Ch02 IM SDP2 KM - Solution Manual & Test Bank Store · two springs, or other systems to show simple harmonic motion. Drop a ping-pong ball and a golf ball simultaneously from 2 meters

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CHAPTER 2

Newton’s Laws

CHAPTER OUTLINE Chapter Introduction: New Horizons—Old Physics 2.1 Force a. Weight b. Friction Physics to Go 2.1 LEARNING CHECK

Engineering Applications: Friction: A Sticky Subject 2.2 Newton’s First Law of Motion

Physics to Go 2.2 a. Centripetal Force

LEARNING CHECK 2.3 Mass LEARNING CHECK 2.4 Newton’s Second Law of Motion a. Force and Acceleration Physics to Go 2.3 b. The International System of Units (SI) LEARNING CHECK 2.5 Examples: Different Forces, Different Motions a. Projectile Motion Revisited Physics to Go 2.4 b. Simple Harmonic Motion c. Falling Body with Air Resistance Physics to Go 2.5 LEARNING CHECK Mathematical Applications: Chaotic Dynamics 2.6 Newton’s Third Law of Motion Physics to Go 2.6 LEARNING CHECK 2.7 The Law of Universal Gravitation

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a. Orbits Physics to Go 2.7 b. Gravitational Field LEARNING CHECK 2.8 Tides LEARNING CHECK Profiles in Physics: Isaac Newton SUMMARY IMPORTANT EQUATIONS MAPPING IT OUT! QUESTIONS PROBLEMS CHALLENGES

CHAPTER OVERVIEW The concept of force is presented and it is argued that a force is needed to cause any change in

motion—in speed or direction. Mass is introduced again and shown also to be involved in changing motions of objects. Newton’s second law brings the above concepts together. Applications of Newton’s second law are given in example calculations. A short section presents the SI system of units. Motion examples from Chapter 1 are now analyzed further in terms of forces. Projectile motion, simple harmonic motion, and motion including air resistance are discussed. The concept of action-reaction forces and Newton’s third law are presented. Newton’s law of universal gravitation, the Cavendish experiment, orbits, and the idea of a gravitational field are explained. The chapter concludes with a discussion of the tides.

LEARNING OBJECTIVES A student who has mastered this material should be able to: 1. Develop a better understanding of the specific meaning of force in mechanics. 2. Understand that gravity causes weight. 3. Explain the mechanisms underlying friction, the laws of friction, and the difference

between static and kinetic friction. 4. State Newton’s first law and explain the meaning of net force. 5. Compare mass, inertia and weight and explain the differences. 6. State Newton’s second law in words as well as mathematically and be able to perform

calculations using it. 7. Be comfortable with the SI system of units and give some examples. 8. Give examples of free fall.

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9. Explain projectile motion, especially the fact that the vertical and horizontal components are independent.

10. Explain simple harmonic motion. 11. Add air resistance to a free-‐fall situation and explain the changes, and explain terminal

speed. 12. State Newton’s third law and give several examples of action-‐reaction force pairs. 13. State Newton’s law of universal gravitation in words and mathematical notation. 14. Describe the Cavendish experiment. 15. Explain how gravity is involved in orbits and relate Newton’s “cannon” idea for a satellite

launch. 16. Make an ellipse with tacks and a loop of string. 17. Picture a gravitational field and explain the benefit of this concept over “action at a

distance.” 18. Explain how tides are created. 19. Give some details of Newton’s life and work. Teaching Suggestions and Lecture Hints

The idea of weight being a force—a pull—is often new to students and must be stressed. You may not wish to dwell on friction to the extent presented in Section 2.1, although kinetic friction is mentioned by name in later chapters. See the interesting article “Soft Matter in a Tight Spot” by Steve Granick in the July 1999 Physics Today about current work on friction and lubrication—he even shows evidence that kinetic friction behaves chaotically.

Question 5 about pressing and pushing on a book resting on a table is a “must do” in class. Have everyone try this, pressing lightly at first, then harder and harder. Then try it again with a sheet of paper between your hand and the book. (When I tried this variation on my computer table using the fourth edition of this book, I couldn’t get the book to slip no matter how hard I pressed!)

Use a weak spring to show how pulling distorts it, and that the amount of stretch can be used to measure the size of a force. Show a spring-type force scale for comparison.

The key idea to bring across in connection with Newton’s first law is that a force is needed to cause any change in motion—in speed or direction. The concept of a net force and an external force may have to be explained. Describe several situations in which a centripetal force keeps something moving along a circular path. Figure 2.12 and Physics to Go 2.2 with the rock should be pointed out and used to dispense with the common misconception that the object would move radially rather than tangentially. If the students don’t get around to trying it right away, tie a rubber stopper to a weak string (strong thread) and swing it in a circle overhead. Hold a sharp knife in front of you and ask the class to predict where the stopper will go after you raise the knife so it will cut the string.

Do the “yank the tablecloth without spilling the wine” demonstration, or some variant. This is as much a demonstration of friction as it is of the third law.

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Attach a weak spring to a loaded dynamics cart and show how it speeds up or slows down when the stretched spring shows that a force is acting.

Swing a bucket of water in a circle about a horizontal axis fast enough to keep the water from falling out at the high point. You could calculate how fast the bucket had to be going.

Use a VCR to show scenes of the rotating space station in the movie 2001: A Space Odyssey. Ask the class if the space station in Star Trek: Deep Space 9 rotated.

Artificial gravity in rotating space stations can be a starting point for discussion of O’Neill’s The High Frontier (see the Suggested Readings at the end of the chapter). Or see his cover article in the September 1974 issue of Physics Today. Research with human subjects suggests that the rotation period would have to be two minutes or larger. With the recent push for human exploration of Mars, ways to manage the effects of long-duration space flight are of renewed interest.

The distinction between mass and weight should be stressed again and the fact that weight is caused by something outside of an object. The concept of weightlessness in an orbiting spacecraft is very tricky to explain to students at this level (see Common Misconceptions below).

Lift, then shove a dynamics cart while describing how weight is involved in the former and mass in the latter.

Hang 1-kg and 2-kg masses from a demonstration force scale to show their weights. Also pass them around the class.

