{2-}\) produce the same ions in solution. We can, therefore, imagine that the solution initially contains 0.15 M KH 2 PO 4 , which gives the following mass balance equation.

\[\left[\mathrm{H}_{3} \mathrm{PO}_{4}\right]+\left[\mathrm{H}_{2} \mathrm{PO}_{4}^-\right]+\left[\mathrm{HPO}_{4}^{2-}\right]+\left[\mathrm{PO}_{4}^{3-}\right]=0.15 \ \mathrm{M} \nonumber\]

The charge balance equation is

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\left[\mathrm{K}^{+}\right]+\left[\mathrm{Na}^{+}\right] =\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]+2 \times\left[\mathrm{HPO}_{4}^{2-}\right] +3 \times\left[\mathrm{PO}_{4}^{3-}\right]+\left[\mathrm{OH}^{-}\right] \nonumber\]

pH of a Monoprotic Weak Acid

To illustrate the systematic approach to solving equilibrium problems, let’s calculate the pH of 1.0 M HF. Two equilibrium reactions affect the pH. The first, and most obvious, is the acid dissociation reaction for HF

\[\mathrm{HF}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{F}^{-}(a q) \nonumber\]

for which the equilibrium constant expression is

\[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]}=6.8 \times 10^{-4} \label{6.3}\]

The second equilibrium reaction is the dissociation of water, which is an obvious yet easily neglected reaction

\[2 \mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q) \nonumber\]

\[K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.00 \times 10^{-14} \label{6.4}\]

Counting unknowns, we find four: [HF], [F – ], [H 3 O + ], and [OH – ]. To solve this problem we need two additional equations. These equations are a mass balance equation on hydrofluoric acid

\[C_{\mathrm{HF}}=[\mathrm{HF}]+\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M} \label{6.5}\]

and a charge balance equation

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]+\left[\mathrm{F}^{-}\right] \label{6.6}\]

With four equations and four unknowns, we are ready to solve the problem. Before doing so, let’s simplify the algebra by making two assumptions.

Assumption One . Because HF is a weak acid, we know that the solution is acidic. For an acidic solution it is reasonable to assume that

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]>>\left[\mathrm{OH}^{-}\right] \nonumber\]

which simplifies the charge balance equation to

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\left[\mathrm{F}^{-}\right] \label{6.7}\]

Assumption Two . Because HF is a weak acid, very little of it dissociates to form F – . Most of the HF remains in its conjugate weak acid form and it is reasonable to assume that

\[[\mathrm{HF}]>>\left[\mathrm{F}^{-}\right] \nonumber\]

which simplifies the mass balance equation to

\[C_{\mathrm{HF}}=[\mathrm{HF}]=1.0 \ \mathrm{M} \label{6.8}\]

For this exercise let’s accept an assumption if it introduces an error of less than ±5%.

Substituting Equation \ref{6.7} and Equation \ref{6.8} into Equation \ref{6.3}, and solving for the concentration of H 3 O + gives us

\[\mathrm{K}_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{F}^{-}\right]}{[\mathrm{HF}]}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{\mathrm{C}_{\mathrm{HF}}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}}{\mathrm{C}_{\mathrm{HF}}}=6.8 \times 10^{-4} \nonumber\]

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{K_{\mathrm{a}} C_{\mathrm{HF}}}=\sqrt{\left(6.8 \times 10^{-4}\right)(1.0)}=2.6 \times 10^{-2} \nonumber\]

Before accepting this answer, we must verify our assumptions. The first assumption is that [OH – ] is significantly smaller than [H 3 O + ]. Using Equation \ref{6.4}, we find that

\[\left[\mathrm{OH}^{-}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{1.00 \times 10^{-14}}{2.6 \times 10^{-2}}=3.8 \times 10^{-13} \nonumber\]

Clearly this assumption is acceptable. The second assumption is that [F – ] is significantly smaller than [HF]. From Equation \ref{6.7} we have

\[\left[\mathrm{F}^{-}\right]=2.6 \times 10^{-2} \ \mathrm{M} \nonumber\]

Because [F – ] is 2.60% of C HF , this assumption also is acceptable. Given that [H 3 O + ] is \(2.6 \times 10^{-2}\) M, the pH of 1.0 M HF is 1.59.

