Chapter 4 FORCES AND NEWTON'S LAWS

Physics Including Human Applications

Chapter 4 Forces and Newton’s Laws

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Chapter 4 FORCES AND NEWTON'S LAWS

GOALS When you have mastered the concepts of this chapter, you will be able to achieve the following goals. Definitions Define each of the following terms, and use each term in an operational definition: force coefficient of friction weight

centripetal force frictional force weightlessness

Newton's Laws State Newton's laws of motion and gravitation. Resolution of Forces Given the force acting on a system, draw a force diagram and/or resolve forces into their components and/or solve for an unknown force. Newton's Second Law Problems Given any two of the three variables in Newton's second law, solve for the third. Centripetal Force Problems Given any three of the four variables in the centripetal force equation, calculate the value of the fourth variable.

PREREQUISITES Before you begin this chapter you should have achieved the goals of Chapter 3, Kinematics, including uniformly accelerated motion and uniform circular motion.



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Chapter 4 FORCES AND NEWTON'S LAWS

4.1 Introduction You have experienced many examples of forces. You are familiar with a muscular exertion producing a push or a pull, which are forces. There are many other examples such as: one solid object acting on another through contact; the gravitational force of the earth on an object; the attraction of a magnet for an iron object; the attraction between a plastic rod that has been rubbed with cat's fur and bits of paper; and the force of a stretched spring. One effect of a force is to alter the state of motion of a body. In this chapter you will study forces. Acceleration, uniformly accelerated motion, projectile motion, and the other kinds of motion require force. If you are to determine these motions, then you must first determine the various forces acting on a body. The study of forces and the associated changes in motion is called dynamics. Of particular interest are the conditions leading to equilibrium. 4.2 Newton's First and Third Laws of Motion In this chapter we will introduce you to the mental constructs devised by Sir Isaac Newton in the seventeenth century. Newton's model seeks to explain how forces acting on an object are related to the motion of an object. This model, sometimes called newtonian mechanics, does not give the correct results in all cases. For example, when objects travel at speeds near the speed of light (3.00 x 108 m/sec) newtonian mechanics is augmented by Einstein's special theory of relativity. In the realm of distances the order of atomic radius (10-10 m) quantum theory must be used to give the proper picture of physical phenomena. Nevertheless, in the world of everyday experiences, Newton's mechanics provides us with a good model for understanding forces and motion. Consider the following experiences you may have had: 1. If you are riding along in an automobile and the brakes are suddenly applied, you

experience a tendency to bang into the windshield. 2. If you are riding a galloping horse and he suddenly stops, you may find yourself on the

ground in front of him. Why? 3. Your book is resting on a sheet of paper on the table. You go away for several hours

and when you return your book is still resting on a sheet of paper on the table. Why? Newton's first law of motion may provide you with some understanding of such experiences.

Every body persists in its state of rest or of uniform linear motion unless it is acted upon by a

unbalanced force. The first law is sometimes spoken of as the inertial law. Recall the concept of inertia introduced in Chapter 2- that is, any object has a tendency to resist change. In mechanics this property of an object to resist a change in motion is called its inertia and is measured by the mass of the object. Newton's first law helps to unify your experiences with objects at rest not suddenly



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moving and with moving objects having difficulty in stopping suddenly. Nature seems to prefer to have things change slowly. The measure of the slowness of change in the motion of an object is its mass. Have you ever faced off to wrestle another person? You grab each others hands and push as hard as you can and for just an instant you are locked in a motionless struggle. You are pushing against your opponent. Your opponent is pushing against you with an equal force. According to Newton's third law that is exactly the same as if you would push against a brick wall (Figure 4.1). Your push against the wall is balanced by the wall's push against you, just as your push was counteracted by that of your wrestling opponent.

Consider the case of your physics book resting on your study desk. The physics book acts on the table with a downward force and the study desk acts on the book with a force in the upward direction. We then have a pair of forces that are equal in magnitude and oppositely directed. One is the book acting on the desk, and the other is the desk acting on the book. One force is called the action force, and the other is called the reaction force. They form an action-reaction pair of forces. These are experiences unified by Newton's third law of motion:

For every action there is an equal and an opposite reaction. Suppose you tie a rope around a tree, and you pull on the rope. You exert a force on the rope and the rope exerts a force on you. This is an action-reaction pair. At the other end of the rope we have another action-reaction pair as the rope acts on the tree and the tree acts on the rope. If one has two interacting bodies, A and B, the action-reaction pair can be described by A acting on B and B acting on A. Consider a tug of war between sides A and B, as shown in Figure 4.2. The rope is acting to the left on group A, RA, and group A is acting to the right on the rope, AR. Group A is acting to the left on the earth, and the earth is acting to the right on the group A. These are two examples of action and reaction. What are the action-reaction pairs for group B? Now suppose group B suddenly wraps their end of the rope around a tree, and group A continues to exert the same force as before. Has the tension in the rope changed? How many pairs of action and reaction forces can you identify around you?

In all of the examples note that each force has magnitude and direction, and thus force is a vector quantity and is so treated. The action-reaction pair of forces are vectors of equal magnitude but opposite directions. A vector representation of the action- reaction forces for the tug of war in Figure 4.2 is given as: FAB = - FBA. The rules of vector addition



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and subtraction can be applied to a force system. Some methods and examples of vector addition were given in Chapter 3. In accordance with the definition of equilibrium, an object at rest experiences no net force.

The vector sum of all forces acting on an object in mechanical equilibrium is zero. Another way of defining equilibrium is that the vector sum of the force components in any direction is equal to zero. Let us look at a particular case. The leg in Figure 4.3 is held in traction by the cord OA and the weight W. Consider the forces acting on the foot at point O. We will call the tension in the cord OA, TA; the tension in the leg, TB, and the tension in the cord OC, TC. The tension TC is equal to the weight, say 10.0 newtons (N), where newtons are the SI units of force. The first step in solving any mechanical equilibrium problem is to draw a force diagram like the one in Figure 4.3b. A force diagram is an idealization of the physical problem. All the forces acting on the system are drawn. Each force is represented by a vector. The length of the arrow is proportional to the magnitude of the force. The direction of the arrow represents the direction in which the force is acting. In many cases, drawing a correct force diagram will essentially solve a force problem for you. The best way to learn how to draw force diagrams is to practice. For each problem and example in this chapter, draw your own force diagram, and then compare it with ours.

