() x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5...

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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 F F T T T x x x x x net N ( ) = = + + = Σ 1 2 3 0 F F T T T y y y y y net N ( ) = = + + = Σ 1 2 3 0 Evaluating the components of the force vectors from the free-body diagram: T T x 1 1 =− T 2x = 0 N T T x 3 3 30 = ° cos T 1y = 0 N T T y 2 2 = T T y 3 3 30 =− ° sin Using Newton’s first law: + °= T T 1 3 30 0 cos N T T 2 3 30 0 °= sin N Rearranging: T T 1 3 30 0 8666 = °= ( )( ) = cos . 100 N 86.7 N T T 2 3 30 05 = °= ( )( ) = sin . 100 N 50.0 N Assess: Since r T 3 acts closer to the x-axis than to the y-axis, it makes sense that T T 1 2 > .

Transcript of () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5...

Page 1: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

F F T T Tx x x x xnet N( ) = = + + =Σ 1 2 3 0 F F T T T

y y y y ynet N( ) = = + + =Σ 1 2 3 0

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

T Tx1 1= − T2x = 0 N T Tx3 3 30= °cos

T1y = 0 N T Ty2 2= T Ty3 3 30= − °sin

Using Newton’s first law:

− + ° =T T1 3 30 0cos N T T2 3 30 0− ° =sin N

Rearranging:

T T1 3 30 0 8666= ° = ( )( ) =cos .100 N 86.7 N T T2 3 30 0 5= ° = ( )( ) =sin .100 N 50.0 N

Assess: Since rT3 acts closer to the x-axis than to the y-axis, it makes sense that T T1 2> .

Page 2: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 onlythree forces are the tension forces shown in the free-body diagram. Note that the diagram defines the angle θ.

Solve: Because the ring is in equilibrium it must obey rFnet N= 0 . This is a vector equation, so it has both x- and

y-components:

F T Txnet N( ) = − =3 2 0cosθ ⇒ =T T3 2cosθ

F T T T Tynet N( ) = − = ⇒ =1 3 3 10sin sinθ θ

We have two equations in the two unknowns T3 and θ. Divide the y-equation by the x-equation:

T

T

T

T3

3

1

2

11 60 1 60 58 0sin

costan . tan . .

θθ

θ θ= = = = ⇒ = ( ) = °−80 N

50 N

Now we can use the x-equation to find

TT

32= =

°=

cosθ50 N

cos58.094.3 N

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

Page 3: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 . sin . .θ θ= = ⇒ = = °−2 m

3 m0 677 0 667 41 81

Newton’s first law for this situation is

F F T T T Tx x x xnet N N( ) = = + = ⇒ − + =Σ 1 2 1 20 0cos cosθ θ

F F T T w T T wy y y y ynet N N( ) = = + + = ⇒ + − =Σ 1 2 1 20 0sin sinθ θ

The x-component equation means T T1 2= . From the y-component equation:

2 1T wsinθ = ⇒ = = =( )( )

°= =T

w mg1 2 2 2 41 8sin sin sin .θ θ

20 kg 9.8 m/s 196 N

1.333147 N

2

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

Page 4: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.4. Model: We can assume that the coach and his sled are a particle being towed at a constant velocity by thetwo 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:

F F T T fx x x x xnet k N( ) = = + + =Σ 1 2 0 F F T T f

y y y y ynet k N( ) = = + + =Σ 1 2 0

From the free-body diagram:

T T f1 2

1

2

1

20cos cosθ θ

+

− =k N T T1 2

1

2

1

20 0sin sinθ θ

+ = N N

From the second of these equations T T1 2= . Then from the first:

2 101T cos ° = 1000 N ⇒ =°

= =T1 2 10

1000 N 1000 N

1.970508 N

cos

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

Page 5: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

aF

mxx=

( )= − =net 24 N 2 N

2 kg1.0 m/s a

F

myy=

( )= − =

net 23 N 3 N

2 kg0 m/s

For the diagram on the right:

aF

mxx=

( )= − =net 24 N 2 N

2 kg1.0 m/s a

F

myy=

( )= − − =

net 23 N 1 N 2 N

2 kg0 m/s

Page 6: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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. Noticethat the angle between the 3 N force and the –y-axis is the same 20° by which the coordinates are tilted. ApplyingNewton’s second law,

aF

mxx=

( )=

− − °( ) =net 25 N 1 N N

2 kg1.49 m/s

3 20sin

aF

myy=

( )=

− °( ) =net 22.82 N N

2 kg0 m/s

3 20cos

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

aF

m

aF

m

xx

yy

=( )

=°( ) + °( ) −

= −

=( )

=+ °( ) − °( ) =

net 2

net 2

N N 3 N

2 kg0.28 m/s

1.414 N N N

2 kg0 m/s

2 15 2 15

2 15 2 15

cos sin

sin cos

Page 7: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 tbetween 0 s and 3 s,

av

txx= = − =∆

∆12 m/s 0 s

3 s4 m/s2

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,

av

txx= = − = −∆

∆0 12 m/s m/s

2 s6 m/s2

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 NAssess: The magnitudes of the forces look reasonable, given the small mass of the object. The positive andnegative signs are appropriate for an object first speeding up, then slowing down.

Page 8: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.8. Visualize: Please refer to Figure Ex5.8. Positive forces result in the object gaining speed and negativeforces 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 tocalculate the accelerations. The acceleration from t = 0 s to t = 3 s is

aF

mxx= = =4 N

2.0 kg2 m/s2

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

aF

mxx= = − = −2 N

2.0 kg1 m/s2

The acceleration from t = 5 s to 8 s is ax = 0 m/s2. In particular, a tx at s m/s2=( ) =6 0 .We can now use one-dimensional kinematics to calculate v at t = 6 s as follows:

v v a t t a t t= + −( ) + −( )= + ( )( ) + −( )( ) = − =

0 1 1 0 2 2 0

0 2 m/s 3 s 1 m/s 2 s 6 m/s 2 m/s 4 m/s2 2

Assess: The positive final velocity makes sense, given the greater magnitude and longer duration of the positive rF1 .

A velocity of 4 m/s also seems reasonable, given the magnitudes and directions of the forces and the mass involved.

