Física para Ciencias e Ingeniería. Volumen I - Douglas C. Giancoli - 4ta edición [Solucionario]

Chapter 4 Dynamics: Newton’s Laws of Motion

Fnet = FTcosθ − mg = 0 → FT = mg →
cosθ
y

( )FH mg
= cosθ sin θ = mg tan θ = (27 kg) 9.80 m s2 tan 2.15° = 9.9 N

( )(b) s2
FT = mg = (27 kg) 9.80 m = 260 N
cosθ
cos 2.15°

GG
46. (a) In the free-body diagrams below, FAB = force on block A exerted by block B, GFBA = force on
G
block B exerted by block A, FBC = force on block B exerted by block C, and FCB = force on
GG
block C exerted by block B. The magnitudes of FBA and FAB are equal, and the magnitudes of
GG
FBC and FCB are equal, by Newton’s third law.

G
FA N G
G FB N G G
G G FBA FBC G FC N
F FAB FCB

mAgG mBgG mCgG

(b) All of the vertical forces on each block add up to zero, since there is no acceleration in the
vertical direction. Thus for each block, FN = mg . For the horizontal direction, we have the
following.

∑ ( )F = F − FAB + FBA − FBC + FCB = F = mA + mB + mC a → a = mA + F + mC
mB

(c) For each block, the net force must be ma by Newton’s second law. Each block has the same

acceleration since they are in contact with each other.

FA net = F mA mA + mC FB net = F mA mB + mC F3 net = F mA mC + mC
+ mB + mB + mB

(d) From the free-body diagram, we see that for mC, FCB = FC net = F mC . And by

mA + mB + mC

Newton’s third law, FBC = FCB = F mC . GG
Of course, F23 and F32 are in opposite
mA + mB + mC

directions. Also from the free-body diagram, we use the net force on mA.

F − FAB = FA net = F mA mA + mC → FAB =F−F mA →
+ mB mA + mB + mC

FAB = F mB + mC
mA + mB + mC

By Newton’s third law, FBC = FAB = F m2 + m3 .

m1 + m2 + m3

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(e) Using the given values, a = m1 + F + m3 = 96.0 N = 3.20 m s2 . Since all three masses
m2 30.0 kg

( )are the same value, the net force on each mass is Fnet = ma = (10.0 kg) 3.20 m s2 = 32.0 N .

This is also the value of FCB and FBC. The value of FAB and FBA is found as follows.

( )FAB = FBA = (m2 + m3 ) a = (20.0 kg) 3.20 m s2 = 64.0 N

To summarize:

FA net = FB net = FC net = 32.0 N FAB = FBA = 64.0 N FBC = FCB = 32.0 N

The values make sense in that in order of magnitude, we should have F > FBA > FCB , since F is the

net force pushing the entire set of blocks, FAB is the net force pushing the right two blocks, and FBC
is the net force pushing the right block only.

47. (a) Refer to the free-body diagrams shown. With the stipulation GG
that the direction of the acceleration be in the direction of motion FN FN +y

for both objects, we have aC = aE = a. a a
m2gG
mEg − FT = mEa ; FT − mCg = mCa +y
(b) Add the equations together to solve them. m1gG

(mEg − FT ) + ( FT − mCg ) = mEa + mCa →

mE g − mCg = mEa + mCa →

mE − mC 1150 kg − 1000 kg
mE + mC 1150 kg + 1000 kg
( )a = 0.68 m s2
= g = 9.80 m s2

⎛ mE − mC ⎞ 2mCmE 2 (1000 kg) (1150 kg)
⎜ mE + mC ⎟ mE + mC
⎝ ⎠ 1150 kg + 1000 kg
( )FT mC (g a) g s2
= + = mC + g = g = 9.80 m

= 10, 483 N ≈ 10,500 N

48. (a) Consider the free-body diagram for the block on the frictionless G
surface. There is no acceleration in the y direction. Use Newton’s
second law for the x direction to find the acceleration. FN y

∑ Fx = mg sinθ = ma → x

( )a = g sinθ = 9.80 m s2 sin 22.0° = 3.67 m s2 θ mθgG

(b) Use Eq. 2-12c with v0 = 0 to find the final speed.

( )v2 − v02 = 2a ( x − x0 ) → v = 2a ( x − x0 ) = 2 3.67 m s2 (12.0 m) = 9.39 m s

49. (a) Consider the free-body diagram for the block on the frictionless G
surface. There is no acceleration in the y direction. Write Newton’s
second law for the x direction. FN y

∑ Fx = mg sinθ = ma → a = g sinθ x

Use Eq. 2-12c with v0 = −4.5m s and v = 0 m s to find the distance θ mθgG
that it slides before stopping.

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100

Chapter 4 Dynamics: Newton’s Laws of Motion

v2 − v02 = 2a ( x − x0 ) →

v2 − v02 0 − (−4.5 m s)2
2a
9.80 m s2 sin 22.0°
( )( x − x0 ) = = 2 = −2.758 m ≈ 2.8 m up the plane

(b) The time for a round trip can be found from Eq. 2-12a. The free-body diagram (and thus the

acceleration) is the same whether the block is rising or falling. For the entire trip, v0 = −4.5m s
and v = +4.5 m s.

t v − v0 (4.5 m s) − (−4.5 m s) = 2.452 s ≈ 2.5 s
a
9.80 m s2 sin 22o
( )v = v0 + at → = =

50. Consider a free-body diagram of the object. The car is moving to the right. The G
acceleration of the dice is found from Eq. 2-12a. FT θ
mgG
v = v0 + = axt → ax = v − v0 = 28 m s − 0 = 4.67 m s2
t 6.0 s

Now write Newton’s second law for both the vertical (y) and horizontal (x)

directions.

∑ Fy = FT cosθ − mg = 0 → FT = mg ∑ Fx = FT sin θ = max
cosθ
Substitute the expression for the tension from the y equation into the x equation.

max = FT sin θ = mg sin θ = mg tan θ → ax = g tanθ
cosθ

θ = tan−1 ax = tan −1 4.67 m s2 = 25.48o ≈ 25o
g 9.80 m s2

51. (a) See the free-body diagrams included. yx G
FT
(b) For block A, since there is no motion in the vertical direction, G
we have FNA = mAg. We write Newton’s second law for the x FNA G

∑direction: FAx = FT = mAaAx. For block B, we only need to FT

∑consider vertical forces: FBy = mBg − FT = mBaBy. Since the mAgG mBgG

two blocks are connected, the magnitudes of their accelerations

will be the same, and so let aAx = aBy = a. Combine the two force equations from above, and

solve for a by substitution.

FT = mAa mBg − FT = mBa → mBg − mAa = mBa →

mAa + mBa = mBg → a = g mB FT = mAa = g mAmB
mA + mB mA + mB

52. (a) From Problem 51, we have the acceleration of each block. Both blocks have the same

acceleration.

mB 5.0 kg
mA + mB
( )a (5.0 kg + 13.0 kg) s2
= g = 9.80 m s2 = 2.722 m ≈ 2.7 m s2

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101

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

(b) Use Eq. 2-12b to find the time.

at 2 2(x − x0 ) 2 (1.250 m ) = 0.96 s

a 2.722 m s2
( )x− x0 = v0t + 1 → t= =
2

(c) Again use the acceleration from Problem 51.

a = g mB = 1 g → mB = 1 → mA = 99mB = 99 kg
mA + mB 100 mA + mB 100

53. This problem can be solved in the same way as problem 51, with the modification that we increase

mass mA by the mass of l A and we increase mass mB by the mass of l B. We take the result from

problem 51 for the acceleration and make these modifications. We assume that the cord is uniform,
and so the mass of any segment is directly proportional to the length of that segment.

mB mB + l lB mC mB + l lB mC
mA + mB A +l A +l
a = g → a=g B = g B

⎛ mA + lA mC ⎞ + ⎛ mB + lB mC ⎞ mA + mB + mC
⎜ A +l ⎟⎠ ⎜⎝ A +l ⎠⎟
⎝ l l
B B

Note that this acceleration is NOT constant, because the lengths l A and l B are functions of time.

Thus constant acceleration kinematics would not apply to this system.

54. We draw a free-body diagram for each mass. We choose UP to be the G
FC
positive direction. The tension force in the cord is found from analyzing
G
the two hanging masses. Notice that the same tension force is applied to FT

each mass. Write Newton’s second law for each of the masses. G
FT
FT − m1g = m1a1 FT − m2 g = m2a2 m1
3.2 kg
Since the masses are joined together by the cord, their accelerations will G m1gG
FT
have the same magnitude but opposite directions. Thus a1 = −a2.
G
Substitute this into the force expressions and solve for the tension force. FT
m2
FT − m1g = −m1a2 → FT = m1g − m1a2 → a2 = m1g − FT 1.2 kg
m1 m2gG

FT − m2 g = m2a2 = m2 ⎛ m1g − FT ⎞ → FT = 2m1m2 g
⎝⎜ m1 ⎟⎠ m1 + m2

Apply Newton’s second law to the stationary pulley.

( )FC − 2FT = 0 s2
→ FC = 2FT = 4m1m2 g = 4 (3.2 kg) (1.2 kg) 9.80 m = 34 N
m1 + m2
4.4 kg

55. If m doesn’t move on the incline, it doesn’t move in the vertical direction, and G
θ FN
so has no vertical component of acceleration. This suggests that we analyze mgG θ

the forces parallel and perpendicular to the floor. See the force diagram for

the small block, and use Newton’s second law to find the acceleration of the

small block.

∑ Fy = FN cosθ − mg = 0 → FN = mg
cosθ

∑ Fx = FN sinθ = ma → a = FN sinθ = mg sinθ = g tanθ
m m cosθ

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102

Chapter 4 Dynamics: Newton’s Laws of Motion

Since the small block doesn’t move on the incline, the combination of both masses has the same
horizontal acceleration of g tanθ . That can be used to find the applied force.

Fapplied = (m + M ) a = (m + M ) g tanθ

Note that this gives the correct answer for the case of θ = 0, , where it would take no applied force to

keep m stationary. It also gives a reasonable answer for the limiting case of θ → 90°, where no

force would be large enough to keep the block from falling, since there would be no upward force to

counteract the force of gravity. G G
56. Because the pulleys are massless, the net force on them must be 0. FTC FTC

Because the cords are massless, the tension will be the same at both aC mC aC
ends of the cords. Use the free-body diagrams to write Newton’s

second law for each mass. We are using the same approach taken in G G mCgG
problem 47, where we take the direction of acceleration to be FTA FTA G
positive in the direction of motion of the object. We assume that mC G FTA
FTA
is falling, mB is falling relative to its pulley, and mA is rising

relative to its pulley. Also note that if the acceleration of mA mA aA mB aB
mAgG mBgG
relative to the pulley above it is aR , then aA = aR + aC. Then, the
acceleration of mB is aB = aR − aC, since aC is in the opposite

direction of aB .

∑ ( )mA :
F = FTA − mA g = mAaA = mA aR + aC

∑ ( )mB : F = mBg − FTA = mBaB = mB aR − aC

∑mC : F = mCg − FTC = mCaC

∑pulley: F = FTC − 2FTA = 0 → FTC = 2FTA

Re-write this system as three equations in three unknowns FTA , aR , aC.

( )FTA − mA g = mA aR + aC → FTA − mAaC − mAaR = mA g

( )mBg − FTA = mB aR − aC → FTA − mBaC + mBaR = mBg

mC g − 2FTA = mCaC → 2FTA + mCaC = mCg

This system now needs to be solved. One method to solve a system of linear equations is by

determinants. We show that for aC.