We can add these vector forces graphically (see Section 3.3) as shown in Figure 4.4. The fact that the three vectors form a closed polygon means the vector sum of the forces is 0. Can you explain why? What are the magnitudes of TB and TA? Let us use the component method to answer this question. First find all the components of forces acting in the horizontal direction:

horizontal components = TA cos 53o- TB = 0

Then sum the components in the vertical direction: vertical components = TA sin 53o - W = 0Substituting values for the sine and cosine of 53 o, 0.600 TA = TB, and 0.800 TA = 10.0. Solving these two equations gives us the values for the magnitudes of TA and TB,



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TA = 12.5 N, and TB = 7.50 N. The directions are shown in Figure 4.3, TB acting horizontally to the left and TA acting to the right and upward 53 o above horizontal. EXAMPLES 1. Given the tension on the teeth bands as 1.00 N (a reasonable tension as applied by an

orthodontist), find the net force, F, that tends to realign the out-of-line tooth as shown in Figure 4.5a. First draw the diagram of the forces acting at the contact between the tooth and the teeth bands. Represent each force by an arrow of about the proper length and pointing in what you assume is the proper direction. See Figure 4.5b. Once you have completed the force diagram you can use the method of vector addition as described in Section 3.3 to form the closed polygon that results when all the vectors add to zero (Figure 4.5c) .

Notice that this problem is a bit tricky. What is asked for is not the force the tooth

exerts on the teeth bands, not the force which keeps the system in equilibrium, but rather the force that tends to realign the tooth. However, from Newton's third law we know that the force of the tooth against the bands and of the bands against the tooth are an action-reaction pair, equal in magnitude, but opposite in direction.

From Figure 4.5c we can see what equations we need to find the numerical value for the magnitude of F. The closed polygon formed by T1, T2 and - F is an isoceles triangle with F the length of its base and T the length of the equal sides. Therefore, in terms of the magnitude of - F and T

F = 2T sin 30o; F = 2(1.00 N)(0.500) = 1 N

So the force that tends to realign the tooth has a magnitude of 1 N and acts inwardly at an angle of 60 o from the teeth bands.

2. Consider the case of a 400 N youth hanging on a rope so that his arms make an angle of 80 o above the horizontal (Figure 4.6).

The rope is horizontal between his hands and makes an angle of 30o above the horizontal beyond his hands. What are the forces exerted by the youth's arms and by the tension in the rope?

First let us draw a force diagram. We know the weight of the youth is a vector pointing vertically downward with a length proportional to 400 N. The only other



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forces acting on the youth are the forces exerted by his right arm FR and by his left arm FL. These two forces are of equal size, as is implied by the statement of the problem, and act at an angle of 80o to the horizontal. The force diagram for the situation is shown in Figure 4.6. Once again use the method of vector addition to draw the closed polygon shown in Figure 4.6c. Now we can use algebraic and trigonometric procedures to solve for the magnitudes of the forces, FR and FL. If we consider just the sum of the horizontal forces, we obtain the equation, FR cos 80o = FL cos 80o, since the 400 N weight of the youth has no horizontal component it does not appear in this equation. The result of the horizontal component equation is that the two forces, FR and FL, are of equal magnitude, FR = FL. From the sum of the vertical components, we obtain

400 N = FR sin 80o + FL sin 80o = 2 FR sin 80o Substituting the value of sin 80o, we obtain

203 N = FR = FL To complete this problem we can draw the force diagram for the forces acting at point B, where the hand of the youth is holding tightly to the rope. Remembering Newton's third law, we know that the force in his arm, which we showed pulling upward to the right in Figure 4.6b, will be pulling downward and to the left on the rope. See Figure 4.6d. From this figure we can use algebraic-trigonometric methods to find the magnitudes of the tensions TBC and TAB. Since TAB is acting only in the horizontal direction, the vertical upward component of TBC must just be balanced by the downward pull of the youth's arm: TBC sin 30o = FL sin 80o = 230 N sin 80o

TBC = 400 N Finally, from the equation for the horizontal components of the forces, we can obtain the value for the magnitude of TAB.

TAB + FL sin 80o = TBC cos 30o TAB + 203 N cos 80o = 400 N cos 30o

TAB = 346 N - 176 N TAB = 170 N

3. Let us examine two different exercise systems used to develop and strengthen muscles in the leg of a patient. First consider the system in which an 80.0 N weight is placed on the seated patient's foot. The patient then straightens his leg to a horizontal position (see Figure 4.7). What is the component of the force that acts perpendicular to the patient's leg so as to oppose the extension of the leg? We will calculate the magnitude of this component for the situations where the lower leg is vertical, 30 o from vertical, 60 o from vertical and 90 o from vertical (horizontal). The force applied to the leg by the 80.0 N weight will always be vertically down. The four different force diagrams are shown in Figure 4.7b through e. In the first position, when the leg is vertical, the applied force is also vertical, and there is no component of the force perpendicular to the extension of the leg.



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In the second position, the force acts at an angle of 30o to the direction of the leg, so the force opposing the horizontal extension of the leg is 80.0 N sin 30o, or 40.0 N. In the third position, the force acts at an angle of 60o to the leg, so the magnitude of the force perpendicular to the line of the leg is given by 80.0 N sin 60o, or 69.3 N. In the horizontal position all 80.0 N of the weight is acting perpendicular to the leg extension. In the second exercise system, an exercise load of 80.0 N is applied to the foot through a pulley system so that the load is maximum when the lower leg is in a vertical position (see Figure 4.8). Once again let us find the force perpendicular to the lower leg opposing the extension of the leg at four positions, lower leg vertical, 30o from vertical, 60o from vertical, and horizontal. Using the geometry given in Figure 4.8, we can calculate the angle between the lower leg and the rope from the exercise apparatus. For the four positions A, B, C, and D the angles Θ between the lower leg and the applied force are 90o, 63o, 40o, and 17o respectively.

4. As a final example, let us consider the forces required to hold you head in a flexed position(see Figure 4.9). The atlanto-occipital joint exerts a force on the skull at an angle of 37o from the vertical and the neck extensor muscles exert a force at an angle of 53o from the vertical. The vertical components of the muscle tension and weight are balanced by the vertical force exerted by the joint.

W + T cos 53o = C cos 37o The horizontal component of the muscle tension is balanced by the horizontal force of the joint.