Page 9: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.9. Model: We assume that the box is a particle being pulled in a straight line. Since the ice is frictionless, thetension 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 toNewton’s first law, the tension in the rope must be zero.(b) For this situation again, ax = 0 m/s2, so F Tnet 0 N= = .(c) Here, the velocity of the box is irrelevant, since only a change in velocity requires a nonzero net force. Sinceax = 5.0 m/s2,

F T maxnet250 kg 5.0 m/s 250 N= = = ( )( ) =

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

Page 10: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.10. Model: We assume that the box is a point particle that is acted on only by the tension in the rope and thepull 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:

F T w T w mgnet2 N 50 kg 9.8 m/s 490 N= − = ⇒ = = = ( )( ) =0

(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/s2 ,

F T w ma

T w

ynet250 kg 5.0 m/s 250 N

250 N 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/s2 and wehave instead

F T w ma

T w

ynet250 kg 5.0 m/s 250 N

250 N 250 N 490 N 240 N

= − = = ( ) −( ) = −

⇒ = − + = − + =Assess: For parts (a) and (b) the zero accelerations immediately imply that the box’s weight must be exactlybalanced by the upward tension in the rope. For part (c) the tension not only has to support the box’s weight butmust also accelerate it upward, hence, T must be greater than w. When the box accelerates downward, the rope neednot support the entire weight, hence, T is less than w.

Page 11: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

mw

g= = =earth

earth2

800 N

9.8 m/s81.6 kg

Therefore, the weight of the astronaut on Mars is

w mgMars Mars281.6 kg 3.76 m/s 307 N= = ( )( ) =

Assess: The smaller acceleration of gravity on Mars reveals that objects are less strongly attracted to Mars than tothe earth, so the smaller weight on Mars makes sense.

Page 12: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

w mgearth earth255 kg 9.8 m/s 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

w mgmoon moon255 kg 1.62 m/s 89 N= = ( )( ) =

Assess: The smaller acceleration due to gravity on the moon reveals that objects are less strongly attracted to themoon than to the earth. Thus the woman’s smaller weight on the moon makes sense.

Page 13: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.13. Model: We assume that the passenger is a particle subject to two vertical forces: the downward pull ofgravity 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

w wa

gw

gmgy

app260 kg 9.8 m/s 590 N= +

= +

= = ( )( ) =1 10

(b) The elevator speeds up from v0y = 0 m/s to its cruising speed at vy = 10 m/s. We need its acceleration before wecan find the apparent weight:

av

ty = = − =∆∆

10 m/s m/s

4.0 s2.5 m/s20

The passenger’s apparent weight is

w wa

gy

app

2

2590 N2.5 m/s

9.8 m/s= +

= ( ) +

1 1 = (590 N)(1.26) = 740 N

(c) The passenger is no longer accelerating since the elevator has reached its cruising speed. Thus, w wapp 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 reasonablethat the floor will have to push up harder on him, increasing his apparent weight.

Page 14: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.14. Model: We assume that the passenger is a particle acted on by only two vertical forces: the downwardpull 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) decreasingvelocity, 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 andthen apply Equation 5.10. The acceleration for the first segment is

av v

t ty = −−

= −−

=1 0

1 0

0

0

8 m/s m/s

2 s s4 m/s2

⇒ = +

= ( ) +

= ( )( ) +

=w w

a

gmgy

app

2

224 m/s

9.8 m/s75 kg 9.8 m/s 1035 N1 1 1

4

9 8.

For the second segment, ay = 0 m/s2 and the apparent weight is

w wg

mgapp

22 m/s

75 kg 9.8 m/s 740 N= +

= = ( )( ) =10

For the third segment,

av v

t t

w w

y = −−

= −−

= −

⇒ = + −

= ( )( ) −( ) =

3 2

3 2

0

1 1 0 2

m/s 8 m/s

10 s 6 s2 m/s

2 m/s

9.8 m/s75 kg 9.8 m/s 590 N

2

app

2

22 .

Assess: As expected, the apparent weight is greater than normal when the elevator is accelerating upward andlower than normal when the acceleration is downward. When there is no acceleration the weight is normal. In allthree cases the magnitudes are reasonable, given the mass of the passenger and the accelerations of the elevator.

Page 15: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.15. Model: We assume that the safe is a particle moving only in the x-direction. Since it is sliding during theentire 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 verticaland horizontal directions:

F F F F f f F Fx xnet B C k k B C N 350 N 385 N 735 N( ) = = + − = ⇒ = + = + =Σ 0

F F n w n w mgy ynet

2 N 300 kg 9.8 m/s 2940 N( ) = = − = ⇒ = = = ( )( ) =Σ 0

Then, for kinetic friction:

f nf

nk k kk 735 N

2940 N= ⇒ = = =µ µ 0 25.

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

Page 16: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 normalforce has no component in the x-direction, so we can ignore it here. For the other two forces:

F F f w f w mgx x x xnet s s

2 N 4000 kg 9.8 m/s 10,145 N( ) = = − = ⇒ = = = ( )( ) °( ) =Σ 0 15sin sinθ

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

Page 17: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 forcesin both the x- and y-directions. However, we know that ay = 0 m/s2. We have

aF

m

f

ma

F

m

n w

m

n mg

mxx

yy=

( )= − = =

( )= − = −net k 2 net

m/s0

We used f fxk k( ) = − because the free-body diagram tells us that

rfk points to the left. The friction model relates

rfk

to rn with the equation f nk k= µ . The y-equation is solved to give n mg= . Thus, the kinetic friction force is

f mgk k= µ . Substituting this into the x-equation yields:

amg

mgx = − = − = −( )( ) = −µ µk

k2 29.8 m/s 5.88 m/s0 6.

The acceleration is negative because the acceleration vector points to the left as the car slows. Now we have aconstant-acceleration kinematics problem. ∆t isn’t known, so use

v v a x xx12

02

2

0 22

= = + ⇒ = − ( )−( ) = m /s40 m/s

5.88 m/s136 m2 2

2∆ ∆

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

Page 18: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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:

av

t= = − =∆

∆82 m/s m/s

35 s2.34 m/s20

F F F f ma F f maxnet thrust r thrust r( ) = = − = ⇒ = +ΣFor rubber rolling on concrete, µr = 0.02 (Table 5.1), and since the runway is horizontal, n w mg= = . Thus:

F w ma mg ma m g athrust r r r

275,000 kg 9.8 m/s 2.34 m/s N

= + = + = +( )= ( ) ( )( ) +[ ] =

µ µ µ

0 02 190 0002. ,

Assess: It’s hard to evaluate such an enormous thrust, but comparison with the plane’s mass suggests that190,000 N is enough to produce the required acceleration.

Page 19: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.19. Model: We treat the train as a particle subject to rolling friction but not to drag (because of its slow speedand 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 thatn w mg= = . Thus,

f mgr r250,000 kg 9.8 m/s 980 N= = ( )( )( ) =µ 0 002.