1 mA −mA

1 mB mB

aC = 2 mC 0 g = −mBmC + mA (2mB ) − mA ( mC − 2mB ) g
1 −mA −mA −mBmC − mA (2mB ) − mA ( mC + 2mB )

1 −mB mB

2 mC 0

= 4mAmB − mAmC − mBmC g = mAmC + mBmC − 4mAmB g
−4mAmB − mAmC − mBmC 4mAmB + mAmC + mBmC

Similar manipulations give the following results.

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103

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

( )aR
= 2 mAmC − mBmC g ; FTA = 4mAmBmC mBmC g
4mAmB + mAmC + mBmC 4mAmB + mAmC +

(a) The accelerations of the three masses are found below.

( )aA
= aR + aC = 2 mAmC − mBmC g + mAmC + mBmC − 4mAmB g
4mAmB + mAmC + mBmC 4mAmB + mAmC + mBmC

= 3mAmC − mBmC − 4mAmB g
4mAmB + mAmC + mBmC

( )aB
= aR − aC = 2 mAmC − mBmC g − mAmC + mBmC − 4mAmB g
4mAmB + mAmC + mBmC 4mAmB + mAmC + mBmC

= mAmC − 3mBmC + 4mAmB g
4mAmB + mAmC + mBmC

aC = mAmC + mBmC − 4mAmB g
4mAmB + mAmC + mBmC

(b) The tensions are shown below.

FTA = 4mAmBmC g ; FTC = 2FTA = 8mAmBmC g

4mAmB + mAmC + mBmC 4mAmB + mAmC + mBmC

57. Please refer to the free-body diagrams given in the textbook for this problem. Initially, treGat the two
boxes and the rope as a single system. Then the only accelerating force on the system is FP. The
mass of the system is 23.0 kg, and so using Newton’s second law, the acceleration of the system is

a = FP = 35.0 N = 1.522 m s2 ≈ 1.52 m s2 . This is the acceleration of each part of the system.
m 23.0 kg

G

Now consider mB alone. The only force on it is FBT , and it has the acceleration found above. Thus
FBT can be found from Newton’s second law.

( )FBT = mBa = (12.0 kg) 1.522 m s2 = 18.26 N ≈ 18.3 N

GG

Now consider the rope alone. The net force on it is FTA − FTB, and it also has the acceleration found

above. Thus FTA can be found from Newton’s second law.

( )FTA − FTB = mCa → FTA = FTB + mCa = 18.26 N + (1.0 kg) 1.522 m s2 = 19.8 N

58. First, draw a free-body diagram for each mass. Notice that the same G G
tension force is applied to each mass. Choose UP to be the positive FT FT
direction. Write Newton’s second law for each of the masses.
m2 m1
FT − m2 g = m2a2 FT − m1g = m1a1 2.2 kg 3.6 kg

Since the masses are joined together by the cord, their accelerations will m2gG m1gG

have the same magnitude but opposite directions. Thus a1 = −a2.

Substitute this into the force expressions and solve for the acceleration by
subtracting the second equation from the first.

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104

Chapter 4 Dynamics: Newton’s Laws of Motion

FT − m1g = −m1a2 → FT = m1g − m1a2

FT − m2 g = m2a2 → m1g − m1a2 − m2 g = m2a2 → m1g − m2 g = m1a2 + m2a2

( )a2m1− m2 3.6 kg − 2.2 kg = 2.366 m s2
= m1 + m2 g = 3.6 kg + 2.2 kg 9.80 m s2

The lighter block starts with a speed of 0, and moves a distance of 1.8 meters with the acceleration

found above. Using Eq. 2-12c, the velocity of the lighter block at the end of this accelerated motion

can be found.

( )v2 − v02 = 2a ( y − y0 ) → v = v02 + 2a ( y − y0 ) = 0 + 2 2.366 m s2 (1.8 m) = 2.918 m s

Now the lighter block has different conditions of motion. Once the heavier block hits the ground,
the tension force disappears, and the lighter block is in free fall. It has an initial speed of 2.918 m/s
upward as found above, with an acceleration of –9.80 m/s2 due to gravity. At its highest point, its
speed will be 0. Eq. 2-12c can again be used to find the height to which it rises.

v2 − v02 0 − (2.918 m s)2
2a
2 −9.80 m s2
( )v2 − v02 = 2a ( y − y0 )→ (y− y0 ) = = = 0.434 m

Thus the total height above the ground is 1.8 m + 1.8 m + 0.43 m = 4.0 m .

G θ G
59. The force F is accelerating the total mass, since it is the only force external to the FT

system. If mass mA does not move relative to mC, then all the blocks have the mBgG
same horizontal acceleration, and none of the blocks have vertical acceleGration. We
solve for the acceleration of the system and then find the magnitude of F from
Newton’s second law. Start with free-body diagrams for mA and mB.

∑mB : Fx = FT sinθ = mBa ; G
∑ Fy = FT cosθ − mBg = 0 → FT cosθ = mBg FN
G
Square these two expressions and add them, to get a relationship between FT and a. FT

FT2 sin2 θ = mB2a2 ; FT2 cos2 θ = mB2 g 2 → mAgG

( ) ( ) ( )FT2 sin2 θ + cos2 θ = mB2 g 2 + a2 → FT2 = mB2 g 2 + a2

Now analyze mA.

Fx = FT = mAa → FT2 = mA2 a2 ; Fy = FN − mA g = 0
∑ ∑mA :

Equate the two expressions for FT2, solve for the acceleration and then finally the magnitude of the
applied force.

( )FT2 = mB2 g 2 + a2 = mA2 a2 a2 = mB2 g 2 → a= mB g
mA2 − mB2
( )→ ( )mA2 − mB2 →

(mA + mB + mC ) mB g

mA2 − mB2
( )F = (mA + mB + mC ) a =

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105

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

60. The velocity can be found by integrating the acceleration function, and the position can be found by
integrating the position function.

∫ ∫F = ma = Ct2 → a = C t2 = dv → dv = C t 2dt → v dv = t C t 2dt → v= C t3
m dt m 0 0 m 3m

C t3 dx C t 3dt x dx t C t 3dt C t4
∫ ∫v= 3m = dt → dx = 3m → 0 = 0 3m → x = 12m

61. We assume that the pulley is small enough that the part of the cable that is touching

the surface of the pulley is negligible, and so we ignore any force on the cable due to

the pulley itself. We also assume that the cable is uniform, so that the mass of a

portion of the cable is proportional to the length of that portion. We then treat the m2
m1
cable as two masses, one on each side of the pulley. The masses are given by

m1 = y M and m2 = l − y M. Free-body diagrams for the masses are shown.
l l
G G
(a) We take downward motion of m1 to be the positive direction for m1, FT FT

and upward motion of m2 to be the positive direction for m2. Newton’s second
law for the masses gives the following.

Fnet 1 = m1g − FT = m1a ; Fnet 2 = FT − m2 g → a = m1 − m2 g m2gG m1gG
m1 + m2

a = yM l −yM g = y − (l − y) g = 2y −l g= ⎛ 2y − 1⎠⎟⎞ g
l − l y + (l − y) l ⎝⎜ l
yM +l −yM
l l

(b) Use the hint supplied with the problem to set up the equation for the velocity. The cable starts

with a length y0 (assuming y0 > l1 ) on the right side of the pulley, and finishes with a length

2

l on the right side of the pulley.

a = ⎛ 2y − 1⎞⎟⎠ g = dv = dv dy = v dv → ⎛ 2y − 1⎠⎟⎞ gdy = vdv →
⎝⎜ l dt dy dt dy ⎜⎝ l

∫ ∫ ( )l⎛2y − 1⎠⎟⎞ gdy = vf vdv → g ⎛ y2 − y ⎞ l = v1 2 vf → gy0 ⎛ 1 − y0 ⎞ = v1 2 →
⎝⎜ l 0 ⎜⎝ l ⎠⎟ y0 ⎜⎝ l ⎠⎟
y0 20 2f

vf = 2 gy0 ⎜⎝⎛1 − y0 ⎞
l ⎟⎠

(c) For y0 = l2 , we have vf = 2 gy0 ⎛⎝⎜1 − y0 ⎞ = 2g ( 2 ) l ⎛⎜⎝1 − 2 l ⎞ = 2 gl .
l ⎠⎟ 3 3 ⎟⎠ 3
3
l

62. The acceleration of a person having a 30 “g” deceleration is a = ( 30" g ") ⎛ 9.80 m s2 ⎞ = 294 m s2 .
⎜⎝ "g" ⎠⎟

( )The average force causing that acceleration is F = ma = (65 kg) 294 m s2 = 1.9 × 104 N . Since

the person is undergoing a deceleration, the acceleration and force would both be directed opposite
to the direction of motion. Use Eq. 2-12c to find the distance traveled during the deceleration. Take

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106

Chapter 4 Dynamics: Newton’s Laws of Motion

the initial velocity to be in the positive direction, so that the acceleration will have a negative value,
and the final velocity will be 0.

v0 = (95 km h) ⎛ 1m s h ⎞ = 26.4 m s
⎜⎝ 3.6 km ⎠⎟

v2 − v02 0 − (26.4 m s)2
2a
2 −294 m s2
( )v2 − v02 = 2a ( x − x0 ) → (x − x0 ) = = = 1.2 m

63. See the free-body diagram for the falling pGurse. Assume that down is the positive G
direction, and that the air resistance force Ffr is constant. Write Newton’s second law for Ffr
the vertical direction. mgG

∑ F = mg − Ffr = ma → Ffr = m ( g − a)

Now obtain an expression for the acceleration from Eq. 2-12c with v0 = 0 , and substitute
back into the friction force.

v2 − v02 = 2a ( x − x0 ) → a = v2 x0 )

2(x −

Ff = m ⎛ g − 2 ( v2 x0 ) ⎞ = (2.0 kg) ⎛ 9.80 m s2 − (27 m s)2 ⎞ = 6.3 N
⎜ x− ⎟ ⎜ 2 (55 m) ⎟
⎝ ⎠ ⎝ ⎠

64. Each rope must support 1/6 of Tom’s weight, and so must have a vertical component of tension

given by Tvert = 1 mg . For the vertical ropes, their entire tension is vertical.
6

( )T1 kg) s2 = 120.9 N ≈ 1.21×102 N
= 1 mg = 1 (74.0 9.80 m 30o
6 6
T2
For the ropes displaced 30o from the vertical, see the first diagram.

T2 vert = T2 cos 30° = 1 mg → T2 = mg = 120.9 N = 1.40 ×102 N
6 6 cos 30° cos 30°

For the ropes displaced 60o from the vertical, see the second diagram.

T3 vert = T3 cos 60° = 1 mg → T3 = mg 120.9 N = 2.42 ×102 N T3
6 6 cos 60° = cos 60°
60o

The corresponding ropes on the other side of the glider will also have the same

tensions as found here.

65. Consider the free-body diagram for the soap block on the frictionless G
surface. There is no acceleration in the y direction. Write Newton’s
second law for the x direction. FN y

∑ Fx = mg sinθ = ma → a = g sinθ x

Use Eq. 2-12b with v0 = 0 to find the time of travel. θ mθgG

x − x0 = v0t + 1 at 2 →
2

2(x − x0 ) 2 ( x − x0 ) 2 (3.0 m) 2.0 s
9.80 m s2 sin (8.5°)
a g sinθ
( )t = = = =

Since the mass does not enter into the calculation, the time would be the same for the heavier bar of
soap.

