FORCE AND NEWTON S LAWS OF MOTION

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1 FORCE AND NEWTON S LAWS OF MOTION 4. ORGANIZE AND PLAN In what ways would a more massive gol ball change its light? We need to thin o what orces will act on the gol ball, and how the ball s light will be aected by these orces. SOLVE Once a gol ball is in light, it experiences orces due to air resistance and wind. Newton s second law o motion ( F = ma ) says that, or a given orce, a more massive gol ball will experience less acceleration than its lighter counterpart. Thus, or example, the drag orce due to air resistance will produce less o a deceleration or a heavy gol ball, so the heavier ball will maintain its speed more eectively than the lighter ball. Forces due to spinning o the gol ball ( slice ) will also produce less acceleration on the heavy ball, so it will ly straighter. REFLECT One might thin that or a given gol-club-swing, a heavier gol ball would experience less acceleration than a lighter ball. I only things were so simple The initial speed o a gol ball depends on a complex relationship between the speed o the gol club swing and the hardness o the ball. Since a heavier ball will liely be harder (more dense), it will react dierently than a lighter ball when you hit it, but there is no guarantee that it will experience more acceleration. There are entire boos written on the subject o the physics o gol. For those interested, more inormation is available with a quic inter search.. ORGANIZE AND PLAN Review the irst section o chapter 4 or a good discussion o mass. SOLVE You can thin o mass in two dierent ways. First, it is a measure o the amount o matter in an object (e.g., the number o sub-atomic particles in an object). This interpretation o mass is oten conused with weight, but weight is the orce applied on a body due to gravity, whereas mass is simply the amount o matter in an object. For this reason, this interpretation o mass is sometimes reerred to as gravitational mass. In outer space, where there is no gravity, you would have zero weight, but your (gravitational) mass would be unchanged. The second interpretation o mass is that it is the resistance all objects give to a change in their velocity. This second interpretation o mass is reerred to as inertial mass, because the term inertia is the quality o an object to resist a change in its velocity. REFLECT I you wonder i gravitational and inertial mass are the same, then you are in good company. Einstein posed this question and spent a large part o his career answering it. The answer, as ar as is presently nown, is yes. 3. ORGANIZE AND PLAN We will consider dierent orces and how they aect the acceleration and velocity o an object. This involves thining about Newton s second law ( F = ma ). Remember that gravity is given by Eq. 4. ( w= mg ), drag orces always act to oppose the velocity o an object, and that the orce on an object is simply the vector sum o all the orces acting on the object. SOLVE Just beore she opens her parachute,. The orce due to gravity is w= mg (toward the center o the Earth).. The drag orce cancels the orce due to gravity. 3. The orce is the sum o gravity and drag, and is zero due to. 4. Because o 3, she is not accelerating (Newton s second law). 5. Because o 4, her velocity (which is oriented downwards) is constant. 4.

2 4. Chapter 4 Just ater she opens her parachute:. The orce due to gravity is w= mg (toward the center o the Earth).. The drag orce increases dramatically, and is oriented upwards. 3. Net orce is the sum o gravity and drag, and is pointing upwards due to. 4. Because o 3, her acceleration is upwards. 5. Because o 4, her velocity (which is oriented downwards) decreases. As she continues now towards the Earth, the orce on her due to gravity will essentially not change. The magnitude o the drag orce will decrease as her velocity is reduced, but will always point upward. Because the drag orce is changing, her orce will change as well, until the drag orce cancels the orce due to gravity. As the drag orce changes, her acceleration will change (Newton s second law). Since the drag orce and hence her acceleration is oriented opposite to her velocity, her velocity will decrease. When she reaches her new terminal velocity (i.e., when drag orce cancels gravity so orce is zero), she will no longer accelerate, and her velocity will be constant until she reaches the Earth. REFLECT The orce due to the Earth s gravity depends on how ar you are rom the center o the Earth, but or the distances considered in this problem, we can ignore this eect. 4. ORGANIZE AND PLAN When object A (the box, in this case) rests on object B (the incline, in this case), a normal orce must act on A to counteract the orce on A due to gravity (notice that we are ignoring the orces acting on B since we are only interested in the orces on A). Some rictional orces may also come into play in this problem. We will draw a diagram to schematically illustrate these orces. SOLVE There are three orces on the slab. Incline F F N Slab F g The irst is the orce due to gravity, which acts downward. The second is the normal orce applied on the slab by the incline on which the slab rests. The normal orce acts in a direction perpendicular to the incline, and hence is not parallel to gravity. The third orce is due to riction between the slab and the incline, and acts to oppose any motion on the part o the slab. Thus this orce will be parallel to the slab, since it can only move on this plane. REFLECT The riction orce may be the static rictional orce, or the iic rictional orce, depending on the nature o the suraces o the slab and the incline (i.e., i they are rough or smooth). The magnitude o the riction orce will depend on the suraces o the slab and the incline. I the static riction coeicient is large enough, the static rictional orce will be large enough to the component o the gravitational orce that acts along the incline as shown in the igure above, and the bloc will not move. I the static coeicient is not suiciently large, the bloc will begin to slide and the coeicient o iic riction will then come into play in determining the riction orce. 5. ORGANIZE AND PLAN Uniorm circular motion is caused a centripetal ( center-seeing ) orce, which, as its name implies, is oriented toward the center o the circle. Review Figure 4.3 in the text or a clear picture o the orces at wor. SOLVE For a car on a lat curve (not baned), the normal orce exerted by the road on the car is oriented straight up, opposing gravity. On a baned curve, the normal orce is still perpendicular to the road, but it is no longer parallel to gravity. The steeper a curve is baned, the larger will be the component o the normal orce toward the center o the curve, and this is the direction o orce needed to mae the car execute circular motion. Thus the normal orce rom a baned curve aids the car in turning. I a car is going too ast on an inner section o the trac, the riction between the tires and the road will not be able to exert the required centripetal orce to execute the turn, and the car will slide outward. Here the trac will be