From Newton’s second law for the decelerating locomotive,

af

mx = − = − = −r 2980 N

50,000 kg0.0196 m/s

Since we’re looking for the distance the train rolls, but we don’t have the time:

v v a x xv v

axx

12

02 1

202 2 2

22

0

2− = ( ) ⇒ = − = ( ) − ( )

−( ) =∆ ∆ m/s 10 m/s

0.0196 m/s2550 m

2

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

Page 20: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

f mgs s2120 kg 9.8 m/s 940 Nmax .= = ( )( )( ) =µ 0 8

The maximum static friction force is greater than the farmer’s maximum pull of 800 N; thus, the farmer will not beable to budge the mule.Assess: The farmer should have known better.

Page 21: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 theskydiver falls feet first, the surface perpendicular to the drag has the cross-sectional area A = 20 cm × 40 cm. Thephysical conditions needed to use Equation 5.16 for the drag force are satisfied. The terminal speed corresponds tothe 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

D Av v v≈ = ×( ) =1

40 25 0 0202 2 2. .0.20 0.40 N N

The skydiver’s weight is w mg= = ( )( ) =75 kg 9.8 m/s 735 N2 . The mathematical form of the condition defining

dynamical equilibrium for the skydiver and the terminal speed isr r rF w Dnet N= + = 0

⇒ − = ⇒ = ≈0 02 2. v vterm term N 735 N 0 N735

0.02192 m/s

Assess: The result of the above simplified physical modeling approach and subsequent calculation, even ifapproximate, shows that the terminal velocity is very high. This result implies that the skydiver will be very badlyhurt at landing if the parachute does not open in time.

Page 22: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 theresultant 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 ofthe motion.Solve: The mathematical equation defining the dynamical equilibrium situation for the falling ball is

r r r rF w Dnet N= + = 0

Since only the vertical direction matters, one can write:ΣF F D wy = ⇒ = − =0 0 N Nnet

When this condition is satisfied, the speed of the ball becomes the constant terminal speed v v= term . Themagnitudes of the weight and drag forces acting on the ball are:

w mg m

D Av R v

= = ( )≈ ( ) = ( ) = ( )( ) ( ) =

9.80 m/s

0.11 m 85 m/s 69 N

2

term2

term21

40 25 0 252 2 2. .π π

The condition for dynamic equilibrium becomes:

9.80 m/s 68.66 N 0 N69 N

.8 m/s7.0 kg2

2( ) − = ⇒ = =m m9

Assess: The value of the mass of the bowling ball obtained above seems reasonable. Such a ball is heavy so that ithas a significant impact on the pins, but not “excessively” heavy so that it can be lifted and rolled by an averagehuman player.

Page 23: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

aF

m

F

mxx= =∑ prop

Table 4.3 gave the typical propulsion force of a car as Fprop N.≈ 5000 The mass of a typical car is m ≈ 1500 N, avalue that has been used in several examples. Thus

ax ≈ ≈5000

15003

N

N m/s2

Assess: One significant figure is appropriate for an estimate. This is about 30% of g, which seems reasonable fora car’s acceleration.

Page 24: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 valuefor µk of cloth sliding on loose dirt. It is probably greater than µk for wood on wood, due to the roughness of bothsurfaces. Let’s guess 0.40. From the free-body diagram, n = w = mg, since there’s no vertical acceleration. Thus,

f mgk k280 kg 9.8 m/s 314 N= = ( )( )( ) =µ 0 40.

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

aF

m= = − ≈ −net 2314 N

80 kg4 m/s

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

Page 25: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.25. Visualize:

We used the force-versus-time graph to draw the acceleration-versus-time graph. The peak acceleration wascalculated as follows:

aF

mmaxmax= = =10 N

5 kg2 m/s2

Solve: The acceleration is not constant, so we cannot use constant acceleration kinematics. Instead, we use themore general result that

v(t) = v0 + area under the acceleration curve from 0 s to tThe object starts from rest, so v0 = 0 m/s. The area under the acceleration curve between 0 s and 6 s is 1

2 (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.

Page 26: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.26. Visualize:

The acceleration is a F mx x= , so the acceleration-versus-time graph has exactly the same shape as the force-versus-time graph. The maximum acceleration is a F mmax max

2 N kg m / s= = ( ) ( ) =6 2 3 .Solve: The acceleration is not constant, so we cannot use constant-acceleration kinematics. Instead, we use themore 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 1

2 3 2 3× × =( ) m/s s m/s2 . Thus vx = 9 m/s at t = 4 s.

Page 27: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.27. Model: You can model the beam as a particle and assume rFnet = 0N to calculate the tensions in the sus-

pension ropes.Visualize:

Solve: The beam attached to the ropes will remain in static equilibrium only if both inequalities T T1 < max andT T2 < max hold, where Tmax is the maximum sustained tension. The equilibrium equations in vector and componentform are:

r r r r rF T T w

F T T w

F T T w

x x x x

y y y y

net

net

net

N

N

N

= + + =

( ) = + + =

( ) = + + =

1 2

1 2

1 2

0

0

0

Using the free-body diagram yields:

− + = + − =T T T T w1 1 2 2 1 1 2 20 0sin sin cos cosθ θ θ θ N 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

− ° + ° = ° + ° =T T T T1 2 1 220 30 0 20 30sin sin cos cos N 9800 N

You can solve this system of equations by simple substitution. The result is T T1 2= =6397 N and 4376 N . The ropeon the left side (rope 1) breaks since the tension in this rope is larger than 6000 N. Once the left rope is broken, theright 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 ropethan in the right rope.

Page 28: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

F T T T T

F T T T w T T T mg

x x x

y y y y y

net

net

N

N

( ) = = + = −

( ) = = + + + = − − −

0

0

1 2 2 2 1 1

1 2 3 3 1 1 2 2

cos cos

sin sin

θ θ

θ θ

Notice how the force components all appear in the second law with plus signs because we are adding forces. Thenegative signs appear only when we evaluate the various components. These are two simultaneous equations in thetwo unknowns T2 and T3. From the x-equation we find

TT

21 1

2

15

25= = ( ) °

°=cos

cos

cos

cos

θθ

500 N533 N

(b) Now we can use the y-equation to findT T T mg3 1 1 2 2= + + =sin sinθ θ 5250 N

Page 29: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 thestring’s tension plus two long-range forces. The equilibrium condition is