How does the calculation change if we require that the error introduced in our assumptions be less than ±1%? In this case we no longer can assume that [HF] >> [F – ] and we cannot simplify the mass balance equation. Solving the mass balance equation for [HF]

\[[\mathrm{HF}]=C_{\mathrm{HF}}-\left[\mathrm{F}^{-}\right]=C_{\mathrm{HF}}-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \nonumber\]

and substituting into the K a expression along with Equation \ref{6.7} gives

\[K_{\mathrm{a}}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}}{C_{\mathrm{HF}}-\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\]

Rearranging this equation leaves us with a quadratic equation

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}+K_{\mathrm{a}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]-K_{\mathrm{a}} C_{\mathrm{HF}}=0 \nonumber\]

which we solve using the quadratic formula

\[x=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a} \nonumber\]

where a , b , and c are the coefficients in the quadratic equation

\[a x^{2}+b x+c=0 \nonumber\]

Solving a quadratic equation gives two roots, only one of which has chemical significance. For our problem, the equation’s roots are

\[x=\frac{-6.8 \times 10^{-4} \pm \sqrt{\left(6.8 \times 10^{-4}\right)^{2}-(4)(1)\left(-6.8 \times 10^{-4}\right)}}{(2)(1)} \nonumber\]

\[x=\frac{-6.8 \times 10^{-4} \pm 5.22 \times 10^{-2}}{2} \nonumber\]

\[x=2.57 \times 10^{-2} \text { or }-2.64 \times 10^{-2} \nonumber\]

Only the positive root is chemically significant because the negative root gives a negative concentration for H 3 O + . Thus, [H 3 O + ] is \(2.57 \times 10^{-2}\) M and the pH is 1.59.

You can extend this approach to calculating the pH of a monoprotic weak base by replacing K a with K b , replacing C HF with the weak base’s concentration, and solving for [OH – ] in place of [H 3 O + ].

Exercise 6.7.4

Calculate the pH of 0.050 M NH 3 . State any assumptions you make in solving the problem, limiting the error for any assumption to ±5%. The K b value for NH 3 is \(1.75 \times 10^{-5}\).

To determine the pH of 0.050 M NH 3 , we need to consider two equilibrium reactions: the base dissociation reaction for NH 3

\[\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q) \nonumber\]

and water’s dissociation reaction.

These two reactions contain four species whose concentrations we need to consider: NH 3 , \(\text{NH}_4^+\), H 3 O + , and OH – . We need four equations to solve the problem—these equations are the K b equation for NH 3

\[K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=1.75 \times 10^{-5} \nonumber\]

the K w equation for H 2 O

\[K_{w}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \nonumber\]

a mass balance equation on ammonia

\[C_{\mathrm{NH}_{3}}=0.050 \ \mathrm{M}=\left[\mathrm{NH}_{3}\right]+\left[\mathrm{NH}_{4}^{+}\right] \nonumber\]

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]+\left[\mathrm{NH}_{4}^{+}\right]=\left[\mathrm{OH}^{-}\right] \nonumber\]

To solve this problem, we will make two assumptions. Because NH 3 is a base, our first assumption is

\[\left[\mathrm{OH}^{-}\right]>>\left[\mathrm{H}_{3} \mathrm{O}^{+}\right] \nonumber\]

\[\left[\mathrm{NH}_{4}^{+}\right]=\left[\mathrm{OH}^{-}\right] \nonumber\]

Because NH 3 is a weak base, our second assumption is

\[\left[\mathrm{NH}_{3}\right]>>\left[\mathrm{NH}_{4}^{+}\right] \nonumber\]

\[C_{\mathrm{NH}_{3}}=0.050 \ \mathrm{M}=\left[\mathrm{NH}_{3}\right] \nonumber\]

Substituting the simplified charge balance equation and mass balance equation into the K b equation leave us with

\[K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{OH}^{-}\right]}{C_{\mathrm{NH}_3}}=\frac{\left[\mathrm{OH}^{-}\right]^{2}}{C_{\mathrm{NH_3}}}=1.75 \times 10^{-5} \nonumber\]

\[\left[\mathrm{OH}^{-}\right]=\sqrt{K_{\mathrm{b}} C_{\mathrm{NH_3}}}=\sqrt{\left(1.75 \times 10^{-5}\right)(0.050)}=9.35 \times 10^{-4} \nonumber\]

