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5.1.

Model: We can assume that the ring is a single massless particle in static equilibrium. Visualize:

Solve:

Written in component form, Newton’s first law is

( Fnet ) x = ΣFx = T1x + T2 x + T3 x = 0 N

( Fnet ) y = ΣFy = T1y + T2 y + T3 y = 0 N

Evaluating the components of the force vectors from the free-body diagram:

T1 x = − T1

T2x = 0 N

T1y = 0 N

T2 y = T2

T3 x = T3 cos 30° T3 y = − T3 sin 30°

Using Newton’s first law:

− T1 + T3 cos 30° = 0 N

T2 − T3 sin 30° = 0 N

Rearranging:

T1 = T3 cos 30° = (100 N )(0.8666) = 86.7 N T2 = T3 sin 30° = (100 N )(0.5) = 50.0 N r Assess: Since T3 acts closer to the x-axis than to the y-axis, it makes sense that T1 > T2 .


5.2.

Model: We can assume that the ring is a particle. Visualize:

This is a static equilibrium problem. We will ignore the weight of the ring, because it is “very light,” so the only three forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle θ. r Solve: Because the ring is in equilibrium it must obey Fnet = 0 N . This is a vector equation, so it has both x- and y-components:

( Fnet ) x = T3 cosθ − T2 = 0 N ⇒ T3 cosθ = T2 ( Fnet ) y = T1 − T3 sin θ = 0 N ⇒ T3 sin θ = T1 We have two equations in the two unknowns T3 and θ. Divide the y-equation by the x-equation:

T3 sin θ T 80 N = tan θ = 1 = = 1.60 ⇒ θ = tan −1 (1.60) = 58.0° T3 cosθ T2 50 N Now we can use the x-equation to find

T3 =

T2 50 N = = 94.3 N cosθ cos58.0°

The tension in the third rope is 94.3 N directed 58.0° below the horizontal.


5.3. Model: We assume the speaker is a particle in static equilibrium under the influence of three forces: gravity and the tensions in the two cables. Visualize:

Solve:

From the lengths of the cables and the distance below the ceiling we can calculate θ as follows:

sin θ =

2m = 0.677 ⇒ θ = sin −1 0.667 = 41.8° 3m

Newton’s first law for this situation is

( Fnet ) x = ΣFx = T1x + T2 x = 0 N ⇒ −T1 cosθ + T2 cosθ = 0 N ( Fnet ) y = ΣFy = T1y + T2 y + w y = 0 N ⇒ T1 sin θ + T2 sin θ − w = 0 N The x-component equation means T1 = T2 . From the y-component equation:

2T1 sin θ = w ⇒ T1 =

(20 kg)(9.8 m/s2 ) 196 N w mg = = = = 147 N 1.333 2 sin θ 2 sin θ 2 sin 41.8°

Assess: It’s to be expected that the two tensions are equal, since the speaker is suspended symmetrically from the two cables. That the two tensions add to considerably more than the weight of the speaker reflects the relatively large angle of suspension.


5.4.

Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by the two ropes, with friction providing the force that resists the pullers. Visualize:

Solve:

Since the sled is not accelerating, it is in dynamic equilibrium and Newton’s first law applies:

( Fnet ) x = ΣFx = T1x + T2 x + fkx = 0 N

( Fnet ) y = ΣFy = T1y + T2 y + fky = 0 N

From the free-body diagram:

1 1 T1 cos θ  + T2 cos θ  − fk = 0 N 2  2 

1 1 T1 sin θ  − T2 sin θ  + 0 N = 0 N 2  2 

From the second of these equations T1 = T2 . Then from the first:

2T1 cos10° = 1000 N ⇒ T1 =

1000 N 1000 N = = 508 N 2 cos10° 1.970

Assess: The two tensions are equal, as expected, since the two players are pulling at the same angle. The two add up to only slightly more than 1000 N, which makes sense because the angle at which the two players are pulling is small.


5.5. Solve:

Visualize: Please refer to the Figure Ex5.5. Applying Newton’s second law to the diagram on the left,

ax =

( Fnet )x m

=

4 N−2 N = 1.0 m /s2 2 kg

ay =

( Fnet ) y m

=

3N−3N = 0 m /s 2 2 kg

For the diagram on the right:

ax =

( Fnet )x m

=

4 N−2 N = 1.0 m /s2 2 kg

ay =

( Fnet ) y m

=

3 N −1 N − 2 N = 0 m /s 2 2 kg


5.6.

Visualize: Please refer to Figure Ex5.6. Solve: For the diagram on the left, three of the vectors lie along the axes of the tilted coordinate system. Notice that the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted. Applying Newton’s second law, ( Fnet ) x 5 N − 1 N − (3sin 20°) N ax = = = 1.49 m /s 2 2 kg m

ay =

( Fnet ) y m

=

2.82 N − (3 cos 20°) N = 0 m /s 2 2 kg

For the diagram on the right, the 2-newton force in the first quadrant makes an angle of 15° with the positive x-axis. The other 2-newton force makes an angle of 15° with the negative y-axis. The accelerations are

ax = ay =

( Fnet ) x (2 cos15°) N + (2 sin15°) N − 3 N = = −0.28 m /s 2 2 kg

m

( Fnet ) y m

=

1.414 N + (2 sin 15°) N − (2 cos15°) N = 0 m /s 2 2 kg


5.7.

Visualize: Please refer to Figure Ex5.7. Solve: We can use the constant slopes of the three segments of the graph to calculate the three accelerations. For t between 0 s and 3 s, ∆v x 12 m /s − 0 s ax = = = 4 m /s 2 ∆t 3s For t between 3 s and 6 s, ∆vx = 0 m/s, so ax = 0 m/s2. For t between 6 s and 8 s, ∆v x 0 m /s − 12 m /s ax = = = −6 m /s 2 ∆t 2s From Newton’s second law, at t = 1 s we have Fnet = max = (2.0 kg)(4 m/s2) = 8 N At t = 4 s, ax = 0 m/s2, so Fnet = 0 N. At t = 7 s, Fnet = max = (2.0 kg)(–6.0 m/s2) = –12 N Assess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive and negative signs are appropriate for an object first speeding up, then slowing down.


5.8.

Visualize: Please refer to Figure Ex5.8. Positive forces result in the object gaining speed and negative forces result in the object slowing down. The final segment of zero force is a period of constant speed. Solve: We have the mass and net force for all the three segments. This means we can use Newton’s second law to calculate the accelerations. The acceleration from t = 0 s to t = 3 s is F 4N ax = x = = 2 m /s 2 m 2.0 kg

The acceleration from t = 3 s to t = 5 s is

ax =

Fx −2 N = = −1 m /s 2 m 2.0 kg

The acceleration from t = 5 s to 8 s is ax = 0 m/s2. In particular, a x (at t = 6 s) = 0 m /s 2 . We can now use one-dimensional kinematics to calculate v at t = 6 s as follows: v = v0 + a1 (t1 − t 0 ) + a2 (t2 − t 0 )

= 0 + (2 m/s 2 )(3 s) + ( −1 m/s 2 )(2 s) = 6 m/s − 2 m/s = 4 m/s

r Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the positive F1 . A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the mass involved.


5.9.

Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, the tension in the rope is the only horizontal force. Visualize:

Solve: (a) Since the box is at rest, ax = 0 m/s2, and the net force on the box must be zero. Therefore, according to Newton’s first law, the tension in the rope must be zero. (b) For this situation again, ax = 0 m/s2, so Fnet = T = 0 N . (c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since ax = 5.0 m/s2,

Fnet = T = ma x = (50 kg)(5.0 m/s 2 ) = 250 N

Assess: For parts (a) and (b), the zero acceleration immediately implies that the rope is exerting no horizontal force on the box. For part (c), the 250 N force (the equivalent of about half the weight of a small person) seems reasonable to accelerate a box of this mass at 5.0 m/s2.


5.10. Model: We assume that the box is a point particle that is acted on only by the tension in the rope and the pull of gravity. Both the forces act along the same vertical line. Visualize:

Solve:

(a) Since the box is at rest, ay = 0 m/s2 and the net force on it must be zero:

Fnet = T − w = 0 N ⇒ T = w = mg = (50 kg)(9.8 m/s 2 ) = 490 N (b) Since the box is rising at a constant speed, again ay = 0 m/s2, Fnet = 0 N, and T = w = 490 N . (c) The velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Since ay = 5.0 m/s 2 ,

Fnet = T − w = ma y = (50 kg)(5.0 m /s 2 ) = 250 N ⇒ T = 250 N + w = 250 N + 490 N = 740 N (d) The situation is the same as in part (c), except that the rising box is slowing down. Thus ay = –5.0 m/s 2 and we have instead

Fnet = T − w = ma y = (50 kg)( −5.0 m /s 2 ) = −250 N ⇒ T = −250 N + w = −250 N + 490 N = 240 N Assess: For parts (a) and (b) the zero accelerations immediately imply that the box’s weight must be exactly balanced by the upward tension in the rope. For part (c) the tension not only has to support the box’s weight but must also accelerate it upward, hence, T must be greater than w. When the box accelerates downward, the rope need not support the entire weight, hence, T is less than w.


5.11. Solve:

Model: The astronaut is treated as a particle. The mass of the astronaut is

m=

wearth 800 N = = 81.6 kg 9.8 m /s 2 gearth

Therefore, the weight of the astronaut on Mars is

wMars = mgMars = (81.6 kg)(3.76 m/s 2 ) = 307 N Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than to the earth, so the smaller weight on Mars makes sense.