3 Force and Newton s Laws o Motion 4.3 more baned, providing an extra orce component to assist the tires in maintaining the turn, and (hopeully) the car will regain its circular trajectory. REFLECT When running or siing around a turn, why do these athletes lean their bodies toward the center o the turn? 6. ORGANIZE AND PLAN See the igure below or a schematic drawing o the situation. Elevator in constant motion (or stationary) N N ma Elevator accelerating upward at a m/s N Elevator accelerating downward at a m/s x x mg mg mg Net acceleration ma -ma Let s consider each situation individually, and apply Newton s second and third laws. SOLVE When the elevator is stationary (or moving at a constant velocity) you are not accelerating so the orce on you must be zero (Newton s second law). The scale must push up on you to counteract the orce on you due to gravity (see let side o igure above). By Newton s third law, you push down on the scale with a orce o equal magnitude but opposite in direction. Thus the scale will read your normal weight. When you accelerate upward in the elevator, Newton s second law states that there is a orce pushing you upward (see center o igure above). This orce must be supplied by the scale, since it s the only thing touching you. By Newton s third law, there will be a orce o the same magnitude but in the opposite direction (i.e., downward) exerted by you on the scale. This will add to the orce due to gravity that is still present, so the scale reading will increase. Now you decelerate in the elevator as it stops at its destination, so by Newton s second law you must experience a orce downward to produce this deceleration. The orces on you are still due only to gravity and to the scale (and gravity has not changed). Thus the scale must push up on you with less orce so that the orce you experience (gravity + scale) is downward (see right side o igure above). Thus i the scale pushes up with less orce, by Newton s third law you will push down on the scale with less orce and it will read something less than your normal weight. When the elevator accelerates downward rom its stationary position, the situation is the same as described in the preceding paragraph all that is important is that you are accelerating downward, not your initial speed. Now we reach constant (downward) velocity, and the scale will again read your normal weight, just as when you were traveling upward at a constant velocity. Finally, the elevator decelerates to come to a stop at a loor. This is just the initial situation (you are experiencing an upward acceleration, which is reducing your downward velocity), so the scale will again read higher than your normal weight. REFLECT This problem highlights the parallels between gravitational and inertial mass. Einstein constructed one o his many amous thought experiments to study the relationship between gravitational and inertial mass in just this way. One consequence, he realized, o gravitational and inertial mass being the same is that light would be bent in a gravitational ield. How did he come to this conclusion? I gravitational and inertial mass are identical, you cannot now i you are in an accelerating elevator in zero gravity, or in a stationary elevator in a gravitational ield. I it s the ormer, then light entering a pinhole on one side o the elevator will exit the elevator at a slightly

4 4.4 Chapter 4 dierent height since the elevator will have moved during the light s passage rom one side o the elevator to the other. And i it s the latter, it s the same as the ormer (i.e., gravitational mass = inertial mass), so the light ray must similarly be bent in a gravitational ield 7. ORGANIZE AND PLAN Apply Newton s third law to the situation. SOLVE Newton s third law says that all orces come in pairs, with each member o the pair being o equal magnitude but opposite direction. Ball B receives a orce oriented at 3 above horizontal, so by Newton s third law, ball A must experience a orce oriented at 3 8 = 5 above horizontal, or 5 below horizontal. REFLECT I ball A experiences a orce at 5 below the horizontal, why is it not moving in that direction? By Newton s second law it is nown that ball A was indeed accelerated in the direction 5 below the horizontal. However, the acceleration lasted only a brie moment, so was not enough to modiy the initial velocity to orient it completely in the new direction. 8. ORGANIZE AND PLAN Review the section on inertial reerence rames. You may want to draw a diagram o a car rounding a turn and draw on the diagram the car s velocity vector or several positions o the car in the turn. SOLVE Reerence rames that move with a constant velocity are called inertial reerence rames. Velocity is a vector consisting o two quantities, magnitude (i.e., speed) and direction. A car executing a turn at a constant speed is not experiencing constant velocity, since the direction o its motion is changing. Thus you are not in an inertial reerence rame, but in a noninertial reerence rame. REFLECT I your reerence rame is accelerating you must experience a orce i you want to remain stationary with respect to your reerence rame. This is indeed the case within a car that is turning. The car seats and walls push you toward the inside o the turn to accelerate you and change your velocity. 9. ORGANIZE AND PLAN The orce due to gravity is always oriented toward the center o the Earth, which will be downward in our drawing. The drag orce always acts opposite to velocity, so it will act at 5 8 = 55 above the horizontal, or 55 below the horizontal. SOLVE 5 drag x REFLECT Oten baseballs will not ly straight, but will hoo and go out o play. What orce(s) cause this? I a ball has spin on it, the cushion o air will create a pressure dierence across the ball, and that will produce a orce that can act transverse to the ball s velocity.. ORGANIZE AND PLAN I there is no air resistance there is no drag, and gravity is the only orce acting on the projectile. By Newton s second law the projectile will experience a constant orce (mg) in the downward direction. The velocity in the ˆx will be constant (no acceleration), since there is no orce acting in this direction. The inematic equations in the ˆx and ŷ directions are thus [Eq.]: x = x + v, xt y= y + v, yt (/ ) gt where v, x and v, y are the initial velocities in each direction. I there is air resistance (i.e., drag) it will act opposite to the projectiles velocity. Thus there will now be a non-zero orce causing a deceleration in both the ˆx and y ˆ. Using Newton s second law again, we can ind the acceleration [Eq. ] drag F = ˆ x ( vx) maxx drag F ( v ) = ma yˆ mg y y y drag Where F ( v ) and drag F ( v ) represent the drag orce in the ˆx and ŷ directions and are unctions o velocity in x x y y