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107

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

66. See the free-body diagram for the load. The vertical component of the tension force must

be equal to the weight of the load, and the horizontal component of the tension

accelerates the load. The angle is exaggerated in the picture. G θ
FT
Fnet = FT sinθ = ma → a = FT sin θ ; Fnet = FT cosθ − mg = 0 →
m
x y

( )FTmg mg sin θ
= cosθ → aH = cosθ m = g tanθ = 9.80 m s2 tan 5.0° = 0.86 m s2 mgG

67. (a) Draw a free-body diagram for each block. Write G G
Newton’s second law for each block. Notice that the G FT FN yA
acceleration of block A in the yA direction will be zero, yB FT
since it has no motion in the yA direction.
mBgG θ xA
∑ FyA = FN − mA g cosθ = 0 → FN = mA g cosθ mAgG θ

∑ FxA = mA g sin θ − FT = mAaxA
∑ ( )FyB = FT − mBg = mBayB → FT = mB g + ayB

Since the blocks are connected by the cord, ayB = axA = a. Substitute the expression for the

tension force from the last equation into the x direction equation for block 1, and solve for the
acceleration.

mA g sinθ − mB ( g + a) = mAa → mA g sinθ − mBg = mAa + mBa

a= g (mA sinθ − mB )
(mA + mB )

(b) If the acceleration is to be down the plane, it must be positive. That will happen if

mA sinθ > mB (down the plane) . The acceleration will be up the plane (negative) if

mA sinθ < mB (up the plane) . If mA sinθ = mB, then the system will not accelerate. It will

move with a constant speed if set in motion by a push.

68. (a) From problem 67, we have an expression for the acceleration.

(mA sinθ − mB ) [(1.00 kg) sin 33.0° − 1.00 kg]
(mA + mB )
( )a 2.00 kg
= g = 9.80 m s2 = −2.23 m s2

≈ −2.2 m s2

The negative sign means that mA will be accelerating UP the plane.
(b) If the system is at rest, then the acceleration will be 0.

a = g (mA sinθ − mB ) = 0 → mB = mA sinθ = (1.00 kg) sin 33.0° = 0.5446 kg ≈ 0.545 kg
(mA + mB )

(c) Again from problem 68, we have FT = mB ( g + a).

( )Case (a): FT = mB ( g + a) = (1.00 kg) 9.80 m s2 − 2.23m s2 = 7.57 N ≈ 7.6 N

( )Case (b): FT = mB ( g + a) = (0.5446 kg) 9.80 m s2 + 0 = 5.337 N ≈ 5.34 N

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108

Chapter 4 Dynamics: Newton’s Laws of Motion

69. (a) A free-body diagram is shown for each block. G G G y
We define the positive x-direction for mA to be FN-A FT G FN-B
FT

up its incline, and the positive x-direction for mB y x
θA θA
to be down its incline. With that definition the x
masses will both have the same acceleration. mAgG θB θB
Write Newton’s second law for each body in the
x direction, and combine those equations to find mBgG
the acceleration.

∑mA : Fx = FT − mA g sinθA = mAa

∑mB : Fx = mBg sinθB − FT = mBa add these two equations

( FT − mAg sinθA ) + (mBg sinθB − FT ) = mAa + mBa → a= mB sinθB − mA sinθA g
mA + mB

(b) For the system to be at rest, the acceleration must be 0.

a = mB sinθB − mA sinθA g = 0 → mB sinθB − mA sinθA →
mA + mB

mB = mA sin θ A = (5.0 kg) sin 32° = 6.8 kg
sin θ B sin 23°

The tension can be found from one of the Newton’s second law expression from part (a).

( )mA : FT − mAg sinθA = 0 → FT = mAg sinθA = (5.0 kg) 9.80 m s2 sin 32° = 26 N

(c) As in part (b), the acceleration will be 0 for constant velocity in either direction.

a = mB sinθB − mA sinθA g = 0 → mB sinθB − mA sinθA →
mA + mB

mA = sin θ B = sin 23° = 0.74
mB sin θ A sin 32°

70. A free-body diagram for the person in the elevator is shown. The scale reading is the
magnitude of the normal force. Choosing up to be the positive direction, Newton’s

∑second law for the person says that F = FN − mg = ma → FN = m ( g + a). The

kg reading of the scale is the apparent weight, FN , divided by g, which gives mgG G
FN
FN-kg = FN = m(g + a).
g
g

( )(a) a = 0 → FN = mg = (75.0 kg) 9.80 m s2 = 7.35 ×102 N

FN-kg = mg =m= 75.0 kg
g

(b) a = 0 → FN = 7.35 ×102 N , FN-kg = 75.0 kg

(c) a = 0 → FN = 7.35 ×102 N , FN-kg = 75.0 kg

( )(d) FN = m ( g + a) = (75.0 kg) 9.80 m s2 + 3.0 m s2 a = 9.60 ×102 N

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109

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

FN-kg = FN = 960 N = 98.0 kg
g 9.80 m s2

( )(e) FN = m ( g + a) = (75.0 kg) 9.80 m s2 − 3.0 m s2 a = 5.1×102 N

FN-kg = FN = 510 N = 52 kg
g 9.80 m s2

71. The given data can be used to calculate the force with which the road pushes G y x
FN
against the car, which in turn is equal in magnitude to the force the car

pushes against the road. The acceleration of the car on level ground is found G
FP
from Eq. 2-12a.

v − v0 = at → a = v − v0 = 21m s − 0 = 1.68 m s2 θmgG θ
t 12.5 s

The force pushing the car in order to have this acceleration is found from

Newton’s second law.

( )FP = ma = (920 kg) 1.68 m s2 = 1546 N

We assume that this is the force pushing the car on the incline as well. Consider a free-body diagram
for the car climbing the hill. We assume that the car will have a constant speed on the maximum
incline. Write Newton’s second law for the x direction, with a net force of zero since the car is not
accelerating.

∑F = FP − mg sinθ = 0 → sin θ = FP
x mg

= sin−1 FP = sin−1 1546 N = 9.9°
mg
(920 kg) 9.80 m
( )θ s2

72. Consider a free-body diagram for the cyclist coasting downhill at a constant G G
speed. Since there is no acceleration, the net force in each direction must be Ffr FN
zero. Write Newton’s second law for the x direction (down the plane).
θ θmgG
∑ Fx = mg sinθ − Ffr = 0 → Ffr = mg sinθ
y G
This establishes the size of the air friction force at 6.0 km/h, and so can be Gx FN
used in the next part. FP G
Ffr
Now consider a free-body diagram for the cyclist climbing the hill. FP is the
force pushing the cyclist uphill. Again, write Newton’s second law for the x θ θmgG
direction, with a net force of 0.
G G
∑ Fx = Ffr + mg sin θ − FP = 0 → Fair FN

FP = Ffr + mg sinθ = 2mg sinθ y

( )= 2 (65 kg) 9.80 m s2 (sin 6.5°) = 1.4 ×102 N x

73. (a) The value of the constant c can be found from the free-body diagram, θ θmgG
knowing that the net force is 0 when coasting downhill at the specified
speed.

∑ Fx = mg sinθ − Fair = 0 → Fair = mg sinθ = cv →

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110

Chapter 4 Dynamics: Newton’s Laws of Motion

( )c
= mg sinθ = (80.0 kg) 9.80 m s2 sin 5.0° = 40.998 N ≈ 41 N
v ms ms
(6.0 km h) ⎛ 1m s ⎞
⎜⎝ 3.6 km h ⎠⎟
G
(b) Now consider the cyclist with an added pushing force FP directed along
G G
the plane. The free-body diagram changes to reflect the additional force Fair FN

the cyclist must exert. The same axes definitions are used as in part (a). G
FP
∑ Fx = FP + mg sin θ − Fair = 0 →
θ θmgG
FP = Fair − mg sin θ = cv − mg sin θ

= ⎛ 40.998 N ⎞ ⎛ (18.0 km h ) ⎛ 1m s h ⎞ ⎞
⎝⎜ ms ⎠⎟ ⎜ ⎜⎝ 3.6 km ⎠⎟ ⎟
⎝ ⎠

( )− (80.0 kg) 9.80 m s2 sin 5.0° = 136.7 N ≈ 140 N

74. Consider the free-body diagram for the watch. Write Newton’s second law for G
θ FT
both the x and y directions. Note that the net force in the y direction is 0 because y
x mgG
there is no acceleration in the y direction.

∑ Fy = FT cosθ − mg = 0 → FT = mg
cosθ

∑ Fx = FT sinθ = ma → mg sin θ = ma
cosθ

( )a = g tanθ = 9.80 m s2 tan 25° = 4.57 m s2

Use Eq. 2-12a with v0 = 0 to find the final velocity (takeoff speed).

( )v − v0 = at → v = v0 + at = 0 + 4.57 m s2 (16s) = 73 m s

75. (a) To find the minimum force, assume that the piano is moving with a constant G
velocity. Since the piano is not accelerating, FT4 = Mg. For the lower pulley, since FT4
the tension in a rope is the same throughout, and since the pulley is not accelerating,

it is seen that FT1 + FT2 = 2FT1 = Mg → FT1 = FT2 = Mg 2.

It also can be seen that since F = FT2 , that F = Mg 2 . MgG
(b) Draw a free-body diagram for the upper pulley. From that

diagram, we see that FT 3 = FT1 + FT 2 +F = 3Mg . G G Upper G
2 FT2 FT1 Pulley FT3

To summarize: G Lower G
FT4 Pulley
FT1 = FT2 = Mg 2 FT3 = 3 Mg 2 FT4 = Mg FT2 G
F
G
FT1

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111

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

76. Consider a free-body diagram for a grocery cart being pushed up an G
G FN
incline. Assuming that the cart is not accelerating, we write Newton’s FP y
x
second law for the x direction.

∑ Fx = FP − mg sinθ = 0 → sin θ = FP
mg

= sin−1 FP sin −1 18 N = 4.2° θθ
mg
(25 kg) 9.80 m mgG
( )θ = s2

77. The acceleration of the pilot will be the same as that of the plane, since the pilot G
Fnet
is at rest with respect to the plane. Consider first a free-body diagram of the
18o
pilot, showing only the net force. By Newton’s second law, the net force MUST
G
point in the direction of the acceleration, and its magnitude is ma . That net force Fnet

is the sum of ALL forces on the pilot. If we assume that the force of gravity and G
Fseat
ttsheereamtfotsorocffeinvodefcFtGthoseeratsc,=oFGcFGnkentept=i−t msmegGagGt+=oFGnmsetaaGht e=− pmilaGo.t are the only forces on the pilot, then in mgG θ
mgG. Solve this equation for the force of the

A vector diagram of that equation is

shown. Solve for the force of the seat on the pilot using components.

( )Fx seat = Fx net = ma cos18° = (75 kg) 3.8 m s2 cos18° = 271.1 N

Fy seat = mg + Fy net = mg + ma sin18°

( ) ( )= (75 kg) 9.80 m s2 + (75 kg) 3.8 m s2 sin18° = 823.2 N

The magnitude of the cockpit seat force is as follows.

F= F + F =2 2 (271.1N)2 + (823.2 N)2 = 866.7 N ≈ 870 N
x seat y seat

The angle of the cockpit seat force is as follows.