5.12. Solve:

Model: Use the particle model for the woman. (a) The woman’s weight on the earth is

wearth = mgearth = (55 kg)(9.8 m/s2 ) = 540 N (b) Since mass is a measure of the amount of matter, the woman’s mass is the same on the moon as on the earth. Her weight on the moon is

wmoon = mgmoon = (55 kg)(1.62 m/s2 ) = 89 N Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to the moon than to the earth. Thus the woman’s smaller weight on the moon makes sense.


5.13. Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull of gravity and the upward push of the elevator floor. We can use one-dimensional kinematics and Equation 5.10. Visualize:

Solve:

(a) The apparent weight is

ay    0 wapp = w1 +  = w1 +  = mg = (60 kg)(9.8 m /s 2 ) = 590 N g g   (b) The elevator speeds up from v0y = 0 m/s to its cruising speed at vy = 10 m/s. We need its acceleration before we can find the apparent weight:

ay =

∆v 10 m /s − 0 m /s = = 2.5 m /s 2 ∆t 4.0 s

The passenger’s apparent weight is

ay    2.5 m /s 2  wapp = w1 +  = (590 N )1 +  = (590 N)(1.26) = 740 N g 9.8 m /s 2    (c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus, wapp = w = 590 N as in part (a). Assess: The passenger’s apparent weight is his normal weight in parts (a) and (c), since there is no acceleration. In part (b), the elevator must not only support his weight but must also accelerate him upward, so it’s reasonable that the floor will have to push up harder on him, increasing his apparent weight.


5.14. Model: We assume that the passenger is a particle acted on by only two vertical forces: the downward pull of gravity and the upward force of the elevator floor. Visualize: Please refer to Figure Ex5.14. The graph has three segments corresponding to different conditions: (1) increasing velocity, meaning an upward acceleration, (2) a period of constant upward velocity, and (3) decreasing velocity, indicating a period of deceleration (negative acceleration). Solve: Given the assumptions of our model, we can calculate the acceleration for each segment of the graph and then apply Equation 5.10. The acceleration for the first segment is ay =

v1 − v0 8 m /s − 0 m /s = 4 m /s 2 = 2s−0 s t1 − t 0

ay    4  4 m /s 2  2  ⇒ wapp = w1 +  = ( mg)1 + = 1035 N  = (75 kg)(9.8 m /s )1 + 9.8  g 9.8 m /s 2    For the second segment, ay = 0 m/s2 and the apparent weight is

 0 m /s 2  2 wapp = w1 +  = mg = (75 kg)(9.8 m /s ) = 740 N g   For the third segment,

ay =

v3 − v2 0 m /s − 8 m /s = = −2 m /s 2 10 s − 6 s t3 − t 2

 −2 m /s 2  2 ⇒ wapp = w1 +  = (75 kg)(9.8 m /s )(1 − 0.2) = 590 N 9.8 m /s 2   Assess: As expected, the apparent weight is greater than normal when the elevator is accelerating upward and lower than normal when the acceleration is downward. When there is no acceleration the weight is normal. In all three cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator.


5.15. Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during the entire problem, we can use the model of kinetic friction. Visualize:

Solve: The safe is in equilibrium, since it’s not accelerating. Thus we can apply Newton’s first law in the vertical and horizontal directions:

( Fnet ) x = ΣFx = FB + FC − fk = 0 N

⇒ fk = FB + FC = 350 N + 385 N = 735 N

( Fnet ) y = ΣFy = n − w = 0 N ⇒ n = w = mg = (300 kg)(9.8 m/s 2 ) = 2940 N Then, for kinetic friction:

fk = µ k n ⇒ µ k =

fk 735 N = = 0.25 2940 N n

Assess: The value of µk = 0.25 is hard to evaluate without knowing the material the floor is made of, but it seems reasonable.


5.16. Model: We assume that the truck is a particle in equilibrium, and use the model of static friction. Visualize:

Solve: The truck is not accelerating, so it is in equilibrium, and we can apply Newton’s first law. The normal force has no component in the x-direction, so we can ignore it here. For the other two forces:

( Fnet ) x = ΣFx =

fs − w x = 0 N ⇒ fs = w x = mg sin θ = ( 4000 kg)(9.8 m/s 2 )(sin 15°) = 10,145 N

Assess: The truck’s weight (mg) is roughly 40,000 N. A friction force that is ≈25% of the truck’s weight seems reasonable.


5.17. Model: We assume that the car is a particle undergoing skidding, so we will use the model of kinetic friction. Visualize:

Solve: We begin with Newton’s second law. Although the motion is one-dimensional, we need to consider forces in both the x- and y-directions. However, we know that ay = 0 m/s2. We have

( Fnet ) y n − w n − mg − fk = = a y = 0 m /s 2 = m m m m m r r We used ( fk ) x = − fk because the free-body diagram tells us that fk points to the left. The friction model relates fk r to n with the equation fk = µ k n . The y-equation is solved to give n = mg . Thus, the kinetic friction force is fk = µ k mg . Substituting this into the x-equation yields: ax =

( Fnet ) x

=

− µ k mg = − µ k g = −(0.6)(9.8 m /s 2 ) = −5.88 m /s 2 m The acceleration is negative because the acceleration vector points to the left as the car slows. Now we have a constant-acceleration kinematics problem. ∆t isn’t known, so use (40 m /s)2 = 136 m v12 = 0 m 2 /s 2 = v02 + 2 a x ∆x ⇒ ∆x = − 2( −5.88 m /s 2 ) ax =

Assess: The skid marks are 136 m long. This is ≈ 430 feet, reasonable for a car traveling at ≈80 mph. It is worth noting that an algebraic solution led to the m canceling out.


5.18. Model: We assume that the plane is a particle accelerating in a straight line under the influence of two forces: the thrust of its engines and the rolling friction of the wheels on the runway. We can use one-dimensional kinematics. Visualize:

Solve:

We can use the definition of acceleration to find a, and then apply Newton’s second law. We obtain: ∆v 82 m /s − 0 m /s a= = = 2.34 m/s 2 ∆t 35 s

( Fnet ) = ΣFx = Fthrust − fr = ma ⇒ Fthrust = fr + ma For rubber rolling on concrete, µr = 0.02 (Table 5.1), and since the runway is horizontal, n = w = mg. Thus: Fthrust = µ r w + ma = µ r mg + ma = m( µ r g + a)

[

]

= (75,000 kg) (0.02)(9.8 m /s 2 ) + 2.34 m /s 2 = 190, 000 N Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that 190,000 N is enough to produce the required acceleration.


5.19. Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow speed and large mass). We can use the one-dimensional kinematic equations. Visualize:

Solve: The locomotive is not accelerating in the vertical direction, so the free-body diagram shows us that n = w = mg. Thus, fr = µ r mg = (0.002)(50,000 kg) 9.8 m/s 2 = 980 N

(

)

From Newton’s second law for the decelerating locomotive, −f −980 N ax = r = = −0.0196 m /s 2 50,000 kg m Since we’re looking for the distance the train rolls, but we don’t have the time:

v 2 − v02 (0 m /s) − (10 m /s) v − v = 2 a x ( ∆x ) ⇒ ∆x = 1 = = 2550 m 2ax 2( −0.0196 m /s 2 ) 2

2 1

2

2 0

Assess: The locomotive’s enormous inertia (mass) and the small coefficient of rolling friction make this long stopping distance seem reasonable.


5.20. Model: We assume that the mule is a particle acted on by two opposing forces in a single line: the farmer’s pull and friction. The mule will be subject to static friction until (and if!) it begins to move; after that it will be subject to kinetic friction. Visualize:

Solve: Since the mule does not accelerate in the vertical direction, the free-body diagram shows that n = w = mg . The maximum friction force is fs max = µ s mg = (0.8)(120 kg) 9.8 m/s 2 = 940 N The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not be able to budge the mule. Assess: The farmer should have known better.

(

)


5.21. Model: We assume that the skydiver is shaped like a box and is a particle. Visualize:

The skydiver falls straight down toward the earth’s surface, that is, the direction of fall is vertical. Since the skydiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm × 40 cm. The physical conditions needed to use Equation 5.16 for the drag force are satisfied. The terminal speed corresponds to the situation when the net force acting on the skydiver becomes zero. Solve: The expression for the magnitude of the drag with v in m/s is 1 D ≈ Av 2 = 0.25(0.20 × 0.40)v 2 N = 0.020v 2 N 4

(

)

The skydiver’s weight is w = mg = (75 kg) 9.8 m/s 2 = 735 N . The mathematical form of the condition defining dynamical equilibrium for the skydiver and the terminal speed is r r r Fnet = w + D = 0 N

735 ≈ 192 m/s 0.02 Assess: The result of the above simplified physical modeling approach and subsequent calculation, even if approximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very badly hurt at landing if the parachute does not open in time. 2 ⇒ 0.02vterm N − 735 N = 0 N ⇒ vterm =


5.22. Model: We will represent the bowling ball as a particle. Visualize:

The bowling ball falls straight down toward the earth’s surface. The bowling ball is subject to a net force that is the resultant of the weight and drag force vectors acting vertically, in the downward and upward directions, respectively. Once the net force acting on the ball becomes zero, the terminal velocity is reached and remains constant for the rest of the motion. Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is r r r r Fnet = w + D = 0 N Since only the vertical direction matters, one can write: ΣFy = 0 N ⇒ Fnet = D − w = 0 N When this condition is satisfied, the speed of the ball becomes the constant terminal speed v = vterm . The magnitudes of the weight and drag forces acting on the ball are: w = mg = m 9.80 m /s 2

(

)

1 2 2 2 2 = (0.25π )(0.11 m ) (85 m /s) = 69 N ( Avterm ) = 0.25(πR2 )vterm 4 The condition for dynamic equilibrium becomes: (9.80 m /s2 )m − 68.66 N = 0 N ⇒ m = 9.869mN/s2 = 7.0 kg Assess: The value of the mass of the bowling ball obtained above seems reasonable. Such a ball is heavy so that it has a significant impact on the pins, but not “excessively” heavy so that it can be lifted and rolled by an average human player. D≈


5.23. Model: Treat the car as a particle. Assume that rolling friction is negligible. Visualize:

Solve:

There’s only one force in the horizontal direction, so the car’s acceleration is given by Newton’s second law as ∑ Fx = Fprop ax = m m Table 4.3 gave the typical propulsion force of a car as Fprop ≈ 5000 N. The mass of a typical car is m ≈ 1500 N, a value that has been used in several examples. Thus 5000 N ax ≈ ≈ 3 m /s 2 1500 N Assess: One significant figure is appropriate for an estimate. This is about 30% of g, which seems reasonable for a car’s acceleration.