Talk about the Apollo astronauts’ experience on the Moon and how they developed their own way of “walking”. A film loop or videotape would be very useful.

I usually work out several examples and/or problems on the board. Incidentally, the momentum form of the second law is presented in Section 3.2.

DHP, page M-18, item Md-2 shows one setup using an air track and photogate timers to demonstrate the second law.

DHP, page M-16, item Mc-2 shows a classic demonstration of inertia, where you hang a heavy ball from a thread and hang another thread from the ball, and you can break the top thread with a gradually increasing downward pull, or break the bottom thread with a quick jerk.

Discuss challenges 1 and 3 at the end of the chapter (the force of a bat hitting a baseball and banked highway curves).

LAB EXERCISE. Attach a rubber stopper to one end of a 1.5-meter piece of strong fishing line. Pass the other end through a 20-cm length of glass tubing and attach it to a weight hanger. The weight of the hanger and masses supplies the centripetal force on the stopper as it is swung in a circle overhead. A small alligator clip can be used as a marker on the string below the tube so that different radii can be used. The period of rotation is measured, and from this, the speed and acceleration of the stopper. Then m times a is compared with the centripetal force. A 10-g stopper, a radius of 1 meter, and a total of 200 g supplying the centripetal force yield reasonable accuracy.

I use the term SI a great deal and try to get the students to automatically use SI units.

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The fact that the states of no motion and of uniform motion are equivalent as far as forces are concerned needs to be pointed out again. Projectile motion, simple harmonic motion, and falling body with air resistance can be treated lightly if desired. The sinusoidal graphs for simple harmonic motion do appear in later chapters.

DHP, pages M-7 to M-11, gives several demonstrations related to projectile motion. Item Mb-14 (a spring gun that launches one ball horizontally and drops one straight down at the same time) is a nice way to show that horizontal motion does not change the effect of the force of gravity.

Use a mass on a spring, a simple pendulum, a metronome, a glider on an air track attached to two springs, or other systems to show simple harmonic motion.

Drop a ping-pong ball and a golf ball simultaneously from 2 meters or more and listen to the former hit the floor later.

Figure 2.26 can lead to a discussion of physics in space. NASA has available many films of physics demonstrations in Skylab and Space Shuttle missions. Page 334 in Sport Science has a table that gives the terminal velocities of several different balls used in sports. It is interesting that a baseball can leave a pitcher’s hand fast enough to equal its terminal velocity when falling, so it is decelerating at 1 g. It must be decelerating at 1 g because the force of air resistance on the ball must be the same as when it is falling at its terminal speed (how could the ball tell the difference between being thrown at its terminal speed vs. falling at that speed?). Here, though, the force of gravity is not present to balance the air resistance force. When falling at terminal speed, the air resistance force is equal to the force of gravity, so with air resistance only (at the same speed), the acceleration is the same magnitude as when you have gravity only.)

LAB EXERCISE. A calibration curve for an inertial balance can be plotted after the periods of known masses are measured. One can plot period versus time, or compute the frequencies and plot them.

Even though chaos is mentioned only briefly in Section 2.5, fun can be had with computer programs dealing with fractals and chaotic dynamical systems. Added fun can be had with actual devices that exhibit chaos—see for example Appendix C, “Chaotic Toys” in Chaotic and Fractal Dynamics, an Introduction for Applied Scientists and Engineers by Francis C. Moon (1992, Wiley) for several interesting ideas.

The fact that a wall or other passive object can exert a force is a new and important concept for students.

In addition to the demonstrations with dynamics carts shown in Figure 2.33, you might look at the many action-reaction and thrust demonstrations in DHP, pages M-17 to M-25.

Discuss Challenge 5 (the tiny gravitational force between two people). The concepts of an inverse square force and a field are important and will be seen again in

Chapter 7. The text relates Newton’s reasoning about gravity being an inverse square law early in Section 2.7 (the Moon is 60 times farther away from the center of the Earth than something on the Earth’s surface is from the center of the Earth, and the acceleration of the Moon is 1/3600th—

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1/60th squared—of the acceleration of, say, a falling apple) but does not go into the more difficult problem of proving that it does not matter that the mass of the Earth is spread throughout the interior of a huge planet. It is quite remarkable that at the Earth’s surface, with the ground right under your feet and nearby soil attracting you much more strongly (in a relative sense) than matter at places inside and on the far side of the Earth—each bit pulling in a different direction—that the overall effect is exactly the same as if all the matter in the Earth were concentrated at its center. (It might be interesting to have students chase down people who can do calculus and are willing to show them the details of the proof.) The Profiles in Physics makes brief mention of Newton’s use of the calculus to prove this, but more clarification is probably needed.

Use a string and a volunteer’s fingers instead of tacks to draw elliptical orbits on the chalkboard as in Physics to Go 2.7. Show that the ellipse is difficult to distinguish from a circle unless the two foci are rather far apart.

Discuss Challenges 6, 7, 8, and 9. Challenge 8 on the stationing of a geosynchronous communication satellite is very interesting and practical. Use the internet to download some “live” (well, 30 minute or so “old”) weather satellite images of your region.

You may have students who have never experienced tides, so a simple description of the phenomenon might be in order. Have some students find (or make) friends who live near the coast and call to ask them about the tides.

The following websites contain valuable simulations you can use for physics exercises:

http://phet.colorado.edu/en/simulations/category/physics http://www.physicslessons.com/iphysics.htm http://physics-‐animations.com/Physics/English/index.htm http://jersey.uoregon.edu/vlab/ http://www.physicslessons.com/iphysics.htm Links to Specific Animations: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/ClassMechanics/VertCircular/VertCircular.html http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/ClassMechanics/TwoBall

sGravity/TwoBallsGravity.html http://www.upscale.utoronto.ca/PVB/Harrison/Flash/ClassMechanics/SHM/TwoSHM.ht

ml http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/ClassMechanics/Damped

SHM/DampedSHM.html

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http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Flash/ClassMechanics/Circular2SHM/Circular2SHM.html

http://physics-animations.com/Physics/English/mech.htm

COMMON MISCONCEPTIONS What?—When there isn’t force, objects can still move with constant velocity? How can the net force of a car be zero if it is moving? Newton’s first law can lead to confusion if students place it in an unintended context. They

might be misled by the need to exert a force on an object to bring it up to speed from rest. But Newton’s first law does not refer to an object’s history, only what is going on right now. If no forces are acting the object simply coasts at constant velocity. They also may be unwittingly thinking of a concept closer to that of energy or momentum when they mistakenly use the term force.