T sin 53o = C sin 37o where W is magnitude of the downward vertical force of your head, T is magnitude of the



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tension in the neck extensor muscles and C is magnitude of the compressive force exerted at the atlanto-occipital joint. The solutions to these equations are T = 2.14W; so C = 2.85 W. So if your head weighs 50.0 N (about 11 lb), your neck muscles will have to exert a force of 107 N when you incline your head at a 37o angle. You may need to practice drawing force diagrams. This is an important problem-solving technique that is based upon the superposition concept. The individual parts of a problem are isolated and solved separately. Then the final result is obtained by adding up the answers obtained for the individual partial solutions. 4.3 Newton's Second Law of Motion Recall your experiences moving furniture. The massive pieces of furniture required great efforts to get them to move. The smaller pieces could be moved more rapidly as you increased the force you exerted against them. These are not quantitative experiences, but they may make you willing to accept the assertion that the force required to give an object a certain acceleration is proportional to its mass, and that the acceleration you are able to give an object is proportional to the force that you exert upon it. The objects that you move have an unbalanced force acting on them. As a result the objects are not at rest and are not moving at constant velocity but are traveling with a changing velocity. An object has an acceleration in the direction of the unbalanced force as described by Newton's second law: The time rate of change in velocity of an object is proportional to the net unbalanced force acting upon it. You can quantify your experiences with moving objects by using a spring balance, a clock, and a low-friction object in the physics laboratory. You will be able to explain your data by assuming that the force required to produce a given acceleration is proportional to the mass:

F ∝ m for constant acceleration For a given mass the acceleration is proportional to the force a ∝ F for a given object of constant mass.

These two proportionalities can be combined into one equation of the form F = kma, where k is a proportionality constant. It is possible to make k unity, if one chooses two of the variable units and defines the third in terms of these two. That is the customary approach.

In SI we define the newton as the magnitude of force that gives a mass of one kilogram an acceleration of one meter per second per second (m/sec2). The vector equation for Newton's second law becomes:

F = m a thus F and a are parallel (4.1)

where F is the net force in newtons, m is the mass in kilograms and a is the acceleration in meters per second per second. The weight of a body in newtons is then given by multiplying the mass of the body in kilograms by g, the acceleration due to gravity in meters per second per second, W = m g. The mass of an object is the same everywhere. Because the acceleration due to gravity depends on position of the mass on the earth, the object's weight varies accordingly. For the numerical calculations in this book, we will use



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a value of 9.80 m/sec2 for g near the surface of the earth. According to Newton's second law the effect of a force is to alter the state of motion of a body. In Chapter 3 we dealt with accelerated motion. Let us now look at some of those problems to see what unbalanced force is acting. EXAMPLES 1. Chapter 3, exercise 3. Assume a bus has a mass of 2000 kg, and it has an acceleration of

4.00 m/sec2. The magnitude of the force needed to accomplish this acceleration becomes Fnet = ma = 2000 kg • 4.00 m/sec2. Then the net force equals 8000 N in the direction of the acceleration.

2. Chapter 3, exercise 6. A 52.3 kg female drops 28.4 m into a freshly plowed garden and is stopped in 15.3 cm. Find the deceleration force. During the fall the force acting is her weight of mg, and the acceleration is that of gravity, 9.80 m/ sec2. In order to get the deceleration, we need to find the velocity at impact, that is, the velocity with which the woman strikes the ground. We can compute this velocity using Equation 3.9.

2gs = vf2 - v0

2 The initial velocity is zero since the woman started from rest, s = 28.4 m, and g = 9.80

m/sec2 . Then we can compute vf , the velocity at impact.

vf2 - v0

2= 2gs vf

2- 0 = 2gs = 2 x 9.80 x 28.4 = 557 m2/sec2 vf = 23.6 m/sec, vertically downward

During deceleration, the woman travels 15.3 cm, and we must now find the force of deceleration. Once again we can use Equation 3.9 where this time the initial velocity, v0 is 23.6 m/sec, since the body has this velocity at the instant before impact. The distance traveled is the 15.3 x 10-2 m she dents the garden dirt. The final velocity vf is zero since the woman is finally at rest.

2as = vf2 - v0

2 2 x 15.3 x 10-2 • a = 02 - (23.6) 2 (3.9)

a = - 1.82 x 103 m/sec2, an acceleration of nearly 200 g! The minus sign indicates an upward acceleration or a downward deceleration. The net retarding force is found using Newton's second law.

F = ma F = (52.3 kg) 1.82 x 103 m/sec2)

An upward impact Fnet = -9.52 x 104 N, a force of more than 10 tons! The data for this problem are from a real case, and the woman survived.



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3. Find the force exerted on the body if a 50.0-kg person jumps stiff- legged from a porch 1.00 m onto a concrete walkway. Assume that the padding on the foot compresses 6.00 mm in stopping the fall (see Figure 4.10). This problem is similar to the previous example. We must use Equation 3.9 twice. First, given the vertical distance of the jump and the acceleration of gravity, we can find the velocity at impact. Then using our computed value for the velocity at impact and the known distance of foot compression, we can calculate the stopping acceleration. Finally, using our calculated value for stopping acceleration, we use Newton's second law to find the force exerted on the body. If we choose the vertical downward as positive, then first,

2gs = 2 x 9.80 m/sec2• 1m = vw2

vw = 4.43 m/sec Second, upon stopping, we find

2 • 6 • 10-3m • a = -(4.43 m/sec)2 where a is the stopping acceleration. Solving for a,

a = -1.63 x 103 m/sec2 To find the force exerted on the body, we substitute this value for a in the equation for Newton's second law.

F = ma = 50 kg • (1.63 • 104 m/sec2) = -8.17 x 104 N This force would be transmitted to the leg joints where it would be potentially dangerous. How does flexing the legs at the knees reduce the danger of such a jump? Let us review the prescription followed in solving these example problems. These

problems asked for the determination of a force that was known to provide a certain acceleration for an object of known mass. This unknown force is equal to the mass of the object times the acceleration specified in the problem. The direction of the force will be in the direction of the acceleration. The acceleration called for in the problem can be determined by using the initial and final velocities and the distance traveled during the acceleration. In this case, we use Equation 3.9,

2as = vf2 - v0

2 to find the acceleration. It is also possible to give the initial and final velocities and the time of such acceleration. In this case we use Equation 3.4

vf = v0+ at to find the acceleration.

