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physics practice problems statics

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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physics practice problems statics

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.

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  • 3.3 Average and Instantaneous Acceleration
  • 3.4 Motion with Constant Acceleration
  • 3.5 Free Fall
  • 3.6 Finding Velocity and Displacement from Acceleration
  • 4.1 Displacement and Velocity Vectors
  • 4.2 Acceleration Vector
  • 4.3 Projectile Motion
  • 4.4 Uniform Circular Motion
  • 4.5 Relative Motion in One and Two Dimensions
  • 5.2 Newton's First Law
  • 5.3 Newton's Second Law
  • 5.4 Mass and Weight
  • 5.5 Newton’s Third Law
  • 5.6 Common Forces
  • 5.7 Drawing Free-Body Diagrams
  • 6.1 Solving Problems with Newton’s Laws
  • 6.2 Friction
  • 6.3 Centripetal Force
  • 6.4 Drag Force and Terminal Speed
  • 7.2 Kinetic Energy
  • 7.3 Work-Energy Theorem
  • 8.1 Potential Energy of a System
  • 8.2 Conservative and Non-Conservative Forces
  • 8.3 Conservation of Energy
  • 8.4 Potential Energy Diagrams and Stability
  • 8.5 Sources of Energy
  • 9.1 Linear Momentum
  • 9.2 Impulse and Collisions
  • 9.3 Conservation of Linear Momentum
  • 9.4 Types of Collisions
  • 9.5 Collisions in Multiple Dimensions
  • 9.6 Center of Mass
  • 9.7 Rocket Propulsion
  • 10.1 Rotational Variables
  • 10.2 Rotation with Constant Angular Acceleration
  • 10.3 Relating Angular and Translational Quantities
  • 10.4 Moment of Inertia and Rotational Kinetic Energy
  • 10.5 Calculating Moments of Inertia
  • 10.6 Torque
  • 10.7 Newton’s Second Law for Rotation
  • 10.8 Work and Power for Rotational Motion
  • 11.1 Rolling Motion
  • 11.2 Angular Momentum
  • 11.3 Conservation of Angular Momentum
  • 11.4 Precession of a Gyroscope
  • 12.1 Conditions for Static Equilibrium
  • 12.3 Stress, Strain, and Elastic Modulus
  • 12.4 Elasticity and Plasticity
  • 13.1 Newton's Law of Universal Gravitation
  • 13.2 Gravitation Near Earth's Surface
  • 13.3 Gravitational Potential Energy and Total Energy
  • 13.4 Satellite Orbits and Energy
  • 13.5 Kepler's Laws of Planetary Motion
  • 13.6 Tidal Forces
  • 13.7 Einstein's Theory of Gravity
  • 14.1 Fluids, Density, and Pressure
  • 14.2 Measuring Pressure
  • 14.3 Pascal's Principle and Hydraulics
  • 14.4 Archimedes’ Principle and Buoyancy
  • 14.5 Fluid Dynamics
  • 14.6 Bernoulli’s Equation
  • 14.7 Viscosity and Turbulence
  • 15.1 Simple Harmonic Motion
  • 15.2 Energy in Simple Harmonic Motion
  • 15.3 Comparing Simple Harmonic Motion and Circular Motion
  • 15.4 Pendulums
  • 15.5 Damped Oscillations
  • 15.6 Forced Oscillations
  • 16.1 Traveling Waves
  • 16.2 Mathematics of Waves
  • 16.3 Wave Speed on a Stretched String
  • 16.4 Energy and Power of a Wave
  • 16.5 Interference of Waves
  • 16.6 Standing Waves and Resonance
  • 17.1 Sound Waves
  • 17.2 Speed of Sound
  • 17.3 Sound Intensity
  • 17.4 Normal Modes of a Standing Sound Wave
  • 17.5 Sources of Musical Sound
  • 17.7 The Doppler Effect
  • 17.8 Shock Waves
  • B | Conversion Factors
  • C | Fundamental Constants
  • D | Astronomical Data
  • E | Mathematical Formulas
  • F | Chemistry
  • G | The Greek Alphabet

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 Equation 12.7 to Equation 12.9 . We introduced a problem-solving strategy in Example 12.1 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 ( + ) ( + ) means that the working direction is the actual direction. A minus sign ( − ) ( − ) 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 Equation 12.7 for force components in the x -direction. (b) Use the free-body diagram to write a correct equilibrium condition Equation 12.11 for force components in the y -direction. (c) Use the free-body diagram to write a correct equilibrium condition Equation 12.9 for torques along the axis of rotation. Use Equation 12.10 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 Example 12.1 .

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.

Example 12.3

The torque balance.

w 1 = m 1 g w 1 = m 1 g is the weight of mass m 1 ; m 1 ; w 2 = m 2 g w 2 = m 2 g is the weight of mass m 2 ; m 2 ;

w = m g w = m g is the weight of the entire meter stick; w 3 = m 3 g w 3 = m 3 g is the weight of unknown mass m 3 ; m 3 ;

F S F S 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 12.10 . 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.

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 + y + y -direction to be parallel to F → S , F → S , 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 m 3 m 3 and F S . F S . In Equation 12.17 , we cancel the g factor and rearrange the terms to obtain

To obtain m 3 m 3 we divide both sides by r 3 , r 3 , so we have

To find the normal reaction force, we rearrange the terms in Equation 12.18 , converting grams to kilograms:

Significance

Check your understanding 12.3.

Repeat Example 12.3 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.

In the next example, we show how to use the first equilibrium condition (equation for forces) in the vector form given by Equation 12.7 and Equation 12.8 . 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.

Example 12.4

Forces in the forearm.

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 sin θ = 0 sin θ = 0 in Equation 12.10 . For the y -components we have θ = ± 90 ° θ = ± 90 ° in Equation 12.10 . 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 T y T y and of w y . w y .

and the y -component of the net force satisfies

Equation 12.21 and Equation 12.22 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

Equation 12.23 is the second equilibrium condition (for torques) for the forearm. The free-body diagram shows that the lever arms are r T = 1.5 in . r T = 1.5 in . and r w = 13.0 in . r w = 13.0 in . At this point, we do not need to convert inches into SI units, because as long as these units are consistent in Equation 12.23 , they cancel out. Using the free-body diagram again, we find the magnitudes of the component forces:

We substitute these magnitudes into Equation 12.21 , Equation 12.22 , and Equation 12.23 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 Equation 12.21 for the x -component is equivalent to Equation 12.22 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 Equation 12.25 :

The force at the elbow is obtained by solving Equation 12.24 :

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

The second equilibrium condition, τ T + τ w = 0 , τ T + τ w = 0 , can be now written as

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

Equation 12.26 is identical to Equation 12.25 and gives the result T = 433.3 lb . T = 433.3 lb . Equation 12.27 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.

Check Your Understanding 12.4

Repeat Example 12.4 assuming that the forearm is an object of uniform density that weighs 8.896 N.

Example 12.5

A ladder resting against a wall.

the net force in the y -direction is

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

where τ w τ w is the torque of the weight w and τ F τ F 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 r F = L = 5.0 m r F = L = 5.0 m and the lever arm of the weight is r w = L / 2 = 2.5 m . r w = L / 2 = 2.5 m . With the help of the free-body diagram, we identify the angles to be used in Equation 12.10 for torques: θ F = 180 ° − β θ F = 180 ° − β for the torque from the reaction force with the wall, and θ w = 180 ° + ( 90 ° − β ) θ w = 180 ° + ( 90 ° − β ) for the torque due to the weight. Now we are ready to use Equation 12.10 to compute torques:

We substitute the torques into Equation 12.30 and solve for F : F :

We obtain the normal reaction force with the floor by solving Equation 12.29 : N = w = 400.0 N . N = w = 400.0 N . The magnitude of friction is obtained by solving Equation 12.28 : f = F = 150.7 N . f = F = 150.7 N . The coefficient of static friction is μ s = f / N = 150.7 / 400.0 = 0.377 . μ s = f / N = 150.7 / 400.0 = 0.377 .

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 Equation 12.10 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 Equation 12.10 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, Equation 12.10 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 Equation 12.10 expresses the rectangular component of this vector product along the axis of rotation.