θ = tan−1 Fy seat = tan −1 823.2 N = 72° above the horizontal
Fx seat 271.1 N

78. (a) The helicopter and frame will both have the same acceleration, and so can be G
treated as one object if no information about internal forces (like the cable Flift
tension) is needed. A free-body diagram for the helicopter-frame
combination is shown. Write Newton’s second law for the combination, (mH + mF ) gG
calling UP the positive direction.
G
∑ F = Flift − ( mH + mF ) g = ( mH + mF ) a → FT
mFgG
( )Flift = (mH + mF ) ( g + a) = (7650 kg +1250 kg) 9.80 m s2 + 0.80 m s2

= 9.43×104 N

(b) Now draw a free-body diagram for the frame alone, in order to find the
tension in the cable. Again use Newton’s second law.

∑ F = FT − mF g = mFa →

( )FT = mF ( g + a) = (1250 kg) 9.80 m s2 + 0.80 m s2 = 1.33×104 N

(c) The tension in the cable is the same at both ends, and so the cable exerts a
force of 1.33×104 N downward on the helicopter.

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112

Chapter 4 Dynamics: Newton’s Laws of Motion

79. (a) We assume that the maximum horizontal force occurs when the train is moving very slowly,

and so the air resistance is negligible. Thus the maximum acceleration is given by the
following.

amax = Fmax = 4 ×105 N = 0.625 m s2 ≈ 0.6 m s2
m 6.4 ×105 kg

(b) At top speed, we assume that the train is moving at constant velocity. Therefore the net force
on the train is 0, and so the air resistance and friction forces together must be of the same

magnitude as the horizontal pushing force, which is 1.5 ×105 N .

80. See the free-body diagram for the fish being pulled upward vertically. From Newton’s G
second law, calling the upward direction positive, we have this relationship. FT
mgG
∑ Fy = FT − mg = ma → FT = m ( g + a)

(a) If the fish has a constant speed, then its acceleration is zero, and so FT = mg. Thus

the heaviest fish that could be pulled from the water in this case is 45 N (10 lb) .

(b) If the fish has an acceleration of 2.0 m/s2, and FT is at its maximum of 45 N, then

solve the equation for the mass of the fish.

m= FT = 9.8 m 45 N s2 = 3.8 kg →
g+a s2 + 2.0 m

( )mg = (3.8 kg) 9.8 m s2 = 37 N (≈ 8.4 lb)

(c) It is not possible to land a 15-lb fish using 10-lb line, if you have to lift the fish vertically. If
the fish were reeled in while still in the water, and then a net used to remove the fish from the
water, it might still be caught with the 10-lb line.

81. Choose downward to be positive. The elevator’s acceleration is calculated by Eq. 2-12c. G
FT
v2 − v02 = 2a ( y − y0 ) → a = v2 − v02 = 0 − (3.5 m s)2 = −2.356 m s2 mgG
2 (2.6 m)
2( y − y0 )

See the free-body diagram of the elevator/occupant combination. Write Newton’s second

law for the elevator.

∑ Fy = mg − FT = ma

( )FT = m ( g − a) = (1450 kg) 9.80 m s2 − −2.356 m s2 = 1.76 ×104 N

82. (a) First calculate Karen’s speed from falling. Let the downward direction be positive, and use Eq.

2-12c with v0 = 0 .

( )v2 − v02 = 2a ( y − y0 ) → v = 0 + 2a ( y − y0 ) = 2 9.8 m s2 (2.0 m) = 6.26 m s

Now calculate the average acceleration as the rope stops Karen, again using Eq. 2-12c, with

down as positive. G
Frope
v2 − v02 = 2a ( y − y0 ) → a= v2 − v02 = 0 − (6.26 m s)2 = −19.6 m s2
2(1.0 m)
2( y − y0 )

The negative sign indicates that the acceleration is upward. Since this is her

acceleration, the net force on Karen is given by Newton’s second law, Fnet = ma . mgG
That net force will also be upward. Now consider the free-body diagram of Karen as

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113

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

she decelerates. Call DOWN the positive direction. Newton’s second law says that

Fnet = ma = mg − Frope → Frope = mg − ma. The ratio of this force to Karen’s weight is

Frope = mg − ma = 1.0 − a = 1.0 − −19.6 m s2 = 3.0. Thus the rope pulls upward on Karen
mg g g 9.8 m s2

with an average force of 3.0 times her weight .

(b) A completely analogous calculation for Bill gives the same speed after the 2.0 m fall, but since
he stops over a distance of 0.30 m, his acceleration is –65 m/s2, and the rope pulls upward on
Bill with an average force of 7.7 times his weight . Thus, Bill is more likely to get hurt.

83. Since the climbers are on ice, the frictional force for the G
lower two climbers is negligible. Consider the free- FN1 G
body diagram as shown. Note that all the masses are G Ffr
the same. Write Newton’s second law in the x direction y FN2 G G
for the lowest climber, assuming he is at rest. x FT1
FT1
∑ Fx = FT2 − mg sinθ = 0 G G
G FT2 FT2
FN3
mgG θ

( )FT2 = mg sinθ = (75 kg) 9.80 m s2 sin 31.0° θ mgG θ
θ
= 380 N
mgG
Write Newton’s second law in the x direction for the
middle climber, assuming he is at rest.

∑ Fx = FT1 − FT2 − mg sinθ = 0 → FT1 = FT2 + mg sinθ = 2FT2g sinθ = 760 N

84. Use Newton’s second law.

( )( )F
= ma = m Δv → Δt = mΔv = 1.0 ×1010 kg 2.0 ×10−3 m s = 8.0 ×106s = 93d
Δt F
(2.5 N)

85. Use the free-body diagram to find the net force in the x direction, and then y
x
find the acceleration. Then Eq. 2-12c can be used to find the final speed at G
the bottom of the ramp. FN

∑ Fx = mg sinθ − FP = ma → G
FP
( )a =
mg sinθ − FP = (450 kg) 9.80 m s2 sin 22° − 1420 N θ mθgG
m 450 kg

= 0.516 m s2

( )v2 = v02 + 2a ( x − x0 ) → v = 2a ( x − x0 ) = 2 0.516 m s2 (11.5 m) = 3.4 m s

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114

Chapter 4 Dynamics: Newton’s Laws of Motion

86. (a) We use the free-body diagram to find the force needed to pull the masses at a y
G
constant velocity. We choose the “up the plane” direction as the positive FN-B Gx
G F
direction for both masses. Then they both have the same acceleration FT

even if it is non-zero. G θB
FT mBgG θB
∑mA : Fx = FT − mA g sinθA = mAa = 0 G
FN-A
∑mb : Fx = F − FT − mBg sinθB = mBa = 0 yx

Add the equations to eliminate the tension force and

solve for F. → θA θA
mAgG
( FT − mAg sinθA ) + ( F − FT − mBg sinθB ) = 0

F = g (mA sinθA + mB sinθB )

( )= 9.80 m s2 [(9.5kg) sin 59° + (11.5kg) sin32°] = 1.40 ×102 N

(b) Since θA > θB, if there were no connecting string, mA would have a larger acceleration than

mB. If θA < θB, there would be no tension. But, since there is a connecting string, there will be
G
tension in the string. Use the free-body diagram from above but ignore the applied force F.

∑ ∑mA : Fx = FT − mA g sinθA = mAa ; mb : Fx = −FT − mBg sinθB = mBa

Again add the two equations to eliminate the tension force.

( FT − mAg sinθA ) + (−FT − mBg sinθB ) = mAa + mBa →

mA sinθA + mB sinθB (9.5kg) sin 59° + (11.5kg) sin 32°
mA + mB
( )a 21.0 kg
= −g = − 9.80 m s2

= −6.644 m s2 ≈ 6.64 m s2 , down the planes

(c) Use one of the Newton’s second law expressions from part (b) to find the string tension. It must
be positive if there is a tension.
FT − mA g sinθA = mAa →

( )FT = mA ( g sinθA + a) = (9.5kg) ⎣⎡ 9.80 m s2 (sin 59°) − 6.644 m s2 ⎤⎦ = 17 N

87. (a) If the 2G-block system is taken as a whole system, then the net force on the system is just the
force F, accelerating the total mass. Use Newton’s second law to find the force from the mass

and acceleration. Take the direction of motion caused by the force (left for the bottom block,
right for the top block) as the positive direction. Then both blocks have the same acceleration.

∑ ( ) ( )Fx = F = m + mtop bottom a = (9.0 kg) 2.5 m s2 = 22.5 N ≈ 23 N

(b) The tension in the connecting cord is the only force acting on the top block, and so must be
causing its acceleration. Again use Newton’s second law.

∑ ( )Fx = FT = mtopa = (1.5 kg) 2.5 m s2 = 3.75 N ≈ 3.8 N

This could be checked by using the bottom block.

∑ ( )Fx = F − FT = m abottom → FT = F − m abottom = 22.5 N − (7.5 kg) 2.5 m s2 = 3.75 N

88. (a) For this scenario, find your location at a time of 4.0 sec, using Eq. 2-12b. The acceleration is

found from Newton’s second law.

a= Fforward = 1200 N →
m 750 kg

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115

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

x − x0 = v0t + 1 at 2 = (15 m s) (4.0s) + 1 1200 N ( 4.0 s )2 = 72.8 m > 65 m
2 2 750 kg

Yes , you will make it through the intersection before the light turns red.

(b) For this scenario, find your location when the car has been fully stopped, using Eq. 2-12c. The

acceleration is found from Newton’s second law.

a = Fbraking = 1800 N → v2 = v02 + 2a ( x − x0 ) →
m − 750 kg

x − x0 = v2 − v02 = 0 − (15m s)2 = 46.9 m > 45 m
2a
2 ⎛ − 1800 N ⎞
⎜⎝ 750 kg ⎠⎟

No , you will not stop before entering the intersection.

89. We take the mass of the crate as m until we insert values. A free-body G
diagram is shown.
(a) (i) Use Newton’s second law to find the acceleration. FN y

∑ Fx = mg sinθ = ma → a = g sinθ x

(ii) Use Eq. 2-12b to find the time for a displacement of l. θ mθgG

x − x0 = v0t + 1 at 2 → l = 1 g (sin θ )t2 →
2 2

t= 2l
g sinθ

(iii) Use Eq. 2-12a to find the final velocity.

v = v0 + at = g sin θ ⎡ 2l ⎤ = 2l g sinθ
⎢ ⎥
⎣ g sin θ ⎦

(iv) Use Newton’s second law to find the normal force.

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ

(b) Using the values of m = 1500 kg , g = 9.80 m s2 , and l = 100 m , the requested quantities
become as follows.

a = (9.80sinθ ) m s2 ; t = 2 (100) s ;

9.80 sin θ

v = 2 (100) (9.80) sinθ m s ; FN = (1500) (9.80) cosθ

Graphs of these quantities as a function of θ are given here.