4.4 Force of Friction In all real motion there is a force that opposes the motion. This is the force of friction. The properties of this frictional force may be complex. The source of the frictional force may be obscure. Try some experiments in which you slide one surface over another. Some of the



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variables you may change are the different kinds of surfaces, the conditions of surfaces, the area of contact, and the velocity. Compare starting force and the force for uniform motion, and tabulate your observations. In doing these experiments you may observe that 1. the starting force is greater than the force required to keep an object moving uniformly

after it is started. 2. the frictional force is independent of area of contact. 3. the frictional force is independent of the velocity (for low velocities only). 4. the frictional force is dependent upon the types of surface in contact. 5. the frictional force is dependent upon the condition of the surfaces in contact. 6. the frictional force is dependent upon the forces pressing the surfaces together. To explain such observations we assume that the friction force can be represented mathematically as a force whose magnitude is a constant number multiplied by the force pushing the two surfaces together. ƒs ≤ µsN (4.2) where ƒ s is the static frictional force, µs is the coefficient of static friction, and N is the normal force, the force perpendicular to the surfaces, that press the surfaces together. The direction of the friction force is opposite to the direction of impending motion. The force of static friction can be any value from zero to µsN, its maximum value just before motion begins. The coefficient of friction is greatest between two surfaces at rest with respect to one another (observation 1 above). When the two surfaces begin to move with respect to each other the friction force is reduced. As soon as sliding begins, a kinetic frictional force opposes the motion. The coefficient of friction for this force is the coefficient of kinetic friction µk. The kinetic frictional force is given by the product of µk and the normal force, N, pressing the surfaces together. ƒk = µk N (4.3) Some approximate values of µs and µk are given in Table 4.1. Notice that each case µk<µs, the kinetic friction is less than the static friction. In many cases it is desirable to reduce the frictional force to a minimum. In mechanical systems the contact surfaces are polished to reduce the frictional forces. Another way to reduce the effects of friction is to lubricate the surfaces. Can you think of a model to explain how oil works as a lubricant? In the case of the membranes and joints in the human body, the problem is also solved by the use of lubricants. Note in Table 4.1 the value of µ in the human body.



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How can the joints in the human body have such a small coefficient of friction? The answer to this question is provided by the synovial joint. In a synovial joint the ends of the bones, each covered by cartilage, rest against an enclosed membrane sack (Figure 4.11). Inside the membrane sack is the synovial fluid, which is similar in physical properties to blood plasma. The synovial fluid helps to support the forces exerted by the bones and provides a medium to reduce the frictional forces. Not all frictional forces are detrimental. Make a list of examples in which frictional forces are absolutely necessary. How could you operate in a frictionless world?

EXAMPLES Consider a patient with neck traction (T) applied by calipers or tongs attached to the skull (see Figure 4.12). Assume that the head W weighs 50.0 N and that the coefficient of friction µ between the back of the head and the hospital bed is 0.200. 1. Assume that the bed and patient are in a horizontal position and the applied traction

force is also horizontal. What is the maximum frictional force that must be overcome before the traction pull will be effective in stretching the neck structures? For a horizontal surface; the normal force N is equal in magnitude to the weight. There is no vertical motion, so the net vertical forceis equal to zero (see Figure 4.13a). W = N

ƒmax = µN = µW. To stretch the neck structures the force T must be greater than the maximum friction

force, T > ƒmax = (0.200) (50.0 N) = 10.0 N.

2. Consider the use of this traction system where a technique is used to stretch the

posterior neck structures by applying the traction at an angle Θ, say 37o, with the horizontal (see Figure 4.13b) . What is the magnitude of the traction T that would just start the head moving toward the end of the bed? For a horizontal surface with force acting at 37o, the downward force of W is opposed in the vertical component of T:

N = W - T sin Θ ƒmax = µN = TcosΘ; horizontal component of T must at least

equal the maximum value of ƒ to start the motion.



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ƒmax = T cos 37o = µ (W - T sin 37o) 0.800 T = 0.200 (50.0 - 0.800 T) 0.800 T + 0.160 T = 10.0 0.96 T = 10.0 N T = 10.4 N. 3. A patient needs neck traction using this system, but his lungs fill with fluid when he is lying horizontally. You solve this by strapping the

patient to the bed and inclining the bed at an angle Φ of 30 o with the horizontal. What is the maximum force your traction system must overcome if it is to pull the patient parallel to the inclined bed (see Figure 4.14)?

In this case the traction will have to pull against both the maximum frictional ƒmax and the component of the weight of the head that is parallel to the inclined plane of the bed. For an incline of 30o with a force parallel to the incline:

N = W cosΦ T = ƒmax + W sin Φ From the equation that defines friction ƒmax = µ N = µ W cos Φ T = µ W cosΦ+ W sinΦ T = 0.200 (50.0) cos30o + 50.0 sin30o = 10.0 (0.866) + (50.0)(0.500) = 8.66 + 25.0 T = 33.7 N

4. Now assume the patient who requires an inclined bed also needs to have the posterior neck structures stretched. What traction T acting

at 37 o to the bed inclined at 30 o will be required to just slide the patient's head (see Figure 4.15). For an incline of 30 o with a force 37 o above the incline: N + T sin Θ= W cos Φ

ƒmax = µ N = µ (W cos Φ- T sinΘ) T cos Θ = W sin Φ+ ƒmax T cos Θ = W sin Φ+ µ (W cos Φ - T sin Θ) 0.800 T = 50.0 (0.500) + 0.200 [50.0 (0.866) - T (0.600)] 0.92T = 33.7; T = 37.7 N

5. If the only unbalanced force acting on a moving body is the frictional force, then it opposes the motion and produces a deceleration. A 15,000 N automobile is traveling



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39.2 km/hr on a horizontal road. The brakes are applied, and it slides to rest with µk = 0.500. What is the force acting? What is the acceleration? How far did the car slide?

F = ƒmax = µk N F = 0.500 • 15,000 N = 7500 N F = ma or a = F/m a = F/(W/g) = 7,500/(15,000/g) = 1/2 • 9.80 = 4.90 m/ sec2 v0 = 39,200 m/3,600 sec = 10.9 m/sec 2as = vf

2 - v02

a = - 4.9 m/ sec2 -2(4.9 m/ sec2 )s = 0 – (10.9 m/sec)2 s = 119 (m/sec)2/9.80 m/sec2= 12.1 m

6. a. A water skier (whose mass is 70.0 kg) is pulled through the water at a constant velocity of 20.0 m/sec by a forward force on the skier of 100 N. Find the frictional drag force on the skier. Since the skier moves at a constant speed (a = 0), the net force on the skier is zero. Thus the drag force must be 100 N.

b. A velocity dependent drag force is common for motion through fluids. Assume that the frictional drag force is proportional to the speed of the skier, and find the frictional drag coefficient. ƒd = k v

Thus k = ƒd / v = 100 N/20.0 m/sec = 5.00 N sec/m 7. Consider the system illustrated in Figure 4.16. The masses are MA = 10.0 kg and MB =

20.0 kg, and the coefficient of sliding friction is 0.200 for the incline. Find the acceleration of each block.