Check Your Understanding 12.5

For the situation described in Example 12.5 , determine the values of the coefficient μ s μ s of static friction for which the ladder starts slipping, given that β β is the angle that the ladder makes with the floor.

Example 12.6

Forces on door hinges.

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 sin β , sin β , we use the geometry of the triangle shown in part (a) of the figure. Now we substitute these torques into Equation 12.32 and compute B x : B x :

Therefore the magnitudes of the horizontal component forces are A x = B x = 100.0 N . A x = B x = 100.0 N . The forces on the door are

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

Check Your Understanding 12.6

Solve the problem in Example 12.6 by taking the pivot position at the center of mass.

Check Your Understanding 12.7

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.

Check Your Understanding 12.8

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.

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Physics library

Welcome to the physics library, unit 1: one-dimensional motion, unit 2: two-dimensional motion, unit 3: forces and newton's laws of motion, unit 4: centripetal force and gravitation, unit 5: work and energy, unit 6: impacts and linear momentum, unit 7: torque and angular momentum, unit 8: oscillations and mechanical waves, unit 9: fluids, unit 10: thermodynamics, unit 11: electric charge, field, and potential, unit 12: circuits, unit 13: magnetic forces, magnetic fields, and faraday's law, unit 14: electromagnetic waves and interference, unit 15: geometric optics, unit 16: special relativity, unit 17: quantum physics, unit 18: discoveries and projects, unit 19: review for ap physics 1 exam.

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Engineering Statics: Open and Interactive

(5 reviews)

physics practice problems statics

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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11: Fluid Statics (Exercises)

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  • Page ID 4195

Conceptual Questions

11.1: what is a fluid.

1. What physical characteristic distinguishes a fluid from a solid?

2. Which of the following substances are fluids at room temperature: air, mercury, water, glass?

3. Why are gases easier to compress than liquids and solids?

4. How do gases differ from liquids?

11.2: Density

5. Approximately how does the density of air vary with altitude?

6. Give an example in which density is used to identify the substance composing an object. Would information in addition to average density be needed to identify the substances in an object composed of more than one material?

7. Figure shows a glass of ice water filled to the brim. Will the water overflow when the ice melts? Explain your answer.

A glass filled to the brim with water and ice cubes.

11.3: Pressure

8. How is pressure related to the sharpness of a knife and its ability to cut?

9. Why does a dull hypodermic needle hurt more than a sharp one?

10. The outward force on one end of an air tank was calculated in Example. How is this force balanced? (The tank does not accelerate, so the force must be balanced.)

11. Why is force exerted by static fluids always perpendicular to a surface?

12. In a remote location near the North Pole, an iceberg floats in a lake. Next to the lake (assume it is not frozen) sits a comparably sized glacier sitting on land. If both chunks of ice should melt due to rising global temperatures (and the melted ice all goes into the lake), which ice chunk would give the greatest increase in the level of the lake water, if any?

13. How do jogging on soft ground and wearing padded shoes reduce the pressures to which the feet and legs are subjected?

14. Toe dancing (as in ballet) is much harder on toes than normal dancing or walking. Explain in terms of pressure.

15. How do you convert pressure units like millimeters of mercury, centimeters of water, and inches of mercury into units like newtons per meter squared without resorting to a table of pressure conversion factors?

11.4: Variation of Pressure with Depth in a Fluid

16. Atmospheric pressure exerts a large force (equal to the weight of the atmosphere above your body—about 10 tons) on the top of your body when you are lying on the beach sunbathing. Why are you able to get up?

17. Why does atmospheric pressure decrease more rapidly than linearly with altitude?

18. What are two reasons why mercury rather than water is used in barometers?

19. Figure shows how sandbags placed around a leak outside a river levee can effectively stop the flow of water under the levee. Explain how the small amount of water inside the column formed by the sandbags is able to balance the much larger body of water behind the levee.

The figure shows a flooding river on the extreme right, with a levee set up on its left, and sandbags are stacked on the left of the levee. The height of the levee and that of the stacked sandbags is greater than the water level of the flooding river, so the water does not flow over their tops, but a leak under the levee allows some water to flow under it and reach the sandbags.

Because the river level is very high, it has started to leak under the levee. Sandbags are placed around the leak, and the water held by them rises until it is the same level as the river, at which point the water there stops rising.

20. Why is it difficult to swim under water in the Great Salt Lake?

21. Is there a net force on a dam due to atmospheric pressure? Explain your answer.

22. Does atmospheric pressure add to the gas pressure in a rigid tank? In a toy balloon? When, in general, does atmospheric pressure not affect the total pressure in a fluid?

23. You can break a strong wine bottle by pounding a cork into it with your fist, but the cork must press directly against the liquid filling the bottle—there can be no air between the cork and liquid. Explain why the bottle breaks, and why it will not if there is air between the cork and liquid.

11.5: Pascal’s Principle

24. Suppose the master cylinder in a hydraulic system is at a greater height than the slave cylinder. Explain how this will affect the force produced at the slave cylinder.

11.6: Gauge Pressure, Absolute Pressure, and Pressure Measurement

25. Explain why the fluid reaches equal levels on either side of a manometer if both sides are open to the atmosphere, even if the tubes are of different diameters.

26. Figure shows how a common measurement of arterial blood pressure is made. Is there any effect on the measured pressure if the manometer is lowered? What is the effect of raising the arm above the shoulder? What is the effect of placing the cuff on the upper leg with the person standing? Explain your answers in terms of pressure created by the weight of a fluid.

27. Considering the magnitude of typical arterial blood pressures, why are mercury rather than water manometers used for these measurements?

11.7: Archimedes’ Principle

28. More force is required to pull the plug in a full bathtub than when it is empty. Does this contradict Archimedes’ principle? Explain your answer.

29. Do fluids exert buoyant forces in a “weightless” environment, such as in the space shuttle? Explain your answer.

30. Will the same ship float higher in salt water than in freshwater? Explain your answer.

31. Marbles dropped into a partially filled bathtub sink to the bottom. Part of their weight is supported by buoyant force, yet the downward force on the bottom of the tub increases by exactly the weight of the marbles. Explain why.

11.8: Cohesion and Adhesion in Liquids: Surface Tension and Capillary Action

32. The density of oil is less than that of water, yet a loaded oil tanker sits lower in the water than an empty one. Why?

33. Is surface tension due to cohesive or adhesive forces, or both?

34. Is capillary action due to cohesive or adhesive forces, or both?

35. Birds such as ducks, geese, and swans have greater densities than water, yet they are able to sit on its surface. Explain this ability, noting that water does not wet their feathers and that they cannot sit on soapy water.

36. Water beads up on an oily sunbather, but not on her neighbor, whose skin is not oiled. Explain in terms of cohesive and adhesive forces.

37. Could capillary action be used to move fluids in a “weightless” environment, such as in an orbiting space probe?

38. What effect does capillary action have on the reading of a manometer with uniform diameter? Explain your answer.

39. Pressure between the inside chest wall and the outside of the lungs normally remains negative. Explain how pressure inside the lungs can become positive (to cause exhalation) without muscle action.

Problems & Exercises

40. Gold is sold by the troy ounce (31.103 g). What is the volume of 1 troy ounce of pure gold?

Solution \(1.610cm^3\)

41. Mercury is commonly supplied in flasks containing 34.5 kg (about 76 lb). What is the volume in liters of this much mercury?

42. (a) What is the mass of a deep breath of air having a volume of 2.00 L?

(b) Discuss the effect taking such a breath has on your body’s volume and density.

Solution (a) 2.58 g (b) The volume of your body increases by the volume of air you inhale. The average density of your body decreases when you take a deep breath, because the density of air is substantially smaller than the average density of the body before you took the deep breath.

43. A straightforward method of finding the density of an object is to measure its mass and then measure its volume by submerging it in a graduated cylinder. What is the density of a 240-g rock that displaces \(89.0cm^3\) of water? (Note that the accuracy and practical applications of this technique are more limited than a variety of others that are based on Archimedes’ principle.)

Solution \(2.70g/cm^3\)

44. Suppose you have a coffee mug with a circular cross section and vertical sides (uniform radius). What is its inside radius if it holds 375 g of coffee when filled to a depth of 7.50 cm? Assume coffee has the same density as water.