10

Acceleration (m/s2) 8

6

4

2

0

0 15 30 45 60 75 90

Angle (degrees)

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116

Chapter 4 Dynamics: Newton’s Laws of Motion

We consider the limiting cases: at Time to bottom (s) 40
an angle of 0° , the crate does not
move, and so the acceleration and 30 75 90
final velocity would be 0. The 75 90
time to travel 100 m would be 20 75 90
infinite, and the normal force
would be equal to the weight of 10

W = mg 0
0 15 30 45 60
= (1500 kg) (9.80 m )s2
Angle (degrees)
= 1.47 ×104 N.
The graphs are all consistent with 50
those results. 40
30
For an angle of 90° , we would Final Velocity (m/s) 20
expect free-fall motion. The 10
0
acceleration should be 9.80 m s2 .
0 15 30 45 60
The normal force would be 0. The
free-fall time for an object Angle (degrees)
dropped from rest a distance of
100 m and the final velocity after 15000
that distance are calculated below. 12000
9000
x − x0 = v0t + 1 at 2 → 6000
2 3000

l = 1 gt 2 → Normal force (N) 0
2 0 15 30 45 60

t= 2l 2 (100 m) = 4.5 s Angle (degrees)
g=
9.80 m s2

v2 = v02 + 2a ( x − x0 ) →
v = 2g ( x − x0 )

= 2 (9.80 m s2 ) (100 m)

= 44 m s

Yes, the graphs agree with these results for the limiting cases.

The spreadsheet used for this problem can be found on the Media Manager, with filename
“PSE4_ISM_CH04.XLS,” on tab “Problem 4.89b.”

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117

CHAPTER 5: Using Newton’s Laws: Friction, Circular Motion, Drag Forces

Responses to Questions

1. Static friction between the crate and the truck bed causes the crate to accelerate.

2. The kinetic friction force is parallel to the ramp and the block’s weight has a component parallel to
the ramp. The parallel component of the block’s weight is directed down the ramp whether the block
is sliding up or down. However, the frictional force is always in the direction opposite the block’s
motion, so it will be down the ramp while the block is sliding up, but up the ramp while the block is
sliding down. When the block is sliding up the ramp, the two forces acting on it parallel to the ramp
are both acting in the same direction, and the magnitude of the net force is the sum of their
magnitudes. But when the block is sliding down the ramp, the friction and the parallel component of
the weight act in opposite directions, resulting in a smaller magnitude net force. A smaller net force
yields a smaller (magnitude) acceleration.

3. Because the train has a larger mass. If the stopping forces on the truck and train are equal, the

i(nnvegerasteivlye)parcocpeolretriaotnioalntoofmthaesstr(aaGin=wFGillmbe).mTuhcehtsraminalwleirllthtaaknethloant goefrthtoe truck, since acceleration is
stop, as it has a smaller

acceleration, and will travel a greater distance before stopping. The stopping force on the train may

actually be greater than the stopping force on the truck, but not enough greater to compensate for the

much greater mass of the train.

4. Yes. Refer to Table 5-1. The coefficient of static friction between rubber and many solid surfaces is
typically between 1 and 4. The coefficient of static friction can also be greater than one if either of
the surfaces is sticky.

5. When a skier is in motion, a small coefficient of kinetic friction lets the skis move easily on the snow
with minimum effort. A large coefficient of static friction lets the skier rest on a slope without
slipping and keeps the skier from sliding backward when going uphill.

6. When the wheels of a car are rolling without slipping, the force between each tire and the road is
static friction, whereas when the wheels lock, the force is kinetic friction. The coefficient of static
friction is greater than the coefficient of kinetic friction for a set of surfaces, so the force of friction
between the tires and the road will be greater if the tires are rolling. Once the wheels lock, you also
have no steering control over the car. It is better to apply the brakes slowly and use the friction
between the brake mechanism and the wheel to stop the car while maintaining control. If the road is
slick, the coefficients of friction between the road and the tires are reduced, and it is even more
important to apply the brakes slowly to stay in control.

7. (b). If the car comes to a stop without skidding, the force that stops the car is the force of kinetic

friction between the brake mechanism and the wheels. This force is designed to be large. If you slam
on the brakes and skid to a stop, the force that stops the car will be the force of kinetic friction
between the tires and the road. Even with a dry road, this force is likely to be less that the force of
kinetic friction between the brake mechanism and the wheels. The car will come to a stop more
quickly if the tires continue to roll, rather than skid. In addition, once the wheels lock, you have no
steering control over the car.

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118

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

8. The forces in (a), (b), and (d) are all equal to 400 N in magnitude.
(a) You exert a force of 400 N on the car; by Newton’s third law the force exerted by the car on you
also has a magnitude of 400 N.
(b) Since the car doesn’t move, the friction force exerted by the road on the car must equal 400 N,
too. Then, by Newton’s third law, the friction force exerted by the car on the road is also 400 N.
(c) The normal force exerted by the road on you will be equal in magnitude to your weight
(assuming you are standing vertically and have no vertical acceleration). This force is not
required to be 400 N.
(d) The car is exerting a 400 N horizontal force on you, and since you are not accelerating, the
ground must be exerting an equal and opposite horizontal force. Therefore, the magnitude of the
friction force exerted by the road on you is 400 N.

9. On an icy surface, you need to put your foot straight down onto the sidewalk, with no component of
velocity parallel to the surface. If you can do that, the interaction between you and the ice is through
the static frictional force. If your foot has a component of velocity parallel to the surface of the ice,
any resistance to motion will be caused by the kinetic frictional force, which is much smaller. You
will be much more likely to slip.

10. Yes, the centripetal acceleration will be greater when the speed is greater since centripetal
acceleration is proportional to the square of the speed. An object in uniform circular motion has an
acceleration, since the direction of the velocity vector is changing even though the speed is constant.

11. No. The centripetal acceleration depends on 1/r, so a sharp curve, with a smaller radius, will generate
a larger centripetal acceleration than a gentle curve, with a larger radius. (Note that the centripetal
force in this case is provided by the static frictional force between the car and the road.)

12. The three main forces on the child are the downward force of gravity (weight), the normal force up
on the child from the horse, and the static frictional force on the child from the surface of the horse.
The frictional force provides the centripetal acceleration. If there are other forces, such as contact
forces between the child’s hands or legs and the horse, which have a radial component, they will
contribute to the centripetal acceleration.

13. As the child and sled come over the crest of the hill, they are moving in an arc. There must be a
centripetal force, pointing inward toward the center of the arc. The combination of gravity (down)
and the normal force (up) provides this centripetal force, which must be greater than or equal to zero.
(At the top of the arc, Fy = mg – N = mv²/r ≥ 0.) The normal force must therefore be less than the
child’s weight.

14. No. The barrel of the dryer provides a centripetal force on the clothes to keep them moving in a
circular path. A water droplet on the solid surface of the drum will also experience this centripetal
force and move in a circle. However, as soon as the water droplet is at the location of a hole in the
drum there will be no centripetal force on it and it will therefore continue moving in a path in the
direction of its tangential velocity, which will take it out of the drum. There is no centrifugal force
throwing the water outward; there is rather a lack of centripetal force to keep the water moving in a
circular path.

15. When describing a centrifuge experiment, the force acting on the object in the centrifuge should be
specified. Stating the rpm will let you calculate the speed of the object in the centrifuge. However, to
find the force on an object, you will also need the distance from the axis of rotation.

16. She should let go of the string at the moment that the tangential velocity vector is directed exactly at
the target.

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

17. The acceleration of the ball is inward, directly toward the pole, and is provided by the horizontal
component of the tension in the string.

18. For objects (including astronauts) on the inner surface of the cylinder, the normal force provides a
centripetal force which points inward toward the center of the cylinder. This normal force simulates
the normal force we feel when on the surface of Earth.
(a) Falling objects are not in contact with the floor, so when released they will continue to move
with constant velocity until the floor reaches them. From the frame of reference of the astronaut
inside the cylinder, it will appear that the object falls in a curve, rather than straight down.
(b) The magnitude of the normal force on the astronaut’s feet will depend on the radius and speed
of the cylinder. If these are such that v²/r = g (so that mv²/r = mg for all objects), then the
normal force will feel just like it does on the surface of Earth.
(c) Because of the large size of Earth compared to humans, we cannot tell any difference between
the gravitational force at our heads and at our feet. In a rotating space colony, the difference in
the simulated gravity at different distances from the axis of rotation would be significant.

19. At the top of bucket’s arc, the gravitational force and normal forces from the bucket provide the
centripetal force needed to keep the water moving in a circle. (If we ignore the normal forces, mg =

mv²/r, so the bucket must be moving with speed v ≥ gr or the water will spill out of the bucket.)

At the top of the arc, the water has a horizontal velocity. As the bucket passes the top of the arc, the
velocity of the water develops a vertical component. But the bucket is traveling with the water, with
the same velocity, and contains the water as it falls through the rest of its path.

20. (a) The normal force on the car is largest at point C. In this case, the centripetal force keeping the
car in a circular path of radius R is directed upward, so the normal force must be greater than the
weight to provide this net upward force.

(b) The normal force is smallest at point A, the crest of the hill. At this point the centripetal force
must be downward (towards the center of the circle) so the normal force must be less than the
weight. (Notice that the normal force is equal to the weight at point B.)

(c) The driver will feel heaviest where the normal force is greatest, or at point C.
(d) The driver will feel lightest at point A, where the normal force is the least.
(e) At point A, the centripetal force is weight minus normal force, or mg – N = mv2/r. The point at

which the car just loses contact with the road corresponds to a normal force of zero. Setting
N = 0 gives mg = mv2/r or v = gr.

21. Leaning in when rounding a curve on a bicycle puts the bicycle tire at an angle with respect to the
ground. This increases the component of the (static) frictional force on the tire due to the road. This
force component points inward toward the center of the curve, thereby increasing the centripetal
force on the bicycle and making it easier to turn.

22. When an airplane is in level flight, the downward force of gravity is counteracted by the upward lift
force, analogous to the upward normal force on a car driving on a level road. The lift on an airplane
is perpendicular to the plane of the airplane’s wings, so when the airplane banks, the lift vector has
both vertical and horizontal components (similar to the vertical and horizontal components of the
normal force on a car on a banked turn). The vertical component of the lift balances the weight and
the horizontal component of the lift provides the centripetal force. If L = the total lift and φ = the
banking angle, measured from the vertical, then L cosϕ = mg and L sinϕ = mv2 r so

( )ϕ = tan−1 v2 gr .

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120

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

23. If we solve for b, we have b = –F/v. The units for b are N·s/m = kg·m·s/(m·s²) = kg/s.
24. The force proportional to v² will dominate at high speed.

Solutions to Problems

1. A free-body diagram for the crate is shown. The crate does not accelerate G
vertically, and so FN = mg. The crate does not accelerate horizontally, and FN G
so FP = Ffr . G FP
Ffr
( )FP = Ffr = μk FN = μk mg = (0.30) ( 22 kg ) 9.80 m s2 = 65 N
mgG

If the coefficient of kinetic friction is zero, then the horizontal force required

is 0 N , since there is no friction to counteract. Of course, it would take a force to START the crate

moving, but once it was moving, no further horizontal force would be necessary to maintain the
motion.

2. A free-body diagram for the box is shown. Since the box does not accelerate G
FN G
vertically, FN = mg. G FP
(a) To start the box moving, the pulling force must just overcome the Ffr

force of static friction, and that means the force of static friction will mgG

reach its maximum value of Ffr = μsFN. Thus we have for the starting

motion,

∑ Fx = FP − Ffr = 0 →

FP 35.0 N = 0.60
mg
kg) 9.80 m
( )FP = Ffr = μsFN = μsmg → μs = = s2
(6.0

(b) The same force diagram applies, but now the friction is kinetic friction, and the pulling force is
NOT equal to the frictional force, since the box is accelerating to the right.

