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Chapter 4 Forces and Newton's Law

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 Law

OVERVIEW - The statements which we call Newton's Laws of Motion are considered to

be a central feature of the study of the motion of objects. These laws generalize much of our common experiences with forces and their influence on the state of motion of objects. As you read and consider the Laws of Motion presented in this chapter, make a point to relate the meaning of the laws to real experiences you have had with bicycles, automobiles, electric trains, etc.

SUGGESTED STUDY PROCEDURE - As you begin your study of this module, begin by

carefully reading each of the Chapter Goals: Definitions, Newton's Laws, Resolution of Forces, Newton's Second Law Problems, and Centripetal Force Problems. Remember that on the next page of this Study Guide you will find a paragraph concerning each of the terms listed under Definitions. Next, read text sections 4.1-4.7. Remember that answers to questions asked in the text are given in the second section of this Study Guide. As you read, be sure to consider the examples given. If you have difficulties with the mathematical treatment of forces and their components, refer to the initial chapter of this Study Guide (Chapter 0).

At the end of the chapter, read the Chapter Summary and complete Summary Exercises 1-17. Now do Algorithmic Problems 1, 2, 5, 6, 7, 8, 9, 10, and 11. Check your answers to each problem carefully. Now complete Problems and Exercises 1, 2, 5, 8, 10, 12, 14, 26, 27, and 28.

Now turn to the Examples portion of this Study Guide and consider the additional problems and questions presented. After you have completed this task, you should be prepared to attempt the Practice Test on Forces and Newton's Laws. This study procedure is outlined below.

--------------------------------------------------------------------------------------------------------------------- Chapter Goals Suggested Summary Algorithmic Exercises Text Readings Exercises Problems & Problems --------------------------------------------------------------------------------------------------------------------- Definitions 4.1, 4.2, 4.3, 1-6 4.4,4.5,4.7 Newton's Laws 4.2, 4.3 7-11 2, 9, 10 8 Resolution of Forces 4.2, 4.4 12, 13 2, 10, 11 1, 2, 5, 10, 12, 26 Newton's Second Law Problems 4.2, 4.3, 4.4, 14-17 1,5,6,7 8,28 4.6 Centripetal Force Problems 4.6 8 14,27

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DEFINITIONS FORCE - A measure of the strength of an interaction; a push or pull; the effect of a force

is to alter the state of motion of a body. A vector quantity with a magnitude measured in newtons (N). The strength of the earth's gravitational field is measured by the weight of an object. Weight is a force acting vertically downward.

WEIGHT - The product of the mass and the acceleration due to gravity.

Another way to think of g is to treat it as the strength of the force of gravity near the surface of the earth. Then the weight is a downward, vertical force.

FRICTIONAL FORCE - Force that opposes the motion of an object.

In all of the cases of motion you commonly encounter there is a force acting to oppose the motion. When you are coasting down a slope on your bicycle, what are the forces that tend to slow you down? All of them can be classified as frictional forces.

COEFFICIENT OF FRICTION - Ratio of the maximum frictional force to the net force

pressing the surface together. Try rubbing two sheets of paper together, then try rubbing two sheets of sandpaper together. Which requires the greater force to slide them along? This illustrates the larger coefficient of friction of sandpaper. If you wet the sandpaper, what happens to the force required to slide the paper along?

CENTRIPETAL FORCE - The unbalanced force acting on the body towards the center

of rotation in circular motion. The radial acceleration of an object going around in a circle is proportional to centripetal force.

WEIGHTLESSNESS - The gravitational attraction of the earth for an object in motion

around the earth is precisely equal to the centrifugal force of the body in its orbit. When you are in such an orbiting spacecraft floating objects and persons are the rule rather than the exception. Ask your physics instructor to show you some of the interesting films made from NASA photographs.

ANSWERS TO QUESTIONS FOUND IN THE TEXT SECTION 4.4 Force of Friction Suppose you think of oil as composed of many smooth planes loosely packed together. Then the oil could slide easily along the planes and would greatly reduce friction in those directions. This is not a very good model for oil. Can you think of a better one? At least this smooth planes model for oil offers an explanation for the reduction of friction in certain directions. Many frictional forces are essential. They enable us to place ourselves in a fixed position and remain there even when subjected to small external forces. If your chair were resting on a frictionless floor, the slightest breeze would blow you away from the table where you are studying physics. That would be too bad, wouldn't it?

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Indeed, it would be difficult to live as you presently do in a frictionless world. If we got started moving, how would we stop?

GENERAL COMMENTS (valuable insights included in the , free of charge) NEWTON'S LAWS OF MOTION as given in Latin in his famous book PRINCIPIA.

Lex. I - Corpus omneperseverare in statu suo quiesendi vel movendi uniformiter in directum, nisi quatenus illuda viriblus impressis cogitur. Lex. II - Mutationem motus, proportionalem esse vi motrici impressae, et fieri secujdum lina lineam rectam qua vis illa imprimitur. Lex. III - Actioni contrarium semper et aequalem esse reactionem; sive corporum duorum actiones in se mutuo semper esse aequales et in partes contrarias dirigi.