Strongly tied to this confusion is what E. (“Joe”) Redish calls Newton’s Zeroth Law of Motion: “At a time t, an object only responds to forces that are exerted on it itself at time t.” (Am. J. Phys., July 1999, p. 571). Students need to see that if a force acts on an object over a time interval, the object will accelerate during that time interval, but if the force stops, so will the acceleration. So a car must have a net force acting on it while it is speeding up, but not after it has reached and maintains cruising speed. Similarly, an object going in a circle must have a net force on it all the time, because it always has a centripetal acceleration. (If the force were to quit, then the object would necessarily instantly quit moving in a circle and head off in a straight line.)

If a skier is going downhill at a constant speed there’s no force exerted because there’s no acceleration, right? The only flaw in the above question is that it should say “no net force exerted”—there are

clearly forces exerted on the skier by the snow surface (partly frictional, partly supportive), gravity, and air resistance, but they add vectorially to zero (I’m assuming the hill has a constant slope—otherwise constant speed wouldn’t imply constant velocity and therefore zero acceleration).

What is the difference between a vector and a force? Will a force produce a vector? A force is one particular example of a vector quantity. I think this questioner is suffering from

attempting to learn the general concept of a vector before seeing some examples. How can a ball have inertia if it is still? Can there be inertia with an object that is standing still? Inertia, just like mass, is a very elusive concept. I see very little distinction between inertia and

mass until relativity is discussed. Mass is an invariant, yet objects get harder and harder to accelerate as their speeds approach the speed of light, so I would like to say their inertia increases at relativistic speeds even though their mass stays constant.

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Thinking of inertia as a “resistance” to changes in motion simply does not satisfactorily address questions regarding what it is. Perhaps spending some time simply struggling with the concept will convey the deep issue it really is. Encourage research into this subject.

I don’t really understand inertia. How can someone hold a brick on their head and have someone hit it with a sledgehammer and have it not hurt? Perhaps this demonstration should be preceded (or followed, if you prefer suspense) by hanging

the brick from a rope and hitting it on the side with the hammer. If the brick is massive enough, the hammer will more or less bounce off, and it will be clear that a person’s head on the far side would be in little danger. You need something a lot bigger than a chimney brick if you want to hit it with a sledgehammer! Experiment carefully.

When I begin discussing Newton’s third law, I always start by pushing hard on the demonstration table (it’s bolted down). The students can easily be convinced that the table pushes back just as hard on me. Then I use a movable chair or desk and push gently on it (not hard enough to move it). I ask the students about the chair’s force on me, and they say it is the same strength as my push. Then I push hard enough to actually move the chair—how do the forces compare now? Most of them confidently report that now I am pushing harder than the chair is pushing back. They are very surprised when I tell them they are wrong—that the forces are still the same size.

I understand Newton’s 3rd law the best, I think. It’s neat how you can be pushing a huge wall and the reason it doesn’t move is basically because it’s pushing on you too. This is wrong. The reason the wall doesn’t move is because its attachments are able to exert

forces on the wall equal and opposite to your push. If you push hard enough (with some sort of help, obviously) the wall will move.

If the Moon pulls on the Earth as hard as the Earth pulls on the Moon why isn’t the Earth’s orbit around the Sun affected by the Moon? It is. The Earth and Moon actually both orbit about the center of mass of the Earth-Moon

system once a month. This point is actually beneath the Earth’s surface, so the Earth’s true motion looks more like a wobble than an orbit, but this motion is fundamentally involved in causing the tides.

I don’t understand the part where the author talks about the third law with the ball falling towards Earth. How can the ball exert a force on the Earth when there is air between the ball and the Earth? How can the Earth exert an equal force as a ball? The force the ball exerts on the Earth is a gravitational force, no more (or less) mysterious than

the gravitational force the Earth exerts on the ball. It does seem strange that a tiny ball can pull on the Earth just as strongly as the enormous Earth pulls on the ball. Is it at all comforting to note that the effect of the Earth’s gravity on the ball (an acceleration of 9.8 m/s2) is much more dramatic than the effect of the ball’s gravity on the Earth (an acceleration, but one that is immeasurably small because of the Earth’s huge mass)?

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G is so small! Does it really make that much of a difference in calculations? I once had my algebra students compare the cost (per ounce) of different kinds of pop in order

to discover which brand was the best deal. One of them reacted to the tiny values (1.28 cents/oz for cheap cola vs. 2.56 cents/oz for brand name colas in convenience stores, for example), saying something like “who cares about pennies?” —totally missing the point that the differences really add up when you purchase in quantity. The same goes for the gravitational constant, G. It needs to emphasized that its small value does imply a kind of “weakness” about gravity (you need a lot of mass to see any effect at all), but on the other hand, one could say that we just happened to choose inconvenient force, mass, and length units when defining newtons, kilograms, and the meter, all of which conspire to require a tiny G value to make Newton’s law of gravitation work out right. New units could be invented that would make a G value of 1 possible (you could leap ahead for a moment and talk about the definition of the ampere). But make the point that the value of G, even though its numerical size depends on the above factors, is a deep universal fact of nature—something worth pondering.

It says that the only force on a spacecraft in orbit is gravity and that the objects inside are weightless. I don’t understand. If there’s gravity, why is everything weightless? We are stuck with two meanings of the term “weight”—weight as the force exerted by gravity

on an object (the one we try to convey in physics) vs. weight as the force supporting an object as it sits on a bathroom scale, for instance. If you are simply standing motionless on a scale, the downward force of gravity is exactly balanced by the equal strength upward force of the scale—there seems little point in fussing a lot about the distinction between the two. We may as well think of the scale force as representing the weight. But this is a very specific situation. In orbit I suppose you could glue your feet to a bathroom scale and push yourself toward a wall of your ship, feet first—the scale would register a force as you made contact with the wall, and the strength and duration of the force would vary depending on whether you bent your knees to soften the blow or had them “locked” as well as whether or not you decide to “kick off” from the wall in some way. Here the scale force is indeed what causes your acceleration during your impact with the wall—it has nothing to do with gravity at all. After you break contact with the wall the scale will read zero—you are “weightless” in the second sense above. Yet all this while gravity has been acting on you—you are definitely not “weightless” in the first sense. The point is that in orbit, gravity gives rise to a centripetal acceleration, whereas on the Earth’s surface, it gives rise to a contact force that counters the gravitational force. It is hard not to think of this contact force as “weight”, and that’s what leads to all the confusion.