Before we accept this answer, we must verify our two assumptions. The first assumption is that the concentration of OH – is significantly greater than the concentration of H 3 O + . Using K w , we find that

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{OH}^{-}\right]}=\frac{1.00 \times 10^{-14}}{9.35 \times 10^{-4}}=1.07 \times 10^{-11} \nonumber\]

Clearly this assumption is acceptable. Our second assumption is that the concentration of NH 3 is significantly greater than the concentration of \(\text{NH}_4^+\). Using our simplified charge balance equation, we find that

\[\left[\mathrm{NH}_{4}^{+}\right]=\left[\mathrm{OH}^{-}\right]=9.35 \times 10^{-4} \nonumber\]

Because the concentration of \(\text{NH}_4^+\) is 1.9% of \(C_{\text{NH}_3}\), our second assumption also is reasonable. Given that [H 3 O + ] is \(1.07 \times 10^{-11}\), the pH is 10.97.

pH of a Polyprotic Acid or Base

A more challenging problem is to find the pH of a solution that contains a polyprotic weak acid or one of its conjugate species. As an example, consider the amino acid alanine, whose structure is shown in Figure 6.7.1 . The ladder diagram in Figure 6.7.2 shows alanine’s three acid–base forms and their respective areas of predominance. For simplicity, we identify these species as H 2 L + , HL, and L – .

pH of 0.10 M Alanine Hydrochloride (H 2 L + )

Alanine hydrochloride is the salt of the diprotic weak acid H 2 L + and Cl – . Because H 2 L + has two acid dissociation reactions, a complete systematic solution to this problem is more complicated than that for a monoprotic weak acid. The ladder diagram in Figure 6.7.2 helps us simplify the problem. Because the areas of predominance for H 2 L + and L – are so far apart, we can assume that a solution of H 2 L + will not contain a significant amount of L – . As a result, we can treat H 2 L + as though it is a monoprotic weak acid. Calculating the pH of 0.10 M alanine hydrochloride, which is 1.72, is left to the reader as an exercise.

pH of 0.10 M Sodium Alaninate (L – )

The alaninate ion is a diprotic weak base. Because L – has two base dissociation reactions, a complete systematic solution to this problem is more complicated than that for a monoprotic weak base. Once again, the ladder diagram in Figure 6.7.2 helps us simplify the problem. Because the areas of predominance for H 2 L + and L – are so far apart, we can assume that a solution of L – will not contain a significant amount of H 2 L + . As a result, we can treat L – as though it is a monoprotic weak base. Calculating the pH of 0.10 M sodium alaninate, which is 11.42, is left to the reader as an exercise.

pH of 0.10 M Alanine (HL)

Finding the pH of a solution of alanine is more complicated than our previous two examples because we cannot ignore the presence of either H 2 L + or L – . To calculate the solution’s pH we must consider alanine’s acid dissociation reaction

\[\mathrm{HL}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{L}^{-}(a q) \nonumber\]

and its base dissociation reaction

\[\mathrm{HL}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{H}_{2} \mathrm{L}^{+}(a q) \nonumber\]

and, as always, we must also consider the dissociation of water

This leaves us with five unknowns—[H 2 L + ], [HL], [L – ], [H 3 O + ], and [OH – ]—for which we need five equations. These equations are K a2 and K b2 for alanine

\[K_{\mathrm{a} 2}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{L}^{-}\right]}{[\mathrm{HL}]} \nonumber\]

\[K_{\mathrm{b} 2}=\frac{K_{\mathrm{w}}}{K_{\mathrm{a1}}}=\frac{\left[\mathrm{OH}^{-}\right]\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]}{[\mathrm{HL}]} \nonumber\]

the K w equation

\[K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right] \nonumber\]

a mass balance equation for alanine

\[C_{\mathrm{HL}}=\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]+[\mathrm{HL}]+[\mathrm{L}^{-}] \nonumber\]

\[\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{OH}^-]+[\mathrm{L^-}] \nonumber\]

Because HL is a weak acid and a weak base, it seems reasonable to assume that little of it will dissociate and that

\[[\mathrm{HL}]>>\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]+[\mathrm{L}^-] \nonumber\]

which allows us to simplify the mass balance equation to

\[C_{\mathrm{HL}}=[\mathrm{HL}] \nonumber\]