5.24. Model: We can treat the sliding player as a particle experiencing kinetic friction. Visualize:

Solve: We can assume a mass of 80 kg (which corresponds to a weight of about 175 pounds). We have no value for µk of cloth sliding on loose dirt. It is probably greater than µk for wood on wood, due to the roughness of both surfaces. Let’s guess 0.40. From the free-body diagram, n = w = mg, since there’s no vertical acceleration. Thus,

fk = µ k mg = (0.40)(80 kg)(9.8 m/s 2 ) = 314 N

Assess: A frictional force of 314 N would produce an acceleration of

a=

Fnet −314 N = ≈ − 4 m /s 2 80 kg m

A player running initially at 8 m/s would thus be brought to a stop in about 2 seconds, which seems somewhat too long. Our estimate of µk is probably a bit low but 8 m/s is a large speed as well.


5.25. Visualize:

We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration was calculated as follows: F 10 N amax = max = = 2 m /s 2 5 kg m Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use the more general result that v(t) = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is 12 (4 s) (2 m/s2) = 4.0 m/s. We’ve used the fact that the area between 4 s and 6 s is zero. Thus, at t = 6 s, vx = 4.0 m/s.


5.26. Visualize:

The acceleration is a x = Fx m , so the acceleration-versus-time graph has exactly the same shape as the forceversus-time graph. The maximum acceleration is amax = Fmax m = (6 N ) (2 kg) = 3 m / s 2 . Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use the more general result that v(t) = v0 + area under the acceleration curve from 0 s to t The object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 4 s is a rectangle (3 m/s2 × 2 s = 6 m/s) plus a triangle 12 × 3 m/s 2 × 2 s = 3 m/s . Thus vx = 9 m/s at t = 4 s.

(

)


r Model: You can model the beam as a particle and assume Fnet = 0 N to calculate the tensions in the suspension ropes. Visualize:

5.27.

Solve: The beam attached to the ropes will remain in static equilibrium only if both inequalities T1 < Tmax and T2 < Tmax hold, where Tmax is the maximum sustained tension. The equilibrium equations in vector and component form are: r r r r r Fnet = T1 + T2 + w = 0 N

( Fnet ) x = T1x + T2 x + w x = 0 N ( Fnet ) y = T1y + T2 y + w y = 0 N Using the free-body diagram yields:

− T1 sin θ1 + T2 sin θ 2 = 0 N

T1 cosθ1 + T2 cosθ 2 − w = 0 N

The mathematical model is reduced to a simple algebraic system of two equations with two unknowns, T1 and T2 . Substituting θ1 = 20°, θ2 = 30°, and w = mg = 9800 N, the simultaneous equations become

− T1 sin 20° + T2 sin 30° = 0 N

T1 cos 20° + T2 cos 30° = 9800 N

You can solve this system of equations by simple substitution. The result is T1 = 6397 N and T2 = 4376 N . The rope on the left side (rope 1) breaks since the tension in this rope is larger than 6000 N. Once the left rope is broken, the right rope will also break because now the whole weight of the beam will be applied to the rope on the right side. Assess: The above approach and result seem reasonable. Intuition indicates there is more tension in the left rope than in the right rope.


5.28. Model: The piano is in static equilibrium and is to be treated as a particle. Visualize:

Solve:

(a) Based on the free-body diagram, Newton’s second law is

( Fnet ) x = 0 N = T1x + T2 x = T2 cosθ 2 − T1 cosθ1 ( Fnet ) y = 0 N = T1y + T2 y + T3 y + w y = T3 − T1 sin θ1 − T2 sin θ 2 − mg Notice how the force components all appear in the second law with plus signs because we are adding forces. The negative signs appear only when we evaluate the various components. These are two simultaneous equations in the two unknowns T2 and T3. From the x-equation we find T cosθ1 (500 N ) cos15° T2 = 1 = = 533 N cosθ 2 cos 25° (b) Now we can use the y-equation to find T3 = T1 sin θ1 + T2 sin θ 2 + mg = 5250 N


5.29. Model: The plastic ball is represented as a particle in static equilibrium. Visualize:

Solve: The electric force, like the weight, is a long-range force. So the ball experiences the contact force of the string’s tension plus two long-range forces. The equilibrium condition is

( Fnet ) x = Tx + ( Felec ) x = T sin θ − Felec = 0 N ( Fnet ) y = Ty + w y = T cosθ − mg = 0 N We can solve the y-equation to get

(0.001 kg)(9.8 m/s 2 ) mg = = 0.0104 N cosθ cos 20° Substituting this value into the x-equation, T=

Felec = T sin θ = (1.04 × 10 −2 N ) sin 20° = 0.0036 N

(b) The tension in the string is 0.0104 N.


5.30. Model: We will represent Henry as a particle. His motion is governed by constant-acceleration kinematic equations. Visualize: Please refer to the Figure Ex5.30. Solve: (a) Henry undergoes an acceleration from 0 s to 2.0 s, constant velocity motion from 2.0 s to 10.0 s, and another acceleration as the elevator brakes from 10.0 s to 12.0 s. The apparent weight is the same as the true weight during constant velocity motion, so Henry’s true weight mg is 750 N. His apparent weight is less than his true weight during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, the elevator’s initial motion is down. (b) Because Henry’s true weight is 750 N, his mass is m = w/g = 76.5 kg. (c) The apparent weight (see Equation 5.10) during vertical motion is given by   wapp  a wapp = wtrue 1 +  ⇒ a = g − 1 g    wtrue During the interval 0 s ≤ t ≤ 2 s, the elevator’s acceleration is

600 N  a = g − 1 = −1.96 m /s 2  750 N  At t = 2 s, Henry’s position is

y1 = y0 + v0 ∆t 0 +

1 1 2 2 a( ∆t 0 ) = a( ∆t 0 ) = −3.92 m 2 2

and his velocity is v1 = v0 + a∆t0 = a∆t0 = –3.92 m/s During the interval 2 s ≤ t ≤ 10 s, a = 0 m/s2. This means Henry travels with a constant velocity v1 = –3.92 m/s. At t = 10 s he is at position y2 = y1 + v1∆t1 = –35.3 m and he has a velocity v2 = v1 = –3.92 m/s. During the interval 10 s ≤ t ≤ 12.0 s, the elevator’s acceleration is 900 N  a = g − 1 = +1.96 m /s 2  750 N  The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t = 12.0 s Henry is at position 1 y3 = y2 + v2(∆t2) + a(∆t2)2 = –39.2 m 2 Thus Henry has traveled distance 39.2 m.


5.31. Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single vertical line: gravity and the supporting force of the elevator. Visualize:

Solve:

(a) Before the elevator starts braking, Zach is not accelerating. His apparent weight (see Equation 5.10) is

  a 0 m /s 2  2 wapp = w1 +  = w1 +  = mg = (80 kg)(9.8 m /s ) = 784 N g g    (b) Using the definition of acceleration, ∆v v1 − v0 0 − ( −10) m /s a= = = 3.33 m /s 2 = ∆t t1 − t 0 3.0 s

  a 3.33 m /s 2  ⇒ wapp = w1 +  = (80 kg)(9.8 m /s 2 )1 +  = (784 N)(1+0.340) = 1050 N g 9.8 m /s 2    Assess: While the elevator is braking, it not only must support Zach’s weight but must also push upward on him to decelerate him, so the apparent weight is greater than his normal weight.


5.32. Model: We can assume your body is a particle moving in a straight line under the influence of two forces: gravity and the support force of the scale. Visualize:

Solve:

The apparent weight (see Equation 5.10) of an object moving in an elevator is

 a  wapp  − 1 g wapp = w1 +  ⇒ a =   w  g  When accelerating upward, the acceleration is 170 lb  a= − 1 9.8 m /s 2 = 1.3 m /s 2  150 lb  When braking, the acceleration is 120 lb  a= − 1 9.8 m /s 2 = −2.0 m /s 2  150 lb  Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire State Building must be. Also note that we did not have to convert the units of the weights from pounds to newtons because the weights appear as a ratio.