∑ F = FP − Ffr = ma → FP − μk FN = ma → FP − μk mg = ma →

( ( ) )μk s2
= FP − ma = 35.0 N − (6.0 kg) 0.60 m = 0.53
mg (6.0 kg)
9.80 m s2

3. A free-body diagram for you as you stand on the train is shown. You do not

accelerate vertically, and so FN = mg. The maximum static frictional force is μsFN , G
Ffr
and that must be greater than or equal to the force needed to accelerate you in order

for you not to slip. mgG G
Ffr ≥ ma → μs FN ≥ ma → μsmg ≥ ma → μs ≥ a g = 0.20g g = 0.20 FN

The static coefficient of friction must be at least 0.20 for you to not slide.

4. See the included free-body diagram. To find the maximum angle, assume G y x
that the car is just ready to slide, so that the force of static friction is a FN θ
maximum. Write Newton’s second law for both directions. Note that for G
both directions, the net force must be zero since the car is not accelerating. Ffr

∑F = FN − mg cosθ = 0 → FN = mg cosθ θmgG
y

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

∑F = mg sinθ − Ffr = 0 → mg sinθ = Ffr = μsFN = μsmg cosθ
x

μs = mg sinθ = tanθ = 0.90 → θ = tan−1 0.90° = 42°
mg cosθ

5. A free-body diagram for the accelerating car is shown. The car does not G
accelerate vertically, and so FN = mg. The static frictional force is the FN G
accelerating force, and so Ffr = ma. If we assume the maximum acceleration,
then we need the maximum force, and so the static frictional force would be its Ffr
maximum value of μsFN. Thus we have mgG
Ffr = ma → μsFN = ma → μsmg = ma →

( )a = μs g = 0.90 9.80 m s2 = 8.8 m s2

6. (a) Here is a free-body diagram for the box at rest on the plane. The GG y
force of friction is a STATIC frictional force, since the box is at rest. Ffr FN
x
(b) If the box were sliding down the plane, the only change is that θ mgG
the force of friction would be a KINETIC frictional force. θ

(c) If the box were sliding up the plane, the force of friction would
be a KINETIC frictional force, and it would point down the
plane, in the opposite direction to that shown in the diagram.

Notice that the angle is not used in this solution.

7. Start with a free-body diagram. Write Newton’s second law for each G G
direction. Ffr FN y

∑ Fx = mg sinθ − Ffr = max

∑ Fy = FN − mg cosθ = may = 0 x
θ mgG θ
Notice that the sum in the y direction is 0, since there is no motion
(and hence no acceleration) in the y direction. Solve for the force of
friction.

mg sin θ − Ffr = max →

( )Ffr = mg sinθ − max = ( 25.0 kg) ⎡⎣ 9.80 m s2 (sin 27°) − 0.30 m s2 ⎤⎦ = 103.7 N ≈ 1.0 ×102 N

Now solve for the coefficient of kinetic friction. Note that the expression for the normal force comes

from the y direction force equation above.

Ffr 103.7 N
mg cosθ 9.80 m s2
( )Ffr = μk FN = μk mg cosθ → μk = = (cos 27°) = 0.48
(25.0 kg)

8. The direction of travel for the car is to the right, and that is also the positive G
FN
horizontal direction. Using the free-body diagram, write Newton’s second law in G mgG
the x direction for the car on the level road. We assume that the car is just on the Ffr

verge of skidding, so that the magnitude of the friction force is Ffr = μsFN.

∑ Fx = −Ffr = ma Ffr = −ma = −μsmg → μs = a = 3.80 m s2 = 0.3878
g 9.80 m s2

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122

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

Now put the car on an inclined plane. Newton’s second law in the x-direction Gy x
for the car on the plane is used to find the acceleration. We again assume FN
the car is on the verge of slipping, so the static frictional force is at its
maximum. G
Ffr
∑ Fx = −Ffr − mg sinθ = ma →

a = − Ffr − mg sinθ = −μsmg cosθ − mg sinθ = −g (μs cosθ + sinθ ) θ
m m θ mgG

( )= − 9.80 m s2 (0.3878 cos 9.3° + sin 9.3°) = −5.3 m s2

9. Since the skier is moving at a constant speed, the net force on the skier must G G
be 0. See the free-body diagram, and write Newton’s second law for both Ffr FN
the x and y directions.
mg sinθ = Ffr = μsFN = μsmg cosθ → y

μs = tanθ = tan 27° = 0.51 x

θ mgG θ

10. A free-body diagram for the bar of soap is shown. There is no motion in G
the y direction and thus no acceleration in the y direction. Write Newton’s G FN
second law for both directions, and use those expressions to find the Ffr
acceleration of the soap. y

∑ Fx = FN − mg cosθ = 0 → FN = mg cosθ x

∑ Fx = mg sinθ − Ffr = ma θθ
mgG
ma = mg sinθ − μk FN = mg sinθ − μkmg cosθ

a = g (sinθ − μk cosθ )

Now use Eq. 2-12b, with an initial velocity of 0, to find the final velocity.

x = x0 + v0t + 1 at 2 →
2

2x 2x 2(9.0 m)
a= (sin 8.0° − (0.060) cos8.0°)
g (sinθ − μk cosθ ) =
( )t = s2 = 4.8 s
9.80 m

11. A free-body diagram for the box is shown, assuming that it is moving to the G G
right. The “push” is not shown on the free-body diagram because as soon as the Ffr FN
box moves away from the source of the pushing force, the push is no longer
mgG
applied to the box. It is apparent from the diagram that FN = mg for the vertical

direction. We write Newton’s second law for the horizontal direction, with

positive to the right, to find the acceleration of the box.

∑ Fx = −Ffr = ma → ma = −μk FN = −μkmg →

( )a = −μk g = −0.15 9.80 m s2 = −1.47 m s2

Eq. 2-12c can be used to find the distance that the box moves before stopping. The initial speed is
4.0 m/s, and the final speed will be 0.

v2 − v02 0 − (3.5 m s)2
2a 2
−1.47 m s2
( )v2 − v02 = 2a ( x − x0 ) → x − x0 = = = 4.17 m ≈ 4.2 m

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

12. (a) A free-body diagram for the car is shown, assuming that it is moving to the G G
right. It is apparent from the diagram that FN = mg for the vertical direction. Ffr FN
Write Newton’s second law for the horizontal direction, with positive to the
right, to find the acceleration of the car. Since the car is assumed to NOT be mgG
sliding, use the maximum force of static friction.

∑ Fx = −Ffr = ma → ma = −μs FN = −μsmg → a = −μs g

Eq. 2-12c can be used to find the distance that the car moves before stopping. The initial speed

is given as v, and the final speed will be 0.

v2 − v02 = 2a ( x − x0 ) → (x − x0 ) = v2 − v02 = 0 − v2 = v2
2a 2μs g
2 (−μs g )

(b) Using the given values:

⎛ 1m s ⎞ v2 (26.38 m s)2
⎝⎜ 3.6 km h ⎟⎠ 2μs g 2 (0.65) 9.80 m
v = (95 km h ) = 26.38 m s ( )(x− x0 ) = = s2 = 55 m

(c) From part (a), we see that the distance is inversely proportional to g, and so if g is reduced by a

factor of 6, the distance is increased by a factor of 6 to 330 m .

13. We draw three free-body diagrams – one for the car, one for the trailer, and G G
FCT FNC G
then “add” them for the combination of car and trailer. Note that since the G
Ffr FCG
car pushes against the ground, the groGund will push against the car with an
equal but oppositely directed force. FCG is the force on the car due to the mC gG
G G G
ground, FTC is the force on the trailer due to the car, and FCT is the force on FTC
G G
the car due to the trailer. Note that by Newton’s rhird law, FCT = FTC .

From consideration of the vertical forces in the individual free-body G mTgG
diagrams, it is apparent that the normal force on each object is equal to its FNT

weight. This leads to the conclusion that Ffr = μk FN T = μkmT g =

( )(0.15) (350 kg) 9.80 m s2 = 514.5 N . G
FCG
Now consider the combined free-body diagram. Write G
Newton’s second law for the horizontal direction, This allows Ffr (mC + mT ) gG
the calculation of the acceleration of the system.
GG
∑ F = FCG − Ffr = (mC + mT ) a → FNT + FNC

a = FCG − Ffr = 3600 N − 514.5 N = 1.893 m s2
mC + mT 1630 kg

Finally, consider the free-body diagram for the trailer alone. Again write Newton’s second law for

the horizontal direction, and solve for FTC.

∑ F = FTC − Ffr = mTa →

( )FTC = Ffr + mTa = 514.5 N + (350 kg) 1.893 m s2 = 1177 N ≈ 1200 N G
G FN
14. Assume that kinetic friction is the net force causing the deceleration. See the Ffr
free-body diagram for the car, assuming that the right is the positive direction,
and the direction of motion of the skidding car. There is no acceleration in the mgG
vertical direction, and so FN = mg . Applying Newton’s second law to the x

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124

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

direction gives the following.

∑ F = −Ff = ma → − μk FN = −μk mg = ma → a = −μk g

Use Eq. 2-12c to determine the initial speed of the car, with the final speed of the car being zero.

v2 − v02 = 2a ( x − x0 ) →

( )v0 = v2 − 2a ( x − x0 ) = 0 − 2 (−μk g ) ( x − x0 ) = 2 (0.80) 9.80 m s2 (72 m) = 34 m s

15. (a) Consider the free-body diagram for the snow on the roof. If the snow G
is just ready to slip, then the static frictional force is at its maximum G FN
Ffr
value, Ffr = μsFN. Write Newton’s second law in both directions, y

with the net force equal to zero since the snow is not accelerating. x

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ θθ
mgG
∑ Fx = mg sinθ − Ffr = 0 →

mg sinθ = Ffr = μsFN = μsmg cosθ → μs = tanθ = tan 34° = 0.67

If μs > 0.67 , then the snow would not be on the verge of slipping.
(b) The same free-body diagram applies for the sliding snow. But now the force of friction is

kinetic, so Ffr = μk FN , and the net force in the x direction is not zero. Write Newton’s second

law for the x direction again, and solve for the acceleration.

∑ Fx = mg sinθ − Ffr = ma

a= mg sinθ − Ffr = mg sinθ − μk mg cosθ = g (sinθ − μk cosθ )
m m

Use Eq. 2-12c with vi = 0 to find the speed at the end of the roof.

v2 − v02 = 2a ( x − x0 )

v = v0 + 2a ( x − x0 ) = 2g (sinθ − μk cosθ ) ( x − x0 )

( )= 2 9.80 m s2 (sin 34° − (0.20) cos 34°) (6.0 m) = 6.802 m s ≈ 6.8 m s

(c) Now the problem becomes a projectile motion problem. The projectile 34o
has an initial speed of 6.802 m/s, directed at an angle of 34o below the
horizontal. The horizontal component of the speed, (6.802 m/s) cos 34o

= 5.64 m/s, will stay constant. The vertical component will change due

to gravity. Define the positive direction to be downward. Then the
starting vertical velocity is (6.802 m/s) sin 34o =3.804 m/s, the vertical acceleration is 9.80 m/s2,

and the vertical displacement is 10.0 m. Use Eq. 2-12c to find the final vertical speed.