COMMENTS ON NEWTON'S FIRST LAW

The inertia of a body may be defined as that property of a body that tends to resist a change in its state of rest or motion. Mass is defined as a quantitative measure of inertia.

COMMENTS ON NEWTON'S SECOND LAW Another way of stating the classical version of Newton's second law is: when a body is acted upon by a constant force, its resulting acceleration is proportional to the force and inversely proportional to its mass; a = F/m

COMMENTS ON NEWTON'S THIRD LAW

It is important to note that the action force and the reaction force in Newton's third law of motion act on different bodies.

EXAMPLES RESOLUTION OF FORCES

1. A beanbag chair is at rest on a driveway inclined at 22ø. The chair has a weight of 68.0 newtons. (a) Draw the diagram of the chair and label all of the forces acting on the chair. (b) Calculate the values for all forces, etc., for this situation. What Data Are Given? The weight of the chair (68.0 N) and the angle of incline of the plane (22ø) are given. The velocity of the chair is zero throughout. What Data Are Implied? The weight is a force acting vertically down in a region where the strength of the earth's gravitational force produces an acceleration of 9.80 m/s2. The chair is not moving because of the presence of friction, so a frictional force is implied in this problem. Likewise, the plane is supporting the chair so a normal force is acting on the chair because the plane is implicit in this problem. What Physics Principles Are Involved? Both the second and third laws of Newton are necessary to understand this problem as well as knowledge of the vector nature of force. Before we go on to

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the algebraic solution of this problem we are ready to use a force diagram to answer part (a) and get us ready to solve part (b). (a) Force Diagram

What Equations Are to be Used? We can make use of Newton's Second Law. We know the velocity remains constant (at zero) so the acceleration is zero; thus, the net force on the chair must be zero. That means the vector sum of the forces on the chair must be zero; W + N + f = 0 Algebraic Solution The vector equation can be written as two separate scalar equations, a vertical component equation and a horizontal component equation. Vertical forces = 0 = W = N cos θ - f sin θ where positive is downward. (1) Horizontal forces = 0 = N sin q - f cos θ where positive is to the right. (2) These equations can be combined to solve for the magnitudes of the two unknown forces f and N. Rearrange Equation (2) and divide by cos θ: f = N sin θ / cos θ) = N tan θ (3)

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Rearrange equation (1), divide by cos θ and substitute the friction for f, N = (W/cos θ) - (f sin θ/cos θ = (W/cos θ) - N tan2 θ (4) Now combine the terms containing the normal force N, so N (1 + tan2 θ) = W/cos θ or N = W / [cos θ (1 + tan2 θ)] (5) Substitute this expression into Equation (3) to obtain f = W tan θ / [cos θ (1 + tan2θ] (6) Numerical Solution W = 68.0 N; θ = 22ø, so cos θ = 0.927; tan θ = 0.404. N = 68.0 N / [0.927(1 + (.404)2)] = 63.1 N perpendicular to the driveway. f = N tan θ = (63.1 N) (.404) = 25.5 N, parallel and up the driveway. Thinking About the Answer Notice that both answers come out in the correct units, newtons. The normal force is much larger than the friction force. This seems reasonable for a low angle driveway. Notice as the angle decreases in Equations 5 and 6 the normal force approaches a value of W and the friction force goes to zero. Is that what you expect?

NEWTON'S SECOND LAW PROBLEMS

2. Assume the beanbag chair given in the previous problem is lying at rest on an icy (friction ÷ zero) driveway. What happens? Give quantitative results. What Data Are Given Same as in the previous problem except now the friction force f is equal to zero. What Data Are Implied? Same as in the previous problem, except the frictional force is zero. Motion along the driveway can be deduced. What Physics Principles Are Involved? As before, Newton's 2nd and 3rd laws. Now Newton's 2nd law shows that the beanbag chair will undergo some acceleration down the driveway since the net force cannot be zero. This is clearly seen by removing the frictional force from the previous force diagram. Then you are left with

It is not possible for the weight and the normal force to add to zero. Hence, the chair must experience some net force and some acceleration. What Equations Are to be Used? The vector equation for the 2nd law in this case is given by ma = W + N (7)

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However, we know the chair can only move along the driveway since there is no force acting to lift the chair. Hence we can transform Equation (7) into a scalar equation by considering only motion and forces that act parallel to the driveway. The normal force has no components parallel to the driveway. The weight has only one component, W sin q, acting down the driveway, so we can write ma = W sin θ {motion down the driveway} (8) Algebraic Solution Remember, only the weight W, g, and q are given, so we need to replace the mass on the left side of Equation (8) by W/g, thus (W/g)a = W sin q or a = g sin θ (9) Numerical Solution g = 9.80 m/s2, sin 22ø = 0.375 a = (9.80)(0.375) = 3.68 m/s2, down the driveway. Thinking About the Answer Note the units are correct, and that if the slope of the driveway is decreased to zero degrees the acceleration goes to zero; i.e., the chair remains at rest, does that seem reasonable to you?