F T F T F

F T w T mg

x x x

y y y

net elec elec

net

N

N

( ) = + ( ) = − =

( ) = + = − =

sin

cos

θ

θ

0

0

We can solve the y-equation to get

Tmg= =

( )( )°

=cos cosθ

0.001 kg 9.8 m/s0.0104 N

2

20Substituting this value into the x-equation,

F Telec N 0.0036 N= = ×( ) ° =−sin . sinθ 1 04 10 202

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

Page 30: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.30. Model: We will represent Henry as a particle. His motion is governed by constant-acceleration kinematicequations.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, andanother acceleration as the elevator brakes from 10.0 s to 12.0 s. The apparent weight is the same as the true weightduring constant velocity motion, so Henry’s true weight mg is 750 N. His apparent weight is less than his trueweight during the initial acceleration, so the acceleration is in a downward direction (negative a). Thus, theelevator’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

w wa

ga g

w

wapp trueapp

true

= +

⇒ = −

1 1

During the interval 0 s ≤ t ≤ 2 s, the elevator’s acceleration is

a g= −

= −600 N

750 N1.96 m/s21

At t = 2 s, Henry’s position is

y y v t a t a t1 0 0 0 0

2

0

21

2

1

2= + + ( ) = ( ) = −∆ ∆ ∆ 3.92 m

and his velocity isv1 = 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. Att = 10 s he is at position

y2 = y1 + v1∆t1 = –35.3 mand he has a velocity v2 = v1 = –3.92 m/s. During the interval 10 s ≤ t ≤ 12.0 s, the elevator’s acceleration is

a g= −

= +900 N

750 N1.96 m/s21

The upward acceleration vector slows the elevator and Henry feels heavier than normal. At t = 12.0 s Henry is atposition

y3 = y2 + v2(∆t2) + 1

2a(∆t2)

2 = –39.2 m

Thus Henry has traveled distance 39.2 m.

Page 31: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.31. Model: We’ll assume Zach is a particle moving under the effect of two forces acting in a single verticalline: 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

w wa

gw

gmgapp

22 m/s

80 kg 9.8 m/s 784 N= +

= +

= = ( )( ) =1 10

(b) Using the definition of acceleration,

av

t

v v

t t= = −

−=

− −( ) =∆∆

1 0

1 0

0 10 m/s

3.0 s3.33 m/s2

⇒ = +

= ( )( ) +

w wa

gapp2

2

280 kg 9.8 m/s3.33 m/s

9.8 m/s1 1 = (784 N)(1+0.340) = 1050 N

Assess: While the elevator is braking, it not only must support Zach’s weight but must also push upward on himto decelerate him, so the apparent weight is greater than his normal weight.

Page 32: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.32. Model: We can assume your body is a particle moving in a straight line under the influence of twoforces: 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

w wa

ga

w

wgapp

app= +

⇒ = −

1 1

When accelerating upward, the acceleration is

a = −

( ) =170 lb

150 lb9.8 m/s 1.3 m/s2 21

When braking, the acceleration is

a = −

( ) = −120 lb

150 lb9.8 m/s 2.0 m/s2 21

Assess: A 10-20% change in apparent weight seems reasonable for a fast elevator, as the ones in the Empire StateBuilding must be. Also note that we did not have to convert the units of the weights from pounds to newtonsbecause the weights appear as a ratio.

Page 33: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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. Wecan also treat the hanging mass as a particle in equilibrium. Since the pulleys are frictionless, the tension is the sameeverywhere 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 58.8 N2

(b) Using Newton’s first law for the vertical direction on the pulley attached to the foot,

F F T T wy ynet foot N( ) = = − ° − =Σ sin sinθ 15 0

⇒ = ° + = ° +sinsin

sinθ T w

T

m g

T

1515foot foot = +

( )( )= + =0 259 0 259 0 667 0 926. . . .

4 kg 9.8 m/s

58.8 N

2

⇒ = = °−θ sin . .1 0 926 67 8

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

F F T T Fx xnet traction N( ) = = + ° − =Σ cos cosθ 15 0

⇒ = + ° = ° + °( )F T T Ttraction cos cos cos . cosθ 15 67 8 15

= + = =(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 downwardpull of the lower segment of the rope, it makes sense that its angle is larger (a more direct upward pull). Themagnitude of the traction force, roughly one-tenth the weight of a human body, seems reasonable.

Page 34: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: (a) In order to use Newton’s second law for the passenger, we’ll need the acceleration. Since we don’thave the stopping time:

v v a x x12

02

02= + −( )1 ⇒ = −−( ) =

− ( )−( )

= −av v

x x12

02

0

2

2

0

2 1 01

2 22 m /s 15 m/s

m m112.5 m/s

⇒ = = = ( ) −( ) = −F F manet260 kg 112.5 m/s 6750 N

The net force is 6750 N to the left.(b) Using the same approach as in part (a),

F ma mv v

x x= = −

−( ) = ( ) − ( )( ) = −1

202

0

2

260

0

21

2 2

kg m /s 15 m/s

0.005 m1,350,000 N

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’sweight. 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 2300times the passenger’s weight, on the other hand, would surely be catastrophic.

Page 35: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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. Thenwe can relate the acceleration to the force with Newton’s second law. (Note that the barrel length is not relevant tothe problem.) The kinematic equation is

v v a x av

x12

02 0

2 2

5 222 2

6 67 10= + ⇒ = − = − ( )( )

= − ×∆∆

400 m/s

0.12 m m/s.

Notice that a is negative, in agreement with the vector ra in the motion diagram. Turning to forces, the wood exerts

two forces on the bullet. First, an upward normal force that keeps the bullet from “falling” through the wood.

Second, a retarding friction force rfk that stops the bullet. The only horizontal force is

rfk , which points to the left

and thus has a negative x-component. The x-component of Newton’s second law is

F f ma f maxnet k k

20.01 kg m/s 6670 N( ) = − = ⇒ = − = −( ) − ×( ) =6 67 105.

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 v v a t1 0= + ∆ as follows:

∆tv

a= − = × =−0 46 00 10. s 600 µs

(c)

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

vx = + − × × =−400 6 667 10 60 10 3605 6m/s m/s s m/s2( . )( )

Page 36: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 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 ballundergoes free fall (a = –g). The initial velocity for part 2 is the final velocity of part 1, as the ball emerges from thetube. The free body diagram for part 1 shows two forces: the air pressure force and the weight force. We need onlythe y-component of Newton’s second law:

a aF

m

F w

m

F

mgy

y= =( )

= − = − = − =net air air 2 22 N

0.05 kg9.8 m/s 30.2 m/s

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

v v a y y v ay12

02

1 0 1 12 2 2= + −( ) ⇒ = = ( )( ) =30.2 m/s 1 m 7.77 m/s2

For part 2, free-fall kinematics v v g y y22

12

2 12= − −( ) gives

y yv

g2 112

2− = = 3.08 m

Page 37: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

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

F w ma F ma mgthrust thrust earth− = ⇒ = + = ( ) +( )0.200 kg 10 m/s 9.8 m/s2 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.