2( )v 2
y − v y0 y = 2a y − y0

( )vy =
v2 + 2a ( y − y0 ) = (3.804 m s)2 + 2 9.80 m s2 (10.0 m) = 14.5 m/s
y0

To find the speed when it hits the ground, the horizontal and vertical components of velocity

must again be combined, according to the Pythagorean theorem.

v= vx2 + v 2 = (5.64 m s)2 + (14.5 m/s)2 = 15.6 m s ≈ 16 m s
y

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125

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

16. Consider a free-body diagram for the box, showing force on the box. When G
G Ffr
FP = 23 N, the block does not move. Thus in that case, the force of friction FP G
FN
is static friction, and must be at its maximum value, given by Ffr = μsFN. 28ο mgG

Write Newton’s second law in both the x and y directions. The net force in
each case must be 0, since the block is at rest.

∑ Fx = FP cosθ − FN = 0 → FN = FP cosθ

∑ Fy = Ffr + FP sinθ − mg = 0 → Ffr + FP sinθ = mg

μs FN + FP sinθ = mg → μs FP cosθ + FP sinθ = mg

FP 23 N
g 9.80 m
( )m
= (μs cosθ + sin θ ) = s2 0.40 cos 28o + sin 28o = 1.9 kg

17. (a) Since the two blocks are in contact, they can be treated as a G
single object as long as no information is needed about internal FP m1 +
forces (like the force of one block pushing on the other block). G m2
Since there is no motion in the vertical direction, it is apparent that Ffr G (m1 + m2 ) gG
FN
FN = (m1 + m2 ) g, and so Ffr = μk FN = μk (m1 + m2 ) g. Write

Newton’s second law for the horizontal direction.

∑ ( )Fx = FP − Ffr = m1 + m2 a →

( )a s2
= FP − Ffr = FP − μk (m1 + m2 ) g 650 N − (0.18) (190 kg) 9.80 m
m1 + m2
m1 + m2 = 190 kg

= 1.657 m s2 ≈ 1.7 m s2 G
F21
(b) To solve for the contact forces between the blocks, an individual block
mGust be analyzed. Look at the free-body diagram for the second block. G m2
F21 is the force of the first block pushing on the second block. Again, it Ffr2 G
FN2 m2gG
is apparent that FN2 = m2 g and so Ffr2 = μk FN2 = μkm2g. Write Newton’s
second law for the horizontal direction.

∑ Fx = F21 − Ffr2 = m2a →

( ) ( )F21 = μkm2g + m2a = (0.18) (125 kg) 9.80 m s2 + (125 kg) 1.657 m s2 = 430 N

By Newton’s third law, there will also be a 430 N force to the left on block # 1 due to block # 2.
G
(c) If the crates are reversed, the acceleration of the system will remain F12
the same – the analysis from part (a) still applies. We can also repeat the
G m1
analysis from part (b) to find the force of one block on the other, if we Ffr1 G m1gG
simply change m1 to m2 in the free-body diagram and the resulting FN1
equations.

∑a = 1.7 m s2 ; Fx = F12 − Ffr1 = m1a →

( ) ( )F12 = μkm1g + m1a = (0.18) (65 kg) 9.80 m s2 + (65 kg) 1.657 m s2 = 220 N

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126

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

18. (a) Consider the free-body diagram for the crate on the surface. There is G y
no motion in the y direction and thus no acceleration in the y direction. G FN
Write Newton’s second law for both directions. Ffr

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ

∑ Fx = mg sinθ − Ffr = ma θθ x
mgG
ma = mg sinθ − μk FN = mg sinθ − μkmg cosθ

a = g (sinθ − μk cosθ )

( )= 9.80 m s2 (sin 25.0° − 0.19 cos 25.0°) = 2.454 m s2 ≈ 2.5 m s2

(b) Now use Eq. 2-12c, with an initial velocity of 0, to find the final velocity.

( )v2 − v02 = 2a ( x − x0 ) → v = 2a ( x − x0 ) = 2 2.454 m s2 (8.15m) = 6.3m s

19. (a) Consider the free-body diagram for the crate on the surface. There is G y
no motion in the y direction and thus no acceleration in the y direction. FN
Write Newton’s second law for both directions, and find the
acceleration. G x
Ffr
∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ θ mgθG

∑ Fx = mg sinθ + Ffr = ma

ma = mg sinθ + μk FN = mg sinθ + μkmg cosθ

a = g (sinθ + μk cosθ )

Now use Eq. 2-12c, with an initial velocity of −3.0 m s and a final velocity of 0 to find the

distance the crate travels up the plane.

v2 − v02 = 2a ( x − x0 ) →

( )x−x0= −v02 = 2 s2 − (−3.0 m s)2 = −0.796 m
2a 9.80 m (sin 25.0° + 0.17 cos 25.0°)

The crate travels 0.80 m up the plane.

(b) We use the acceleration found above with the initial velocity in Eq. 2-12a to find the time for
the crate to travel up the plane.

( )v = v0 + at → tup = − v0 =− (−3.0 m s) = 0.5308 s
aup 9.80 m s2 (sin 25.0° + 0.17 cos 25.0°)

The total time is NOT just twice the time to travel up the plane, because G y
the acceleration of the block is different for the two parts of the motion. G FN
The second free-body diagram applies to the block sliding down the Ffr
plane. A similar analysis will give the acceleration, and then Eq. 2-12b
with an initial velocity of 0 is used to find the time to move down the θ mgθG x
plane.

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ

∑ Fx = mg sinθ − Ffr = ma

ma = mg sinθ − μk FN = mg sinθ − μkmg cosθ

a = g (sinθ − μk cosθ )

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127

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

x − x0 = v0t + 1 at 2 →
2

( )2 x − x0 2(0.796 m) = 0.7778s
(sin 25.0° − 0.17 cos 25.0°)
adown
( )tdown = =
9.80 m s2

t = tup + tdown = 0.5308s + 0.7778s = 1.3s

It is worth noting that the final speed is about 2.0 m/s, significantly less than the 3.0 m/s original
speed.

20. Since the upper block has a higher coefficient of friction, that y GG
block will “drag behind” the lower block. Thus there will be FG NB FfrB
tension in the cord, and the blocks will have the same x FT
acceleration. From the free-body diagrams for each block, we
write Newton’s second law for both the x and y directions for G G θ
each block, and then combine those equations to find the FNA FT
acceleration and tension.
(a) Block A: G mBgG
FfrA
∑ FyA = FNA − mA g cosθ = 0 → FNA = mA g cosθ
θ
θ mAgG

∑ FxA = mA g sinθ − FfrA − FT = mAa

mAa = mA g sinθ − μAFNA − FT = mA g sinθ − μAmA g cosθ − FT
Block B:

∑ FyB = FNB − mBg cosθ = 0 → FNB = mBg cosθ

∑ FxB = mA g sinθ − FfrA + FT = mBa

mBa = mBg sinθ − μBFNB + FT = mBg sinθ − μBmBg cosθ + FT
Add the final equations together from both analyses and solve for the acceleration.

mAa = mA g sinθ − μAmA g cosθ − FT ; mBa = mBg sinθ − μBmBg cosθ + FT

mAa + mBa = mA g sinθ − μAmA g cosθ − FT + mBg sinθ − μBmBg cosθ + FT →

a = g ⎡ mA ( sin θ − μA cosθ ) + mB (sinθ − μB cosθ )⎤
⎢ ( mA + mB )
⎣ ⎥
⎦

( )= ⎡ (5.0 kg) (sin 32° − 0.20cos 32°) + (5.0 kg) (sin 32° − 0.30cos 32°) ⎤
9.80 m s2 ⎢ ⎥
⎣ (10.0 kg) ⎦

= 3.1155 m s2 ≈ 3.1m s2

(b) Solve one of the equations for the tension force.
mAa = mA g sinθ − μAmA g cosθ − FT →

FT = mA ( g sinθ − μA g cosθ − a)

( )= (5.0 kg) ⎣⎡ 9.80 m s2 (sin 32° − 0.20 cos 32°) − 3.1155 m s2 ⎤⎦ = 2.1N

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128

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

21. (a) If μA < μB, the untethered acceleration of mA would be greater than that of mB. If there were
no cord connecting the masses, mA would “run away” from mB. So if they are joined together,

mA would be restrained by the tension in the cord, mB would be pulled forward by the tension
in the cord, and the two masses would have the same acceleration. This is exactly the situation
for Problem 20.

(b) If μA > μB, the untethered acceleration of mA would be less than that of mB. So even if there

is a cord between them, mB will move ever closer to mA , and there will be no tension in the

cord. If the incline were long enough, eventually mB would catch up to mA and begin to push
it down the plane.
(c) For μA < μB , the analysis will be exactly like Problem 20. Refer to that free-body diagram and
analysis. The acceleration and tension are as follows, taken from the Problem 20 analysis.

a = g ⎡ mA ( sin θ − μA cosθ ) + mB (sinθ − μB cosθ ) ⎤
⎢ ( mA + mB ) ⎥
⎣ ⎦

mAa = mA g sinθ − μAmA g cosθ − FT →

FT = mA g sinθ − μAmA g cosθ − mAa

= mA g sin θ − μAmA g cosθ − mA g ⎡ mA ( sin θ − μA cosθ ) + mB (sinθ − μB cosθ )⎤
⎢ ( mA + mB )
⎣ ⎥
⎦

= mA mB g cosθ ( μB − μA )

(mA + mB )

For μA > μB, we can follow the analysis of Problem 20 but not include the tension forces.

Each block will have its own acceleration. Refer to the free-body diagram for Problem 20.
Block A:

∑ FyA = FNA − mA g cosθ = 0 → FNA = mA g cosθ

∑ FxA = mA g sinθ − FfrA = mAaA

mAaA = mA g sinθ − μAFNA = mA g sinθ − μAmA g cosθ →

aA = g (sinθ − μA cosθ )

Block B:

∑ FyB = FNB − mBg cosθ = 0 → FNB = mBg cosθ

∑ FxB = mA g sinθ − FfrA = mBaB

mBaB = mBg sinθ − μBFNB = mBg sinθ − μBmBg cosθ →

aB = g (sinθ − μB cosθ )

Note that since μA > μB, aA > aB as mentioned above. And FT = 0 .

22. The force of static friction is what decelerates the crate if it is not sliding on the G G
truck bed. If the crate is not to slide, but the maximum deceleration is desired, Ffr FN

then the maximum static frictional force must be exerted, and so Ffr = μsFN. mgG

The direction of travel is to the right. It is apparent that FN = mg since there is

no acceleration in the y direction. Write Newton’s second law for the truck in

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

the horizontal direction.

∑ ( )Fx = −Ffr = ma → − μsmg = ma → a = −μs g = − (0.75) 9.80 m s2 = −7.4 m s2

The negative sign indicates the direction of the acceleration – opposite to the direction of motion.

23. (a) For mB to not move, the tension must be equal to mBg, and so mBg = FT. For mA to not

move, the tension must be equal to the force of static friction, and so FS = FT. Note that the

normal force on mA is equal to its weight. Use these relationships to solve for mA.

mBg = FT = Fs ≤ μsmA g → mA ≥ mB = 2.0 kg = 5.0 kg → mA ≥ 5.0 kg
μs 0.40

(b) For mB to move with constant velocity, the tension must be equal to mBg . For mA to move
with constant velocity, the tension must be equal to the force of kinetic friction. Note that the

normal force on mA is equal to its weight. Use these relationships to solve for mA .

mBg = Fk = μkmA g → mA = mB = 2.0 kg = 6.7 kg
μk 0.30

24. We define f to be the fraction of the cord that G y G
is handing down, between mB and the pulley. G FN xG FT
Ffr
Thus the mass of that piece of cord is fmC. FT x

We assume that the system is moving to the mBgG + fmCgG

right as well. We take the tension in the cord mAgG (1 − f ) mCgG
to be FT at the pulley. We treat the hanging
mass and hanging fraction of the cord as one

mass, and the sliding mass and horizontal part

of the cord as another mass. See the free-body

diagrams. We write Newton’s second law for each object.