CENTRIPETAL FORCE PROBLEMS

3. An 80.0 kg earth satellite circles the earth once every 94.2 minutes at a mean altitude of 522 km. Use 6.38 x 106 m as the radius of the earth. (a) What is the gravitational force acting on the satellite while in orbit? (b) What fraction of its force at the surface of the earth is it? (c) If you were riding inside the satellite what forces would you experience? What Data Are Given? The mass of the satellite is 80.0 kg. Data which can be used to calculate its velocity and the radius of its circular path are given. What Data Are Implied? The problem intends for you to assume the path is a circle with the center of the earth as the center of the path, hence the radius of the path is 6.38 x 106 m + 522 km = 6.38 x 106 + 0.52 x 106 = 6.90 x 106 m. What Physics Principles Are Involved? You need to recognize that the gravitational force must have the same magnitude as the centripetal force. Furthermore, for an observer inside the satellite the gravitational force will appear to be canceled by the centrifugal force and the condition of weightlessness will be experienced. What Equations Are to be Used? The equation for centripetal force is needed. F = mv2/r (4.6) In order to calculate the velocity v we need to divide the circumference of the orbit by the time required for one pass around it. v = 2π(radius)/time for one orbit (10) Algebraic Solutions

(a) In orbit Fo = mv2/r but v = 2πr/t Fo = m4π2r2) / rt2 = 4π2mr /t2 (11)

(b) On the surface Fe = mg Ratio of orbit force to surface force is given by F0/Fe = 4π2r/gt2 (12)

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Numerical Solution r = 6.90 x 106 m t = 94.2 minutes = 5.65 x 103 seconds m = 80.0 kg; g = 9.80 m/s2 (a) Putting these values in Equation (11) Fo = (80.0 kg)4π2(6.90 x106) / (5.65 x 103)2 = Fo = 6.82 x 102 kg/s2 Fo = 6.82 x 102 N (b) Then Fo/mg = 0.87 which is 87%. In summary, the force on the satellite in orbit is 682 N. Since its weight on earth is 784 N, the orbital force is 87% of its weight on the earth. As an observer inside the satellite you would observe the effects of the gravitational force being balanced by a centrifugal force and you would experience weightlessness. Thinking About the Answers Notice again that the units come out correctly, 1 newton is a kilogram meter per second squared. According to Equation (12) the ratio of the orbit force to the surface of the earth force is independent of the mass of the object since the symbol (m) does not appear in Equation (12). Does that seem reasonable to you?

PRACTICE TEST

1. Newton's Universal Law of Gravitation applies to all massive objects, including planets, automobiles, and people. However, many people misrepresent the gravitational law. Use your knowledge of Newton's Law and write a short correction for the following statements.

a. Because of its larger mass, the sun attracts the earth with a greater force than the earth attracts the sun.

b. The earth's rotation produces our gravitation. If the rotation were to stop, we would drift into space.

c. A body taken to an altitude of about 100 miles has no weight because the gravitational force does not extend out into space beyond the atmosphere.

2. The Fig. shown below shows a modified Russell traction apparatus for femoral

fixation.

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a. Construct a force diagram isolating point B showing all the forces which are at that point.

b. Find the total horizontal force applied to the leg at B by the apparatus when a 4 kilogram mass is suspended at A.

3. A 500 kilogram subcompact car is parked on top of a long straight incline

whose inclination is 30ø from the horizontal. a. Draw a free body diagram isolating the car and showing all the forces holding

the car in equilibrium. b. Assume that the brakes are released and the car rolls (without friction) down

the plane. Using Newton's second law, predict the car's acceleration.

4. Astronaut Owen Garriott (mass = 65 kilograms) is placed aboard a spinning device used to determine his ability to withstand high "g" forces.

a. If the apparatus is revolving at 1 revolution per second, what force does Garriott experience due to this rotation motion?

b. How many times his weight is this

force? (how many "g" 's does he experience?)

ANSWERS:

1. a) Newton's Universal Law states that these forces must be exactly equal in magnitude, opposite in direction.

b) Rotation of the earth has little effect on our weight. Actually, if rotation were to cease, our apparent weight (as measured by a spring balance on the surface of the earth) would increase slightly.

c) The gravitational force extends indefinitely into space. In a space ship which is circling the earth, we say an astronaut is weightless because in the condition of "falling" around the earth, he exerts no force on a spring balance in his ship. Actually, his actual weight (attraction to the earth) is 95% of the force at the earth's surface.

2. 69.6 N 3. 5 m/s2 4. 25,600 N, 39 g's

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