(

)

(

)


5.33. Model: We can assume the foot is a single particle in equilibrium under the combined effects of gravity, the tensions in the upper and lower sections of the traction rope, and the opposing traction force of the leg itself. We can also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the tension is the same everywhere in the rope. Because all pulleys are in equilibrium, their net force is zero. So they do not contribute to T. Visualize:

Solve:

(a) From the free-body diagram for the mass, the tension in the rope is

T = w = mg = (6 kg)(9.8 m/s 2 ) = 58.8 N (b) Using Newton’s first law for the vertical direction on the pulley attached to the foot,

( Fnet ) y = ΣFy = T sin θ − T sin15° − wfoot = 0 N ⇒ sin θ =

(4 kg)(9.8 m/s 2 ) T sin 15° + wfoot m g = 0.259 + 0.667 = 0.926 = sin 15° + foot = 0.259 + 58.8 N T T ⇒ θ = sin −1 0.926 = 67.8°

(c) Using Newton’s first law for the horizontal direction,

( Fnet ) x = ΣFx = T cosθ + T cos15° − Ftraction = 0 N ⇒ Ftraction = T cosθ + T cos15° = T (cos 67.8° + cos15°)

= (58.8 N)(0.3778 + 0.9659) = (58.8 N)(1.344) = 79.0 N Assess: Since the tension in the upper segment of the rope must support the foot and counteract the downward pull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). The magnitude of the traction force, roughly one-tenth the weight of a human body, seems reasonable.


5.34. Model: We can assume the person is a particle moving in a straight line under the influence of the combined decelerating forces of the air bag and seat belt or, in the absence of restraints, the dashboard or windshield. Visualize:

Solve: (a) In order to use Newton’s second law for the passenger, we’ll need the acceleration. Since we don’t have the stopping time:

v12 = v02 + 2 a( x1 − x 0 ) ⇒ a =

0 m 2 /s 2 − (15 m /s) v12 − v02 = = −112.5 m /s 2 2( x1 − x 0 ) 2(1 m − 0 m ) 2

⇒ Fnet = F = ma = (60 kg)( −112.5 m/s 2 ) = −6750 N The net force is 6750 N to the left. (b) Using the same approach as in part (a),

0 m 2 /s 2 − (15 m /s) v12 − v02 = (60 kg) = −1,350,000 N 2( x1 − x 0 ) 2(0.005 m ) 2

F = ma = m

The net force is 1,350,000 N to the left. (c) The passenger’s weight is mg = (60 kg)(9.8 m/s2) = 588 N. The force in part (a) is 11.5 times the passenger’s weight. The force in part (b) is 2300 times the passenger’s weight. Assess: An acceleration of 11.5g is well within the capability of the human body to withstand. A force of 2300 times the passenger’s weight, on the other hand, would surely be catastrophic.


5.35. Model: We will represent the bullet as a particle. Visualize:

Solve: (a) We have enough information to use kinematics to find the acceleration of the bullet as it stops. Then we can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant to the problem.) The kinematic equation is

v02 (400 m /s) = −6.67 × 10 5 m /s 2 =− 2 ∆x 2(0.12 m ) r Notice that a is negative, in agreement with the vector a in the motion diagram. Turning to forces, the wood exerts two forces on the bullet. First, anrupward normal force that keeps the bullet from “falling” through the wood. r Second, a retarding friction force fk that stops the bullet. The only horizontal force is fk , which points to the left and thus has a negative x-component. The x-component of Newton’s second law is 2

v12 = v02 + 2 a∆x ⇒ a = −

( Fnet ) x = − fk = ma ⇒ fk = − ma = −(0.01 kg)(−6.67 × 10 5 m/s 2 ) = 6670 N Notice how the signs worked together to give a positive value of the magnitude of the force. (b) The time to stop is found from v1 = v0 + a∆t as follows:

∆t = −

v0 = 6.00 × 10 −4 s = 600 µs a

(c)

Using the above kinematic equation, we can find the velocity as a function of t. For example at t = 60 µs,

v x = 400 m /s + ( −6.667 × 10 5 m /s 2 )(60 × 10 −6 s) = 360 m /s


5.36. Model: The ball is represented as a particle that obeys constant-acceleration kinematic equations. Visualize:

Solve: This is a two-part problem. During part 1 the ball accelerates upward in the tube. During part 2 the ball undergoes free fall (a = –g). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from the tube. The free body diagram for part 1 shows two forces: the air pressure force and the weight force. We need only the y-component of Newton’s second law:

ay = a =

( Fnet ) y m

=

Fair − w Fair 2N = −g= − 9.8 m /s 2 = 30.2 m /s 2 m m 0.05 kg

We can use kinematics to find the velocity v1 as the ball leaves the tube:

v12 = v02 + 2 a( y1 − y0 ) ⇒ v1 = 2 ay1 = 2(30.2 m /s 2 )(1 m ) = 7.77 m /s For part 2, free-fall kinematics v22 = v12 − 2 g( y2 − y1 ) gives

y2 − y1 =

v12 = 3.08 m 2g


5.37. Model: We assume the rocket is a particle moving in a vertical straight line under the influence of only two forces: gravity and its own thrust. Visualize:

Solve:

(a) Using Newton’s second law and reading the forces from the free-body diagram,

Fthrust − w = ma ⇒ Fthrust = ma + mgearth = (0.200 kg)(10 m/s 2 + 9.8 m/s 2 ) = 3.96 N (b) Likewise, the thrust on the moon is (0.200 kg)(10 m/s2 + 1.62 m/s2) = 2.32 N. Assess: The thrust required is smaller on the moon, as it should be, given the moon’s weaker gravitational pull. The magnitude of a few newtons seems reasonable for a small model rocket.


5.38. Model: Represent the rocket as a particle that follows Newton’s second law. Visualize:

Solve:

(a) The y-component of Newton’s second law is

ay = a =

( Fnet ) y m

=

Fthrust − mg 300,000 N = − 9.8 m /s 2 = 5.20 m /s 2 m 20,000 kg

(b) At 5000 m the acceleration has increased because the rocket mass has decreased. Solving the equation of part (a) for m gives

m5000 m =

Fthrust 300,000 N = = 18,990 kg a5000 m + g 6.00 m /s 2 + 9.8 m /s 2

The mass of fuel burned is mfuel = minitial – m5000 = 1010 kg.


5.39.

Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of friction µs = 0.80 and a kinetic coefficient of friction µk = 0.60. Visualize:

Solve:

(a) While the block is at rest, Newton’s second law is

( Fnet ) x = T − fs = 0 N ⇒ T = fs ( Fnet ) y = n − w ⇒ n = mg The static friction force has a maximum value ( fs ) max = µ s mg . The string tension that will cause the block to slip and start to move is

T = µ s mg = (0.80)(2 kg)(9.8 m/s 2 ) = 15.7 N

Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension for motion. (b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the x-component of Newton’s second law as follows: T− f ( Fnet ) x = T − fk = max ⇒ ax = m k The kinetic friction force fk = µkmg. The acceleration of the block is

ax =

20 N − (0.60)(2 kg)(9.8 m /s 2 ) 2 kg

= 4.12 m /s 2

Using kinematics, the block’s speed after moving 1.0 m will be

v12 = 0 m 2 /s 2 + 2( 4.12 m/s 2 )(1.0 m ) ⇒ v1 = 2.87 m/s

(c) The only difference in this case is the coefficient of kinetic friction whose value is 0.05 instead of 0.60. The acceleration of the block is 20 N − (0.05)(2 kg) 9.8 m /s 2 ax = = 9.51 m /s 2 2 kg The block’s speed after moving 1.0 m will be

(

)

v12 = 0 m 2 /s 2 + 2(9.51 m/s 2 )(1.0 m ) ⇒ v1 = 4.36 m/s


5.40. Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of two forces: the thrust of his jet skis and the resisting force of friction on the skis. We can use one-dimensional kinematics. Visualize:

Solve: (a) The friction force of the snow can be found from the free-body diagram and Newton’s first law, since there’s no acceleration in the vertical direction:

n = w = mg = (75 kg)(9.8 m/s 2 ) = 735 N ⇒ fk = µ k n = (0.1)(735 N ) = 73.5 N

Then, from Newton’s second law:

( Fnet ) x = Fthrust − fk = ma0 ⇒ a0 = From kinematics:

Fthrust − fk 200 N − 73.5 N = = 1.687 m /s 2 75 kg m

v1 = v0 + a0 t1 = 0 m/s + (1.687 m/s 2 )(10 s) = 16.87 m/s

(b) During the acceleration, Sam travels to

x1 = x 0 + v0 t1 +

1 2 1 2 a0 t1 = (1.687 m /s 2 )(10 s) = 84 m 2 2

After the skis run out of fuel, Sam’s acceleration can again be found from Newton’s second law: F −73.5 N ( Fnet ) x = − fk = −73.5 N ⇒ a1 = mnet = 75 kg = −0.98 m /s 2 Since we don’t know how much time it takes Sam to stop:

v22 = v12 + 2 a1 ( x 2 − x1 ) ⇒ x 2 − x1 =

2 2 v22 − v12 0 m /s − (16.87 m /s) = = 145 m 2 a1 2( −0.98 m /s 2 ) 2

The total distance traveled is ( x 2 − x1 ) + x1 = 145 m + 84 m = 229 m. Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coasting distance of nearly 150 m also seems possible, starting from a high speed, given that we’re neglecting air resistance.