5 Force and Newton s Laws o Motion 4.5 each direction. Note that they act to oppose the velocity in each direction. This complicates the inematic equations, giving [Eq.3] drag x = x + v, xt+ (/ )( F x/ m) t drag y= y + v t+ (/ )( g F / m) t, y SOLVE Without the explicit orm o the drag orce, we cannot solve this equation, but we can see that the drag orce will act to reduce the displacement in both the ˆx and ŷ directions. I we assume the drag orce is given by [Eq. 4]: drag Fx ( vx) = γvx drag F ( v ) = γv y y y where γ is a constant, then the trajectory o our projectile or dierent γ s will be as shown in the igure below. y Trajectory with no drag Trajectory with drag With no air resistance (γ = ), the projectile will ollow a parabolic path. I we include air resistance by setting γ >, the projectile trajectory will be reduced, as shown in the igure above. REFLECT Notice that or greater values o γ, the trajectory is more severely reduced. This is because the drag orce is greater, causing a greater deceleration in both the ˆx and ŷ directions. We have assumed a drag orce that is linear to the velocity. Can you imagine other orms or the drag orce? How would that modiy the trajectory o the projectile?. ORGANIZE AND PLAN This problem involves the interaction o two objects: a person (object ) waling on the Earth (object ). We will have to thin about Newton s third law to resolve this problem. Regarding the dierence between static and iic riction, remember that static riction requires that the interace between the two objects is static (i.e., not changing) during the time that the static riction orce is being applied. For iic riction, the interace is dynamic (i.e., changing), as with a sate sliding over ice, or a tire sidding on the pavement. SOLVE When you wal you push against the Earth with your eet to propel you orward. From Newton s third law, the Earth exerts a orce o equal magnitude but opposite direction to the sole o your shoe. It is the riction between the soles o your shoes and the ground that actually transmits the orce to your eet to move you orward. As you step orward, the oot you re pushing with does not move with respect to the ground, so it is static riction that is at wor. I you re on a slippery surace and your oot begins to slide as you push o with it, then it is iic riction that comes into play. REFLECT When your oot slips out rom beneath you unexpectedly, what happens to your body? Does it lurch orward? Fall straight down?. ORGANIZE AND PLAN Force is a vector quantity, so it is composed o a magnitude and a direction. We will loo at the orce acting on the box. Draw a diagram o a box with all the orces acting on it and decompose the orces into their horizontal and vertical components. X Friction Box Normal orce Force applied to pull box Component o orce pulling box orward Component o orce liting box mg Sum the orces in the horizontal and vertical directions to see how pulling upward on the box aects the situation.

6 4.6 Chapter 4 SOLVE I you pull the box slightly in an upwards direction with respect to the loor, a component o the orce you apply will be liting the box up, counteracting gravity. Thus the normal orce exerted by the loor will decrease, and with it the rictional orce (see Equation 4.6). Less orce will thereore be required to slide your box orward. I there is no riction, then there is no reason to apply a orce upwards on the box since it will not change the orce required to move the box in the horizontal direction. REFLECT Note that i the box does not move in the vertical direction, then by Newton s second law there must be zero orce acting in that direction. This is how it is nown that the normal orce must be reduced when an upwards orce is applied on the box. O course, i that upward orce is greater than the orce due to gravity, the box will accelerate in the vertical direction. What happens to the normal orce in this case? 3. ORGANIZE AND PLAN Draw a orce diagram and loo at the orces acting in the vertical direction. SOLVE I the sydiver is alling with a constant velocity, then she is not accelerating and Newton s second law tells us that there is no orce acting on her. Thereore, the orce exerted by her parachute must exactly cancel the orce on her due to gravity. This means her parachute will exert a orce o magnitude equal to her weight, but in the upward direction. REFLECT When the sydiver irst opens her parachute, is the orce exerted by the parachute greater than, less then, or the same as the orce due to gravity? Since the sydiver s velocity decreases, she must be undergoing an acceleration in the upward direction, thus by Newton s second law the parachute exerts an upward orce on her but the magnitude o that orce is unnown since the magnitude o her acceleration is unnown. 4. ORGANIZE AND PLAN Review Newton s second law o motion. Note that there are two aspects to it,. The object s acceleration is proportional to the orce acting on it. The object s acceleration is inversely proportional to its mass The junior high student has invoed only the irst o these two aspects, so we should remind him o the second. SOLVE You can tell your junior high student that acceleration may well be proportional to orce, but it is also inversely proportional to mass. In other words, a=f/ m, so i you double the numerator and the denominator, the quotient remains unchanged The acceleration due to gravity is thus the same in both cases. REFLECT Have you ever tested this theory? Tae two similar sized objects, but o dierent weights, hold them above your head, and let them drop. Do you believe Newton s laws? 5. ORGANIZE AND PLAN Draw a orce diagram or the roller coaster at the top o a hill and at the bottom o a valley, and consider in which direction the roller coaster must be accelerating at each point. Now apply Newton s second and third laws. SOLVE When the roller coaster is going over a hill, its velocity is changing rom an upward direction beore the summit o the hill to a downward direction ater the summit, so you are experiencing an overall downward acceleration. By Newton s second law, there must be a downward orce on the roller coaster. Since the orce due to gravity has not changed, the only other orce acting on the system (the normal orce rom the trac) must have been reduced. This is shown in the igure below, where the normal orce at the top o the hill is noticeably weaer than the orce due to gravity. N = Force exerted by trac N top N bot Roller coaster Trac mg mg Net orce