I don’t understand why if gravity is the only force acting upon a spacecraft, you are weightless. Wouldn’t gravity pull you down? Gravity does pull you down—it is causing the centripetal acceleration that both you and your

craft experience. “Down” must be thought of as “toward the center of the Earth,” however, not “pulled out of orbit” or something.

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In section 2.8 it talks about shooting a cannonball into orbit. Why do things stay in orbit? Why don’t they go like this? They can! To achieve a perfectly circular orbit

the launch velocity has to be a very precise value—if it’s slower than that the cannonball will fall to the ground (as shown in Figure 2.40). Any faster and the trajectory will curve outward as shown here. But even then, there are two possibilities: the orbit can curve back around following an ellipse or (if the cannonball is launched fast enough) it can continue outward forever following a hyperbolic path.

Is the gravitational force on a person in space zero because of weightlessness? The premise is in error—the gravitational force

is not zero on a person in space (though it might be vanishingly small way out in deep space, far from any other masses). Also, weightlessness is not a cause of anything, just a characterization of a particular situation.

When an object reaches terminal speed does the object’s speed and air resistance equal out? No. The forces of gravity and of air resistance upon the object are equal in strength. It doesn’t

make sense to compare a speed with air resistance (a force). Terminal velocity is basically your highest speed. Not quite. If you fall from rest, your terminal velocity will be the highest speed you attain, but

it is possible to throw something downward fast enough so that the air resistance force is larger than the force of gravity at the start, slowing the fall to terminal velocity.

I’m a little confused on how the tides work. When the water is closer to the Moon and is “heaping up,” does that mean the tide is out? “Out” as in “the tide is out” means that the water has flowed oceanward—“out” away from the

beach, meaning low tide. “Out” as in “pulled out into a bulge” means upward, away from the Earth’s surface. At the location of a bulge, the tide would high, or “in.”

Does the space shuttle shut off its engines in orbit? Yes. But often movies show spaceships with engines blazing continuously. I enjoy finding

physics flubs in science fiction shows that demonstrate misunderstandings such as this one (some of my students thought this rather weird).

What exactly moves a rocket forward in space? What does the thrust push against?

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The thrust pushes against the rocket! Of course, the question may imply a mistaken belief that the thrust needs a way to get “leverage” or a sort of “foothold” on some external medium, which is not the case. This is a good place to discuss the idea of a system again. When the rocket and its exhaust are considered together, the system as a whole doesn’t get anywhere—it just spreads out—the rocket going forward and the exhaust particles going backward (and the forces are all internal). Thinking of the rocket by itself is actually more complicated.

CONSIDER THIS— Where does gravity come from and why is it there? How is the Earth pulling on the Moon? I don’t think questions of this type should be ducked. Instead, point out that this sort of curiosity

and wonder is exactly what drives physicists. (Good luck finding the answers.) How come gravity pulls harder on objects with greater mass? This is actually a very deep question. It is fascinating that gravity pulls harder in exact

proportion to how much more massive the object is—that’s why everything free falls (near the Earth’s surface) with an acceleration of 9.8 m/s2.

If gravity affects a person with a larger mass more, and the force of gravity works on them harder, why can’t smaller people consistently jump higher and farther than big people? Are there maybe too many variables in people for this to hold true? Investigate this directly with a bunch of large and small student volunteers.

Since we’re on the subject of forces, what would happen if all the people in China jumped at the same time and landed at the same exact moment? This invites experimentation: How high can people jump?; calculation—what force do they

exert on the Earth during a jump?; and a bit of research—how many people are there in China? Collect the answers, then do F = ma for the Earth. Getting the force exerted during a jump really brings together a lot of material: Use the free-fall equations to calculate the jumper’s initial velocity from their jump height. Then time the jump—more specifically, time from when they first start pushing up with their legs to when their feet leave the floor (videotaping the jump and counting video frames works well). Use this time plus the initial velocity to get the average acceleration. From that get the average net force on the jumper from Newton’s second law. This net force is how much the force of their legs exceeds their weight, so it must be added to their weight to get the force generated by their legs. Newton’s third law then says the same strength force acts on the Earth. Multiplying by the number of people in China will give the force on the Earth, and F = ma can again be applied using the Earth’s mass this time. The result will be tiny, of course—a good demonstration of the enormity of the Earth, but hopefully also a demonstration that physics can provide explicit quantitative answers to many questions.

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Several students pointed out to me that even landing with a parachute is not exactly a soft landing (the text mentions a landing speed of 10 mph, another I checked said 25 mph). Urge your students to make some phone calls to skydiving outfits to get a fuller story. To get a real feel for how hard such landings are, compute the height of a table you might jump from in order to hit the floor with these speeds (it is a little scary). Chapter 3 shows how to do this calculation using energy conservation.

Consider two satellites in orbit around Earth. Both are in the same orbit, but on opposite sides of the Earth. Both satellites have a limited amount of maneuvering fuel available. What kinds of maneuvers would be done to get these satellites close enough to each other to “link up”? When the maneuvering is done, they will be in the same orbit as the beginning. The operation must be done as quickly as possible, with a minimum of fuel use. What would be a likely approach? I have a feeling the requirements “as quickly as possible” and “minimum of fuel use” are

incompatible (and too vague). This is a very difficult problem. Nevertheless, it could be used to motivate detailed study of real space missions. Request a press kit for a space shuttle mission—I found it a trivial matter to get the phone number for Johnson Space Center in Houston from Directory Assistance and got a lot of free information with just a phone call (OK, two phone calls).