5.41. Model: We assume Sam is a particle moving in a straight line down the slope under the influence of gravity, the thrust of his jet skis, and the resisting force of friction on the snow. Visualize:

Solve:

From the height of the slope and its angle, we can calculate its length: h h 50 m = sin θ ⇒ x1 − x 0 = = = 288 m x1 − x 0 sin θ sin10°

Since Sam is not accelerating in the y-direction, we can use Newton’s first law to calculate the normal force:

( Fnet ) y = ΣFy = n − w cosθ = 0 N

⇒ n = w cosθ = mg cosθ = (75 kg)(9.8 m/s2)(cos10°) = 724 N

One-dimensional kinematics gives us Sam’s acceleration:

v12 = v02 + 2 a x ( x − x 0 ) ⇒ a x =

v12 − v02 (40 m /s) − 0 m 2 /s 2 = 2.78 m /s 2 = 2( x1 − x 2 ) 2(288 m ) 2

Then, from Newton’s second law and the equation fk = µ k n :

( Fnet ) x = ΣFx = w sin θ + Fthrust − fk = max ⇒ µk =

=

mg sin θ + Fthrust − ma n

(75 kg)(9.8 m/s 2 )(sin10°) + 200 N − (75 kg)(2.78 m/s 2 )

724 N Assess: This coefficient seems a bit high for skis on snow, but not impossible.

= 0.165


5.42. Model: We will represent the crate as a particle. Visualize:

r r (a) When the belt runs at constant speed, the crate has an acceleration a = 0 m/s 2 and is in dynamic r r equilibrium. Thus Fnet = 0. It is tempting to think that the belt exerts a friction force on the crate. But if it did, there would be a net force because there are no other possible horizontal forces to balance a friction force. Because there is no net force, there cannot be a friction force. The only forces are the upward normal force and the crate’s weight. (A friction force would have been needed to get the crate moving initially, but no horizontal force is needed to keep it moving once it is moving with the same constant speed as the belt.) (b) If the belt accelerates gently, the crate speeds up without slipping on the belt. Because it is accelerating, the crate must have a net horizontal force. So now there is a friction force, and the force points in the direction of the crate’s motion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative to the belt. Thus, it is a static friction force that accelerates the crate so that it moves without slipping on the belt. (c) The static friction force has a maximum possible value (fs)max = µsn. The maximum possible acceleration of the crate is ( fs )max = µs n amax = m m If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from the free-body diagram that n = w = mg . Thus, amax = µsg = (0.5)(9.8 m/s2) = 4.9 m/s2 Solve:


5.43. Model: We assume the suitcase is a particle accelerating horizontally under the influence of friction only. Visualize:

Solve: Because the conveyor belt is already moving, friction drags your suitcase to the right. It will accelerate until it matches the speed of the belt. We need to know the horizontal acceleration. Since there’s no acceleration in the vertical direction, we can apply Newton’s first law to find the normal force:

n = w = mg = (10 kg)(9.8 m/s 2 ) = 98 N

The suitcase is accelerating, so we use µk to find the friction force

fk = µ k mg = (0.3)(98 N ) = 29.4 N We can find the horizontal acceleration from Newton’s second law:

( Fnet ) x = ΣFx =

fk = ma ⇒ a =

fk 29.4 N = = 2.94 m /s 2 10 kg m

From one of the kinematic equations:

v12 = v02 + 2 a( x1 − x 0 ) ⇒ x1 − x 0 =

v12 − v02 (2.0 m /s) − (0 m /s) = = 0.68 m 2a 2(2.94 m /s 2 ) 2

2

The suitcase travels 0.68 m before catching up with the belt and riding smoothly. Assess: If we imagine throwing a suitcase at a speed of 2.0 m/s onto a motionless surface, 0.68 m seems a reasonable distance for it to slide before stopping.


5.44. Model: We will represent Johnny as a particle, and use the model of kinetic friction. Visualize:

Use a tilted coordinate system to make this a one-dimensional motion problem. Solve: The free-body diagram shows that the hill exerts both a normal force and a kinetic friction force on r Johnny. The friction force is up the hill, opposite his direction of motion, while n is perpendicular to the hill. It is r important to note that the angle between w and the –y-axis is the same angle as the tilt of the x-axis. Solve: Working from the free-body diagram, we can write Newton’s second law as

ax = a =

( Fnet ) x m

=

w sin θ − fk mg sin θ − fk = m m

( Fnet ) y

n − w cosθ n − mg cosθ = m m m The friction model is fk = µ k n . First solve the y-equation to give n = mgcosθ. Use this in the friction model to get fk = µ k mg cosθ . Now substitute this result for fk into the x-equation: a y = 0 m /s 2 =

ax =

=

mg sin θ − µ k mg cosθ = g(sin θ − µ k cosθ ) = (9.8 m /s 2 )(sin 20° − 0.5 cos 20°) = −1.253 m /s 2 m

Notice that ax came out negative, as it should. Now we can do constant-acceleration kinematics:

v12 = v02 + 2 a x ( x1 − x 0 ) ⇒ v0 = −2 a x x1 = −2( −1.253 m /s 2 )(3.5 m ) = 2.96 m /s


5.45. Model: We will represent the wood block as a particle, and use the model of kinetic friction and kinematics. Assume w sin θ > fs, so it does not hang up at the top. Visualize:

The block ends where it starts, so x2 = x0 = 0 m. We expect v2 to be negative, because the block will be moving in the –x-direction, so we’ll want to take |v2| as the final speed. Because of friction, we expect to find |v2| < v0. r → Solve: (a) The friction force is opposite to v , so fk points down the slope during the first half of the motion and up r r the slope during the second half. w and n are the only other forces. Newton’s second law for the upward motion is

a x = a0 =

( Fnet ) x m

a y = 0 m /s 2 =

=

− w sin θ − fk − mg sin θ − fk = m m

( Fnet ) y m

=

n − w cosθ n − mg cosθ = m m

The friction model is fk = µ k n . First solve the y-equation to give n = mgcosθ. Use this in the friction model to get fk = µ k mg cosθ . Now substitute this result for fk into the x-equation:

a0 =

− mg sin θ − µ k mg cosθ = − g(sin θ + µ k cosθ ) = −(9.8 m /s 2 )(sin 30° + 0.20 cos 30°) = −6.60 m /s 2 m

Kinematics now gives

v12 = v02 + 2 a0 ( x1 − x 0 ) ⇒ x1 =

2 2 v12 − v02 0 m /s − (10 m /s) = = 7.58 m 2 a0 2( −6.60 m /s 2 ) 2

The block’s height is then h = x1sinθ = (7.58 m)sin30° = 3.79 m. r (b) For the return trip, fk points up the slope, so the x-component of the second law is

a x = a1 =

( Fnet ) x m

=

− w sin θ + fk − mg sin θ + fk = m m

Note the sign change. The y-equation and the friction model are unchanged, so we have

a1 = − g(sin θ − µ k cosθ ) = −3.20 m / s 2 The kinematics for the return trip are

v22 = v12 + 2 a1 ( x 2 − x1 ) ⇒ v2 = −2 a1 x1 = −2( −3.20 m /s 2 )(7.58 m ) = −6.97 m /s Notice that we used the negative square root because v2 is a velocity with the vector pointing in the –x-direction. The final speed is |v2| = 6.97 m/s.


5.46. Model: We will model the box as a particle, and use the models of kinetic and static friction. Visualize:

The pushing force is along the +x-axis, but the force of friction acts along the –x-axis. A component of the box’s weight acts along the –x-axis as well. The box will move up if the pushing force is at least equal to the sum of the friction force and the component of the weight in the x-direction. Solve: Let’s determine how much pushing force you would need to keep the box moving up the ramp at steady speed. Newton’s second law for the box in dynamic equilibrium is (Fnet)x = ΣFx = nx + wx + (fk)x + (Fpush)x= 0 N – mg sinθ – fk + Fpush = 0 N (Fnet)y = ΣFy = ny + wy + (fk)y + (Fpush)y = n – mg cosθ + 0 N + 0 N = 0 N The x-component equation and the model of kinetic friction yield: Fpush = mg sinθ + fk = mg sinθ + µkn Let us obtain n from the y-component equation as n = mg cosθ, and substitute it in the above equation to get Fpush = mg sinθ + µk mg cosθ = mg (sinθ + µk cosθ) = (100 kg)(9.80 m/s2)(sin 20° + 0.60 cos20°) = 888 N The force is less than your maximum pushing force of 1000 N. That is, once in motion, the box could be kept moving up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of static friction applies. The analysis is the same except that the coefficient of static friction is used and we use the maximum value of the force of static friction. Therefore, we have Fpush = mg (sinθ + µs cosθ) = (100 kg)(9.80 m/s2)(sin 20° + 0.90 cos 20°) = 1160 N Since you can push with a force of only 1000 N, you can’t get the box started. The big static friction force and the weight are too much to overcome.


5.47. Model: We will model the sled and friend as a particle, and use the model of kinetic friction because the sled is in motion. Visualize:

The net force on the sled is zero (note the constant speed of the sled). That means the component of the pulling force along the +x-direction is equal to the magnitude of the kinetic force of friction in the –x-direction. Also note that (Fnet)y = 0 N, since the sled is not moving along the y-axis. Solve: Newton’s second law is (Fnet)x = ΣFx = nx + wx + (fk)x + (Fpull)x= 0 N + 0 N – fk + Fpull cosθ = 0 N (Fnet)y = ΣFy = ny + wy + (fk)y + (Fpull)y = n – mg + 0 N + Fpull sinθ = 0 N The x-component equation using the kinetic friction model fk = µkn reduces to

µkn = Fpull cosθ The y-component equation gives n = mg – Fpull sinθ We see that the normal force is smaller than the weight because Fpull has a component in a direction opposite to the direction of the weight. In other words, Fpull is partly lifting the sled. From the x-component equation, µk can now be obtained as

µk =

Fpull cosθ mg − Fpull sin θ

=

(75 N)(cos 30°)

(60 kg)(9.8 m/s 2 ) − (75 N)(sin 30°)

= 0.12

Assess: A quick glance at the various µk values in Table 5.1 suggests that a value of 0.12 for µk is reasonable.