The first step in solving such problems is to draw a free-body diagram (seeFigure 4.16b) to show clearly all forces acting on each body.

The bodies are connected by a rope; if the rope doesn't stretch or break, both bodies will have the same acceleration. Therefore, the net force on each body must equal its mass times the acceleration. Let us solve the problem: For mass A: There is no motion perpendicular to the plane, thus, the magnitude of the normal force N must equal the component of the weight acting vertically: N - MA g cos Θ = 0



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or N = MA g cos Θ = 10.0 kg • 9.80 m/sec2 • 0.800 = 78.4 N. For mass A to move up the plane, we must have a net positive force acting parallel to the plane: T - (MA g sinΘ + µ N) = MA a where a is positive for acceleration up the plane.

For mass B, the tension in the rope acts to oppose its weight, thus the net force equation is MB g - T = MB a where a is positive for weight B falling down.

We must now solve the equations for the magnitude of a. This can be done by solving the MB equation for T and substituting this into equation for mass A. Thus we get: MB g - MA g sinΘ - µMA g cos Θ = (MA + MB)a a = (MB g - MA g sinΘ - µMA g cos Θ)/(MA + MB) a = (20.0 • 9.80 - 10.0 • 9.80 • 0.600 - 0.200 • 10 • 9.80 • 0.800)/30.0 a = (196 N - 118.1 N - 15.7 N)/30.0 kg = 2.09 m/sec2 Thus, T = MB (g- a) = 154 N

4.5 The Law of Attraction Between Two Bodies In Chapter 3 on kinematics you learned that a freely falling body has an acceleration g. In accordance with Newton's second law there must be an unbalanced force acting on the falling body. This force is the weight of the body and results from the attraction between the earth and the falling body. Sir Isaac Newton stated the law of gravitational attraction between two idealized point masses: any two point masses attract each other by a force that is proportional to the product of their masses and inversely proportional to the square of the distance between the masses. The direction of this force is along the line between the point masses. This is called the universal law of gravitation because it is assumed to apply to point mass interactions throughout the universe. We can write this law as an equation using a proportionality constant which we will signify by G:

F = G m1m2/r2 (4.4) where G is a constant known as the universal gravitation constant. It has the same value for any pair of masses, and its value in SI units is G = 6.6732 x 10-11 N- m2/kg2. The universal law of gravitation can be extended to all spherical bodies that have mass distributions that depend only on distance from the center of the body. The superposition principle applied to these bodies shows us that such a body may be treated as if its entire mass were concentrated at its center. Consider a mass, m, freely falling near the surface of the earth. The force of gravity



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produces the acceleration g. Thus we have: (Me = mass of earth and re = earth radius) mg = GMem/re

2 or g = GMe/re 2 (4.5)

Gravitation is another example of interaction-at-a-distance since an object does not have to be in contact with the earth to be acted upon by the force of gravity. It is reasonable to construct the concept of a gravitational field that surrounds the earth. At any point a distance r from the center of the earth, the gravitational field can be defined to have the magnitude of GMe/r 2 newtons and be directed toward the center of the earth. If we place an object of mass m at that point, we will have its weight, or the gravitational force acting upon it, of GmMe/r 2 newtons. We can use the change in gravitational field strength with distance as a way of explaining the changing weight of an object as it recedes from the earth. Furthermore, we can ascribe gravitational fields to other objects such as the sun and moon, and we can use the properties of these fields to explain such phenomena as tides and planetary orbits.

4.6 Centripetal Force Probably you have tied an object to a string and whirled it around. If you have not, do the experiment now. What did you experience? In Chapter 3 on kinematics you learned that in uniform circular motion, there is an acceleration directed toward the center. Newton's second law states that if there is an acceleration of a body, there is an unbalanced force acting upon the body. The vectors of the acceleration and the unbalanced force are directed toward the center of the circle. In uniform circular motion the magnitude of the unbalanced force acting on the body toward the center of rotation is given by: F = m ar = mv2/r (4.6) since the radial acceleration is given by Equation 3.21. The speed of an object traveling in a circle of radius r is equal to 2πrn, where n is the number of revolutions per sec. This expression can be substituted for v in Equation 4.6 to yield, F = 4π2 n2 mr (4.7) This force is called the centripetal force. In Equation 4.7 the expression 4π2 n2 is the square of 2πn, where 2π is the number of radians in one revolution (2π radians = 360o ), and n is the number of revolutions per second. The quantity 2πn is called the angular velocity. It is measured in radians per second and is designated by ω. ω = 2πn So Equation 4.7 can be rewritten as F = mrω2 (4.8) In the space age we are aware of the concept of weightlessness as experienced by astronauts in orbiting satellites. For these people and for the entire contents of the spacecraft, the gravitational attraction of the earth is precisely equal to the centripetal force of the body in its orbit, F = G Mem/r 2 = m v2/r (4.9) where Me is the mass of the earth, G is the gravitation constant and r is the radius of the orbit. Hence the speed of the spacecraft in orbit is given by



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v = √(G Me/r) (4.10) Notice the difference in experience between the spacecraft observers on the ground and the orbiting astronauts. To us on the earth the astronauts and all the orbiting equipment are falling around the earth acted upon by a centripetal force whose force is given by Equation 4.8. However, in the orbiting spacecraft, all objects are at rest with respect to one another. The astronaut, when he steps away from the dinner table, does not fall toward the earth but continues in an orbit around the earth. To the astronaut the force of gravity on objects is canceled by an imaginary force, called centrifugal force, which arises from the orbital motion of spacecraft. Actually, the astronauts are only a few hundred kilometers above the earth. The gravitational force at that altitude is reduced by only about ten percent. Hence, the experience of weightlessness occurs in the frame of reference of the astronauts who can account for the objects that float through the spacecraft by assuming these objects are acted upon by a fictitious outward force equal in magnitude to the gravitational force given by Equation 4.8. EXAMPLES