If you were in the space shuttle and some outside force pushed the shuttle (you, floating in midair inside, still), would you stay in the same place as the shuttle moved until some part in the interior hit you, or would you float in the same spot in relation to the interior? The wall would move over and hit you. I often wonder if some television reporters mistakenly

think the other answer makes sense—I have seen them act as if they expect astronauts doing untethered EVA’s in the shuttle’s cargo bay to be suddenly whisked away if they drift a bit too far out. Impossible. Now, if one of their little maneuvering thrusters got stuck on—that would be another matter.

How does a probe that is in orbit fall out of orbit and then tumble to Earth? Air resistance in low Earth orbit is slight, but not zero. Over time the orbit slowly decays lower

into gradually thicker atmosphere where the air resistance is greater, speeding up the rate of decay until the reentry. Communication satellites in much higher geosynchronous orbits do not suffer this fate. By the way, the fact that geosynchronous orbits have to be above the equator of a planet is often missed (especially on sci fi shows). Have the students try to see why there is no other possibility. Are weather satellites geosynchronous? If they are, and therefore orbit above the equator, can we not get weather photos of the polar regions of the Earth, then?

Since there is always air resistance why do physicists ignore it when they study projectile motion? Wouldn’t whatever they find be wrong because they didn’t take air resistance into account? The book keeps wanting to “ignore air resistance”, or “ignore gravity” or “ignore friction.” Well, those things do exist and do play a part in daily life. Why calculate

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things or talk about things that don’t really happen on Earth? I don’t care to talk about imaginary probabilities. This point was addressed back in Chapter 1. To get a handle on the more difficult real-life

problems, we have to start simple. Also, more fundamental behavior may be hidden by real-life complexity. Stripping that away may lead to better understanding. The first question is correct, though—ignore air resistance in a real-life situation and you will go wrong.

Find out more about the history of friction and lubrication. Research friction, wear, and lubricants in modern engineering and biomechanics. Look into how car braking systems work. Try searches for anti-lock and ABS brakes and tribology.

Investigate the conflict between Newton and others like Hooke and Leibniz.

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ANSWERS TO MAPPING IT OUT! 1. Universal gravitation concepts:

• Every object exerts a gravitational force on every other object via the gravitational field it creates.

• Gravitational forces between masses are • proportional to each object’s mass, • inversely proportional to the square of the distance between them,

• given by the formula 1 22 .GmmFd

=

• G was measured by Cavendish using a torsion balance. • G = 6.67 ⋅ 10–11 N·∙m2/kg2.

• Earth’s mass (6 ⋅ 1024 kg) can be computed from G and g. • Every planet has a different “g.” • Orbits of planets and comets (and everything else) around the Sun are ellipses. • Achieving Earth orbit is like throwing a projectile horizontally at just the right speed

(about 7,900 m/s). Here’s an attempt at prioritizing— Law of universal gravitation; mass creates a gravitational field; the field obeys an inverse

square law; gravity is one of the four fundamental forces. The gravity force between a pair of objects (weight is an example) is proportional to each

mass in the pair; gravity can be measured with a Cavendish torsion balance; it reveals G = 6.67 ⋅ 10–11 N·∙m2/kg2. G and g are related; every planet has its own “g.”

Orbits: satellite orbits imagined in Newton’s ‘cannon’ thought-‐experiment; orbital velocities can be calculated; orbits of planets, comets, etc., are ellipses with the Sun at one focus.

Tides: on Earth, primarily due to Moon’s gravity and the fact it is different on the near and far sides of the Earth; neap and spring tides are due to Sun’s added influence.

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2.

ANSWERS TO QUESTIONS 1. Force is a push or a pull that usually causes distortion and/or acceleration. Examples:

weight (force of gravity) pulling down on everything, force of friction between you and your seat, force of friction holding nails and screws in desks, cabinets, walls, etc., and so on.

2. Weight is the force of gravity acting on a body. An object is truly weightless only if there is no other body around to exert a gravitational force on it.

3. The two forces acting on the book are the force of air resistance acting toward the rear of the car and a force of static friction between the car’s roof and the book acting forward.

Net force

which may be

yields

Simple Harmonic

Mot ion

Constant

and

Opposite to velocity

yields

Straight line motion

wi th

Pe rpendicular to velocity

yields

Pa rallel to velocity

yields

Decreasing accelerat ion

and

Terminal speed

Decreasing speed

Proport ional to and opposite to displacement

Circular motion

Proport ional to and opposite to velocity

yields

Straight line motion

Incr easing speed

wi th

2-‐16


4. Static friction is the type of friction that acts when there is no relative motion between two bodies. Kinetic friction acts when there is relative motion between two bodies in contact. One example of both acting at the same time is a moving car: static friction acts between the road and the tires, and kinetic friction acts between the air and the outside of the car. (See page 49.)

5. There are friction forces at both contact surfaces—between the hand and the book on top, and between the book and the table underneath. If the friction force between the hand and the book is larger than that between the table and the book, the book will be dragged along by the hand. In this case static friction acts between the book and the hand, and kinetic friction acts between the table and the book.

In the case where the book stays put and the hand slips, it’s very tempting to think that the friction between the book and the hand is less than that between the book and the table. But that is not true—the two friction forces must be equal! The book is motionless, so it has zero acceleration and therefore zero net force. The two friction forces (acting in opposite directions) must exactly cancel each other. Here static friction acts between the book and the table, and kinetic friction acts between the hand and the book.

Do several trials with your hand and a real book. Press gently at first so your hand slips, then gradually increase the pressure. In successive trials where the book does not move, the static friction between the book and table simply grows in strength to match the increasing kinetic friction force applied by your hand. When you reach the pressure where the book begins to slide, your hand is exerting a friction force larger than the maximum possible static friction force between the book and table at that pressure.

6. An external force is one caused by something outside of the body under consideration. An internal force can’t accelerate an object because, by Newton’s third law, any internal force acting in some direction on one part of a body would produce an equal but opposite force acting on another part. The two forces would cancel each other.

7. Because the applied forces are of the same magnitude (for this question a magnitude of 1), the combined forces of Player A and the other teammate on Player B will the same as a single force 1.41 (the square root of 2) times as strong on B toward the southeast. What B’s subsequent motion will be depends on whether he was already moving (how fast and in what direction) or not moving, how massive he is, whether other forces in addition to the two mentioned also act on B, and how long the forces continue to act. Only in the simple case of B being at rest when the other players hit him can we say he will go toward the southeast. Newton’s second law can be applied to solve for the motion in all the other cases, with more work involved.