Page 38: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

a aF

m

F mg

myy= =

( )= − = − =

net thrust 2 2300,000 N

20,000 kg9.8 m/s 5.20 m/s

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

mF

a g50005000

mthrust

m2 2

300,000 N

6.00 m/s 9.8 m/s18,990 kg=

+=

+=

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

Page 39: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.39. Model: The steel block will be represented by a particle. Steel-on-steel has a static coefficient of frictionµs = 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

F T f T f F n w n mgx ynet s s net0 N ( ) = − = ⇒ = ( ) = − ⇒ =

The static friction force has a maximum value f mgs max s( ) = µ . The string tension that will cause the block to slipand start to move is

T mg= = ( )( )( ) =µs2 kg m/s N0 80 2 9 8 15 7. . .

Any tension less than this will not be sufficient to cause the block to move, so this is the minimum tension formotion.(b) As the block is moving with a tension of 20 N in the string, we can find its acceleration from the x-componentof Newton’s second law as follows:

F T f ma aT f

mx x xnet kk( ) = − = ⇒ = −

The kinetic friction force fk = µkmg. The acceleration of the block is

ax =− ( )( )( )

=20 0 60 2 9 8

24 12

N kg m/s

kg m/s

2

2. .

.

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

v v12

10 2 4 12 1 0 2 87= + ( )( ) ⇒ = m /s m/s m m/s2 2 2. . .

(c) The only difference in this case is the coefficient of kinetic friction whose value is 0.05 instead of 0.60. Theacceleration of the block is

ax =− ( )( )( )

=20 0 05 2 9 8

29 51

N kg m/s

kg m/s

2

2. .

.

The block’s speed after moving 1.0 m will be

v v12

10 2 9 51 1 0 4 36= + ( )( ) ⇒ = m /s m/s m m/s2 2 2. . .

Page 40: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.40. Model: We assume that Sam is a particle moving in a straight horizontal line under the influence of twoforces: 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, sincethere’s no acceleration in the vertical direction:

n w mg= = = ( )( ) =75 kg 9.8 m/s 735 N2 ⇒ = = ( )( ) =f nk k 735 N 73.5 Nµ 0 1.

Then, from Newton’s second law:

F F f ma aF f

mxnet thrust kthrust k 2200 N 73.5 N

75 kg1.687 m/s( ) = − = ⇒ = − = − =0 0

From kinematics:

v v a t1 0 0 1 0= + = + ( )( ) = m/s 1.687 m/s 10 s 16.87 m/s2

(b) During the acceleration, Sam travels to

x x v t a t1 0 0 1 0 12 21

2

1

21 687 10= + + = ( )( ) =. m/s s 84 m2

After the skis run out of fuel, Sam’s acceleration can again be found from Newton’s second law:

F f aF

mxnet knet 273.5 N

73.5 N

75 kg0.98 m/s( ) = − = − ⇒ = = − = −1

Since we don’t know how much time it takes Sam to stop:

v v a x x22

12

1 2 12= + −( ) ⇒ − = − =− ( )

−( ) =x xv v

a2 122

12

1

2

2

0

2

m /s 16.87 m/s

0.98 m/s145 m

2 2

2

The total distance traveled is x x x2 1 1 145 84−( ) + = + m m = 229 m.

Assess: A top speed of 16.9 m/s (roughly 40 mph) seems quite reasonable for this acceleration, and a coastingdistance of nearly 150 m also seems possible, starting from a high speed, given that we’re neglecting air resistance.

Page 41: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.41. Model: We assume Sam is a particle moving in a straight line down the slope under the influence ofgravity, 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

x xx x

h

1 01 0−

= ⇒ − = =°

=sinsin

θθ

50 m

sin10288 m

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

F F n wy ynet N( ) = = − =Σ cosθ 0 ⇒ = =n w mgcos cosθ θ = (75 kg)(9.8 m/s2)(cos10°) = 724 N

One-dimensional kinematics gives us Sam’s acceleration:

v v a x xx12

02

02= + −( ) ⇒ = −−( ) = ( ) −

( )=a

v v

x xx12

02

1 2

2

2

0

2

40 m/s m /s

288 m2.78 m/s

2 22

Then, from Newton’s second law and the equation f nk k= µ :

F F w F f ma

mg F ma

n

x x xnet thrust k

kthrust

( ) = = + − =

⇒ = + −

Σ sin

sin

θ

µ θ

=( )( ) °( ) + − ( )( )

=75 kg 9.8 m/s 200 N 75 kg 2.78 m/s

724 N

2 2sin.

100 165

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

Page 42: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: (a) When the belt runs at constant speed, the crate has an acceleration r ra = 0 m/s2 and is in dynamic

equilibrium. Thus r rFnet .= 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 thereis 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 keepit 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 cratemust have a net horizontal force. So now there is a friction force, and the force points in the direction of the crate’smotion. Is it static friction or kinetic friction? Although the crate is moving, there is no motion of the crate relative tothe 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 thecrate is

af

m

n

ms

maxmax=

( )= µs

If the belt accelerates more rapidly than this, the crate will not be able to keep up and will slip. It is clear from thefree-body diagram that n w mg= = . Thus,

amax = µsg = (0.5)(9.8 m/s2) = 4.9 m/s2

Page 43: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 accelerateuntil it matches the speed of the belt. We need to know the horizontal acceleration. Since there’s no acceleration inthe vertical direction, we can apply Newton’s first law to find the normal force:

n w mg= = = ( )( ) =10 kg 9.8 m/s 98 N2

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

f mgk k 98 N 29.4 N= = ( )( ) =µ 0 3.

We can find the horizontal acceleration from Newton’s second law:

F F f max xnet k( ) = = =Σ ⇒ = = =a

f

mk 229.4 N

10 kg2.94 m/s

From one of the kinematic equations:

v v a x x12

02

1 02= + −( ) ⇒ − = − = ( ) − ( )( ) =x x

v v

a1 012

02 2

2

0

2

2.0 m/s m/s

2.94 m/s0.68 m

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 areasonable distance for it to slide before stopping.

Page 44: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 onJohnny. The friction force is up the hill, opposite his direction of motion, while

rn is perpendicular to the hill. It is

important to note that the angle between rw 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

a aF

m

w f

m

mg f

m

aF

m

n w

m

n mg

m

xx

yy

= =( )

= − = −

= =( )

= − = −

net k k

2 net m/s

sin sin

cos cos

θ θ

θ θ0

The friction model is f nk k= µ . First solve the y-equation to give n = mgcosθ. Use this in the friction model toget f mgk k= µ θcos . Now substitute this result for fk into the x-equation:

amg mg

mgx = − = −( ) = ( ) ° − °( ) = −sin cos

sin cos sin . cosθ µ θ θ µ θk

k2 29.8 m/s 1.253 m/s20 0 5 20

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

v v a x x v a xx x12

02

1 0 0 12 2 2= + −( ) ⇒ = − = − −( )( ) =1.253 m/s 3.5 m 2.96 m/s2

Page 45: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.45. Model: We will represent the wood block as a particle, and use the model of kinetic friction andkinematics. 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 inthe –x-direction, so we’ll want to take |v2| as the final speed. Because of friction, we expect to find |v2| < v0.