7 Force and Newton s Laws o Motion 4.7 From Newton s third law the trac must experience a orce equal in magnitude to the normal orce, but in the opposite (i.e., downward) direction. Thus, the trac going over a summit experiences a orce that is less than the weight o the roller coaster. When the roller coaster is going through a valley, the opposite logic applies. The orce on the roller coaster must be upward. This upward orce must be exerted by the trac, which is added to the normal orce needed to counter the weight. (Note the length o the normal orce vector compared to the weight vector in the igure above). Newton s third law states that the trac will experience a orce o the same magnitude but in the opposite direction. Thus the trac going through a valley experiences a orce that is greater than the weight o the roller coaster. REFLECT At the very top o the summit and at the very bottom o the valley, the trac is horizontal. The acceleration, however, is not. It is oriented toward the center o the curve, in other words downward at the summit and upward in the valley. 6. ORGANIZE AND PLAN The weight o an object is given by equation 4., w=mg, where g is the acceleration due to Earth s gravity ( ~ 9.8 m/s toward the center o the Earth). Known: m = 5 g w= mg = 5 g 9.8 m/s = 47 g m/s, or 47 Newtons (N). The correct response is thus choice (c). REFLECT What is this in pounds? Using the conversion actor =. lb/g, we ind that SOLVE For a mass o 5 g, the magnitude o the weight is 5 g = ( 5 g)(. lb/g) = 33 lb. Not a very big dog. 7. ORGANIZE AND PLAN We can approach this in two ways. Either we can calculate the acceleration due to both orces individually and then do a vector sum o the accelerations, or we can calculate the orce irst and then ind the acceleration due to that orce. We will tae the second approach. To calculate the orce, do a vector sum o all the orces acting on the box, then use Newton s second law to ind the acceleration o the box. Note also that the responses contain vector and scalar accelerations, so we will have to calculate both. Known: F = 5iˆ 87 ˆj, m =.4 g SOLVE To get the acceleration due to each orce individually, we use Newton s second law F =m a. The only orce acting on the box is F, so F = F. We thus ind an acceleration o [Eq. ] a=f / m= 5 N /.4 g i ˆ + 87 N /.4 g ˆj = 75 m/s iˆ 6 m/s ˆj This corresponds to response (a). We will now calculate the magnitude o the orce, and then insert it into Newton s second law to ind the magnitude o the acceleration. Inserting the components o F into the Pythagorean theorem, we ind that the magnitude o the orce is [Eq. ] F = = 36.4 N Using Newton s second law to ind the magnitude o the acceleration gives [Eq. 3] a=f / m= 36.4 N /.4 g = 97.4 m /s which is response (d). REFLECT Which answer gives more inormation? The irst approach tells us the box is accelerated more in the î direction than in the ĵ direction, but the second approach tells us the overall magnitude o the acceleration without urther calculation. I we want to chec our math, we can veriy that the two approaches give the same magnitude or the acceleration. Using the acceleration components rom Eq. (), we ind that magnitude o the acceleration is [Eq. 4] which again is response (d). ( ) a= 75 m/s + 6 m/s = 97.4 m/s

8 4.8 Chapter 4 8. ORGANIZE AND PLAN We can use Newton s second law F =m ato solve this problem. Speciically, we can apply it to the g mass and to the 3 g mass, and then tae the ratio o the resulting equations, which will give us the ratio o the accelerations. SOLVE For the g mass, Newton s second law gives us [Eq. ] F= g a For the 3 g mass, Newton s second law gives us [Eq. ] F= 3g a 3g where the orce F is the same in both equations, but the acceleration a 3 g Taing the ratio o the two equations gives [Eq. 3] F ( g) a = F (3 g) a3 g a = (/3) a which is response (d). 3 g is unnown or the moment. REFLECT The acceleration o the larger mass is less than the acceleration o the smaller mass, which maes sense as we are applying the same orce to both. From this logic alone, we could have chosen the correct answer, since only response (d) its the bill. 9. ORGANIZE AND PLAN This problem involves adding orces. How many ways can we add two orces together? I we assume the irst orce deines one axis (let s call it the î axis), we can add the other orce to it using any angle θ between the two orces. Let s express this mathematically and see what the upper and lower limits are or the magnitude o the resulting orce. Known: F= 3 N ; F= 5 N SOLVE Assuming F is along the î axis, and F maes an angle θ with respect to F, then the orce is F = F +F cos θ ( i ˆ) +Fsin ( θ )( ˆj) [Eq. ]. The magnitude o the orce is [Eq. ] ( ) ( ) ( ) θ θ θ θ F = F +F cos + F sin = 3+ 5cos + 5sin N The maximum orce will occur when θ = (i.e., the two orces are acting in the same direction), and the minimum orce will act when θ = 8 (i.e., the two orces are acting in the opposite direction). In the ormer case we ind [Eq. 3] and in the latter case we ind [Eq. 4] F = 3+ 5 = 83 N ( ) F = 3 5 = 9 N Thus the or must all between 9 and 83 N, or 9N F 83 N [Eq. 5] which is response (c). REFLECT Notice that the maximum orce attainable is greater than either o the two orces alone, which is reasonable. In this case, the minimum orce attainable is less than either o the two orces alone, but is this always so?. ORGANIZE AND PLAN This problem involves Newton s second law, F =m a. In both cases the masses are accelerating under the orce o gravity at g= 9.8m / s. We can apply the same approach as or the previous problem. Known: m =. g; m = 5. g SOLVE Applying Newton s second law to each mass, we have F=mg[Eq. ] and F=mg [Eq. ]. Taing the ratio o the two equations gives F/ F =m / m [Eq. 3] or, solving or F in terms o F, F=Fm/ m=f 5. g /. g = 5F [Eq. 4], which is response (c).