Diagram to support the answer:

2-‐17


Force by player A (FA) green vector Force by player B (FB) red vector FA (east) FR (Southeast) FB(south)

8. A steady centripetal force causes an object to move along a circular path. If the force

disappears, the object moves in a straight line with constant speed, its direction being that in which it was traveling at the instant the force disappeared. (See page 44.)

9. Constant speed and heading together imply constant velocity and therefore zero acceleration. Newton’s second law then says the net force on the train car is zero.

2-‐18


10. Mass is an intrinsic property of a body that determines its acceleration when a net force acts on it. Weight is a force acting on a body caused by something outside the body, like the Earth. Weight depends on where an object is—mass does not. (See page 48.)

11. To accelerate the ball during the act of throwing, the astronaut’s hand will have to apply the same force to the ball in orbit as on the ground, because Newton’s second law says this force depends only on the mass of the ball and the acceleration, both of which will be the same in orbit as on the ground for an identical horizontal throw. The same goes for the force during the catching process, so throwing and catching will “feel” the same in orbit as on the ground.

What’s missing in orbit are the friction and support forces of the ground on the astronauts, so repeated throws or catches in orbit will make each astronaut move away from the other at an increasing speed. They will start spinning head-‐over-‐heels slowly as well, if they catch and throw at roughly head level in the usual way.

12. The track is shaped in such a way that it is accelerating the roller coaster downward. Riders would feel “negative g's.” One shape that would cause this is a steep hill: the track slants upward, then quickly slants downward.

13. If the propeller exerts the same forward force on the craft in both cases, it won’t matter whether it pushes from behind or pulls from the front—the motion of the plane or boat will be the same.

14. The rocket’s mass decreases as its fuel is consumed. The same net force acting on a smaller mass results in a larger acceleration.

15. The SI is a system of units within the metric system that is internally consistent. The units in the answer of any legitimate calculation are SI, provided the units of the input quantities are also SI.

16. The watch and the arrow hit the ground at the same time. The acceleration of the arrow is the same as that of the watch because horizontal motion does not affect vertical motion.

17. For a mass oscillating vertically, as in Figure 2.24, the force varies continuously from the maximum (directed downward) at the high point of the motion, to zero at the equilibrium point, to the maximum (directed upward) at the low point. The size of the force is proportional to the distance from the equilibrium position. By Newton’s second law, the acceleration undergoes the same variation.

18. If a 0.5-‐kg object hanging from a spring stretches it by 0.30 m, we know the weight (force) F = mg. We can use the equation on page 60, F kd= to solve for the spring constant k.

20.5 kg 9.8 m/s 4.9 N4.9 N 16.33 N/m

0.30 m

F mgFF kd kd

= = × =

= ⇒ = = =

Now, how much will the spring be stretched if a 1-‐kg object is suspended from it?

2-‐19


9.8 N 0.6 m16.33 N/m

FF kd dk

= ⇒ = = =

19. As the speed of a falling body increases due to the force of gravity, the size of the force of

air resistance (acting upward) increases as well. Therefore the net force decreases. This continues as the speed increases until the force of air resistance is the same size as the force of gravity and the net force is zero. Then the acceleration is zero, and the speed is constant, equal to the terminal speed.

20. The ping-‐pong ball decelerates initially because the force of air resistance is greater than the weight. After it slows down to its terminal speed (20 mph), it falls at that speed until it hits the ground.

21. When sitting, the two forces on you are the force of gravity (weight) pulling downward and the force of the chair pushing upward. The equal and opposite force to that of the Earth’s gravity on you is the pull (also due to gravity) that your body exerts on the Earth. For the chair, it is your downward push (caused by contact) that is the equal and opposite force.

22. The road exerts a force upward (supporting the car so gravity doesn't accelerate it downward) and a force forward (the equal and opposite force to the force of the tires pushing backward).

23. Your legs exert a downward force on the ground. The ground exerts an equal and opposite force on you and that force accelerates you upward.

24. They move the same in the two cases. By the third law of motion, the forces are always equal and opposite, no matter which skater initiates a force.

25. Cavendish balanced two masses on a rod suspended by a thin wire. Two larger masses were placed next to them so that the gravitational forces acting on the two smaller masses caused the wire to twist. Cavendish used the amount of twist to measure the size of the gravitational force acting on each mass. (See Figure 2.39.)

26. (a) The first part of the question deals only with Newton’s law of universal gravitation, which states that the force is proportional to the masses of both objects. Because the container of marbles weighs more than the container of Styrofoam beads, the container with the marbles has a greater mass than the container of Styrofoam beads. Therefore the gravitational force between the container of marbles and Earth is greater than the gravitational force between the container of Styrofoam beads and Earth. However, the gravitational force between the two containers is the same (equal and opposite) for each container. Therefore, the answer to this part of the question depends on your frame of reference.

2-‐20


(b) The containers both experience the same acceleration due to gravity (g = a constant). When the two containers are released, they will reach the ground at the same time, as demonstrated by Galileo when he dropped the two objects from the leaning tower in Pisa.

27. The weight of everything on Earth would increase a billion times, so buildings would collapse, people would be pulled to the ground and be killed as internal organs would suddenly weigh tons, satellites would be pulled into much lower orbits, the air pressure would increase dramatically and crush many things. Overall it would be a catastrophe.

28. Events that involve jumping or throwing things for distance would be greatly enhanced and records for distance would be broken! Swimming, bicycling (on level terrain), and similar races would not be affected very much. Running races might be adversely affected because the runner would go high in the air with each step and perhaps would not be able to keep up a fast pace.

29. Tides are caused by the gravitational pull of the Moon on the Earth and its oceans. Because the water on the side of the Earth closest to the Moon experiences a stronger force and the water on the opposite of the Earth experiences a weaker force, tidal bulges appear. As the Earth rotates, parts of its surface are alternately in a tidal bulge (resulting in high tide) and between the bulges (low tide).