2. A merry-go-round is turning at the rate of 10.0 revolutions per minute. A 50.0 kg youth is located 5.00 m from the axis of rotation. What is the centripetal force acting on him? From Equations 4.6 and 4.7 we note that

F = mv2/r = 4π2 n2 mr Since n must be in revolutions per second, n = 10.0 rpm/60 sec/min = 1/6.00 rps Then, F = 4π2 /(6.00)2• 50.0 • 5.00 = 1/9.00 • π2 • 250 = 6.85 • 104 N 2. An ultracentrifuge is used to separate parts of biological samples. It has become an

important tool in biochemistry and molecular biology. The rotors in these machines rotate at rates as high as 60,000 rpm. We wish to determine the centripetal force on a microgram sample at a radius of 10.0 cm from the axis of the rotor. Letting Fc represent the force and ac the centripetal acceleration. Fc = m ac and ac = rω2 Substituting r = 0.100 m and the given rate of rotation, ω= 2πn = (60,000 x 2π/60) rad/sec Thus Fc = 10-9 kg • (2π • 103 rad/sec) 2 • 0.100 m = 4.00 π2• 10-4 N = 3.95 • 10-3 N Note that this force is about 4000 times the weight of the sample. It is said that the

sample is subjected to a 4000 g centripetal force. The tangential velocity of the sample at a 10.0-cm radius is given by v= rω = 2π • 103 rad/sec • 0.100 m = 2.00 π • 102 m/sec = 628 m/sec This speed is greater than the velocity of sound in air. In order to avoid shock waves due to supersonic speeds, the rotor of the ultracentrifuge spins inside a



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vacuum chamber. A diagram of an ultracentrifuge is shown in Figure 4.17.

4.7 Inertial and Gravitational Mass Equivalence A fundamental question concerning the inertia of an object was raised by Newton. He wondered if the property determining the weight of an object was the same as the inertial parameter that determines the acceleration produced by force applied to the object. This can also be stated as follows: Is the gravitational mass the same as the inertial mass of an object? In freely falling object experiments, is the acceleration a constant, or does it depend on the properties of the falling object (such as temperature and chemical composition)? If the inertial mass equals the gravitational mass, we should expect all objects to fall with the same acceleration. Likewise if a nongravitational force equal to the weight of the object is applied, the acceleration of the object should be equal to the acceleration due to gravity. Such experiments and many others have been carried out since the seventeenth century, and all experiments show that within experimental error the gravitational mass and the inertial mass have the same value. Most recent experiments have shown this equivalence to be accurate to within one part in 1012.

ENRICHMENT 4.8 Escape Velocity From the Earth If the mass is constant, the force is the product of mass and acceleration. In general, an equation of motion is given by Fnet = ma = m dv/dt = m d2r/dt2 (4.11) where the acceleration a is given by the time rate of change of the velocity a = dv/dt, and where the velocity v is given by the time rate of change of the distance, v = dr/dt.

If we know the expression for F for a given situation we can equate it to m d2r/dt2, and solve the differential equation that results. F may have various functions of distance r or time t. For example, F may be given by the law of gravitational attraction of the earth,

G Mem/r 2 = m d2r/dt2 (4.12) To solve this differential equation, the acceleration is expressed as a function of the

distance derivative of the velocity by multiplying by dr/dr and rearranging the variables,



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a = dv/dt = dv/dt • dr/dr = dr/dt • dv/dr = v (dv/dr) (4.13) For constant mass, Newton's second law may be rewritten as F = m v dv/dr (4.14)

This expression can be set equal to the gravitational attraction of the earth, G Mem/r 2 = m v dv/dr (4.15) where Me is the mass of the earth. We can integrate this equation to obtain an expression for the minimum vertical velocity needed for an object to escape from the earth. We can evaluate the constant for the escape velocity by requiring the velocity of the object to be zero when the object is an infinite distance from the earth. Then the constant is zero. We find that the vertical velocity the object must have at the surface of the earth is vescape = √(2GMe/re) (4.18) where re is the radius of the earth. Questions 1. Assume an asteroid has the same density as the earth, 5.50 x 10-3 kg/m3. Calculate the

size of the asteroid from which you could launch yourself into space by running with the escape velocity on the surface of the asteroid. (Use your estimated maximum running speed on earth for the escape velocity.)

SUMMARY Use these questions to evaluate how well you have achieved the goals of this chapter. The answers to these questions are given at the end of this summary with the number of the section where you can find the related content material. Definitions 1. There must be no unbalanced force on a body

a. at rest b. with constant velocity c. with no friction d. with constant acceleration e. all of these



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2. The weight of a body may vary with its location because thea. mass varies b. temperature varies c. acceleration due to gravity varies d. pressure varies e. all of these

3. The coefficient of friction isa. inversely proportional to normal force b. greater for kinetic than static case c. greater than 1 d. theoretically determined e. experimentally determined

4. The force due to friction alwaysa. is greater than normal force b. opposes motion c. is an unbalanced force d. is zero for bodies at rest e. equals body weight

5. Centripetal force in circular motion accounts for accelerationa. toward the center b. away from the center c. tangent to the path d. opposing gravity e. equal to zero

6. Weightlessness in orbiting satellites is equivalent to motion ina. a centrifuge b. free fall c. a vacuum d. rocket launching e. none of these

Newton's Laws 7. According to Newton's law of gravitational interaction the gravitational force between

particles is_____(attractive / repulsive) with a magnitude proportional to the product of_____ and the square of the ________________.

8. Newton's third law postulates the existence of_________ pairs of forces which are_________and ___________to one another.

9. Newton's second law helps us understand the motion of objects because it relates_________ the of an object to the ________acting on the_____ object with the inertial property of the object called _________as the proportionality constant.

10. The SI unit of force is a_________ and is equivalent to a mass of_________ being accelerated at a constant rate________________________ .

11. Newton's first law may help explain the persistence of the status quo because it states that an object at rest tends to __________ and moving objects tend to __________ unless acted upon by ______________.



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Resolution of Forces 12. A force of 13.0 N acting upward at an angle of 23o to the horizontal has an upward

vertical component of___________ and a horizontal component of ___________. 13. A football sled of downward weight 500 N is pushed horizontally by a force of 400N.