Solve: (a) The friction force is opposite to v→

, sorfk points down the slope during the first half of the motion and up

the slope during the second half. rw and

rn are the only other forces. Newton’s second law for the upward motion is

a aF

m

w f

m

mg f

m

aF

m

n w

m

n mg

m

xx

yy

= =( )

= − − = − −

= =( )

= − = −

0

0

net k k

2 net m/s

sin sin

cos cos

θ θ

θ θ

The friction model is f nk k= µ . First solve the y-equation to give n = mgcosθ. Use this in the friction model toget f mgk k= µ θcos . Now substitute this result for fk into the x-equation:

amg mg

mg0 30 0 20 30= − − = − +( ) = −( ) ° + °( ) = −sin cos

sin cos sin . cosθ µ θ θ µ θk

k2 29.8 m/s 6.60 m/s

Kinematics now gives

v v a x x xv v

a12

02

0 1 0 112

02

0

2

22

0

2= + −( ) ⇒ = − =

− ( )−( ) =

m /s 10 m/s

6.60 m/s7.58 m

2 2

2

The block’s height is then h = x1sinθ = (7.58 m)sin30° = 3.79 m.

(b) For the return trip, rfk points up the slope, so the x-component of the second law is

a aF

m

w f

m

mg f

mxx= =

( )= − + = − +

1net k ksin sinθ θ

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

a g1 = − −( ) = −sin cosθ µ θk23.20 m / s

The kinematics for the return trip are

v v a x x v a x22

12

1 2 1 2 1 12 2 2= + −( ) ⇒ = − = − −( )( ) = −3.20 m/s 7.58 m 6.97 m/s2

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.

Page 46: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 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’sweight acts along the –x-axis as well. The box will move up if the pushing force is at least equal to the sum of thefriction 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 steadyspeed. 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 NThe 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 keptmoving up the ramp. However, if you stop on the ramp and want to start the box from rest, the model of staticfriction applies. The analysis is the same except that the coefficient of static friction is used and we use themaximum 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 theweight are too much to overcome.

Page 47: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.47. Model: We will model the sled and friend as a particle, and use the model of kinetic friction because thesled 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 pullingforce along the +x-direction is equal to the magnitude of the kinetic force of friction in the –x-direction. Also notethat (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 cosθThe y-component equation gives

n = mg – Fpull sinθWe see that the normal force is smaller than the weight because Fpull has a component in a direction opposite to thedirection of the weight. In other words, Fpull is partly lifting the sled. From the x-component equation, µk can now beobtained as

µk = F

mg Fpull

pull

cos

sin

θθ−

= 75 N

60 kg 9.8 m/s 75 N2

( ) °( )( )( ) − ( ) °( )

cos

sin

30

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.

Page 48: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

af

m

n

m

mg

mgmax = = = =s max s s

s

µ µ µ

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 sledwas zero, one could not pull at all without causing the box to slide.

Page 49: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: The shortest stopping distance is the distance for which the static friction force has its maximum valuefs 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 givesv1

2 = v02 + 2a(x1– x0) and µs = 0.8 (Table 5.1), we find

x xv v

a

v

g1 012

02

02 2

2 2 2 0 80−( ) = − = −

−( ) = ( )( )( )( ) =

µs2

15 m/s

9.8 m/s14.3 m

.

Assess: The truck was moving at a speed of 15 m/s or at approximately 34 mph. A stopping distance withoutmaking the contents slide of about 14.3 m or approximately 47 feet looks reasonable.

Page 50: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.50. Model: The antiques (mass = m) in the back of your pickup (mass = M) will be treated as a particle. Theantiques touch the truck’s steel bed, so only the steel bed can exert contact forces on the antiques. The pickup-antiques 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 tothe known coefficients for rubber on concrete. For the pickup-antiques system, with mass m + M, Newton’s secondlaw 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. Theseequations can be combined to yield a = –µg. Since constant-acceleration kinematics gives v1

2 = v02 + 2a(x1 + x0), we

find

av v

x x

v

g x x= −

−( ) ⇒ =−( ) = ( )

( )( )( )=1

202

1 0

02

1 0

2

2 2 2 9 80 58µmin 2

25 m/s

m/s 55 m..

The truck cannot stop if µ is smaller than this. But both the static and kinetic coefficients of friction, 1.00 and 0.80respectively (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 tothe California Highway Patrol website, the stopping distance (with zero reaction time) for a passenger vehicletraveling 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 stopthe truck.

Page 51: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.51. Model: The box will be treated as a particle. Because the box slides down a vertical wood wall, we willalso 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

down the wall at constant speed, so r ra = 0 and the box is in dynamic equilibrium. Thus,

rFnet =

r0 . 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

⇒ =° + °

=( )( )

° + °=F

mgpush

k

22 kg 9.8 m/s23.1 N

µ cos sin . cos sin45 45 0 20 45 45

Page 52: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 thepushing force is within the proper range, the box will remain stuck in place. First, let’s evaluate the sum of all theforces 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. j N . This means the static friction force will be able to prevent the box from moving if f js N= +3 8. ˆ .Using the x-equation and the friction model we get

fs max = µsn = µsFpush cos30° = 5.2 N

where we used µs = 0.5 for wood on wood. The static friction force rfs needed to keep the box from moving is less

than fs max. Thus the box will stay at rest.

Page 53: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.53. Model: We will model the skier along with the wooden skis as a particle of mass m. The snow exerts a contactforce 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

friction rfk and drag

rD 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 +x-direction cancel out the forces along the –x-direction.Solve: Newton’s second law and the models of kinetic friction and drag are

(Fnet)x = ΣFx = nx + wx + (fk)x + (D)x = 0 N + mg sinθ – fk–1

4Av2 = max = 0 N

(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

⇒ = −

v mgAterm

ksin cosθ µ θ14

1 2

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

vterm2

14

3 280 kg 9.8 m/s

kg m 0.72 m 51 m/s= ( )( ) ° − °

( )( )

=sin . cos40 0 06 401 2

Assess: A terminal speed of 51 m/s corresponds to a speed of ≈100 mph. This speed is reasonable but high due tothe steep slope angle of 40° and a small coefficient of friction.