45. (a) A rectangular gasoline tank can hold 50.0 kg of gasoline when full. What is the depth of the tank if it is 0.500-m wide by 0.900-m long?

(b) Discuss whether this gas tank has a reasonable volume for a passenger car.

Solution (a) 0.163 m (b) Equivalent to 19.4 gallons, which is reasonable

46. A trash compactor can reduce the volume of its contents to 0.350 their original value. Neglecting the mass of air expelled, by what factor is the density of the rubbish increased?

47. A 2.50-kg steel gasoline can holds 20.0 L of gasoline when full. What is the average density of the full gas can, taking into account the volume occupied by steel as well as by gasoline?

Solution \(7.9×10^2kg/m^3\)

48. What is the density of 18.0-karat gold that is a mixture of 18 parts gold, 5 parts silver, and 1 part copper? (These values are parts by mass, not volume.) Assume that this is a simple mixture having an average density equal to the weighted densities of its constituents.

Solution \(15.6g/cm^3\)

49. There is relatively little empty space between atoms in solids and liquids, so that the average density of an atom is about the same as matter on a macroscopic scale—approximately \(10^3kg/m^3\). The nucleus of an atom has a radius about \(10^{−5}\) that of the atom and contains nearly all the mass of the entire atom.

(a) What is the approximate density of a nucleus?

(b) One remnant of a supernova, called a neutron star, can have the density of a nucleus. What would be the radius of a neutron star with a mass 10 times that of our Sun (the radius of the Sun is \(7×10^8m\))?

Solution (a) \(10^{18}kg/m^3\) (b) \(2×10^4m\)

50. As a woman walks, her entire weight is momentarily placed on one heel of her high-heeled shoes. Calculate the pressure exerted on the floor by the heel if it has an area of \(1.50cm^2\) and the woman’s mass is 55.0 kg. Express the pressure in Pa. (In the early days of commercial flight, women were not allowed to wear high-heeled shoes because aircraft floors were too thin to withstand such large pressures.)

Solution \(3.59×10^6Pa\); or \(521lb/in^2\)

51. The pressure exerted by a phonograph needle on a record is surprisingly large. If the equivalent of 1.00 g is supported by a needle, the tip of which is a circle 0.200 mm in radius, what pressure is exerted on the record in \(N/m^2\)?

52. Nail tips exert tremendous pressures when they are hit by hammers because they exert a large force over a small area. What force must be exerted on a nail with a circular tip of 1.00 mm diameter to create a pressure of \(3.00×10^9N/m^2\)?(This high pressure is possible because the hammer striking the nail is brought to rest in such a short distance.)

Solution \(2.36×10^3N\)

53. What depth of mercury creates a pressure of 1.00 atm?

Solution 0.760 m

54. The greatest ocean depths on the Earth are found in the Marianas Trench near the Philippines. Calculate the pressure due to the ocean at the bottom of this trench, given its depth is 11.0 km and assuming the density of seawater is constant all the way down.

55. Verify that the SI unit of \(hρg\) is \(N/m^2\).

\((hρg)_{units}=(m)(kg/m^3)(m/s^2)=(kg⋅m^2)/(m^3⋅s^2)\) \(=(kg⋅m/s^2)(1/m^2)\) \(=N/m^2\)

56. Water towers store water above the level of consumers for times of heavy use, eliminating the need for high-speed pumps. How high above a user must the water level be to create a gauge pressure of \(3.00×10^5N/m^2\)?

57. The aqueous humor in a person’s eye is exerting a force of 0.300 N on the \(1.10-cm^2\) area of the cornea.

(a) What pressure is this in mm Hg?

(b) Is this value within the normal range for pressures in the eye?

Solution (a) 20.5 mm Hg (b) The range of pressures in the eye is 12–24 mm Hg, so the result in part (a) is within that range

58. How much force is exerted on one side of an 8.50 cm by 11.0 cm sheet of paper by the atmosphere? How can the paper withstand such a force?

59. What pressure is exerted on the bottom of a 0.500-m-wide by 0.900-m-long gas tank that can hold 50.0 kg of gasoline by the weight of the gasoline in it when it is full?

Solution \(1.09×10^3N/m^2\)

60. Calculate the average pressure exerted on the palm of a shot-putter’s hand by the shot if the area of contact is \(50.0cm^2\) and he exerts a force of 800 N on it. Express the pressure in \(N/m^2\) and compare it with the \(1.00×10^6Pa\) pressures sometimes encountered in the skeletal system.

61. The left side of the heart creates a pressure of 120 mm Hg by exerting a force directly on the blood over an effective area of \(15.0cm^2\). What force does it exert to accomplish this?

Solution 24.0 N

62. Show that the total force on a rectangular dam due to the water behind it increases with the square of the water depth. In particular, show that this force is given by \(F=ρgh^2L/2\), where \(ρ\) is the density of water, \(h\) is its depth at the dam, and \(L\) is the length of the dam. You may assume the face of the dam is vertical. (Hint: Calculate the average pressure exerted and multiply this by the area in contact with the water. (See Figure.)

A two-dimensional view of a dam with dimensions L and h is shown. Force F at h is shown by a horizontal arrow. The force F exerted by water on the dam is F equals average pressure p bar into area A and pressure in turn is average height h bar into density rho into acceleration due to gravity g.

63. How much pressure is transmitted in the hydraulic system considered in Example? Express your answer in pascals and in atmospheres.

Solution \(2.55×10^7Pa\); or 251 atm

64. What force must be exerted on the master cylinder of a hydraulic lift to support the weight of a 2000-kg car (a large car) resting on the slave cylinder? The master cylinder has a 2.00-cm diameter and the slave has a 24.0-cm diameter.

65. A crass host pours the remnants of several bottles of wine into a jug after a party. He then inserts a cork with a 2.00-cm diameter into the bottle, placing it in direct contact with the wine. He is amazed when he pounds the cork into place and the bottom of the jug (with a 14.0-cm diameter) breaks away. Calculate the extra force exerted against the bottom if he pounded the cork with a 120-N force.

Solution \(5.76×10^3N\) extra force

66. A certain hydraulic system is designed to exert a force 100 times as large as the one put into it.

(a) What must be the ratio of the area of the slave cylinder to the area of the master cylinder?

(b) What must be the ratio of their diameters?

(c) By what factor is the distance through which the output force moves reduced relative to the distance through which the input force moves? Assume no losses to friction.

67. (a) Verify that work input equals work output for a hydraulic system assuming no losses to friction. Do this by showing that the distance the output force moves is reduced by the same factor that the output force is increased. Assume the volume of the fluid is constant.

(b) What effect would friction within the fluid and between components in the system have on the output force? How would this depend on whether or not the fluid is moving?

Solution (a) \(V=d_iA_i=d_oA_o⇒d_o=d_i(\frac{A_i}{A_o})\).

Now, using equation:

\(\frac{F_1}{A_1}=\frac{F_2}{A_2}⇒F_o=F_i(\frac{A_o}{A_i})\).

\(W_o=F_od_o=(\frac{F_iA_o}{A_i})(\frac{d_iA_i}{A_o})=F_id_i=W_i\).

In other words, the work output equals the work input.

(b) If the system is not moving, friction would not play a role. With friction, we know there are losses, so that \(W_{out}=W_{in}−W_f\); therefore, the work output is less than the work input. In other words, with friction, you need to push harder on the input piston than was calculated for the nonfriction case.

68. Find the gauge and absolute pressures in the balloon and peanut jar shown in Figure, assuming the manometer connected to the balloon uses water whereas the manometer connected to the jar contains mercury. Express in units of centimeters of water for the balloon and millimeters of mercury for the jar, taking \(h=0.0500 m\) for each.

Solution Balloon: Pg=5.00 cmH2O, Pabs=1.035×103cmH2O. Jar: Pg=−50.0 mm Hg, Pabs=710 mm Hg.

69. (a) Convert normal blood pressure readings of 120 over 80 mm Hg to newtons per meter squared using the relationship for pressure due to the weight of a fluid (\(P=hρg\)) rather than a conversion factor.

(b) Discuss why blood pressures for an infant could be smaller than those for an adult. Specifically, consider the smaller height to which blood must be pumped.