The resultant force is ________in a direction _________ from horizontal. Newton's Second Law 14. If a body is acted on by an unbalanced horizontal force of 40.0 N and experiences an

acceleration of 2.00 m/sec2, its mass isa. 80.0 kg b. 0.050 kg c. 20.0 kg

d. 2.00 kg e. 40.0 kg

15. A body with a weight of 10.0 N is pulled across a horizontal surface with coefficient of kinetic friction µk = 0.200 by a force of 5.00 N making an angle of 37 o with the horizontal. The accelertion of the body will bea. 1.00 g b. 0.140 g c. 0.200 g

d. 2.60 g e. 0.260 g

16. A mass of 2.00 kg moves in a circular path (radius 1.00 m) with an angular velocity of

10.0 rad/sec. The centripetal force on the mass is a. 2.00 g b. 20.0 N c. 10.0 N

d. 200 N e. 50.0 N

17. Friction between a block at a distance d from the center of a turntable rotating at n

rad/sec keeps the block on the turntable. The coefficient of friction between the block and turntable must be at leasta. nd/g b. g c. n2d/g

d. gn2d e. g/ n2d

Answers 1. a, b (Section 4.2) 2. c (Section 4.3) 3. e (Section 4.4) 4. b (Section 4.4) 5. a (Section 4.6) 6. b (Section 4.6) 7. attractive, masses of the particles,

inverse of the distance between the particles (Section 4.6)

8. action- reaction, equal, opposite (Section 4.2)

9. acceleration, unbalanced forces, mass (Section 4.3)

10. newton, 1 kg, 1 m/sec2 (Section 4.3) 11. stay at rest,continue moving in a

straight line of constant speed, unbalanced forces (Section 4.2)

12. 5.00 N, 12.0 N (Section 4.2) 13. 640 N, 59 o down (Section 4.2) 14. c (Section 4.3) 15. e (Section 4.3) 16. d (Section 4.6) 17. c (Section 4.4 and 4.6)



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ALGORITHMIC PROBLEMS Listed below are the important equations from this chapter. The problems following the equations will help you learn to translate words into equations and to solve single concept problems. Equations

2a • s = vf2 – vo

2 (3.9) F = m a (4.1)

ƒk = µk N (4.3) F = Gm1m2/r2 (4.4)

g = GMe/re 2 (4.5)

F = m ar = mv2/r = 4π2 n2 mr = mrω2 (4.6, 4.7, 4.8) G = 6.6732 x 10-11 N- m2/kg2

Problems 1. Calculate the braking force needed to give a 2.00 m/sec2 negative acceleration to a

1000-kg car. 2. Compute the force needed to move a 1.00-kg block across the floor at a constant speed

if the coefficient of friction is 0.300. 3. Find the centripetal force on a satellite of mass m in a circular orbit at a distance D from

the center of the earth. The velocity of the satellite is C. 4. The centripetal force in problem 3 is equal to the gravitational attraction between the

earth and the satellite. If the mass of the earth is M,show that the value of C must be (GM/D)½. (Hint: Use the answer to problem 3.)

5. Given that the force of friction on a sled of mass 100 kg is 50.0 N, find the acceleration of the sled when a 550 N force is applied horizontally in an easterly direction (see

Figure 4.18). 6. Find the force necessary to accelerate a 1000-kg elevator at 1.00 m/ sec2 upward. 7. Assume you have a mass of 60.0 kg and are riding in an elevator which is accelerating

upward with an acceleration of 2.20 m/sec2. What force must your legs withstand? 8. What is the force necessary to hold a 100-g mass in a 0.100-m circle at 1200 rpm? How

does this compare with its weight? 9. What is the force of attraction between the earth and a 2.00-kg mass at the surface of the

earth? 10. Find the coefficient of friction that gives a force of friction of 5.00 N between a box and

the floor if the box weighs 50.0 N. 11. Two boys are holding a box weighing 400 N above the floor. One boy exerts a 225 N

force upward. Find the upward force exerted by the other boy.



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Answers 1. 2000 N 2. 2.92 N 3. mC2/D 5. 5.00 m/sec2 6. 10,800 N

7. 720 N 8. 78.9 N, 80.6 times its weight 9. 19.6 N 10. 0.100 11. 175 N

EXERCISES These exercises are designed to help you apply the ideas of a section to physical situations. Then appropriate numerical answers are given at the end of each exercise. Section 4.2 1. A picture weighing 10.0 N is hung on a hook at the midpoint of a supporting wire.

Each side of the wire makes an angle of 10 o below the horizontal. What is the tension on the wire? [29 N]

2. A 120-N child in a swing is held by a horizontal force so that the ropes make an angle of 37 o with the vertical. What is the tension on each rope and the horizontal force? [75 N, 90 N]

3. On a field trip, a geologist gets his automobile into a soft spot. He attaches a taut cable to a tree 30.0 m from the car. He then hangs a 50.0-kg rock at the midpoint of the cable, and it sags 30.0 cm. Assume there is no stretch in the cable. What is the force acting on the car? [12,300 N]

4. Consider the foot and ankle as an isolated body, and apply the principle of statics at the moment the ball of one foot is in contact with the ground. Use your weight as W. The tibia C makes an angle of 15 o with the vertical, and the Achilles tendon T makes an angle of 21

o with the vertical Figure 4.19. Calculate the magnitude of the compressional force in the tibia and of the tension in the Achilles tendon. [C= 3.42w, T = 2.48w]

5. A weight W is supported by cervical neck traction apparatus consisting of a boom AB and a cable BC as shown in Figure 4.20. Neglecting the weight of the boom, what is the compression of the boom and the tension of the cable? [T = 0.600W, C = 0.800W]

Section 4.3 6. A 50.0-kg student is riding in an elevator. What is the compressive

force in her legs, if the elevator is movinga. upward at a constant rate?

b. with an upward acceleration of 3.20 m/sec2? c. with a deceleration of 2.40 m/sec2? [490 N, 650 N, 370 N]

7. In walking you have to accelerate and decelerate your foot. If you walk at a constant speed of about 6.00 km/hr, the average acceleration of your foot is about 40.0 m/ sec2. What is the force necessary to give each kilogram of your foot an acceleration of this amount? What would this force be if you were running a 100-meter dash in 10.0 sec? [40.0 N, 240 N]