5.48. Model: As long as the static force of friction between the box and the sled is sufficiently large for the box not to slip, the acceleration a of the box is the same as the acceleration of the sled. We will therefore model the box as a particle and use the model of static friction. Visualize:

The force on the box that is responsible for its acceleration is the force of static friction provided by the sled because the box would slide backward without this force. Solve: Newton’s second law for the box is (Fnet)x = ΣFx = nx + wx + (fs)x = 0 N + 0 N + fs max = max = mamax (Fnet)y = ΣFy = ny + wy + (fs)y = n – mg + 0 N = may = 0 N The above equations can be combined along with the friction model to get

amax =

fs max µ s n µ s mg = = = µs g m m m

To find Tmax, we write Newton’s second law for the box-sled system, with total mass m + M, as (Fnet)x = ΣFx = Nx + Wx + (fs)x + (Tmax)x = 0 N + 0 N + 0 N + Tmax = (m + M)amax (Fnet)y = ΣFy = Ny + Wy + (fs)y + (Tmax)y = N – (m + M)g + 0 N + 0 N = 0 N Referring to Table 5.1 for the coefficient of friction, the x-component equation for the box-sled system yields Tmax = (m + M)amax = (m + M)µsg = (0.50)(9.8 m/s2)(5.0 kg + 10 kg) = 74 N That is, the largest tension force for which the box does not slip is 74 N. Assess: If the box was glued to the sled, there would be no resistance to motion because the ice is frictionless, and the tension force could be as large as one wants. On the other hand, if the friction between the box and the sled was zero, one could not pull at all without causing the box to slide.


5.49. Model: We will model the steel cabinet as a particle. It touches the truck’s steel bed, so only the steel bed can exert contact forces on the cabinet. As long as the cabinet does not slide, the acceleration a of the cabinet is equal to the acceleration of the truck. Visualize:

Solve: The shortest stopping distance is the distance for which the static friction force has its maximum value fs max. Newton’s second law for the box and the model of static friction are (Fnet)x = ΣFx = nx + wx + (fs)x = 0 N + 0 N – fs max = max = ma ⇒ –fs max = ma (Fnet)y = ΣFy = ny + wy + (fs)y = n − mg + 0 N = 0 N ⇒ n = mg –fs max = –µsn = –µsmg = ma These three equations can be combined together to get a = –µsg. Because constant-acceleration kinematics gives v12 = v02 + 2a(x1– x0) and µs = 0.8 (Table 5.1), we find

( x1 − x0 ) =

v12 − v02 − v02 (15 m /s) = = = 14.3 m 2a (−2 µs g) (2)(0.80)(9.8 m /s 2 ) 2

Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance without making the contents slide of about 14.3 m or approximately 47 feet looks reasonable.


5.50. Model: The antiques (mass = m) in the back of your pickup (mass = M) will be treated as a particle. The antiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickupantiques system will also be treated as a particle, and the contact force on this particle will be due to the road. Visualize:

Solve: (a) We will find the smallest coefficient of friction that allows the truck to stop in 55 m, then compare that to the known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M, Newton’s second law is (Fnet)x = ΣFx = Nx + Wx + ( f ) x = 0 N + 0 N – f = (m + M)ax = (m + M)a (Fnet)y = ΣFy = Ny + Wy + ( f ) y = N – (m + M)g + 0 N = 0 N The model of static friction is f = µN, where µ is the coefficient of friction between the tires and the road. These equations can be combined to yield a = –µg. Since constant-acceleration kinematics gives v12 = v02 + 2a(x1 + x0), we find

v12 − v02 v02 (25 m /s) ⇒ µ min = = = 0.58 2( x1 − x 0 ) 2 g( x1 − x 0 ) (2)(9.8 m /s 2 )(55 m ) 2

a=

The truck cannot stop if µ is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80 respectively (see Table 5.1), are larger. So the truck can stop. (b) The analysis of the pickup-antiques system applies to the antiques, and it gives the same value of 0.58 for µmin. This value is smaller than the given coefficient of static friction (µs = 0.6) between the antiques and the truck bed. Therefore, the antiques will not slide as the truck is stopped over a distance of 55 m. Assess: The analysis of parts (a) and (b) are the same because mass cancels out of the calculations. According to the California Highway Patrol website, the stopping distance (with zero reaction time) for a passenger vehicle traveling at 25 m/s or 82 ft/s is approximately 43 m. This is smaller than the 55 m over which you are asked to stop the truck.


5.51. Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we will also use the model of kinetic friction. Visualize:

Solve: The normal force due to the wall, which is perpendicular to the wall, is here to the right. The box slides r r r r down the wall at constant speed, so a = 0 and the box is in dynamic equilibrium. Thus, Fnet = 0 . Newton’s second law for this equilibrium situation is (Fnet)x = 0 N = n – Fpush cos45° (Fnet)y = 0 N = fk + Fpush sin45° – w = fk + Fpush sin45° – mg The friction force is fk = µkn. Using the x-equation to get an expression for n, we see that fk = µk Fpush cos45°. Substituting this into the y-equation and using Table 5.1 to find µk = 0.20 gives, µk Fpush cos45° + Fpush sin45° – mg = 0 N

⇒ Fpush =

(2 kg)(9.8 m /s 2 ) mg = = 23.1 N µ k cos 45° + sin 45° 0.20 cos 45° + sin 45°


5.52. Model: Use the particle model for the block and the model of static friction. Visualize:

Solve: The block is initially at rest, so initially the friction force is static friction. If the 12 N push is too strong, the box will begin to move up the wall. If it is too weak, the box will begin to slide down the wall. And if the pushing force is within the proper range, the box will remain stuck in place. First, let’s evaluate the sum of all the forces except friction: ΣFx = n – Fpush cos30° = 0 N ⇒ n = Fpush cos30° ΣFy = Fpush sin30° – w = Fpush sin30° – mg = (12 N)sin30° – (1 kg)(9.8 m/s2) = –3.8 N In the first equation we utilized the fact that any motion is parallel to the wall, so ax = 0 m/s2. These three forces add up to −3.8 jˆ N . This means the static friction force will be able to prevent the box from moving if fs = +3.8 jˆ N . Using the x-equation and the friction model we get fs max = µsn = µsFpush cos30° = 5.2 N r where we used µs = 0.5 for wood on wood. The static friction force fs needed to keep the box from moving is less than fs max. Thus the box will stay at rest.


5.53. Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contact force and the wind exerts a drag force on the skier. We will therefore use the models of kinetic friction and drag. Visualize:

We choose a coordinate system such that the skier’s motion is along the +x-direction. While the forces of kinetic r r friction fk and drag D act along the –x-direction opposing the motion of the skier, the weight of the skier has a component in the +x-direction. At the terminal speed, the net force on the skier is zero as the forces along the +xdirection cancel out the forces along the –x-direction. Solve: Newton’s second law and the models of kinetic friction and drag are 1 (Fnet)x = ΣFx = nx + wx + (fk)x + (D)x = 0 N + mg sinθ – fk– Av2 = max = 0 N 4 (Fnet)y = ΣFy = ny + wy + (fk)y + (D)y = n – mg cosθ + 0 N + 0 N = 0 N fk = µkn These three equations can be combined together as follows: (1/4)Av2 = mg sinθ – fk = mg sinθ – µkn = mg sinθ – µk mg cosθ

 sin θ − µ k cosθ  ⇒ vterm =  mg  1   4 A

12

Using µk = 0.06 and A = 1.8 m × 0.40 m = 0.72 m2, we find 12

  sin 40° − 0.06 cos 40°    = 51 m /s vterm = (80 kg)(9.8 m /s 2 ) 1 3 2    ( 4 kg m )(0.72 m )    Assess: A terminal speed of 51 m/s corresponds to a speed of ≈100 mph. This speed is reasonable but high due to the steep slope angle of 40° and a small coefficient of friction.


5.54.

Model: Treat the Ping Pong ball as a particle with a drag force equal to with vterm as the terminal velocity. Visualize:

1 4

Av 2 , as modeled in the text

Solve: (a) Imagine the ball falling at its terminal speed. The ball’s weight is directed down and the resistive drag force is directed up. The net force is zero because the magnitude of the drag force is equal to the magnitude of the weight, D = w. When the ball is shot upwards at twice the terminal speed, the drag force is four times the terminal drag force. That is, D′ = 4 D = 4 w . Since all the forces are down, the y-component of Newton’s second law is ΣFy = − D′ − w = −4 w − w = −5mg = ma ⇒ a = −5g (b) The ball is initially shot downward. Therefore D′′ is upward but w is down. Again D′′ = 4 D and the ycomponent of Newton’s second law is ΣFy = D′′ − w = 4 w − w = 3mg = ma ⇒ a = 3g (c)

The ball initially decelerates at 3g but as v become smaller, the drag force approaches the weight so the deceleration goes to zero and v approaches vterm. Assess: D′ is very large and with w yields a large initial deceleration when the ball is shot up. When the ball is shot down w opposes D′′ so the ball decelerates at a lesser rate.