Next we solve K b2 for [H 2 L + ]

\[\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]=\frac{K_{\mathrm{w}}[\mathrm{HL}]}{K_{\mathrm{a1}}\left[\mathrm{OH}^{-}\right]}=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right][\mathrm{HL}]}{K_{\mathrm{a1}}}=\frac{C_{\mathrm{HL}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a1}}} \nonumber\]

and solve K a2 for [L – ]

\[[\mathrm{L^-}]=\frac{K_{a2}[\mathrm{HL}]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{K_{a2} C_{\mathrm{HL}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\]

Substituting these equations for [H 2 L + ] and [L – ], and the equation for K w , into the charge balance equation give us

\[\frac{C_{\mathrm{HL}}\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}{K_{\mathrm{a1}}}+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\frac{K_{\mathrm{w}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}+\frac{K_{a2} C_{\mathrm{HL}}}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]} \nonumber\]

which we simplify to

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left(\frac{C_{\mathrm{HL}}}{K_{\mathrm{a1}}}+1\right)=\frac{1}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}\left(K_{\mathrm{w}}+K_{a2} C_{\mathrm{HL}}\right) \nonumber\]

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]^{2}=\frac{\left(K_{\mathrm{a} 2} C_{\mathrm{HL}}+K_{\mathrm{w}}\right)}{\frac{C_{\mathrm{HL}}}{K_{\mathrm{a1}}}+1}=\frac{K_{\mathrm{a1}}\left(K_{\mathrm{a2}} C_{\mathrm{HL}}+K_{\mathrm{w}}\right)}{C_{\mathrm{HL}}+K_{\mathrm{a1}}} \nonumber\]

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{\frac{\left(K_{\mathrm{a1}} K_{a2} C_{\mathrm{HL}}+K_{\mathrm{a1}} K_{\mathrm{w}}\right)}{C_{\mathrm{HL}}+K_{\mathrm{a1}}}} \nonumber\]

We can further simplify this equation if K a1 K w << K a1 K a2 C HL , and if K a1 << C HL , leaving us with

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{K_{\mathrm{a1}} K_{\mathrm{a} 2}} \nonumber\]

For a solution of 0.10 M alanine the [H 3 O + ] is

\[\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=\sqrt{\left(4.487 \times 10^{-3}\right)\left(1.358 \times 10^{-10}\right)}=7.806 \times 10^{-7} \ \mathrm{M} \nonumber\]

or a pH of 6.11.

Exercise 6.7.5

Verify that each assumption in our solution for the pH of 0.10 M alanine is reasonable, using ±5% as the limit for the acceptable error.

In solving for the pH of 0.10 M alanine, we made the following three assumptions: (a) [HL] >> [H 2 L + ] + [L – ]; (b) K a1 K w << K a1 K a2 C HL ; and (c) K a1 << C HL . Assumptions (b) and (c) are easy to check. The value of K a1 (\(4.487 \times 10^{-3}\)) is 4.5% of C HL (0.10), and K a1 K w (\(4.487 \times 10^{-17}\)) is 0.074% of K a1 K a2 C HL (\(6.093 \times 10^{-14}\)). Each of these assumptions introduces an error of less than ±5%.

To test assumption (a) we need to calculate the concentrations of H 2 L + and L – , which we accomplish using the equations for K a1 and K a2 .

\[\left[\mathrm{H}_{2} \mathrm{L}^{+}\right]=\frac{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right][\mathrm{HL}]}{K_{\mathrm{a1}}}=\frac{\left(7.807 \times 10^{-7}\right)(0.10)}{4.487 \times 10^{-3}}=1.74 \times 10^{-5} \nonumber\]

\[\left[\mathrm{L}^{-}\right]=\frac{K_{a 2}[\mathrm{HL}]}{\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]}=\frac{\left(1.358 \times 10^{-10}\right)(0.10)}{7.807 \times 10^{-7}}=1.74 \times 10^{-5} \nonumber\]

Because these concentrations are less than ±5% of C HL, the first assumption also is acceptable.

Effect of Complexation on Solubility

One method for increasing a precipitate’s solubility is to add a ligand that forms soluble complexes with one of the precipitate’s ions. For example, the solubility of AgI increases in the presence of NH 3 due to the formation of the soluble \(\text{Ag(NH}_3)_2^+\) complex. As a final illustration of the systematic approach to solving equilibrium problems, let’s calculate the molar solubility of AgI in 0.10 M NH 3 .