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8. A girl of mass 45.0 kg (weight ≈100 lb) escapes a fire by sliding down an improvised rope made of sheets tied together. The maximum upward force the "rope" can stand is 270 N ( ≈ 60 lb). a. Can she slide down the rope at a constant speed? If not, find the least acceleration

with which she can slide down the “rope.” b. If the window is 5 m above the ground, how long will it take her to reach the side of

the anxious young man below her window. [no, 3.80 m/sec2, 1.62 sec] 9. Compare the kinetic frictional forces for the same normal force pressing the following

surfaces together: a. two dry iron surfaces b. two wood surfaces c. lubricated metal surfaces d. two human body joint surfaces [a:b:c:d = 2:3: 0.4:0.05]10. A 50.0-N box is sliding down a 30o inclined plane at a constant rate. What is the coefficient of kinetic friction? What force would be required to move this box up the inclined plane at a constant rate? [0.580, 50 N]

11. A floor polishing block weighing 10.0 is being pushed at a constant rate over the floor by a 10.0-N force exerted on the handle which makes an angle of 60o with the horizontal. What is the coefficient of kinetic friction? [0.273]

12. Find the magnitude of force F for each case shown in Figure 4.21. All motion is at a uniform rate. Body A weighs 10.0 N. Body B weighs 20.0 N. The coefficient of kinetic friction is 0.300 for all surfaces. [a. 9.00 N; b. 12.0 N; c. 15.0 N]

Section 4.5 13. How far from the center of the earth must a rocket be for

its weight to be 9/10 its weight on earth? [1.05re]Section 4.6

14. A centrifuge is revolving at the rate 10,000 rpm. The radius of revolution for a 0.100-g particle being separated is 8.00 cm. What is the centripetal force acting on it? [8.77 N]

PROBLEMS Each problem may involve more than one physical concept. A problem that requires

material from the enrichment section is marked by a dagger †. The numerical answer to the problem is given in brackets at the end of the problem.

15. A train of total mass M is traveling with velocity v when the throttle is suddenly

opened and kept open for a distance d. If the accelerating force is F, find the velocity at the end of the distance d.[√(v2 + 2dF/m)]



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16. In Figure 4.22, block A has a mass of 10.0 kg, and block B has a mass of 4.00 kg. The coefficient of kinetic friction is 0.300. The pulley is massless and frictionless. Neglecting the mass of the cord, what is the acceleration of the system and the tension in the cord? [0.700 m/sec2, 36.4 N]

17. A weightless, frictionless pulley supports two masses of 90.0 g and 100 g connected by a massless cord (Figure 4.23). The masses are released. What is the acceleration of the system? What is the tension in the cord? [0.516 m/sec, 0.928 N]

18. An 80.0-kg high jumper just clears the bar at 1.80 m. What vertical velocity did he have at the instant he left the ground? If he acquired this velocity in 0.500 sec, what was his acceleration? What vertical force must his leg have experienced? [5.94 m/sec, 11.9 m/sec2, 1740 N]

19. A 1500-kg automobile climbs a hill at a speed of 12.0 m/sec. If the hill rises 1.00 m in 10.0 m and there is a frictional resisting force of 500 N, how much force must the driving wheels exert? [1970 N]

20. In Figure 4.24, object A weighs 50.0 N, and object B weighs 100 N. The coefficient of friction between object A and the plane is 0.500, and the coefficient of friction between object B and the plane is 0.200. A and B are connected by a rigid rod. (Neglect its mass.) What is the acceleration of the system, and the compression in the rod connecting A and B. [3.53 m/sec2, 8.00 N]

21. A student is swinging a bucket partially filled with water in a vertical circle. Her arm is 0.700 m long, and the surface of the water is 0.100 m from the bucket's handle. What is the minimum velocity that the bucket must have to avoid spilling the water? [2.80 m/sec or 3.50 rad/sec]

22. What angle of bank should a road curve of 80.0-m radius have for safety at a speed of 100 km/hr? (Assume no friction between the tires and the road.) [44.5o]

23. A 30.0-kg child is sitting on a horizontal rotating platform at a distance of 2.00 m from the axis of rotation. The coefficient of static friction between the surfaces in contact is 0.200. What is the maximum rate at which the platform can rotate without the child sliding off? [0.990 rad/sec]

24. A track meet is theoretically scheduled on an asteroid which has a radius of 100 km. Assume that the density of the earth and asteroid are the same and that the radius of the earth is 6700 km. How high could a jumper jump on the asteroid if he could jump 2.00 m on the earth? What other assumptions have you made in the solution? [134 m]

25. The mass of the earth is 81.0 times that of the moon. The radius of the earth is 3.75 times that of the moon. On the surface of the earth the weight of an unknown mass as determined by a spring balance and also by an equal arm balance with a standard set of weights is 100 N. All of this equipment is taken to the moon. What is the weight of the unknown mass on the moon as measured by both the spring balance and the equal arm balance? [17.4 N, 100N]



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26. The Russell traction apparatus (Figure

4.25) is used for a fracture of a femur. The mass of the leg is 4.00 kg, and the tension required is 60.0 N. Find the values of masses m1 and m2. [3.06kg, 0.940 kg]

27. The force exerted on the pilot by his seat is 4.00 mg (m = the pilot's mass) when his plane pulls out of a dive at 480 km/hr. Assume that the plane makes a circular arc at the bottom of the dive as shown in Figure 4.26. Find the radius of the arc and the time required to go through a π/2 rad arc. [605 m, 7.15 sec]

28. Many of the data available concerning human impact studies have been gathered from rocket and track experiments. If a rocket is brought to rest on 0.250 sec from a speed of 100 m/sec, find the average acceleration and force on an 80.0-kg subject. How far does the rocket travel while braking? [400 m/sec2, -3.20 x 103 N, 12.5 m]

29. Consider the two exercise systems described in Example 3 of Section 4.2. Suppose that both systems were simultaneously used on the same patient. Use the superposition principle to find the total force perpendicular to the lower leg so as to oppose the extension of the leg. Draw a graph of the total force as a function of the angle Θ that the lower leg makes with vertical. What is the angle of maximum total force? [Fp = 80 N[cos (Θ - φ) + sinΘ] where tan φ = (1 - cosΘ)/(2 + sinΘ) and Θ = leg angle with vertical]

30. Assume Evil Knievel, whose mass is 80 kg, wishes to launch himself into space from a 60 o incline riding on a motorcycle of 170 kg mass. What must be the speed of the motorcycle as it leaves the sea level ramp? Me = 5.97 x 1024 kg; re = 6.37 x 106m. [12.9 km/sec]

Chapter 4 FORCES AND NEWTON'S LAWS

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