30. What matters is the difference in the gravitational forces on the near and far sides of the Earth—the size of the Earth is a much more substantial fraction of the Earth-‐Moon distance than it is of the Earth-‐Sun distance, so it makes a bigger difference (due to the 1/r2 dependence) for the Moon’s gravity than the Sun’s gravity. The Moon’s gravity is about 7% weaker on the far side of the Earth compared to the near side, while the Sun’s gravity only varies about 0.02% over this range. But this is only part of the answer because this effect is largely compensated for by the Sun being much more massive than the Moon (7% of the Moon’s gravity is roughly only double 0.02% of the Sun’s gravity). A full explanation of the tides must include the motion of the Earth about the Earth-‐Moon center of mass. See the special topic on pp. 273-‐75 of University Physics by Harris Benson for a good discussion.

31. (a) Newton’s second law is best for calculating the net force on a car as it slows down. (b) The law of universal gravitation is best for calculating the force exerted on a satellite by

Earth. (c) Newton’s second law (F = ma ⇒ W = mg) shows the mathematical relationship

between mass and weight. (d) Newton’s first law explains the direction that a rubber stopper takes after the string

that was keeping it moving in a circle overhead is cut. (e) Newton’s third law explains why a gun recoils when fired. (f) Newton’s third law can be used to explain why a wing on an airplane is lifted upward as

it moves through the air.

2-‐21


32. For this question, we use Newton’s second law, F = ma. Because the slope of a force versus acceleration graph is defined as the rise (force) divided by the run (acceleration), we know the slope of each graph is the mass (m = F/a). You are then asked to rank the figures according the mass of the object, from smallest to largest. By inspection, you can see the slope of Figure 2 is largest, which thus corresponds to the largest mass. Also by inspection, we see that Figures 1 and 4 have the same slope, which is less than that of Figures 2 and 3, and therefore have the smallest mass. Figure 3 has a slope greater than Figures 1 and 3, but less than the slope of Figure 2.

Figure Number Mass Rank

1 Smallest 1 2 Largest 3 3 In between largest and smallest 2 4 Smallest 1

33. The key to answering the question is the statement that the car/trailer systems are moving “at constant, albeit different” speeds. This means for each car/trailer combination there is no change in speed and therefore no acceleration. If there is no acceleration, then there is no net force acting on the car/trailer combination. If there were a net force, the car/trailer combination would be speeding up or slowing down. Here are two quotes from the text that support this conclusion: “Upon first reading Newton’s first law does not seem to be terribly profound. Obviously, an object will remain stationary unless a net force causes it to move. But the law also states that anything that is already moving with a certain velocity will not speed up, slow down, or change direction unless a net force acts on it….A car traveling at a constant velocity has zero net force acting on it because the various forces (including air resistance and gravity) cancel each other…” (page 50); “The acceleration of a stationary object or of one moving with uniform motion (constant velocity) is zero. By the second law, this means that the net force is also zero (see Figure 2.20). It is that simple. When the net force on a body is zero, we say that it is in equilibrium, whether it is stationary or moving with constant velocity…” (page 57). 34. An object with mass m is attached to a spring with spring constant k and is stretched to the right by 0.20 m and released. Rank the oscillation periods of the mass-‐spring combinations shown in the Table below from shortest to longest. We use the following equation for the frequency of oscillation for a mass-‐spring combination.

2-‐22


1 0.1592

k kfm mπ

= =

Combination Spring Constant k Mass m Frequency f

Period T Rank

(a) 0.5 N/m 0.25 kg 0.22486 s-‐1 4.447212 s 2 (b) 0.5 N/m 0.50 kg 0.159 s-‐1 6.289308 s 3 (c) 0.5 N/m 1.00 kg 0.11243 s-‐1 8.894425 s 4 (d) 1.0 N/m 0.25 kg 0.318 s-‐1 3.144654 s 1 (e) 1.0 N/m 0.50 kg 0.22486 s-‐1 4.447212 s 2

ANSWERS TO EVEN NUMBERED PROBLEMS 2. The mass m of a child who weighs 300 N is 30.6 kg, or 2.1 slugs. 4. The mass of a 1,130-‐kg elephant has a weight in newtons and pounds of (a) 11,074 N (b) 2,491.65 lb 6. The net force on the motorcycle is F = -‐1,500 N. 8. The mass m of an object subjected to a force of 60 N with a resulting acceleration of 4 m/s2

is m = 15 kg. 10. Passengers on the Kingda Ka roller coaster are accelerated uniformly to a speed of 57 m/s

(128 mph) in just 3.5 s. (a) The acceleration experience by the passengers is a = 16.3 m/s2, or 1.7 g’s. (b) The force on a 65-‐kg passenger is F = 1,059.5 N, or 238.2 lb. 12. A jet aircraft with a mass of 4,500 kg has an engine that exerts a thrust of 60,000 N. (a) The jet’s acceleration at take-‐off is a = 13.3 m/s2. (b) The jet’s speed after it accelerates for 8 s is v = 106.4 m/s. (c) The jet travels a distance d = 425.6 m in 8 s.

2-‐23


14. An 80-‐kg sprinter who accelerates uniformly from 0 m/s to 9 m/s in 3 s (a) has acceleration a = 3 m/s2 (b) experiences a force of F = 240 N (c) travels a distance d = 13.5 m 16. A catapult accelerates an 18,000-‐kg jet from 0 to 70 m/s in 2.5 s. (a) The acceleration of the jet is a = 28 m/s2, or 2.9 g’s. (b) The jet travels a distance d = 87.5 m while accelerating. (c) The catapult exerts a force F = 504,000 N on the jet. 18. An airplane built to withstand a maximum acceleration of 6g has a mass of 1,200 kg. A

force of 70,560 N is necessary to cause this acceleration. 20. A 600-‐kg race car rounds a curve of radius 400 m at 60 m/s. (a) The car’s centripetal acceleration is a = 9 m/s2, or 0.918 g's. (b) The centripetal force acting on the car is F = 5,400 N. 22. A 0.1-‐kg ball is attached to a string and whirled around in a circle overhead. The string

breaks if the force on it exceeds 60 N. When the radius of the circle is 1 m, the maximum speed of the ball is v = 24.5 m/s.