We begin by writing the relevant equilibrium reactions, which includes the solubility of AgI, the acid–base chemistry of NH 3 and H 2 O, and the metal‐ligand complexation chemistry between Ag + and NH 3 .

\[\begin{array}{c}{\operatorname{AgI}(s)\rightleftharpoons\operatorname{Ag}^{+}(a q)+\mathrm{I}^{-}(a q)} \\ {\mathrm{NH}_{3}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{OH}^{-}(a q)+\mathrm{NH}_{4}^{+}(a q)} \\ {2 \mathrm{H}_{2} \mathrm{O}(l)\rightleftharpoons \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{OH}^{-}(a q)} \\ {\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q)}\end{array} \nonumber\]

This leaves us with seven unknowns—[Ag + ], [I – ], [NH 3 ], [\(\text{NH}_4^+\) ], [OH – ], [H 3 O + ], and [\(\text{Ag(NH}_3)_2^+\)]—and a need for seven equations. Four of the equations we need to solve this problem are the equilibrium constant expressions

\[K_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]\left[\mathrm{I}^{-}\right]=8.3 \times 10^{-17} \label{6.9}\]

\[K_{\mathrm{b}}=\frac{\left[\mathrm{NH}_{4}^{+}\right]\left[\mathrm{OH}^{-}\right]}{\left[\mathrm{NH}_{3}\right]}=1.75 \times 10^{-5} \label{6.10}\]

\[K_{\mathrm{w}}=\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]\left[\mathrm{OH}^{-}\right]=1.00 \times 10^{-14} \label{6.11}\]

\[\beta_{2}=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]}{\left[\mathrm{Ag}^{+}\right]\left[\mathrm{NH}_{3}\right]^{2}}=1.7 \times 10^{7} \label{6.12}\]

We still need three additional equations. The first of these equations is a mass balance for NH 3 .

\[C_{\mathrm{NH}_{3}}=\left[\mathrm{NH}_{3}\right]+\left[\mathrm{NH}_{4}^{+}\right]+2 \times\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \label{6.13}\]

In writing this mass balance equation we multiply the concentration of \(\text{Ag(NH}_3)_2^+\) by two since there are two moles of NH 3 per mole of \(\text{Ag(NH}_3)_2^+\). The second additional equation is a mass balance between iodide and silver. Because AgI is the only source of I – and Ag + , each iodide in solution must have an associated silver ion, which may be Ag + or \(\text{Ag(NH}_3)_2^+\) ; thus

\[\left[\mathrm{I}^{-}\right]=\left[\mathrm{Ag}^{+}\right]+\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \label{6.14}\]

Finally, we include a charge balance equation.

\[\left[\mathrm{Ag}^{+}\right]+\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]+\left[\mathrm{NH}_{4}^{+}\right]+\left[\mathrm{H}_{3} \mathrm{O}^{+}\right]=[\mathrm{OH}^-]+[\mathrm{I}^-] \label{6.15}\]

Although the problem looks challenging, three assumptions greatly simplify the algebra.

Assumption One. Because the formation of the \(\text{Ag(NH}_3)_2^+\) complex is so favorable (\(\beta_2\) is \(1.7 \times 10^7\)), there is very little free Ag + in solution and it is reasonable to assume that

\[\left[\mathrm{Ag}^{+}\right]<<\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \nonumber\]

Assumption Two . Because NH 3 is a weak base we may reasonably assume that most uncomplexed ammonia remains as NH 3 ; thus

\[\left[\mathrm{NH}_{4}^{+}\right]<<\left[\mathrm{NH}_{3}\right] \nonumber\]

Assumption Three. Because K sp for AgI is significantly smaller than \(\beta_2\) for \(\text{Ag(NH}_3)_2^+\), the solubility of AgI probably is small enough that very little ammonia is needed to form the metal–ligand complex; thus

\[\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]<<\left[\mathrm{NH}_{3}\right] \nonumber\]

As we use these assumptions to simplify the algebra, let’s set ±5% as the limit for error.