9 Force and Newton s Laws o Motion 4.9 REFLECT We ind that the orce required to accelerate the 5. g mass is ive times the orce required to impart the same acceleration to the. g mass. This is consistent with Newton s second law.. ORGANIZE AND PLAN A drawing o the situation will clariy things. N N mg sin (u)(ı) N mg u = mg cos (u)( ) N Note that the coordinate system chosen is aligned to the incline. To ind the acceleration o the bloc o ice we need to ind the orce, and then we can use Newton s second law to ind the acceleration. Known: angle o incline θ = SOLVE The normal orce exerted on the bloc must cancel the ĵ component o the orce due to gravity, otherwise the bloc would move through or away rom the inclined surace. Thus the orce is the just the F =mgsin θ i ˆ. î component o the orce due to gravity, or [Eq. ] From Newton s second law, we can ind the acceleration due to this orce [Eq. ], F =mgsin( θ )( iˆ ) = m a a = gsin( ) iˆ = 3.4 m/s iˆ. which is response (c). REFLECT The acceleration must be less than that due to gravity, since the orce exerted on the bloc is less than the orce due to gravity. We have also ound more inormation than ased or, since we now the direction o the acceleration as well as its magnitude.. ORGANIZE AND PLAN This problem is a continuation o Problem. Use the same drawing and just add the = iˆ. additional orce due to riction [Eq. ], The riction orce must be oriented in the î direction because riction always acts to counter the motion o the object, and the object only has the possibility to move in the î direction. Calculate the orce on the box, and then use Newton s second law to ind the acceleration. Known: iic coeicient o riction μ =.5; angle o incline θ = SOLVE The acceleration will be in the î direction, so ind the orce in this direction [Eq. ]. F = ( mgsin( θ) μn)( iˆ ) = mgsin θ μ mgcos θ iˆ ( ) F =ma= mg mg i [Eq. 3], or ( ˆ ) ( sin( θ) μ ( θ) )( ˆ) ( ˆ cos. / ) Newton s second law gives sin( θ) μ cos( θ) a= g g i = m s i [Eq. 4], which is response (c). REFLECT Because riction acts to counter the bloc s acceleration, the acceleration will be less than what we ound in the rictionless case (Problem ), which is what we ound. 3. ORGANIZE AND PLAN For this problem we can use Eq. 4.9, which gives the orce required or an object to execute uniorm circular motion, F=mv r / R. This orce is directed toward the center o the circle and will be provided by the rictional orce, = μsn= μsmg, so we can equate these two expressions and solve or the coeicient o static riction. Known: R = 75 m; v = m/s SOLVE Equating the rictional orce to the orce needed to execute centripetal motion gives [Eq. ]: μsmg = mv / R ( m/s) μ s =v /( Rg ) = =.6 (75 m)(9.8 m/s ) which is response (a).

10 4. Chapter 4 REFLECT We label the coeicient o riction as static because the part o the tire that is in contact with the road is stationary while it touches the road. Is this a reasonable static coeicient o riction or a tire on a concrete road? Looing at Table 4., we see that a tire on a dry concrete road has a static coeicient o riction o., which is reduced to.3 on wet concrete. Thus the static coeicient o riction o.6 needed to execute this turn is well within the normal range, and we should have no problem executing this turn, even on wet concrete. 4. ORGANIZE AND PLAN To accelerate the rocet ship upward, it must experience a orce in the upward direction. The orce needed can be ound using Newton s second law, since the mass o the rocet and the desired acceleration are nown. Use a coordinate system where ĵ indicates the upward vertical direction. a =.5 m/s ˆj Known: mass o rocet m = g; acceleration rocet SOLVE The orce on the rocet is the orce due to gravity, which is acting downward, and the orce o the engine, which is acting upward. Summing these and using Newton s second law gives [Eq. ]: F =F mg ˆj =ma engine rocet ( ( ˆ rocet )) ( g)(.5 m/s 9.8 m/s )( ˆ),583 N( ˆ) F =m a +g j = + j = j engine This is response (c). REFLECT Note that since the orces in this problem are all acting in a single direction we could have used scalar quantities (i.e., the vector magnitudes) to describe the orces. We would have had to be consistent in our sign convention (positive sign or upward orces, negative sign or downward orces). 5. ORGANIZE AND PLAN Find the horizontal orces acting on the puc, and then use Newton s second law to ind the horizontal acceleration o the puc. Use this acceleration in the inematic equations rom Chapter to ind the initial speed needed to travel 6 m. Choose a coordinate system where the initial velocity o the puc is in the î direction. Also, note that the puc s inal velocity is m/s. Known: μ =.5; distance d = 6 m; inal puc speed v = m/s. SOLVE The horizontal orce acting on the puc is due to iic riction, since the puc is moving with respect to the ice. The magnitude o the iic riction orce is given by Eq. 4.6, = μn, where n is the magnitude o the normal orce. The direction o the iic riction orce will be opposite to the velocity o the puc, or in the î direction. The normal orce n must have the same magnitude as the orce due to gravity, otherwise the puc would experience a vertical acceleration, so n = mg. Using Newton s second law to ind the acceleration due to iic riction, we have F = = ma [Eq. ]. Using Eq. 4.6 to express the iic riction orce in terms o the normal orce, and recalling that the iic riction orce acts in the î direction, we have [Eq. ] μni μmg( i ) ˆ = ma, ˆ = ma, a = μ g i To ind the initial speed needed or the puc to travel 6 meters, use Eq..9 (reproduced here as Eq. 3) x =x +vt+ /at Since the distance traveled d = x x, we have [Eq. 4] d=vt+ /at This equation has two unnowns, v and t, so another equation is needed relating these two quantities. Use Eq..8, v=v +at [Eq. 5], where v is the inal puc speed (i.e., zero). Using Eq. 5 to eliminate t in Eq. 4, we ind [Eq. 6] or ( ˆ) v v d=v + /a a a v = ad