70. How tall must a water-filled manometer be to measure blood pressures as high as 300 mm Hg?

Solution 4.08 m

71. Pressure cookers have been around for more than 300 years, although their use has strongly declined in recent years (early models had a nasty habit of exploding). How much force must the latches holding the lid onto a pressure cooker be able to withstand if the circular lid is \(25.0 cm\) in diameter and the gauge pressure inside is 300 atm? Neglect the weight of the lid.

72. Suppose you measure a standing person’s blood pressure by placing the cuff on his leg 0.500 m below the heart. Calculate the pressure you would observe (in units of mm Hg) if the pressure at the heart were 120 over 80 mm Hg. Assume that there is no loss of pressure due to resistance in the circulatory system (a reasonable assumption, since major arteries are large).

Solution \(ΔP=38.7 mm Hg\), Leg blood pressure\(=\frac{159}{119}.\)

73. A submarine is stranded on the bottom of the ocean with its hatch 25.0 m below the surface. Calculate the force needed to open the hatch from the inside, given it is circular and 0.450 m in diameter. Air pressure inside the submarine is 1.00 atm.

74. Assuming bicycle tires are perfectly flexible and support the weight of bicycle and rider by pressure alone, calculate the total area of the tires in contact with the ground. The bicycle plus rider has a mass of 80.0 kg, and the gauge pressure in the tires is \(3.50×10^5Pa\).

Solution \(22.4cm^2\)

75. What fraction of ice is submerged when it floats in freshwater, given the density of water at 0°C is very close to \(1000 kg/m^3\)?

Solution \(91.7%\)

76. Logs sometimes float vertically in a lake because one end has become water-logged and denser than the other. What is the average density of a uniform-diameter log that floats with 20.0% of its length above water?

77. Find the density of a fluid in which a hydrometer having a density of \(0.750 g/mL\) floats with \(92.0%\) of its volume submerged.

Solution \(815 kg/m^3\)

78. If your body has a density of \(995 kg/m^3\), what fraction of you will be submerged when floating gently in:

(a) Freshwater?

(b) Salt water, which has a density of \(1027 kg/m^3\)?

79. Bird bones have air pockets in them to reduce their weight—this also gives them an average density significantly less than that of the bones of other animals. Suppose an ornithologist weighs a bird bone in air and in water and finds its mass is \(45.0 g\) and its apparent mass when submerged is \(3.60 g\) (the bone is watertight).

(a) What mass of water is displaced?

(b) What is the volume of the bone?

(c) What is its average density?

Solution (a) \(41.4 g\) (b) \(41.4cm^3\) (c) \(1.09 g/cm^3\)

80. A rock with a mass of 540 g in air is found to have an apparent mass of 342 g when submerged in water.

(b) What is the volume of the rock?

(c) What is its average density? Is this consistent with the value for granite?

81. Archimedes’ principle can be used to calculate the density of a fluid as well as that of a solid. Suppose a chunk of iron with a mass of 390.0 g in air is found to have an apparent mass of 350.5 g when completely submerged in an unknown liquid.

(a) What mass of fluid does the iron displace?

(b) What is the volume of iron, using its density as given in [link]

(c) Calculate the fluid’s density and identify it.

Solution (a) 39.5 g (b) \(50cm^3\) (c) \(0.79g/cm^3\) It is ethyl alcohol.

82. In an immersion measurement of a woman’s density, she is found to have a mass of 62.0 kg in air and an apparent mass of 0.0850 kg when completely submerged with lungs empty.

(a) What mass of water does she displace?

(b) What is her volume?

(c) Calculate her density.

(d) If her lung capacity is 1.75 L, is she able to float without treading water with her lungs filled with air?

83. Some fish have a density slightly less than that of water and must exert a force (swim) to stay submerged. What force must an 85.0-kg grouper exert to stay submerged in salt water if its body density is \(1015kg/m^3\)?

Solution 8.21 N

84. (a) Calculate the buoyant force on a 2.00-L helium balloon.

(b) Given the mass of the rubber in the balloon is 1.50 g, what is the net vertical force on the balloon if it is let go? You can neglect the volume of the rubber.

85. (a) What is the density of a woman who floats in freshwater with \(4.00%\) of her volume above the surface? This could be measured by placing her in a tank with marks on the side to measure how much water she displaces when floating and when held under water (briefly).

(b) What percent of her volume is above the surface when she floats in seawater?

Solution (a) \(960kg/m^3\) (b) \(6.34%\) She indeed floats more in seawater.

86. A certain man has a mass of 80 kg and a density of \(955kg/m^3\) (excluding the air in his lungs).

(a) Calculate his volume.

(b) Find the buoyant force air exerts on him.

(c) What is the ratio of the buoyant force to his weight?

87. A simple compass can be made by placing a small bar magnet on a cork floating in water.

(a) What fraction of a plain cork will be submerged when floating in water?

(b) If the cork has a mass of 10.0 g and a 20.0-g magnet is placed on it, what fraction of the cork will be submerged?

(c) Will the bar magnet and cork float in ethyl alcohol?

Solution (a) \(0.24\) (b) \(0.68\) (c) Yes, the cork will float because \(ρ_{obj}<ρ_{\text{ethyl alcohol}}(0.678g/cm^3<0.79g/cm^3)\)

88. What fraction of an iron anchor’s weight will be supported by buoyant force when submerged in saltwater?

89. Scurrilous con artists have been known to represent gold-plated tungsten ingots as pure gold and sell them to the greedy at prices much below gold value but deservedly far above the cost of tungsten. With what accuracy must you be able to measure the mass of such an ingot in and out of water to tell that it is almost pure tungsten rather than pure gold?

Solution The difference is 0.006%.

90 . A twin-sized air mattress used for camping has dimensions of 100 cm by 200 cm by 15 cm when blown up. The weight of the mattress is 2 kg. How heavy a person could the air mattress hold if it is placed in freshwater?

91. Referring to Figure, prove that the buoyant force on the cylinder is equal to the weight of the fluid displaced (Archimedes’ principle). You may assume that the buoyant force is \(F_2−F_1\) and that the ends of the cylinder have equal areas \(A\). Note that the volume of the cylinder (and that of the fluid it displaces) equals \((h_2−h_1)A\).

Solution \(F_{net}=F_2−F_1=P_2A−P_1A=(P_2−P_1)A\) \(=(h_2ρ_{fl}g−h_1ρ_{fl}g)A\) \(=(h_2−h_1)ρ_{fl}gA\)

where \(ρ_{fl}\) = density of fluid. Therefore,

\(F_{net}=(h_2−h_1)Aρ_{fl}g=V_{fl}ρ_{fl}g=m_{fl}g=w_{fl}\)

where is \(w_{fl}\) the weight of the fluid displaced.

92. (a) A 75.0-kg man floats in freshwater with \(3.00%\) of his volume above water when his lungs are empty, and \(5.00%\) of his volume above water when his lungs are full. Calculate the volume of air he inhales—called his lung capacity—in liters.

(b) Does this lung volume seem reasonable?

93. What is the pressure inside an alveolus having a radius of \(2.50×10^{−4}m\) if the surface tension of the fluid-lined wall is the same as for soapy water? You may assume the pressure is the same as that created by a spherical bubble.

Solution \(592N/m^2\)

94. (a) The pressure inside an alveolus with a \(2.00×10^{−4}\)-m radius is \(1.40×10^3Pa\), due to its fluid-lined walls. Assuming the alveolus acts like a spherical bubble, what is the surface tension of the fluid?

(b) Identify the likely fluid. (You may need to extrapolate between values in Table.)

95. What is the gauge pressure in millimeters of mercury inside a soap bubble 0.100 m in diameter?

Solution \(2.23×10^{−2}mm Hg\)

96. Calculate the force on the slide wire in Figure if it is 3.50 cm long and the fluid is ethyl alcohol.

97. Figure(a) shows the effect of tube radius on the height to which capillary action can raise a fluid.

(a) Calculate the height \(h\) for water in a glass tube with a radius of 0.900 cm—a rather large tube like the one on the left.

(b) What is the radius of the glass tube on the right if it raises water to 4.00 cm?