24. A centripetal force of 200 N acts on a 1,000-‐kg satellite moving with a speed of 5,000 m/s in a circular orbit around a planet. The radius of its orbit is 125,000,000 m.

26. At a distance of 10,000 miles from Earth’s center, the gravitational force on a space probe is 600 lb. The force on the probe

(a) at 20,000 miles from Earth’s center is 150 lb (b) at 30,000 miles from Earth’s center is 66.7 lb (c) at 100,000 miles from Earth’s center is 6 lb 28. A mass of m=0.75 kg is attached to a relaxed spring with k=2.5 N/m. The mass rests on a

horizontal, frictionless surface. If the mass is displaced by d=0.33 m, what is the magnitude of the force exerted on the mass by the spring?

(2.5 N/m)(0.33 m) 0.825 NF kd= = = If the mass is released to execute simple harmonic motion along the surface, with what

frequency will it oscillate?

-12.5 N/m0.159 0.290 s 0.290 Hz0.75 kg

f = = =

2-‐24


ANSWERS TO CHALLENGES 1. The force on a baseball as it collides with a bat is not transmitted from the person swinging

the bat. This is a collision between two objects moving in opposite directions. As soon as their surfaces are in contact, the forces arise because of Newton’s first law—they can’t occupy the same space, so they have to change their speed, which requires forces. As the ball and bat deform each other, their stiffness causes equal and opposite forces to act.

2. Two forces, one equal to 15 N pulling to the left and another equal to 40 N pulling to the right, act on a 50-‐kg crate resting on a horizontal surface.

(a) The net force acting on the crate is 40 N – 15N = 25 N.

(b) The horizontal acceleration of the crate is 2

2net force 25 kg m/s 0.5 m/smass 50 kg

a ⋅= = = .

c) The horizontal speed of the crate after 5 s if it starts from rest is ( )( )20.5 m/s 5 s =2.5 m/s.at= =v

(d) The crate travels distance ( )( )( )22 212 0.5 0.5 m/s 5 s = 6.25 md at= = during this time.

3. The centripetal force necessary to keep a car on a flat road is supplied by the static friction between the tires and the road (Figure 2.19), and the force acting downward on the road is the weight of the car (W = mg). On a banked curve, there is a component of the weight acting to hold the car on the curve (similar to the force acting on a block on an inclined plane in Figure 2.5).

4. Two equal and opposite forces cancel each other if they are acting on the same object. In the carriage-‐horse example, one 400-‐N force acts on the carriage and the other acts on the horse. The actual motion of the horse and carriage depends on the net force exerted on each. The frictional forces acting between the road and the horse and between the road and the carriage have to be taken into account. In this case, the latter force is less than 400 N so there is a net force on the carriage and it accelerates.

5. The force acting between two 70-‐kg people standing 1 m apart is given by the following.

( )( )( )

( )

11 2 2

2

7

6.67 10 N m /kg 70 kg 70 kg

1 m

3.3 10 N

F−

−

× ⋅=

= ×

6. Because of the Earth’s rotation, objects on its surface (except at the poles) are moving to the east. The closer they are to the equator, the higher their speed. So objects launched towards the east from near the equator require a bit less fuel to reach orbital speed since they have a running start. For example, rockets sitting on the launch pad in Florida are already moving 400 m/s towards the east.

2-‐25


7. Why are the rockets used to put satellites and spacecraft into orbit usually launched from the equator? Two factors are involved here, both caused by the Earth’s rotation. The main factor is that part of the gravitational force acting on objects is needed as a centripetal force because objects are moving in circles around the Earth’s axis. The closer an object is to the Earth’s equator, the higher its speed and the greater the centripetal force required. The remaining gravitational force causes objects to accelerate while falling. The second factor is that the rotation makes the Earth bulge at the equator so that objects successively closer to the equator are farther away from the Earth’s center. In particular, objects at the Earth’s equator are about 13 miles farther from its center than objects at the poles. By the universal law of gravitation, the gravitational force is therefore less strong near the equator so that falling bodies have lower acceleration there.

8. A 200-‐kg communications satellite is placed into a circular orbit around Earth with a radius of 74.23 10 m× (26,300 miles)

(a) The gravitational force on the satellite is

( )( )( )

( )11 2 2 24

27

6.67 10 N m /kg 200 kg 6 10 kg

4.23 10 m

44.7 N.

F−× ⋅ ×

=×

=

(b) The speed v of the satellite is calculated from the centripetal force F.

( )( )

22

72

2 6 2 2

44.7 N 4.23 10 m200 kg

9.46 10 m /s3,075 m/s

m FrFr m

= ⇒ =

×=

= ×=

v v

v

vv

(c) Show that the period of the satellite is 1 day.

( )( )( )7

8

3

circumference

2 3.14 4.23 10 m23,075 m/s

2.66 10 m 86,388 s3.075 10 m/s

1 day 24 h 60 min 60 s 86,400 s

d tdt

rt π

=

= =

×= =

×= =×

= × × =

v

v v

v

(The discrepancy is due to round off error.)

2-‐26


9. Calculate the mass of the Earth as outlined in Section 2.7.

( )( )

2

2

22 6

11 2 2

14 2 224

11 2 2

9.8 m/s 6.4 10 kg6.67 10 N m /kg

4.014 10 m kg /s 6.0 10 kg6.67 10 N m /kg

GM gRg MR G

M −

−

= ⇒ =

×=

× ⋅× ⋅= ≅ ×

× ⋅

10. Is there a contradiction between the law of universal gravitation and the statement that the gravitational force acting on a falling body near Earth’s center—its weight—is constant? Yes, there is a contradiction, but as a practical matter it can be ignored. When we consider real life free-‐fall problems we usually think of motion near the surface of the Earth over vertical distances on the order of meters. The r in the law of universal gravitation is the distance from the falling body to the center of the Earth, which is thousands of miles. The range of variation of r during the free fall is so small compared to r itself that we can treat r as constant in our approximate calculations and introduce very little error.

The exact calculation can be done using more sophisticated mathematics or a computer program.

Documents

Ch02 IM SDP2 KM - Solution Manual & Test Bank Store · two springs, or other systems to show simple harmonic motion. Drop a ping-pong ball and a golf ball simultaneously from 2 meters