5.55. Model: Take the coin to be a particle held against the palm of your hand (which is like a vertical wall) by friction. The friction needs to be vertical and equal to the weight of the coin to hold the coin in place. Visualize:

The hand pushes on the coin giving a normal force to the coin and causing a friction force. Solve: (a) We need a force (push) to create a normal force. Since the force accelerates the coin, a minimum acceleration amin is needed. (b) Newton’s second law and the model of friction are ΣFy = ( fmin ) y − ( w ) y = fmin − mg = 0 N ΣFx = nmin = mamin fmin = µnmin Since you do not want the coin to slip down the hand you need µ s . Combining the above three equations yields g 9.8 m /s 2 = = 12.3 m /s 2 fmin = mg ⇒ µ s nmin = mg ⇒ µ s mamin = mg ⇒ amin = 0.8 µs


5.56. Model: Take the ball to be a particle with a force (P) pushing down on it. Visualize: (a)

r r r r r (b) The net force is Fnet = P + w. You can see from the free-body diagram that P and w point in the same direction, hence Fnet > w. r (c) The apparent weight of an object is the magnitude of the contact force against the object. P is the contact force against the ball, so the apparent weight is wapp = P. Newton’s second law for the ball is ΣFy = − P − w = − P − mg = ma y Thus the apparent weight is

  ay   ay wapp = P = − mg − ma y = mg − − 1 = w − − 1  g   g  If ay = –g, then wapp = 0. If ay = –1.5g, then wapp = 12 w = 0.5 N. If ay = –2g, then wapp = w = 1.0 Ν. Assess: ay = –g is simply free fall. The hand is not pushing at all, merely following behind the ball. As we’ve seen, an object in free fall has wapp = 0. The harder the hand pushes against the ball, causing the ball to accelerate downward faster than free fall, the larger the ball’s apparent weight.


5.57. Model: We will model the container as a particle of mass m. The steel cable of the crane will be assumed to have zero mass. Visualize:

Solve: As long as the container is stationary or it is moving with a constant speed (zero acceleration), the net force on the container is zero. In these cases, the tension in the cable is equal to the weight of the container: T = mg = 44,000 N The cable should safely lift the load. More tension is required to accelerate the load. Newton’s second law is (Fnet)y = ΣFy = wy + (T)y = –mg + T = may The crane’s maximum acceleration is amax = 1.0 m /s 2 . So the maximum cable tension is

Tmax = mg + mamax = 48, 600 N This is less than the cable’s rating, so the cable must have been defective.


5.58. Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will be assumed to have zero mass. Visualize:

Solve:

Newton’s first law in component form is (Fnet)x = ΣFx = T1x + T2x + wx = – T1 sin 30° + T2 sin 60° + 0 N = 0 N (Fnet)y = ΣFy = T1y + T2y + wy = T1 cos 30° + T2 cos 60° – w = 0 N

Using the x-component equation to obtain an expression for T1 and substituting into the y-component equation yields: w 500 lbs T2 = = = 250 lbs (sin 60°)(cos 30°) 2 + cos 60° sin 30° Substituting this value of T2 back into the x-compoent equation,

T1 = T2

sin 60° sin 60° = 250 lbs = 433 lbs sin 30° sin 30°

We will now find a rope size for a tension force of 433 lbs, that is, the diameter of a rope with a safety rating of 433 lbs. Since the cross-sectional area of the rope is 14 πd 2 , we have 12

  4( 433 lbs) d=  = 0.371 inch 2  π ( 4,000 lbs/ inch )  Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to a diameter of 3/8 of an inch will therefore be appropriate. Assess: If only a single rope was used to hang the sculpture, the rope would have to support a weight of 500 lbs. The diameter of the rope for a safety rating of 500 lbs is 0.399 inches, and the rope size jumps from a diameter of 3/8 to 4/8 of an inch. Also note that the weight of the sculpture is distributed in the two ropes. It is the sum of the ycomponents of the tensions in the ropes that will equal the weight of the sculpture.


5.59. Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will also use the model of kinetic friction for the motion of the shuttle along the square steel rail. Visualize:

Solve: The upward tension component Ty = T sin 45° = 14.1 N is larger than the weight of the shuttle. Consequently, the elastic cord pulls the shuttle up against the rail and the rail’s normal force pushes downward. Newton’s second law in component form is (Fnet)x = Σ Fx = T x + ( fk ) x + (n)x + w x = T cos 45° – fk + 0 N + 0 N = ma x = ma x (Fnet)y = ΣFy = Ty + ( fk ) y + (n)y + wy = T sin 45° + 0 N − n – mg = may = 0 N The model of kinetic friction is fk = µkn. We use the y-component equation to get an expression for n and hence fk. Substituting into the x-component equation and using the value of µk in Table 5.1 gives us

ax = =

T cos 45° − µ k (T sin 45° − mg) m

(20 N) cos 45° − (0.60)[+(20 N) sin 45° − (0.800 kg)(9.8 m /s 2 )] 0.800 kg

= 13.0 m /s 2

Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in the x-direction is max = (0.800 kg)(13.0 m/s2 ) = 10.4 N. This value for ma is reasonable since a part of the 14.1 N tension force is used up to overcome the force of kinetic friction.


5.60. Model: The ball hanging from the ceiling of the truck by a string is represented as a particle. Visualize:

Solve: (a) You cannot tell from within the truck. Newton’s first law says that there is no distinction between “at rest” and “constant velocity.” In both cases, the net force acting on the ball is zero and the ball hangs straight down. (b) Now you can tell. If the truck is accelerating, then the ball is tilted back at an angle. (c) The ball moves with the truck, so its acceleration is 5 m/s2 in the forward direction. r (d) The free-body diagram shows that the horizontal component of T provides a net force in the forward direction. This is the net force that causes the ball to accelerate in the forward direction along with the truck. (e) Newton’s second law for the ball is

ax =

( Fnet ) x m

=

Tx T sin 10° = m m

a y = 0 m 2 /s 2 =

( Fnet ) y m

=

Ty + w y m

We can solve the second equation for the magnitude of the tension:

T=

(1 kg)(9.8 m/s 2 ) mg = = 9.95 N cos10° cos10°

Then the first equation gives the acceleration of the ball and truck:

T sin 10° (9.95 N ) sin 10° = = 1.73 m /s 2 m m The truck’s velocity cannot be determined. ax =

=

T cos10° − mg m


5.61. Model: You will be treated as a particle. You will experience your weight force and the force (P) of the scale pulling up. Visualize:

Solve: The apparent weight of an object is the magnitude of the contact force supporting it. Here, the contact r force is P, so the apparent weight is wapp = P. Newton’s second law is

 a Fnet = ΣFy = Py + w y = ma y ⇒ Fnet = P − w = ma ⇒ wapp = P = mg + ma = mg1 +  g 


5.62. Solve: (a) A 1 kg block is pulled across a level surface by a string, starting from rest. The string has a tension of 20 N, and the block’s coefficient of kinetic friction is 0.50. How long does it take the block to move 1 m? (b) Newton’s second law for the block is ax = a =

( Fnet ) x m

=

T − fk T − µ k n = m m

a y = 0 m /s 2 =

( Fnet ) y m

=

n − w n − mg = m m

where we have incorporated the friction model into the first equation. The second equation gives n = mg. Substituting this into the first equation gives

a=

T − µ k mg 20 N − 4.9 N = = 15.1 m /s 2 1 kg m

Constant acceleration kinematics gives

x1 = x 0 + v0 ∆t +

1 1 2 2 a( ∆t ) = a( ∆t ) ⇒ ∆t = 2 2

2(1 m ) 2 x1 = = 0.364 s a 15.1 m /s 2


5.63. Solve: (a) A 15,000 N truck starts from rest and moves down a 15° hill with the engine providing an 12,000 N force in the direction of the motion. Assume the frictional force between the truck and the road is very small. If the hill is 50 m long, what will be the speed of the truck at the bottom of the hill? (b) Newton’s second law is ΣFy = n y + w y + f y + E y = ma y = 0

ΣFx = n x + w x + f x + E x = ma x ⇒ 0 N + w sin θ + 0 N + 12, 000 N = ma

⇒a=

w sin θ + 12, 000 N (15, 000 N ) sin 15° + 12, 000 N = = 10.4 m/s2 m (15, 000 N/9.8 m/s2 )

where we have calculated the mass of the truck from its weight. Using the constant-acceleration kinematic equation v x2 − v02 = 2 ax ,

v x2 = 2 a x x = 2(10.4 m/s 2 )(50 m ) ⇒ vx = 32.2 m/s


5.64. Solve: (a) A 1 kg box is pushed along an assembly line by a mechanical arm. The arm exerts a 20 N force at a downward angle of 30°. The box has a coefficient of kinetic friction on the assembly line of 0.25. What is the speed of the box after traveling 50 cm, starting from rest? (b) Newton’s second law for the box is ax = a =

( Fnet ) x m

=

Fpush cos 30° − fk m

=

Fpush cos 30° − µ k n m

( Fnet ) y

n − w − Fpush sin 30° n − mg − Fpush sin 30° = = m m m The second equation is solved to give an expression for n. Substituting into the first equation: a y = 0 m /s 2 =

a=

(

Fpush cos 30° − µ k Fpush sin 30° + mg m

) = (20 N) cos 30° − 4.9 N = 12.4 m /s 1 kg

Using kinematics,

v12 = v02 + 2 a∆x = 2 a∆x ⇒ v1 = 2 a∆x = 2(12.4 m /s 2 )(0.5 m ) = 3.5 m /s

2


5.65. Solve: (a) A 61.2 kg skier starts down a 20° slope. Her skies have a coefficient of friction of 0.06, and she is facing into a wind that exerts a 100 N force against her. How far does she travel before reaching a speed of 20 m/s? (b) Newton’s second law for the skier is ( Fnet ) x Fwind + fk − w sin 20° Fwind + µ k n − mg sin 20° ax = a = = = m m m