Solution (a) \(1.65×10^{−3}m\) (b) \(3.71×10^{–4}m\)

98. We stated in Example that a xylem tube is of radius \(2.50×10^{−5}m\). Verify that such a tube raises sap less than a meter by finding h for it, making the same assumptions that sap’s density is \(1050kg/m^3\), its contact angle is zero, and its surface tension is the same as that of water at \(20.0º C\).

99. What fluid is in the device shown in Figure if the force is \(3.16×10^{−3}N\) and the length of the wire is 2.50 cm? Calculate the surface tension \(γ\) and find a likely match from Table.

Solution \(6.32×10^{−2}N/m\) Based on the values in table, the fluid is probably glycerin.

100. If the gauge pressure inside a rubber balloon with a 10.0-cm radius is 1.50 cm of water, what is the effective surface tension of the balloon?

101. Calculate the gauge pressures inside 2.00-cm-radius bubbles of water, alcohol, and soapy water. Which liquid forms the most stable bubbles, neglecting any effects of evaporation?

Solution \(P_w=14.6N/m^2,\) \(P_a=4.46N/m^2,\) \(P_{sw}=7.40N/m^2.\) Alcohol forms the most stable bubble, since the absolute pressure inside is closest to atmospheric pressure.

102. Suppose water is raised by capillary action to a height of 5.00 cm in a glass tube.

(a) To what height will it be raised in a paraffin tube of the same radius?

(b) In a silver tube of the same radius?

103. Calculate the contact angle \(θ\) for olive oil if capillary action raises it to a height of 7.07 cm in a glass tube with a radius of 0.100 mm. Is this value consistent with that for most organic liquids?

Solution \(5.1º\) This is near the value of \(θ=0º\) for most organic liquids.

104. When two soap bubbles touch, the larger is inflated by the smaller until they form a single bubble.

(a) What is the gauge pressure inside a soap bubble with a 1.50-cm radius?

(b) Inside a 4.00-cm-radius soap bubble?

(c) Inside the single bubble they form if no air is lost when they touch?

105. Calculate the ratio of the heights to which water and mercury are raised by capillary action in the same glass tube.

Solution \(−2.78\) The ratio is negative because water is raised whereas mercury is lowered.

106. What is the ratio of heights to which ethyl alcohol and water are raised by capillary action in the same glass tube?

11.9: Pressures in the Body

107. During forced exhalation, such as when blowing up a balloon, the diaphragm and chest muscles create a pressure of 60.0 mm Hg between the lungs and chest wall. What force in newtons does this pressure create on the \(600cm^2\) surface area of the diaphragm?

Solurion 479 N

108. You can chew through very tough objects with your incisors because they exert a large force on the small area of a pointed tooth. What pressure in pascals can you create by exerting a force of \(500 N\) with your tooth on an area of \(1.00mm^2\)?

109. One way to force air into an unconscious person’s lungs is to squeeze on a balloon appropriately connected to the subject. What force must you exert on the balloon with your hands to create a gauge pressure of 4.00 cm water, assuming you squeeze on an effective area of \(50.0cm^2\)?

Solution 1.96 N

110. Heroes in movies hide beneath water and breathe through a hollow reed (villains never catch on to this trick). In practice, you cannot inhale in this manner if your lungs are more than 60.0 cm below the surface. What is the maximum negative gauge pressure you can create in your lungs on dry land, assuming you can achieve \(−3.00 cm\) water pressure with your lungs 60.0 cm below the surface?

Solution \(−63.0 cm\) \(H_2O\)

111. Gauge pressure in the fluid surrounding an infant’s brain may rise as high as 85.0 mm Hg (5 to 12 mm Hg is normal), creating an outward force large enough to make the skull grow abnormally large.

(a) Calculate this outward force in newtons on each side of an infant’s skull if the effective area of each side is \(70.0cm^2\).

(b) What is the net force acting on the skull?

112. A full-term fetus typically has a mass of 3.50 kg.

(a) What pressure does the weight of such a fetus create if it rests on the mother’s bladder, supported on an area of \(90.0cm^2\)?

(b) Convert this pressure to millimeters of mercury and determine if it alone is great enough to trigger the micturition reflex (it will add to any pressure already existing in the bladder).

Solution (a) \(3.81×10^3N/m^2\) (b) \(28.7 mm Hg\), which is sufficient to trigger micturition reflex

113. If the pressure in the esophagus is \(−2.00 mm Hg\) while that in the stomach is \(+20.0 mm Hg\), to what height could stomach fluid rise in the esophagus, assuming a density of 1.10 g/mL? (This movement will not occur if the muscle closing the lower end of the esophagus is working properly.)

114. Pressure in the spinal fluid is measured as shown in Figure. If the pressure in the spinal fluid is 10.0 mm Hg:

(a) What is the reading of the water manometer in cm water?

(b) What is the reading if the person sits up, placing the top of the fluid 60 cm above the tap? The fluid density is 1.05 g/mL.

Diagram of a person lying face-down on a table hooked up to a medical apparatus. A needle attached to a tube is inserted between the patient's vertebrae in the lower back area. The tube, which appears to be filled with fluid, is connected to an upright tube containing an unknown amount of water. The height of the water in the tube is labeled question-mark centimeters H 2 O. A label pointing to the patient's head reads P equals ten millimeters H g.

A water manometer used to measure pressure in the spinal fluid. The height of the fluid in the manometer is measured relative to the spinal column, and the manometer is open to the atmosphere. The measured pressure will be considerably greater if the person sits up.

Solution (a) 13.6 m water (b) 76.5 cm water

115. Calculate the maximum force in newtons exerted by the blood on an aneurysm, or ballooning, in a major artery, given the maximum blood pressure for this person is 150 mm Hg and the effective area of the aneurysm is \(20.0cm^2\). Note that this force is great enough to cause further enlargement and subsequently greater force on the ever-thinner vessel wall.

116. During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force.

(a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius?

(b) What deformation is produced if the disk is 0.800 cm thick and has a Young’s modulus of \(1.5×10^9N/m^2\) ?

Solution (a) \(3.98×10^6Pa\) (b) \(2.1×10^{−3}cm\)

117. When a person sits erect, increasing the vertical position of their brain by 36.0 cm, the heart must continue to pump blood to the brain at the same rate.

(a) What is the gain in gravitational potential energy for 100 mL of blood raised 36.0 cm?

(b) What is the drop in pressure, neglecting any losses due to friction?

(c) Discuss how the gain in gravitational potential energy and the decrease in pressure are related.

118. (a) How high will water rise in a glass capillary tube with a 0.500-mm radius?

(b) How much gravitational potential energy does the water gain?

(c) Discuss possible sources of this energy.

Solution (a) 2.97 cm (b) \(3.39×10^{−6}J\) (c) Work is done by the surface tension force through an effective distance \(h/2\) to raise the column of water.

119. A negative pressure of 25.0 atm can sometimes be achieved with the device in Figure before the water separates.

(a) To what height could such a negative gauge pressure raise water?

(b) How much would a steel wire of the same diameter and length as this capillary stretch if suspended from above?

When the piston is raised the liquid separates and results in negative pressure.

(a) When the piston is raised, it stretches the liquid slightly, putting it under tension and creating a negative absolute pressure \(P=−F/A\) (b) The liquid eventually separates, giving an experimental limit to negative pressure in this liquid.

120. Suppose you hit a steel nail with a 0.500-kg hammer, initially moving at \(15.0 m/s\) size 12{"15" "." 0`"m/s"} {} and brought to rest in 2.80 mm. (a) What average force is exerted on the nail? (b) How much is the nail compressed if it is 2.50 mm in diameter and 6.00-cm long? (c) What pressure is created on the 1.00-mm-diameter tip of the nail?

Solution (a) \(2.01×10^4N\) (b) \(1.17×10^{−3}m\) (c) \(2.56×10^{10}N/m^2\)

121. Calculate the pressure due to the ocean at the bottom of the Marianas Trench near the Philippines, given its depth is \(11.0 km\) and assuming the density of sea water is constant all the way down.

(b) Calculate the percent decrease in volume of sea water due to such a pressure, assuming its bulk modulus is the same as water and is constant.

(c) What would be the percent increase in its density? Is the assumption of constant density valid? Will the actual pressure be greater or smaller than that calculated under this assumption?