11 Using Eq. or the acceleration, we ind that [Eq. 7] ( μ ) Force and Newton s Laws o Motion 4. v = g d = m/s 6 m = 4. m/s which is response (c). REFLECT Looing at the units under the radical in the last expression or v, we see that they are m /s, so that when we tae the square root we are let with m/s, which are the correct units or a velocity. Also notice that we had to be careul about the directions o the velocity and the acceleration, otherwise we may have ound a negative value under the radical in Eq ORGANIZE AND PLAN This problem involves the centripetal (i.e., center-seeing) acceleration o an object. We can use Eq. 4.9, F=mv r / R, and Newton s second law, F = ma, to ind the centripetal acceleration o both vehicles. Then we can just tae the ratio o the accelerations to see how they compare. Known: m cc = 8 g; v cc = 5 m/s; m suv = 43 g; v suv =.5 m/s SOLVE Using Eq. 4.9, the centripetal orce acting on the compact car is [Eq.] F =m v R cc / r cc cc From Newton s second law, we ind its acceleration is [Eq. ] cc a =F / m =v / R cc r cc cc Repeating the calculation or the SUV, we have [Eq. 3] SUV F =m v R v SUV SUV / The centripetal acceleration o the SUV is [Eq. 4] SUV a =F / m =v / R Taing the ratio o the Eq. 4 to Eq. gives [Eq. 5] SUV r SUV SUV SUV cc SUV cc a / a =v / v =.5/ 5 = / 4 Thus the SUV s acceleration is one-ourth that o the compact car, so the response is (e). REFLECT We see that the mass o an object does not aect its centripetal acceleration. This only depends on the object s speed and the radius o the circular trajectory. Since the centripetal acceleration depends on the speed squared, i you cut the speed in hal, you will reduce the centripetal acceleration by a actor o our, as we ound. 7. ORGANIZE AND PLAN We can tae the ratio o the Eqs. and 3 rom the previous problem to answer this question. SOLVE Taing the ratio o centripetal orces o the SUV to the compact car, we have [Eq. ] SUV F ( 43 g)(.5 m/s) v msuvvsuv / R 3 = = = cc Fv mccvcc / R ( 8 g)( 5 m/s) 4 which is response (c). Thus the centripetal orce acting on the SUV is only 75% o that acting on the compact car. REFLECT Notice that we did not need to now the radius o the turn, since we are only interested in the ratio o the orces. Does the answer mae sense? Compared with the compact car, the SUV has three times the mass (increasing the centripetal orce by a actor o three), but only ¼ the speed-squared (reducing the centripetal orce by a actor o 4). Thus it is reasonable that we get an overall actor o ¾ or the ratio o the centripetal orce applied to the SUV to that applied to the compact car. 8. ORGANIZE AND PLAN As explained in section 4.4, drag orces on an object always act in the direction opposite to the velocity o the object. SOLVE As the ball travels straight upward, the drag or will oppose the velocity, so the drag orce will be straight down. The orce due to gravity will also be straight down, so gravity and drag are acting in the same direction or the ball s upward trip. As the ball travels straight down, the same rules apply. The velocity is now downward, so the drag orce will be upward. The orce due to gravity acts downward, as always. Thus or the downward trip the orces due to drag and gravity act in the opposite direction. The response is thereore (b), same/opposite.

12 4. Chapter 4 REFLECT What would happen i the ball were not thrown straight up, but thrown with some horizontal velocity component? In this case the velocity would have a non-zero horizontal component, which means the drag orce would also have a non-zero horizontal component. Thus the drag orce and the gravitational orce would not be aligned, but would have an angle between them that depends on the height o the ball. When the ball reaches the pea o its trajectory, the vertical velocity component would be zero, meaning the drag orce would also have no vertical component. At that instant, the drag orce would either be zero, or perpendicular to gravity. 9. ORGANIZE AND PLAN To ind the orce acting on an object, we perorm a vector addition o all the orces acting on the object. In this case there are only two orces, and they are anti-parallel (i.e., parallel but in opposite directions), so the vector addition is quite simple. We chose a coordinate system where the î direction is toward the right. N F r = 3.7 Nı N F l =.5 N Setch Known: F= ( ˆ r i) 3.7 N ; F=.5 N i l ( ˆ) F l =.5 Nı N F r = 3.7 N Force diagram SOLVE The orce (or total orce) acting on the crate is simply the vector sum o all the orces acting on it, or F =F +F = 3.7 N i ˆ +.5 N i ˆ =. N i ˆ [Eq. ], which is. N toward the right. r l REFLECT Although or this quite simple problem we did not necessarily need to mae a drawing showing the orces, it is oten helpul to do so to avoid maing mistaes or omissions. 3. ORGANIZE AND PLAN Recall that the orce is the vector sum o all orces acting on an object. Two non-zero orces are acting on the box, and we must add a third orce to cancel these orces and render the orce zero. I we call the third orce F 3, we just insert it into the expression or the orce and solve or it. Known: F= 47 N( iˆ) 65 N( ˆj ); F= 57 N( iˆ) 49 N( ˆj ) SOLVE The orce, which we want to be zero, is given by [Eq.]: F = F + F + F 3 =. Inserting the now values or F and F, we can solve or F 3 [Eq. ]: F = =F +F + F3 F3 = F F F3 = ( 47 N + 57 N) ( iˆ) + ( 65 N 49 N) ( ˆj) F = 5 N iˆ + 69 N ˆj 3 We thereore need to apply a orce 3 5 N( ˆ) 69 N( ˆ) F=+ i + j to reduce the orce to zero. REFLECT Note that to construct an anti-parallel vector o equal magnitude to a nown vector, it suices to multiply each component o the nown vector by. 3. ORGANIZE AND PLAN We can use Newton s second law to ind the acceleration o the rocet. The orce on the rocet will be the vector sum o all the orces acting on it, which are the orce due to gravity (straight down) and the orce rom the rocet engine (straight up). We will use a coordinate system where ĵ represents the upward vertical direction. F = 95.3 N ˆj Known: m = 3.5 g; engine SOLVE The orce acting on the rocet is [Eq. ] F =F +mg= 95.3 N ˆj g 9.8 m/s ( ˆj ) = 6 N ˆj. engine From Newton s second law, the acceleration o the toy rocet is [Eq. ] ˆ a=f / m= 6 N j /3.5 g = 7.4 m/s ˆj.