Page 54: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.54. Model: Treat the Ping Pong ball as a particle with a drag force equal to 14

2Av , as modeled in the textwith vterm as the terminal velocity.Visualize:

Solve: (a) Imagine the ball falling at its terminal speed. The ball’s weight is directed down and the resistive dragforce is directed up. The net force is zero because the magnitude of the drag force is equal to the magnitude of theweight, D = w. When the ball is shot upwards at twice the terminal speed, the drag force is four times the terminaldrag force. That is, ′ = =D D w4 4 . Since all the forces are down, the y-component of Newton’s second law is

ΣF D w w w mg may = − ′ − = − − = − =4 5 ⇒ a g= −5

(b) The ball is initially shot downward. Therefore ′′D is upward but w is down. Again ′′ =D D4 and the y-component of Newton’s second law is

ΣF D w w w mg may = ′′ − = − = =4 3 ⇒ a g= 3

(c)

The ball initially decelerates at 3g but as v become smaller, the drag force approaches the weight so thedeceleration 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 isshot down w opposes D′′ so the ball decelerates at a lesser rate.

Page 55: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.55. Model: Take the coin to be a particle held against the palm of your hand (which is like a vertical wall) byfriction. 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 minimumacceleration amin is needed.(b) Newton’s second law and the model of friction are

ΣF f w f mgy y y= − = − =( ) ( )min min 0 N ΣF n max = =min min f nmin min= µ

Since you do not want the coin to slip down the hand you need µs . Combining the above three equations yields

f mgmin = ⇒ µsn mgmin = ⇒ µsma mgmin = ⇒ ag

min = = =µs

229.8 m/s

0.812.3 m/s

Page 56: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

(b) The net force is r r rF P wnet = + . You can see from the free-body diagram that

r rP w and point in the same direction,

hence F wnet > .

(c) The apparent weight of an object is the magnitude of the contact force against the object. rP is the contact force

against the ball, so the apparent weight is w Papp = . Newton’s second law for the ball is

ΣF P w P mg may y= − − = − − =

Thus the apparent weight is

w P mg ma mga

gw

a

gyy y

app = = − − = − −

= − −

1 1

If ay = –g, then wapp = 0. If ay = –1.5g, then wapp = 12 0 5w = . 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’veseen, an object in free fall has wapp = 0. The harder the hand pushes against the ball, causing the ball to acceleratedownward faster than free fall, the larger the ball’s apparent weight.

Page 57: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: As long as the container is stationary or it is moving with a constant speed (zero acceleration), the netforce 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 amax21.0 m/s= . So the maximum cable tension is

T mg mamax ,= + =max N48 600

This is less than the cable’s rating, so the cable must have been defective.

Page 58: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.58. Model: We will model the sculpture as a particle of mass m. The ropes that support the sculpture will beassumed 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:

Tw

2 60 30

3060

500=°( ) °( )

°+ °

=sin cos

sincos

lbs

2= 250 lbs

Substituting this value of T2 back into the x-compoent equation,

T T1 2

60

30= °

°sin

sin = °

°=250

60

30 lbs 433 lbs

sin

sin

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 433lbs. Since the cross-sectional area of the rope is 1

42πd , we have

d = ( )( )

=4

40 371

1 2

433 lbs

,000 lbs/ inch inch

2π.

Any diameter larger than 0.371 inch will ensure a safety rating of at least 433 lbs. The rope size corresponding to adiameter 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 of3/8 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 y-components of the tensions in the ropes that will equal the weight of the sculpture.

Page 59: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.59. Model: We will model the shuttle as a particle and assume the elastic cord to be massless. We will alsouse 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 lawin component form is

(Fnet)x = Σ Fx = T x + ( )fk x + (n)x + wx = T cos 45° – fk + 0 N + 0 N = max = max

(Fnet)y = ΣFy = Ty + ( )f yk + (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

aT T mg

mx =° − ° −( )

=( ) ° − ( ) +( ) ° − ( )( )[ ]

=

cos sin

cos . sin

45 45

45 0 60 45

µk

2

220 N 20 N 0.800 kg 9.8 m/s

0.800 kg13.0 m/s

Assess: The x-component of the tension force is 14.1 N. On the other hand, the net force on the shuttle in thex-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 Ntension force is used up to overcome the force of kinetic friction.

Page 60: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 “atrest” 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.(d) The free-body diagram shows that the horizontal component of

rT 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

aF

m

T

m

T

ma

F

m

T w

m

T mg

mxx x

yy y y=

( )= = ° = =

( )=

+= ° −net 2 2 net

m /ssin cos10

010

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

Tmg=

°=

( )( )°

=cos cos10 10

1 kg 9.8 m/s9.95 N

2

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

aT

m mx = ° = ( ) °sin . sin10 9 95 10 N= 1.73 m/s2

The truck’s velocity cannot be determined.

Page 61: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: The apparent weight of an object is the magnitude of the contact force supporting it. Here, the contactforce is

rP, so the apparent weight is wapp = P. Newton’s second law is

F F P w may y y ynet = = + =Σ ⇒ = − =F P w manet ⇒ = = + = +

w P mg ma mga

gapp 1

Page 62: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.62. Solve: (a) A 1 kg block is pulled across a level surface by a string, starting from rest. The string has atension of 20 N, and the block’s coefficient of kinetic friction is 0.50. How long does it take the block to move1 m?(b) Newton’s second law for the block is

a aF

m

T f

m

T n

ma

F

m

n w

m

n mg

mxx

yy= =

( )= − = − = =

( )= − = −net k k 2 net

m/sµ

0

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

aT mg

m= − = − =µk 220 N 4.9 N

1 kg15.1 m/s

Constant acceleration kinematics gives

x x v t a t a t tx

a1 0 02 2 11

2

1

2

2 2 1= + + ( ) = ( ) ⇒ = = ( )∆ ∆ ∆ ∆

m

15.1 m/s= 0.364 s2

Page 63: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.63. Solve: (a) A 15,000 N truck starts from rest and moves down a 15° hill with the engine providing an12,000 N force in the direction of the motion. Assume the frictional force between the truck and the road is verysmall. 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

ΣF n w f E may y y y y y= + + + = = 0

ΣF n w f E max x x x x= + + + = x ⇒ + + + =0 0 12 000 N N Nw masin ,θ

⇒ = + = ( ) ° +( )a

w

m

sin , , sin ,

, .