122. The hydraulic system of a backhoe is used to lift a load as shown in Figure.

(a) Calculate the force \(F\) the slave cylinder must exert to support the 400-kg load and the 150-kg brace and shovel.

(b) What is the pressure in the hydraulic fluid if the slave cylinder is 2.50 cm in diameter?

(c) What force would you have to exert on a lever with a mechanical advantage of 5.00 acting on a master cylinder 0.800 cm in diameter to create this pressure?

Diagram of the arm and shovel of a backhoe lifting a load of dirt. The weight of the arm, w sub arm, is depicted as a vector extending vertically downward from the arm one point one zero meters from the top of the arm; w sub arm forms a thirty degree angle with the arm of the shovel. The weight of the load, w sub load, is depicted as a vector extending downward from the middle of the shovel one point seven zero meters from the top of the arm. Force F is a vector pushing the arm of the shovel zero point three zero meters from the top of the arm and perpendicular to the arm.

Hydraulic and mechanical lever systems are used in heavy machinery such as this back hoe.

Solution (a) \(1.38×10^4N\) (b) \(2.81×10^7N/m^2\) (c) 283 N

123. Some miners wish to remove water from a mine shaft. A pipe is lowered to the water 90 m below, and a negative pressure is applied to raise the water.

(a) Calculate the pressure needed to raise the water.

(b) What is unreasonable about this pressure?

(c) What is unreasonable about the premise?

124. You are pumping up a bicycle tire with a hand pump, the piston of which has a 2.00-cm radius.

(a) What force in newtons must you exert to create a pressure of \(6.90×10^5Pa\)

(b) What is unreasonable about this (a) result?

(c) Which premises are unreasonable or inconsistent?

Solution (a) 867 N (b) This is too much force to exert with a hand pump. (c) The assumed radius of the pump is too large; it would be nearly two inches in diameter—too large for a pump or even a master cylinder. The pressure is reasonable for bicycle tires.

125. Consider a group of people trying to stay afloat after their boat strikes a log in a lake. Construct a problem in which you calculate the number of people that can cling to the log and keep their heads out of the water. Among the variables to be considered are the size and density of the log, and what is needed to keep a person’s head and arms above water without swimming or treading water.

126. The alveoli in emphysema victims are damaged and effectively form larger sacs. Construct a problem in which you calculate the loss of pressure due to surface tension in the alveoli because of their larger average diameters. (Part of the lung’s ability to expel air results from pressure created by surface tension in the alveoli.) Among the things to consider are the normal surface tension of the fluid lining the alveoli, the average alveolar radius in normal individuals and its average in emphysema sufferers.

Contributors and Attributions

Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a  Creative Commons Attribution License (by 4.0) .

Gurumuda Networks

Fluid statics – problems and solutions

Liquid pressure

1. What is the d ifference between the hydrostatic pressure of blood betwee n the brain and the sole s of the feet of a person whose height 165 cm (suppose the density of blood = 1.0 × 10 3 kg/m 3 , acceleration due to gravity = 10 m/s 2 )

Height (h) = 165 cm = 165/100 m = 1.65 meters

Density of bloods (ρ) = 1.0 × 10 3 kg/m 3

Acceleration due to gravity (g) = 10 m/s 2

Wanted: liquid pressure

P = (1.0 × 10 3 )(10)(1.65)

P = (1.0 × 10 4 )(1.65)

P = 1.65 x 10 4 N/m 2

2. A U pipe is initially filled with water than on one pipe filled with oil, as shown in the figure below. The density of water is 1000 kg/m 3 . If the height of oil is 8 cm and the height of the water is 5 cm, what is the density of oil?

Fluid statics – problems and solutions 1

Density of water = 1000 kg.m -3

The height of water (h 2 ) = 5 cm

The height of oil (h 1 ) = 8 cm

Wanted : density of oil

ρ 1 g h 1 =ρ 2 g h 2

ρ 1 h 1 =ρ 2 h 2

(1000)(5) = (ρ 2 )(8)

5000 = (ρ 2 )(8)

ρ 2 = 625 kg.m -3

3. A U pipe was first filled with kerosene then added water. If the mass of kerosene is 0.8 grams/cm 3 and the density of water is 1 gram/cm 3 and the cross sectional area is 1.25 cm 2 . Determine how much water should be added so that the height difference of the kerosene surface is 15 cm

Fluid statics – problems and solutions 11

Density of kerosene (ρ 1 ) = 0.8 gram/cm 3

Density of water (ρ 2 ) = 1 gram/cm 3

Sectional area of the pip e = 1.25 cm 2

The height difference of the surface of kerosene (h 1 ) = 15 cm

Wanted : Volume of water

The height of water (h 2 ) :

ρ 1 g h 1 = ρ 2 g h 2

(0,8)(15)(1)(h 2 )

h 2 = 12 cm

Volume of water :

V = ( Sectional area of the pip e )(height of water)

V = (1.25 cm 2 )(12 cm)

V = 15 cm 3

1 liter = 1 dm 3 = 10 3 cm 3

1 mililiter = 10 -3 liter s = (10 -3 )(10 3 ) cm 3 = 1 cm 3

Volume of water is 15 cm 3 = 15 mililiters

The correct answer is C.

4. A pipe U filled with water with density of 1000 kg/m 3 . One column of pipe U filled with glyserin with density of 1200 kg/m 3 . If the height of glyserin is 4 cm, determine the height difference of both columns of the pipe.

Density of water (ρ 1 ) = 1000 kg/m 3

Density of glycerin (ρ 2 ) = 1200 kg/m 3

Height of glycerin (h 2 ) = 4 cm

Wanted: The height difference of both columns of the pipe.

The height of the column of the pipe (h 1 ) :

ρ 1 h 1 = ρ 2 h 2

(1000)(h 1 ) = (1200)(4)

(1000)(h 1 ) = 4800

h 1 = 4.8 cm

The height difference of both columns of the pipe U = h 1 – h 2 = 4.8 cm – 4 cm = 0.8 cm

The correct answer is A.

5. A pipe U has two ends are open filled with water with a mass of 1 g/cm 3 . The sectional area along the pipe is the same, that is 1 cm 2 . Someone blows on one end of the foot of the pipe so that the surface of the water at the other foot rises 10 cm from its original position. If the acceleration due to gravity i s 10 m/s2 then determine the force acted by that person.

A. 20 kilodyne

B. 10 kilodyne

C. 2 kilodyne

D. 1 kilodyne

Change all units to the International system.

Density of water (ρ 1 ) = 1 gr/cm 3 = 10 -3 kg / 10 -6 m 3 = 10 3 kg/m 3

Cross sectional area of pipe (A) = 1 cm 2 = 10 -4 m 2

The change of column of pipe (h) = 10 cm = 1 dm = 10 -1 m

Acceleration due to gravity (g) = 10 m.s -2 = 10 1 m.s -2

Volume of moved water (V) = (A)(h) = (1 cm 2 )(10 cm) = 10 cm 3 = (10 1 )(10 -6 m 3 ) = 10 -5 m 3

Wanted : Force (F) acted by the person.

The force that acted by that person = weight of water with a height of 10 cm

F = m g —–> Equation of density : m = ρ V

F = (10 3 )(10 -5 )(10 1 )

F = (10 4 )(10 -5 )

F = 10 -1 Newton —–> 1 Newton = 10 5 dyne

F = (10 -1 )(10 5 dyne)

F = 10 4 dyne

F = 10 kilodyne

The correct answer is B.

6. A Y-shaped tube is inserted upside down so that the left foot and right foot are immersed in two kinds of liquid. After both feet are immersed in the liquid, then the top of the Y pipe is closed with the finger and pulled upwards, so that the two legs of the Y pipe are filled with a column of different high-density liquids. If the density of the first liquid is 0.80 gram.cm -3 and the second density is 0.75 gram.cm -3 , and the lower liquid column is 8 cm, then determine the height difference between the two liquid columns on U pip e.