13 Force and Newton s Laws o Motion 4.3 REFLECT The rocet is accelerating upward at almost twice the acceleration due to gravity. This is a signiicant acceleration, compared to what we experience on a daily basis. To appreciate the acceleration due to gravity, try to catch a stic that alls rom a stationary position, with your hand initially positioned above the stic. It can be done, but it s not easy. Imagine now the rocet accelerating at almost twice that rate 3. ORGANIZE AND PLAN This problem involves a straightorward application o Newton s second law. The phrasing o the problem implies that gravity is not to be included calculating the acceleration. We will chose a coordinate system in which the orce exerted by the molecule acts in the î direction. 8 Known: m= 3. g F = 6. N iˆ ; applied SOLVE Neglecting gravity, the orce on the molecular complex is [Eq. ] F =F =ma or [Eq. ] applied applied a=f / m= 6. N / 3. g i = m/s i 8 ˆ 6 ˆ REFLECT This acceleration is much, much greater than that due to gravity, so we were justiied in neglecting gravity. 33. ORGANIZE AND PLAN I the orce acting on the air trac glider is constant, then by Newton s second law the acceleration it generates must be constant as well. This means the acceleration generated is just the average acceleration, so use Eq..6 or average acceleration (reproduced here as Eq.), a ave = ( v vi )/( t ti ) Knowing the average acceleration, use Newton s second law to ind the orce that acted on the air trac glider to produce this acceleration. Choose a coordinate system where î indicates right ( î indicates let). s;. m/s ˆ t= v= i; t= 4. s,v=. m/s i ˆ ;m =.3 g Known: i i SOLVE Inserting the nown values into Eq. gives [Eq. ] (. m/s. m/s)( iˆ ) ( 4. s s) ( ˆ) a ave = =.8 m/s i We now use Newton s second law to ind the orce needed to produce this acceleration [Eq. 3]. F =ma =.3 g.8 m/s i ˆ =.8 N iˆ Since the orce is acting in the ave î direction, it is acting to the let. REFLECT It is reasonable that the orce is acting to the let, since we needed to reverse the puc s velocity rom rightward to letward. How much orce is this in pounds? Since pound is approximately 4.5 N, (.8 N)( lb / 4.5 N) =.4 lb, which is not very much. 34. ORGANIZE AND PLAN Using Newton s second law, we can ind the orce needed to accelerate the.4 g boo the given amount. We can then use the same law to ind the acceleration o the 3.6 g boo subject to the same orce. Known: m= ˆ ˆ.4 g;m= 3.6 g;a= 3.4 m/s i.8 m/s j SOLVE The orce needed to generate a is F =ma [Eq.]. Applying this orce to the 3.6 g boo results in an acceleration o [Eq. ] m.4 g a ( ) ˆ ˆ ˆ ˆ =F/ m = a = 3.4 m/s i +.8 m/s j =.3 m/s i +.9 m/s j m 3.6 g REFLECT Notice that we did not have to explicitly calculate the orce, we only needed to express it in terms o nown quantities.

14 4.4 Chapter 4 We can also chec that a/ a =m/ m (because by Newton s second law, acceleration is inversely proportional to mass) [Eq. 3]. Taing the ratio, we ind [Eq. 4] which is the same as the ratio o m to m [Eq. 5] a = 3.4 m/s +.8 m/s = 4.6 m/s a =.3 m/s +.9 m/s = 3. m/s a 3. m/s a = 4.6 m/s = m.4 g m = 3.6 g = 35. ORGANIZE AND PLAN This problem involves the application o Newton s second law in vector orm, F=ma. ˆ a =.55 m/s i +.65 m/s ˆj Known: m = g =. g; SOLVE Inserting the nown quantities (with the correct units) into Newton s second law, we have [Eq. ] ˆ F = ma =. g.55 m/s i +.65 m/s ˆj =.55 N iˆ +.65 N ˆj.7.7 This is the orce needed to generate the desired acceleration. The direction o this orce is [Eq. ].55 θ = atan = 68.6, From Eq. () we now the orce vector is in the second quadrant, so we the direction o the orce must be.4 above the horizontal. The magnitude o the orce is [Eq. 3] F =.55 N +.65 N =.698 N REFLECT Note that we had to convert the mass rom grams to ilograms so that the result o our calculation would be in SI units (i.e., Newtons). 36. ORGANIZE AND PLAN Weight is the orce exerted by gravity on an object, and or objects on the Earth, it can be approximated quite well using Eq. (4.) w=mg, where g is the acceleration due to Earth s gravitational ield near the surace o the Earth. Since g is a vector, we will deine or this problem a coordinate system where ĵ points g g ˆj (the symbol means is deined as ). vertically upward rom the surace o the Earth, so Known: m = 7. g; g 9.8 m/s ˆj SOLVE Inserting the nown quantities into Eq. (4.), we get (Eq.) w= mg = 7. g 9.8 m/s ˆj = 7.6 N ˆj. REFLECT Many textboos treat weight as a scalar quantity instead o a vector quantity, but that goes with their implicit assumption that the orce o weight is always oriented toward the center o the Earth. I we treat weight as a vector, that assumption is spelled out clearly, hopeully avoiding possible conusion. 37. ORGANIZE AND PLAN Let s draw a schematic representation o the situation and label all the points o interest. To avoid conusion, we can mae a separate drawing or the upward and downward trip. SOLVE

15 Force and Newton s Laws o Motion 4.5 y = H mg v = y = H/ v mg mg v y = v mg mg v Upward trip Downward trip REFLECT Note that the orce acting on the ball during its entire trajectory does not change; it is simply the orce due to gravity. This orce generates an acceleration on the ball, so the velocity o the ball changes, in both magnitude and direction (compare the upward and downward trips). Note also that we do not now the relationship between the velocity and the orce due to gravity, since we do not now the mass o the ball, so beware o assigning undue meaning to the length o the velocity vector compared with that o the orce vector. However, since there is no air resistance, we do now that the magnitude o the velocity is the same at the equal heights on the upward and downward trips (note that the length o the velocity vectors are the same at y = and y = H/). 38. ORGANIZE AND PLAN Let s call the conversion actor we re looing or x. This variable must satisy the equation ( dyne) x = N [Eq. ]. All we need to do is to solve this equation or x, and we ll have our answer. Known: dyne = gm cm/s ; N = g m/s ; g = gm; m = cm SOLVE Executing our plan, we have [Eq. ] ( dyne) x = N N g m/s g m x = = = dyne gm cm/s gm cm Now, using the relations between g, gm, m, and cm, to eliminate cm and gm rom our ormula, we have [Eq. 3] g m 5 x= = (/ g)(/ m) 5 Inserting the value we ond or x into Eq., we have dyne = N [Eq. 4]. REFLECT Note that the conversion actor is dimensionless (i.e., it has no units). This is expected since both units (dyne and Newton) represent a orce, so they should have the same units and any conversion actor between them should be dimensionless. Also note that we treated the units just as you do algebraic symbols; i.e., we wrote equations between them, used these equations to eliminate unwanted quantities, and canceled the units when they appeared in both the numerator and denominator. In other words, all the regular rules o algebra apply to units. This was an exercise in dimensional analysis, irst developed by no less than Joseph Fourier. Checing that the dimensions mae sense is a useul technique to veriy calculations, and is a very useul tool or scientists and engineers in all ields. 39. ORGANIZE AND PLAN This problem involves a twist on the usual problem o inding the orce on an object due to Earth s gravitational ield. But, as the hint suggests, i we irst calculate the orce on the object we can calculate the orce on the Earth by applying Newton s third law F AB = FBA. We can then use Newton s second law F=ma to ind the acceleration o the Earth. We will deine our coordinate system so that ĵ is pointing radially away rom the center o the Earth. The section on Newton s third law indicates that the mass o the Earth is 6 4 g. g = 9.8 m/s ˆj Known: m object = g; m earth = 6 4 g;