( Fnet ) y

n − w cos 20° n − mg cos 20° = = m m m The second equation gives n = mgcos20°. Substituting into the first equation, F + µ k mg cos 20° − mg sin 20° 100 N + 34 N − (600 N ) sin 20° = −1.16 m /s 2 a = wind = 61.2 kg m r The acceleration is negative, indicating that a points down the hill as it should. Using kinematics, a y = 0 m /s 2 =

(20 m /s) = −172 m v12 = 2 a 2( −1.16 m /s 2 ) ∆x is negative because she travels in the negative x-direction. The distance traveled is 172 m. 2

v12 = v02 + 2 a∆x = 2 a∆x ⇒ ∆x =


5.66. Solve: (a) A driver traveling at 40 m/s in her 1500 kg auto slams on the brakes and skids to rest. How far does the auto slide before coming to rest? (b)

(c) Newton’s second law is ΣFy = n y + w y = n − mg = ma y = 0 N

(

ΣFx = −0.8n = ma x

)

The y-component equation gives n = mg = (1500 kg) 9.8 m/s . Substituting this into the x-component equation yields

(1500 kg)ax

2

= −0.8(1500 kg)(9.8 m/s 2 ) ⇒ a x = ( −0.8)(9.8 m/s 2 ) = −7.8 m/s 2

Using the constant-acceleration kinematic equation v12 = v02 + 2 a∆x, we find

v02 (40 m /s) = 102 m =− 2a 2( −7.8 m /s 2 ) 2

∆x = −


5.67. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20° wooden incline by a rope that is connected to an electric motor. The crate’s acceleration is measured to be 2 m/s2. The coefficient of kinetic friction between the crate and the incline is 0.2. Find the tension T in the rope. (b)

(c) Newton’s second law for this problem in the component form is (Fnet)x = ΣFx = T – 0.2n – (20 kg)(9.80 m/s2) sin 20° = (20 kg)(2 m/s2) (Fnet)y = ΣFy = n – (20 kg)(9.80 m/s2) cos 20° = 0 N Solving the y-component equation, n = 184.18 N. Substituting this value for n in the x-component equation yields T = 144 N.


Solve: (a) You wish to pull a 20 kg wooden crate across a wood floor (µk = 0.2) by pulling on a rope attached to the crate. Your pull is 100 N at an angle of 30° above the horizontal. What will be the acceleration of the crate? (b)

5.68.

(c) Newton’s equations and the model of kinetic friction are

ΣFx = n x + Px + w x + f x = 0 N + (100 N ) cos 30° + 0 N − fk = (100 N ) cos 30° − fk = ma x ΣFy = n y + Py + w y + f y = n + (100 N ) sin 30° − mg − 0 N = ma y = 0 N fk = µ k n From the y-component equation, n = 150 N . From the x-component equation and using the model of kinetic friction with µk = 0.2,

(100 N) cos 30° – (0.2)(150 N) = (20 kg)ax ⇒ a x = 2.8 m /s 2


Solve: (a) A wooden crate with a mass of 20.0 kg is seen to slide down a wooden ramp (µk = 0.2) at a constant speed of 10 m/s. Find the angle θ of the ramp. (b)

5.69.

(c) Newton’s second law is (Fnet)x = ΣFx = – 0.2 n + (20 kg)(9.80 m/s2) sinθ = 0 N (Fnet)y = ΣFy = n – (20 kg)(9.80 m/s2) cosθ = 0 N Using the expression for n obtained from the y-component equation in the x-component equation yields

−0.2(20 kg)(9.8 m/s 2 ) cosθ + (20 kg)(9.8 m/s 2 ) sin θ = 0 N ⇒ tanθ = 0.2 ⇒ θ = 11.3°


5.70.

Model: Assume the ball is a particle on a slope, and that the slope increases as the x-displacement increases. Assume that there is no friction and that the ball is being accelerated to the right so that it remains at rest on the slope. Visualize: Although the ball is on a slope, it is accelerating to the right. Thus we’ll use a coordinate system with horizontal and vertical axes.

Solve:

Newton’s second law is

ΣFx = n sin θ = ma x

ΣFy = n cosθ − w = ma y = 0 N

Combining the two equations, we get

ma x =

w sin θ = mg tan θ ⇒ a x = g tan θ cosθ

The curve is described by y = Ax2, where A = 1 m−1. Its slope is tanθ which is the derivative of the curve. Hence,

dy = tan θ = 2 Ax ⇒ a x = g(2 Ax ) dx

(

) (

)

(b) The acceleration at x = 0.2 m is 9.8 m/s 2 (2) 1 m −1 (0.2 m ) = 3.9 m/s 2 .


5.71. Model: We will model the skier as a particle, and use the model of kinetic friction. Visualize:

Solve:

r r r r Your best strategy, if it’s possible, is to travel at a very slow constant speed a = 0 so Fnet = 0 . Alternatively,

(

)

you want the smallest positive ax. A negative ax would cause you to slow and stop. Let’s find the value of µk that gives r r Fnet = 0. Newton’s second law for the skier and the model of kinetic friction are (Fnet)x = ΣFx = nx + wx + ( fk ) x + (D)x = 0 + mg sinθ – fk – D cosθ = 0 N (Fnet)y = ΣFy = ny + wy + ( fk ) y + (D)y = n – mg cosθ + 0 N – D sinθ = 0 N fk = µk n The x- and y-component equations are fk = + mg sinθ − D cosθ

n = mg cosθ + D sinθ

From the model of kinetic friction,

µk =

2 fk mg sin θ − D cosθ 75 kg(9.8 m /s ) sin 15° − (50 N ) cos15° = = = 0.196 n mg cosθ + D sin θ 75 kg(9.8 m /s 2 ) cos15° + (50 N ) sin 15°

Yellow wax with µk = 0.20 applied to skis will make the skis stick and hence cause the skier to stop. The skier’s next choice is to use the green wax with µk = 0.15.


5.72. Model: We will represent the widget as a particle. Visualize:

Please refer to Figure CP5.72. Solve: (a) There are only two forces on the widget: the normal force of the table and its weight force. (b) Newton’s second law along the y-axis is

( Fnet ) y = ny + w y = ny − mg = may ⇒ ny = m(ay + g) We know what the ay-vs-t graph looks like. To get the ny-vs-t graph, we need to add gy to the graph, which amounts to shifting the whole graph up by 9.8 m/s2, and multiply by m = 5 kg. (c) The normal force is negative for t > 0.75 s. Physically, this means that the normal force is pointed in the downward direction. In other words, the table is pulling down on the widget rather than pushing up on the widget. It can do this because the widget is glued to the table rather than simply sitting on the table. (d) The apparent weight is a maximum at t = 0 s, when the upward acceleration is maximum. (e) The apparent weight is zero at t = 0.75 s when ay = –9.8 m/s2 = –g. (f) If not glued down, the widget will fly off the table at t = 0.75 s, the instant at which its apparent weight becomes zero. The table is accelerating in the negative direction so quickly after t = 0.75 s that the widget can stay on only if the table pulls downward on it. That is the significance of the negative value for ny. If the widget is not glued down, the largest downward acceleration it can achieve is the free fall acceleration.


5.73. Model: We will model the object as a particle, and use the model of drag. Visualize:

Solve: (a) We cannot use the constant-acceleration kinematic equations since the drag force causes the acceleration to change with time. Instead, we must use ax = dvx/dt and integrate to find vx. Newton’s second law for the object is (Fnet)x = ΣFx = nx + wx + (fk)x + (D)x = 0 N + 0 N + 0 N −

dv 1 2 Av x = ma x = m x 4 dt

This can be written

dv x A = dt 2 4m vx We can integrate this from the start (v0x at t = 0) to the end (vx at t):

vx

v0 x

dv x 1 1 A t A = = dt ⇒ − + t 2 4 m ∫0 vx v x v0 x 4 m

Solving for vx gives

vx =

v0 x 1 + Av0 x t 4 m

(b) Using A = (1.6 m)(1.4 m) = 2.24 m2, v0x = 20 m/s, and m = 1500 kg, we get

1 20 m / s 20   20 m /s − 1 ⇒t= =   0.007467   v x 2.24t (20) 1 + 0.007467t  1+ 4 × 1500 where t is in seconds. We can now obtain the time t for v = 10 m/s: vx =

1   20 m /s − 1 = 133.93 20 − 1 = 134 s t=  0.007467   10 m /s   10  When vx = 5 m/s, then t = 402 s. (c) If the only force acting on the object was kinetic friction with µk = 0.05, the force would be (0.05)(1500 kg) 2 (9.8 m/s2) = 735 N. The drag force at an average speed of 10 m/s is D = 14 (2.24)(10) N = 56 N. We conclude that it is not reasonable to neglect the kinetic force of friction.


Chapter 05  

°= ( )( ) = Assess: Since r T 3 acts closer to the x-axis than to the y-axis, it makes sense that T T cos .100N 86.7 N T T =− T 2x = 0 N T T...

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