Fluid statics – problems and solutions 12

B. 0.9375 cm

C. 0.3533 cm

D. 0.5333 cm

Density of first liquid (ρ 1 ) = 0,80 gram.cm -3

Density of second liquid (ρ 2 ) = 0,75 gram.cm -3

The height of the lower liquid (h 1 ) = 8 cm

Wanted : T he height difference between the two liquid columns on U pip e

T he height of the higher liquids (h 2 ) :

(0.80)(8) = (0.75)(h 2 )

6.4 = 0.75 (h 2 )

h 2 = 6.4 / 0.75

h 2 = 8.5 cm

The height difference of liquids = h 2 – h 1 = 8.5333 cm – 8 cm = 0.5333 cm

The correct answer is D.

Buoyant force

7. A stone with the volume of 0.5 m 3 placed in a liquid with the density of 1.5 gr cm –3 . Acceleration due to gravity is 10 m s -2 . What is the buoyant force?

Volume of stone (V) = 0.5 m 3

Density of water (ρ) = 1.5 gr cm –3 = 1500 kg m -3

Acceleration due to gravity (g) = 10 m s -2

Wanted: buoyant force (F A )

The equation of the buoyant force :

F A = ρ g V = (1500 kg m -3 )(10 m s -2 )(0.5 m 3 ) = 7500 kg m/s 2 = 7500 Newton

8. A block of ice float in the sea as shown in the figure below. The density of sea is 1.2 gr cm –3 and density of ice is 0.9 gr c –3 . The volume of ice in sea water = ……. x the volume of ice in the air.

Fluid statics – problems and solutions 2

Density of sea (ρ sea ) = 1.2 gr cm –3

Density of ice (ρ ice ) = 0.9 gr c –3

Wanted: The volume of ice in sea water = ……. x the volume of ice in the air.

Fluid statics – problems and solutions 3

The volume of ice in sea = 0.75

The volume of ice in air = 0.25

The volume of ice in sea water = 3 x the volume of ice in air (3 x 0.25 = 0.75).

9. An object float in a liquid where 2/3 of the object in the liquid. If the density of the object is 0.6 gr cm 3 , then what is the density of water.

The part of the object in liquid = 2/3

Density of object = 0.6 gr cm 3 = 600 kg m 3

Wanted: the density of the liquid (x)

Fluid statics – problems and solutions 4

The density of the liquid is 900 kg m 3

10. A wood float in water, where 3/5 part of wood in the water. If the density of water is 1 × 10 3 kg/m 3 , what is the density of wood?

Part of object in water = 3/5

Density of water = 1×10 3 kg/m 3 = 1000 kg/m 3

Wanted : The density of wood (x)

Fluid statics – problems and solutions 5

The density of wood is 600 kg/m 3 = 6 x 10 2 kg/m 3

  • Answer: Fluid statics, also known as hydrostatics, is the branch of fluid mechanics that studies fluids at rest and the forces exerted by static fluids on immersed objects and container walls.
  • Answer: In a static fluid, pressure increases linearly with depth due to the weight of the fluid column above any given depth. The change in pressure with depth is given by Δ P = ρ g h , where ρ is the fluid density, g is the gravitational acceleration, and ℎ h is the depth.
  • Answer: Pascal’s principle states that a change in pressure applied to an enclosed fluid is transmitted undiminished to all portions of the fluid and to the walls of its container.
  • Answer: A hydraulic lift utilizes Pascal’s principle. When a small force is applied to a small piston, it creates a pressure in the fluid. This pressure is transmitted undiminished throughout the fluid, exerting a much larger force on a larger piston, enabling the lift to raise heavy objects with relatively little effort.
  • Answer: The buoyant force is the upward force exerted by a fluid on any immersed object. According to Archimedes’ principle, the buoyant force on an object is equal to the weight of the fluid displaced by the object.
  • Answer: Whether an object floats or sinks depends on the relationship between the buoyant force and the object’s weight. If the buoyant force (due to the displaced fluid) is greater than the object’s weight, it will float. If the object’s weight is greater, it will sink.
  • Answer: Hydrostatic pressure is the pressure exerted by a fluid at rest due to the force of gravity. It increases linearly with depth in the fluid, and is calculated as P = P 0 ​ + ρ g h , where P 0 ​ is the pressure at the surface, ρ is the fluid density, g is the gravitational acceleration, and ℎ h is the depth.
  • Answer: The atmosphere can be thought of as a fluid. Atmospheric pressure is the pressure exerted by the weight of the air above a given point. It decreases with altitude, similar to how pressure in a liquid decreases as one moves upward in the fluid column.
  • Answer: In fluid statics, the pressure at a given depth depends only on the height of the fluid column above that depth, not on the shape of the container. Thus, pressure at a specific depth is the same regardless of the container’s shape.
  • What is the significance of the hydrostatic paradox?
  • Answer: The hydrostatic paradox highlights that in fluid statics, the force exerted by a static fluid on the bottom of a container depends only on the height of the fluid column, not its volume or the shape of the container. Thus, very different containers with the same fluid height exert the same pressure at their base, even if they hold different amounts of fluid.

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practice problem 1

  • A forklift driver decides to push it without lifting it. What force must be applied to just get the pallet moving?
  • After a bit of time, the pallet begins to slide. How fast is the pallet moving after sliding under the same force you calculated in part a. for half a second?
  • If the forklift stops pushing, how far does the pallet slide before coming to a stop?

Four forces are acting on the pallet: the downward pull of Earth's gravity, the normal force of the floor pushing up, the forward push of the forklift, and the backward resistance of friction. Weight and normal are equal throughout this example since the floor is level. Friction changes from static to kinetic — static friction initially since the pallet isn't moving initially, then kinetic friction once the pallet gets going. The push also changes from nothing to the value needed to get the pallet moving, then back to nothing after 0.5 seconds of motion.

To get the pallet started, the driver must push it with a force equal to the maximum static friction.

Once the pallet starts moving, the coefficient of friction drops from its static value to its kinetic value.

But the forklift is still pushing with 1,650 N of force. Thus we have a nonzero net force.

A net force causes acceleration.

Acceleration goes with a change in velocity.

Once the forklift stops pushing, kinetic friction becomes the net force. This net force will cause an acceleration opposite the direction of motion. When one vector is opposite another, one of the two needs to be negative. The convenient thing to do for this problem is to let friction be the negative one.

Pick the appropriate equation of motion

v 2  =  v 0 2  + 2 a ∆ s

Eliminate the zero term (final velocity), solve for distance, substitute, and calculate. Watch how the negative signs disappear. This has to happen. An object moving forward should be displaced forward.

practice problem 2

  • Determine the car's maximum starting acceleration with and without "burning rubber". How do these two methods of starting a car compare?
  • Determine the car's minimum braking distance with normal brakes and antilock brakes as a function of initial speed. How do these two methods of stopping a car compare?

The net external force propelling a car comes from the friction force between tires and pavement. When a driver starts a car by "flooring it" (pressing the accelerator to the floor) the tires grind on the road producing a smoke of burning rubber and pavement. Since the tires are slipping, the coefficient of kinetic friction determines the maximum acceleration. Under normal circumstances, however, most drivers are not willing to subject their tires to such extreme punishment. Typical car tires rotate over the surface of the road without slipping, thus the coefficient of static friction determines a car's maximum acceleration in most situations.

To solve this problem, set the frictional force on level ground equal to the net force of the second law of motion.

Contrary to popular belief, flooring the accelerator is not an effective method of starting a car. Burning rubber is only about 90% as effective as accelerating a car normally from rest.

The net external force stopping a car comes from the friction force between tires and pavement. Stopping a car with ordinary brakes may result in wheel lock; that is, the wheels lock in position and are not able to rotate. When this happens, the tires skid and the coefficient of kinetic friction determines the braking distance. Cars equipped with an antilock braking system (ABS) have a sensor that releases the brake pads the instant the wheel locks up. After a brief pause the brakes are then quickly re-engaged. If they don't lock up again, all is well. If they do, the ABS releases the brake pads again. This processes can repeat many times a second. In any case, the tires are not allowed to lock for more than a few milliseconds. The car is then stopped using the force of static friction alone.