16 4.6 Chapter 4 SOLVE Solving irst or the orce on the object gives [Eq. ] F earth-object =mg= g 9.8 m/s ˆj F = 98 N ˆj. earth-object In view o the coordinate system, this orce is pulling the object toward the center o the Earth. Newton s third law indicates that [Eq. ] Fobject-Earth = FEarth-object F = 94 N j object-earth We have ound that the orce exerted on the Earth by the g object is 98 N radially away rom the center o the Earth (i.e., up). To ind the Earth s acceleration, we apply Newton s second law [Eq. 3], Fobject-earth = maearth 4 a = (98 N ˆj)/(6 g) = 9.8/ 6 m/s ˆj.6 m/s ˆj. earth REFLECT Note that we simpliied our calculation in the last step by using the rules or dividing quantities with exponents. Had we simply thrown the numbers into our calculator, we might have had trouble deciphering the results due to all the leading zeros. This acceleration seems very small. Let s see how long it would tae or the Earth to move mm (= -3 m) i it accelerated at this rate rom a stationary start [Eq. 4]. x=x +vt+ /at x= + + /.6 m/s t ( ) 3 ( ) ( ) t= x/ a= m /.6 m/s t= 9 3. s 3 years 4. ORGANIZE AND PLAN We will deine a coordinate system in which the astronaut s velocity is in the î direction. From Newton s third law, the motion o both objects will be in the ± î direction, so the vector notation is unnecessary. Use scalar quantities, eeping in mind that a positive (negative) quantity represents the î ( î ) direction. Beore the astronaut starts pushing on the spacecrat, the two have zero velocity with respect to each other ast sc ( v = v = ). Their inal velocity is thereore [Eq. ] ast ast ast v =v + /a t ast ast v = /a t or the astronaut, and [Eq. ] sc sc sc sc v =v + /a t = /a t or the spacecrat. We can now tae the ratio o Eqs. and to get a relationship between the astronaut s and the spacecrat s velocities [Eq. 3]. sc a sc ast v =v ast a Using Newton s second and third laws, obtain a relationship between the astronaut s and spacecrat s accelerations. ast sc ast Known: m = 6 g;m = 3 g;v.35 / ( ˆ = m s i ) SOLVE Newton s third law gives, in scalar orm, [Eq. 4] F = F Applying Newton s second law (in scalar orm) to the astronaut and to the spacecrat, gives [Eq. 5] ast ast F =m a ast-sc sc-ast sc-ast

17 and (Eq.6) sc sc F ast-sc =m a respectively. Taing the ratio o Eqs. 6 to 5, and using Eq. (4), we have [Eq. 7] sc sc Fast-sc m a = ast ast Fast-ast m a sc ast a m = ast sc a m Force and Newton s Laws o Motion 4.7 Inserting this result into Eq. () rom above, we ind that [Eq. 9] ast m sc ast v = v = (.35 m/s)(6 g)/(3 g) =.7 m/s sc m v sc =.7 m/s i ˆ [Eq. 9]. In view o our coordinate system, the complete vector velocity is REFLECT We ind the spacecrat is moving to the right at.7 m/s (perhaps the astronaut is James Bond). Note that we did not need to now the interaction time between the astronaut and the spacecrat, nor the acceleration o each. All we needed to now was the ratio o these quantities. We may also have used conservation o momentum to solve this problem, but we have not yet introduced this concept. Those who are interested may loo ahead to section ORGANIZE AND PLAN This problem is the same as the previous one, with the astronaut replaced by the cannon ball and the spacecrat replaced by the cannon, and with the caveat that we are ased or speed instead o velocity. We will use the same approach to solve this problem. cb Known: m cb = 3.5 g; m cannon = 9 g; v = 95 m/s SOLVE In view o the similarity between this problem and the previous one, we can simply map the variables o Problem 4 to their counterparts in this problem [Eq. ]: sc cannon v v ast cb v v sc cannon m m ast cb m m Using this mapping and Eq. 8 rom Problem 4, we ind [Eq. ] cb m cannon cb v = v = ( 95 m/s)( 3.5 g )/( 9 g) =.36 m/s cannon m Since we are only interested in the speed o the recoiling cannon, we tae the magnitude o the velocity, which is.36 m/s. REFLECT Note that we reported the result to the same number o signiicant igures () as the lowest-precision datum used in calculating the result. 4. ORGANIZE AND PLAN To express the acceleration o the arrow when it hits the target, use the inematic equations.8 and.9. Since the problem ass or vector quantities (in this case, a orce), deine a coordinate system in which the arrow travels in the î direction. Thus, the average acceleration o the arrow is given by [Eq. ] v v a A = t and the displacement o the arrow ater it hits the target is given by [Eq. ] x =x ave +vt+ /a A t Solving Eq. () or t and inserting it into Eq. (), and using the act that the inal arrow velocity is zero gives [Eq. 3] v a ˆ A = i x x With this expression or the average acceleration, use Newton s second law to ind the average orce exerted on the arrow by the target, and Newton s third law to ind the average orce exerted on the target by the arrow.