Assumption two and assumption three suggest that the concentration of NH 3 is much larger than the concentrations of either \(\text{NH}_4^+\) or \(\text{Ag(NH}_3)_2^+\), which allows us to simplify the mass balance equation for NH 3 to

\[C_{\mathrm{NH}_{3}}=\left[\mathrm{NH}_{3}\right] \label{6.16}\]

Finally, using assumption one, which suggests that the concentration of \(\text{Ag(NH}_3)_2^+\) is much larger than the concentration of Ag + , we simplify the mass balance equation for I – to

\[\left[\mathrm{I}^{-}\right]=\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right] \label{6.17}\]

Now we are ready to combine equations and to solve the problem. We begin by solving Equation \ref{6.9} for [Ag + ] and substitute it into \(\beta_2\) (Equation \ref{6.12}), which leaves us with

\[\beta_{2}=\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right][\mathrm{I}^-]}{K_{\mathrm{sp}}\left[\mathrm{NH}_{3}\right]^{2}} \label{6.18}\]

Next we substitute Equation \ref{6.16} and Equation \ref{6.17} into Equation \ref{6.18}, obtaining

\[\beta_{2}=\frac{\left[\mathrm{I}^{-}\right]^{2}}{K_{\mathrm{sp}}\left(C_{\mathrm{NH}_3}\right)^{2}} \label{6.19}\]

Solving Equation \ref{6.19} for [I – ] gives

\[\left[\mathrm{I}^{-}\right]=C_{\mathrm{NH}_3} \sqrt{\beta_{2} K_{s p}} = \\ (0.10) \sqrt{\left(1.7 \times 10^{7}\right)\left(8.3 \times 10^{-17}\right)}=3.76 \times 10^{-6} \ \mathrm{M} \nonumber\]

Because one mole of AgI produces one mole of I – , the molar solubility of AgI is the same as the [I – ], or \(3.8 \times 10^{-6}\) mol/L.

Before we accept this answer we need to check our assumptions. Substituting [I – ] into Equation \ref{6.9}, we find that the concentration of Ag + is

\[\left[\mathrm{Ag}^{+}\right]=\frac{K_{\mathrm{p}}}{[\mathrm{I}^-]}=\frac{8.3 \times 10^{-17}}{3.76 \times 10^{-6}}=2.2 \times 10^{-11} \ \mathrm{M} \nonumber\]

Substituting the concentrations of I – and Ag+ into the mass balance equation for iodide (Equation \ref{6.14}), gives the concentration of \(\text{Ag(NH}_3)_2^+\) as

\[\left[\operatorname{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]=[\mathrm{I}^-]-\left[\mathrm{Ag}^{+}\right]=3.76 \times 10^{-6}-2.2 \times 10^{-11}=3.76 \times 10^{-6} \ \mathrm{M} \nonumber\]

Our first assumption that [Ag + ] is significantly smaller than the [\(\text{Ag(NH}_3)_2^+\)] is reasonable.

Substituting the concentrations of Ag + and \(\text{Ag(NH}_3)_2^+\) into Equation \ref{6.12} and solving for [NH 3 ], gives

\[\left[\mathrm{NH}_{3}\right]=\sqrt{\frac{\left[\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\right]}{\left[\mathrm{Ag}^{+}\right] \beta_{2}}}=\sqrt{\frac{3.76 \times 10^{-6}}{\left(2.2 \times 10^{-11}\right)\left(1.7 \times 10^{7}\right)}}=0.10 \ \mathrm{M} \nonumber\]

From the mass balance equation for NH3 (Equation \ref{6.12}) we see that [\(\text{NH}_4^+\)] is negligible, verifying our second assumption that \([\text{NH}_4^+]\) is significantly smaller than [NH 3 ]. Our third assumption that [\(\text{Ag(NH}_3)_2^+\)] is significantly smaller than [NH 3 ] also is reasonable.

Did you notice that our solution to this problem did not make use of Equation \ref{6.15}, the charge balance equation? The reason for this is that we did not try to solve for the concentration of all seven species. If we need to know the reaction mixture’s complete composition at equilibrium, then we will need to incorporate the charge balance equation into our solution.

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  3. 12.2: Conditions for Static Equilibrium

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  4. 12.2 Examples of Static Equilibrium

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  17. 12.2 Examples of Static Equilibrium

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  18. Chapter 15.3: Solving Equilibrium Problems

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  21. 8.4: Solving Statics Problems

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