To solve this problem, determine acceleration using the displacement-velocity formula of kinematics. Set this equation equal to the formula for acceleration due to friction derived above.

v 0 2  = 2 a Δ s  = 2μ g Δ s

Antilock brakes need 90% of the distance of regular brakes to stop a car traveling at the same speed. This decrease in distance is certainly significant, but doesn't really seem all that great given the high cost of an ABS. In addition to reduced braking distance, however, antilock braking systems also increase performance during extreme braking. Locked brakes are useless for steering. ABS ensures that the wheels retain their static frictional grip on the road, which allows for maneuvering while braking in an emergency.

practice problem 3

Practice problem 4.

Start with Newton's second law of motion.

∑ F  =  m a

A stopping car is acted upon by three forces: weight pointing down, normal pointing up (we'll have to assume the test track is level), and (as long as the wheels don't lock and the car doesn't skid) static friction. Of course, there's also aerodynamic drag, but worrying about that force in this problem would be a waste of time. Weight and normal cancel out since the car is neither accelerating up nor down. Static friction is therefore the net force acting on a braking car.

f s  =  ma

Replace f s with its classical formula.

μ s N  =  ma

Earlier, we assumed (quite sensibly) that the test track would be level, which means that normal equals weight ( W  =  mg ).

μ s mg  =  ma

Work the magic of algebra and solve for the goal of this problem — the coefficient of friction.

Great, but what is a ? Go back to the good old days when you learned the equations of motion. Pick the one that doesn't involve time and solve it for acceleration.

Substitute this expression into the previous one.

Take all the numbers in road-test-summary.txt and run them through this final equation. These results are given in road-test-summary-solution.txt . (Note: I used g  = 9.8 m/s 2 , but one could also use the value of standard gravity g  =  9.80665 m/s 2 .) Using the mean of these 246 trials as the value and the standard deviation as the uncertainty yields the following answer.

μ s  =  0.91 ± 0.10

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  5. Statics: Practice Exam #1

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  1. Statics

    For a static crate on an incline, the static friction force equals the parallel component of the crate's weight. fs = W∥. fs = 965 N. 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 ...

  2. 9: Statics and Torque (Exercises)

    Find the direction and magnitude of the force exerted by the kneecap on the upper leg bone (the femur). The knee joint works like a hinge to bend and straighten the lower leg. It permits a person to sit, stand, and pivot. Solution. 1.1 ×103N 1.1 × 10 3 N θ = 190º θ = 190 º ccw from positive x axis. 48.

  3. Statics

    Determine the normal force of the ground on the lawn mower. A child is pushing a 9.5 kg wagon up a 10° ramp at a constant velocity. His arms are parallel to the ramp as he pushes. The effective coefficient of kinetic friction of the bearings against the axle is 0.21. Draw a free body diagram of the wagon.

  4. Statics Problems

    The required equations and background reading to solve these problems is given on the equilibrium page. Problem # 1 A ball of mass 10 kg is hanging vertically from a string. What is the tension in the string? (Answer: 98 N) Problem # 2 Two boxes are stacked on the floor, one on top of the other. The top box has a mass of 10 kg and the bottom ...

  5. Equilibrium and Statics

    Equilibrium and Statics. When all the forces that act upon an object are balanced, then the object is said to be in a state of equilibrium. The forces are considered to be balanced if the rightward forces are balanced by the leftward forces and the upward forces are balanced by the downward forces. This however does not necessarily mean that ...

  6. 9.4: Applications of Statics, Including Problem ...

    The pole is uniform and has a mass of 5.00 kg. In Figure 9.4.1 9.4. 1, 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 (netF = 0) ( n e t F = 0).

  7. 8.4: Solving Statics Problems

    Formulate and apply six steps to solve static problems. Statics is the study of forces in equilibrium. Recall that Newton's second law states: ∑ F = ma (8.4.1) (8.4.1) ∑ F = m a. Therefore, for all objects moving at constant velocity (including a velocity of 0 — stationary objects), the net external force is zero.

  8. Rotational Statics

    When the bridge is closed the moveable span is balanced so that there is no normal force on the far end. Answer the first two parts by stating the equilibrium conditions when the span is down. left axis. ∑τ counterclockwise =. ∑τ clockwise. (1 L ) ( Wc ) (sin 90°) =. (√2 L ) ( T1 ) (sin 90°) 1 Wc =. √2 T1.

  9. 9.4 Applications of Statics, Including Problem-Solving Strategies

    Learning Objectives. 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.

  10. Statics problems and solutions

    When solving a Statics problem, follow these steps: Read carefully the problem statement. Draw a picture of the physical situation described in the problem. Write in your notebook the givens in the problem statement. Identify the elements that make up the system that should be in static equilibrium: if it is a single body or if on the contrary ...

  11. Static and kinetic friction (practice)

    Static and kinetic friction. Google Classroom. You might need: Calculator. A 7.0 kg box is at rest on a table. The static friction coefficient μ s between the box and table is 0.40 , and the kinetic friction coefficient μ k is 0.10 . Then, a 30 N horizontal force is applied to the box.

  12. 12.2 Examples of Static Equilibrium

    All examples in this chapter are planar problems. Accordingly, we use equilibrium conditions in the component form of Equation 12.7 to Equation 12.9.We introduced a problem-solving strategy in Example 12.1 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.

  13. Static Equilibrium

    This physics video tutorial explains the concept of static equilibrium - translational & rotational equilibrium where everything is at rest and there's no mo...

  14. Statics

    It's a page about solving a particular (and common) kind of problem in mechanics. Informally, statics is the study of forces without motion. More formally, statics is the branch of mechanics that deals with forces in the absence of changes in motion. In contrast, dynamics is the study of forces and motion; or more formally, the branch of ...

  15. Physics library

    Physics library 19 units · 12 skills. Unit 1 One-dimensional motion. Unit 2 Two-dimensional motion. Unit 3 Forces and Newton's laws of motion. Unit 4 Centripetal force and gravitation. Unit 5 Work and energy. Unit 6 Impacts and linear momentum. Unit 7 Torque and angular momentum.

  16. Engineering Statics: Open and Interactive

    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.

  17. 11: Fluid Statics (Exercises)

    The fluid density is 1.05 g/mL. A water manometer used to measure pressure in the spinal fluid. The height of the fluid in the manometer is measured relative to the spinal column, and the manometer is open to the atmosphere. The measured pressure will be considerably greater if the person sits up.

  18. PDF AP Physics Practice Test: Static Equilibrium, Gravitation ...

    48. Draw a free-body diagram of the horizontal beam. cm m. 2m. 3m. A weight of mass 2m is now firmly attached to the left end of the beam, and a mass of 3m attached to the right end of the beam. Calculate the tension T in the support line. Calculate the x and y components of force acting on the pivot point.

  19. Rotational Statics

    The Physics Teacher has published several articles containing free body diagram worksheets. They are available free to members of the American Association of Physics Teachers (AAPT). Everyone else has to pay. Free-body diagrams revisited—II. James E. Court. The Physics Teacher. Vol 37 No. 8 (1999): 490-495. RE1-RE16: Rotational Equilibrium.

  20. Statics practice problems

    100. 0. Okay straight from my physics B statics test: 1. A uniform 20.0 kg, 10.0m long beam is supported by two posts. Post A is 1.00m from an end of the beam and post B is 4.00m from the other end. A 15.00 kg cat sits on the beam directly above post B. a)Calculate the force exerted by post A on the beam.

  21. Fluid statics

    Solution : The volume of ice in sea = 0.75. The volume of ice in air = 0.25. The volume of ice in sea water = 3 x the volume of ice in air (3 x 0.25 = 0.75). 9. An object float in a liquid where 2/3 of the object in the liquid. If the density of the object is 0.6 gr cm3, then what is the density of water.

  22. Dynamics

    practice problem 1. A person stands in an elevator weighing a cheeseburger with a kitchen scale. (It could happen.) The mass of the cheeseburger is 0.150 kg. The scale reads 1.14 N. Draw a free body diagram showing all the forces acting on the cheeseburger. Determine the weight of the cheeseburger. Determine the magnitude and direction of the ...

  23. Friction

    When one vector is opposite another, one of the two needs to be negative. The convenient thing to do for this problem is to let friction be the negative one. a = ∑ F / m = fk / m. a = (−1,000 N)/ (600 kg) a = −1.67 m/s2. Pick the appropriate equation of motion. v2 = v02 + 2 a ∆ s.