θ 12 000 15 000 15 12 000

15 000 8

N N N

N/9 m/s2 = 10.4 m/s2

where we have calculated the mass of the truck from its weight. Using the constant-acceleration kinematic equationv v axx

202 2− = ,

v a xx x2 2 2= = ( )( )10.4 m/s 50 m2 ⇒ vx = 32.2 m/s

Page 64: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 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 Nforce at a downward angle of 30°. The box has a coefficient of kinetic friction on the assembly line of 0.25. What isthe speed of the box after traveling 50 cm, starting from rest?(b) Newton’s second law for the box is

a aF

m

F f

m

F n

m

aF

m

n w F

m

n mg F

m

xx

yy

= =( )

=° −

=° −

= =( )

=− − °

=− − °

net push k push k

2 net push push m/s

cos cos

sin sin

30 30

030 30

µ

The second equation is solved to give an expression for n. Substituting into the first equation:

aF F mg

m=

° − ° +( )= ( ) ° −

=push k push 220 N 4.9 N

1 kg12.4 m/s

cos sin cos30 30 30µ

Using kinematics,

v v a x a x v a x12

02

12 2 2 2= + = ⇒ = = ( )( ) =∆ ∆ ∆ 12.4 m/s 0.5 m 3.5 m/s2

Page 65: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.65. Solve: (a) A 61.2 kg skier starts down a 20° slope. Her skies have a coefficient of friction of 0.06, andshe is facing into a wind that exerts a 100 N force against her. How far does she travel before reaching a speed of20 m/s?(b) Newton’s second law for the skier is

a aF

m

F f w

m

F n mg

m

aF

m

n w

m

n mg

m

xx

yy

= =( )

= + − ° = + − °

= =( )

= − ° = − °

net wind k wind k

2 net m/s

sin sin

cos cos

20 20

020 20

µ

The second equation gives n = mgcos20°. Substituting into the first equation,

aF mg mg

m= + ° − ° =

+ − ( ) °= −wind k 2100 N 34 N 600 N

61.2 kg1.16 m/s

µ cos sin sin20 20 20

The acceleration is negative, indicating that ra points down the hill as it should. Using kinematics,

v v a x a x xv

a12

02 1

2 2

2 22 2

= + = ⇒ = = ( )−( ) = −∆ ∆ ∆20 m/s

1.16 m/s172 m

2

∆x is negative because she travels in the negative x-direction. The distance traveled is 172 m.

Page 66: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 fardoes the auto slide before coming to rest?(b)

(c) Newton’s second law isΣ ΣF n w n mg ma F n may y y y x x= + = − = = = − =0 0 8 N .

The y-component equation gives n mg= = ( )( )1500 kg 9.8 m/s2 . Substituting this into the x-component equation yields

1500 kg 1500 kg 9.8 m/s2( ) = − ( )( )ax 0 8. ⇒ = −( )( ) = −ax 0 8. 9.8 m/s 7.8 m/s2 2

Using the constant-acceleration kinematic equation v v a x12

02 2= + ∆ , we find

∆xv

a= − = − ( )

−( ) =02 2

2 2102

40 m/s

7.8 m/s m

2

Page 67: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

5.67. Solve: (a) A 20.0 kg wooden crate is being pulled up a 20° wooden incline by a rope that is connected toan electric motor. The crate’s acceleration is measured to be 2 m/s2. The coefficient of kinetic friction between thecrate 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 NSolving the y-component equation, n = 184.18 N. Substituting this value for n in the x-component equation yieldsT = 144 N.

Page 68: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

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

Σ

Σ

F n P w f f f ma

F n P w f n mg ma

f n

x x x x x x

y y y y y y

= + + + = + ( ) ° + − = ( ) ° − =

= + + + = + ( ) ° − − = =

=

0 100 30 0 100 30

100 30 0 0

N N N N

N N N

k k

k k

cos cos

sin

µ

From the y-component equation, n = 150 N . From the x-component equation and using the model of kineticfriction with µk = 0.2,

100 N( ) cos 30° – (0.2)(150 N) = (20 kg)ax ⇒ =ax 2.8 m/s2

Page 69: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

(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 9 8 20 9 8 0. . cos . sin kg m/s kg m/s N tan = 0.2 = 11.32 2θ θ θ θ

Page 70: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: Newton’s second law is

Σ ΣF n ma F n w max x y y= = = − = =sin cosθ θ N0

Combining the two equations, we get

maw

mg a gx x= = ⇒ =cos

sin tan tanθ

θ θ θ

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

dxAx a g Axx= = ⇒ = ( )tanθ 2 2

(b) The acceleration at x = 0.2 m is 9.8 m/s 1 m m 3.9 m/s2 2( )( )( )( ) =−2 0 21 . .

Page 71: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

Solve: Your best strategy, if it’s possible, is to travel at a very slow constant speed r r r ra F= =( )0 0 so net . 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 givesr rFnet = 0.Newton’s second law for the skier and the model of kinetic friction are

(Fnet)x = ΣFx = nx + wx + ( )f xk + (D)x = 0 + mg sinθ – fk – D cosθ = 0 N

(Fnet)y = ΣFy = ny + wy + ( )f yk + (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 sinθFrom the model of kinetic friction,

µ θ θθ θk

k

2

2

75 kg 9.8 m/s 50 N

75 kg 9.8 m/s 50 N= = −

+=

( ) ° − ( ) °

( ) ° + ( ) °=f

n

mg D

mg D

sin cos

cos sin

sin cos

cos sin.

15 15

15 150 196

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

Page 72: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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

F n w n mg ma n m a gy y y y y y ynet( ) = + = − = ⇒ = +( )

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 amountsto 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 thedownward 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 becomeszero. The table is accelerating in the negative direction so quickly after t = 0.75 s that the widget can stay on only ifthe 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.

Page 73: () x xx x x y yy y y - Engineering HeroApplying Newton’s second law, a F x m = x = −− () net 5 N 1 N N = 2 2 kg 1.49 m/s 320sin a F y m = y = − () net 2.82 N N = 2 2 kg 0 m/s

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 accelerationto 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 0 01

42 N N N Av ma m

dv

dtx xx

This can be written

dv

v

A

mdtx

x2 4

=

We can integrate this from the start (v0x at t = 0) to the end (vx at t):

dv

v

A

mdt

v v

A

mtx

xv

v

x

t

xx

x

2 00

0 4

1 1

4∫ ∫= ⇒ − + =

Solving for vx gives

vv

Av t mxx

x

=+

0

01 4

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

vx = 20 m / s2.24 20

1500 1

4+ ( )

×t

=+

20

1 0 007467. t⇒ =

tvx

1

0 007467

201

.

m/s

where t is in seconds. We can now obtain the time t for v = 10 m/s:

t =

= −

=1

0 007467

20

101 133 93

20

101

..

m/s

m/s134 s

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)(9.8 m/s2) = 735 N. The drag force at an average speed of 10 m/s is D = ( )( ) =1

422.24 10 N 56 N. We conclude that

it is not reasonable to neglect the kinetic force of friction.