∑ FyA = FN − (mA + (1 − f ) mC ) g = 0

∑ ( )( )FxA = FT − Ffr = FT − μk FN = mA + 1 − f mC a

∑ FxB = (mB + fmC ) g − FT = (mB + fmC ) a

Combine the relationships to solve for the acceleration. In particular, add the two equations for the
x-direction, and then substitute the normal force.

a= ⎡ mB + fmC − μk (mA + (1 − f ) mC ) ⎤ g
⎢ ⎥
⎣ mA + mB + mC ⎦

25. (a) Consider the free-body diagram for the block on the surface. There is G y
no motion in the y direction and thus no acceleration in the y direction. FN
Write Newton’s second law for both directions, and find the
acceleration. G x
Ffr
∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ θ mgθG

∑ Fx = mg sinθ + Ffr = ma

ma = mg sinθ + μk FN = mg sinθ + μkmg cosθ

a = g (sinθ + μk cosθ )

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130

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

Now use Eq. 2-12c, with an initial velocity of v0 , a final velocity of 0, and a displacement of
−d to find the coefficient of kinetic friction.

v2 − v02 = 2a ( x − x0 ) → 0 − v02 = 2g (sinθ + μk cosθ ) (−d ) →

μk = v02 − tanθ
2gd cosθ

(b) Now consider the free-body diagram for the block at the top of its G y
motion. We use a similar force analysis, but now the magnitude of the G FN
friction force is given by Ffr ≤ μsFN , and the acceleration is 0. Ffr

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ

∑ Fx = mg sinθ − Ffr = ma = 0 → Ffr = mg sinθ θ mgθG x

Ffr ≤ μsFN → mg sinθ ≤ μsmg cosθ → μs ≥ tanθ

26. First consider the free-body diagram for the snowboarder on the incline. G
Write Newton’s second law for both directions, and find the acceleration. FN

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ G y
Ffr
∑ Fx = mg sinθ − Ffr = ma
x
ma = mg sinθ − μk1FN = mg sinθ − μk1mg cosθ
θ mgG θ
( )aslope = g (sinθ − μk1 cosθ ) = 9.80 m s2 (sin 28° − 0.18 cos 28°)

= 3.043 m s2 ≈ 3.0 m s2

Now consider the free-body diagram for the snowboarder on the flat surface. G
Again use Newton’s second law to find the acceleration. Note that the normal FN
force and the frictional force are different in this part of the problem, even
though the same symbol is used. G
Ffr
∑ ∑Fy = FN − mg = 0 → FN = mg Fx = −Ffr = ma
mgG
maflat = −Ffr = −μk 2FN = −μk1mg →

( )aflat = −μk2g = − (0.15) 9.80 m s2 = −1.47 m s2 ≈ −1.5 m s2

Use Eq. 2-12c to find the speed at the bottom of the slope. This is the speed at the start of the flat
section. Eq. 2-12c can be used again to find the distance x.

v2 − v02 = 2a ( x − x0 ) →

( )vend of = ( )v02 + 2aslope x − x0 = (5.0 m s)2 + 2 3.043 m s2 (110 m) = 26.35 m s
slope

v2 − v02 = 2a ( x − x0 ) →

v2 − v02 0 − (26.35 m s)2
2aflat
2 −1.47 m s2
( )( x − x0 ) = = = 236 m ≈ 240 m

27. The belt is sliding underneath the box (to the right), so there will be a force of G G
kinetic friction on the box, until the box reaches a speed of 1.5 m/s. Use the free- FN Ffr
body diagram to calculate the acceleration of the box.
mgG
∑(a) Fx = Ffr = ma = μkFN = μkmg → a = μk g

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131

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

∑ Fx = Ffr = ma = μkFN = μkmg → a = μk g

t v − v0 v−0 1.5 m s = 0.22 s
a μk g
(0.70) 9.80 m
( )v = v0 + at → = = = s2

x x0 v2 − v02 v2 (1.5m s)2 = 0.16 m
2a 2μk g
2(0.70) 9.80 m
( )(b)− = = = s2

∑28. We define the positive x direction to be the direction y G G
FT FNB
of motion for each block. See the free-body diagrams. G G x G y
Write Newton’s second law in both dimensions for FNA FT FfrB
both objects. Add the two x-equations to find the
acceleration. G x
Block A: FfrA

FyA = FNA − mA g cosθA = 0 → FNA = mA g cosθA θA θA θB mBθgGB
mAgG

∑ FxA = FT − mA g sinθ − FfrA = mAa

Block B:

∑ FyB = FNB − mBg cosθB = 0 → FNB = mBg cosθB

∑ FxB = mBg sinθ − FfrB − FT = mBa

Add the final equations together from both analyses and solve for the acceleration, noting that in
both cases the friction force is found as Ffr = μFN.

mAa = FT − mA g sinθA − μAmA g cosθA ; mBa = mBg sinθB − μBmBg cosθB − FT

mAa + mBa = FT − mA g sinθA − μAmA g cosθA + mBg sinθB − μBmBg cosθB − FT →

a = g ⎡ −mA ( sin θ A + μA cosθA ) + mB ( sin θ − μB cosθ ) ⎤
⎢ (mA + ⎥
⎣ mB ) ⎦

( )= ⎡ − (2.0 kg) (sin 51° + 0.30 cos 51°) + (5.0 kg) (sin 21° − 0.30 cos 21°) ⎤
9.80 m s2 ⎢⎣ (7.0 kg) ⎦⎥

= −2.2 m s2

29. We assume that the child starts from rest at the top of the slide, and then slides G G
Ffr FN
a distance x − x0 along the slide. A force diagram is shown for the child on
y
the slide. First, ignore the frictional force and so consider the no-friction case.

All of the motion is in the x direction, so we will only consider Newton’s x
second law for the x direction.
θ mθgG
∑ Fx = mg sinθ = ma → a = g sinθ

Use Eq. 2-12c to calculate the speed at the bottom of the slide.

( ) ( ) ( )v2 − v02 = 2a x − x0 → v =(No friction) v02 + 2a x − x0 = 2g sinθ x − x0

Now include kinetic friction. We must consider Newton’s second law in both the x and y directions
now. The net force in the y direction must be 0 since there is no acceleration in the y direction.

∑ Fy = FN − mg cosθ = 0 → FN = mg cosθ

∑ Fx = ma = mg sinθ − Ffr = mg sinθ − μk FN = mg sinθ − μk mg cosθ

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132

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

a= mg sinθ − μk mg cosθ = g (sinθ − μk cosθ )
m

With this acceleration, we can again use Eq. 2-12c to find the speed after sliding a certain distance.

v2 − v02 = 2a ( x − x0 ) → v =(friction) v02 + 2a ( x − x0 ) = 2g (sinθ − μk cosθ ) ( x − x0 )

Now let the speed with friction be half the speed without friction, and solve for the coefficient of
friction. Square the resulting equation and divide by g cosθ to get the result.

( ) ( ) ( ) ( )v = v →(friction)
1 2g sinθ − μk cosθ x − x0 = 1 2g sinθ x − x0
2
2 ( No friction)

2g ( sin θ − μk cosθ )(x − x0 ) = 1 2g ( sin θ )(x − x0 )
4

μk = 3 tan θ = 3 tan 34° = 0.51
4 4

30. (a) Given that mB is moving down, mA must be moving G G
up the incline, and so the force of kinetic friction on G FT FN
mA will be directed down the incline. Since the FT
blocks are tied together, they will both have the G xA yA
same acceleration, and so ayB = axA = a. Write mBGg Ffr

Newton’s second law for each mass. yB m gGB θ
mAgG θ
∑ FyB = mBg − FT = mBa → FT = mBg − mBa

∑ FxA = FT − mA g sinθ − Ffr = mAa

∑ FyA = FN − mA g cosθ = 0 → FN = mA g cosθ

Take the information from the two y equations and substitute into the x equation to solve for the

acceleration.

mBg − mBa − mA g sinθ − μkmA g cosθ = mAa →

a = mBg − mA g sinθ − mA gμk g cosθ = 1 g (1 − sin θ − μk g cosθ )
2
(mA + mB )

( )= (1 − sin 34° − 0.15cos 34°) = 1.6 m s2
1 9.80 m s2
2

(b) To have an acceleration of zero, the expression for the acceleration must be zero.

a = 1 g (1 − sin θ − μk cosθ ) = 0 → 1 − sinθ − μk cosθ = 0 →
2

μk = 1 − sinθ = 1 − sin 34° = 0.53
cosθ cos 34°

31. Draw a free-body diagraGm for each block. G G G
G FNA −Ffr AB FNB −FNA
FT G
Ffr AB G G
mAgG GFT F
Block A (top) Ffr B mBgG

Block B (bottom)

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133

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

G is the force of friction between the two blocks, G is the normal force of contact between the
Ffr AB G FNA G

two blocks, Ffr B is the force of friction between the bottom block and the floor, and FNB is the

normal force of contact between the bottom block and the floor.

Neither block is accelerating vertically, and so the net vertical force on each block is zero.

top: FNA − mA g = 0 → FNA = mA g

bottom: FNB − FNA − mBg = 0 → ( )FNB = FNA + mBg = mA + mB g

Take the positive horizontal direction to be the direction of motion of each block. Thus for the

bottom block, positive is to the right, and for the top block, positive is to the left. Then, since the
blocks are constrained to move together by the connecting string, both blocks will have the same
acceleration. Write Newton’s second law for the horizontal direction for each block.

top: FT − Ffr AB = mAa bottom: F − FT − Ffr AB − Ffr B = mBa
(a) If the two blocks are just to move, then the force of static friction will be at its maximum, and so

the frictions forces are as follows.

( )Ffr AB = μs FNA = μsmA g ; Ffr B = μs FNB = μs mA + mB g

Substitute into Newton’s second law for the horizontal direction with a = 0 and solve for F .

top: FT − μsmA g = 0 → FT = μsmA g

bottom: F − FT − μsmAg − μs (mA + mB ) g = 0 →

F = FT + μsmA g + μs (mA + mB ) g = μsmA g + μsmA g + μs (mA + mB ) g

( )= μs (3mA + mB ) g = (0.60) (14 kg) 9.80 m s2 = 82.32 N ≈ 82 N

(b) Multiply the force by 1.1 so that F = 1.1(82.32 N) = 90.55 N. Again use Newton’s second law

for the horizontal direction, but with a ≠ 0 and using the coefficient of kinetic friction.

top: FT − μkmA g = mAa

bottom: F − FT − μkmAg − μk (mA + mB ) g = mBa

sum: F − μkmAg − μkmAg − μk (mA + mB ) g = (mA + mB ) a →

a = F − μkmA g − μkmA g − μk ( mA + mB ) g = F − μk (3mA + mB ) g
(mA + mB )
(mA + mB )

( )90.55 N − (0.40) (14.0 kg) 9.80 m s2 = 4.459 m s2 ≈ 4.5 m s2

= (8.0 kg)

32. Free-body diagrams are shown for both blocks. There is a force of friction G
between the two blocks, which acts to the right on the top block, and to the left FN
on the bottom block. They are a Newton’s third law pair of forces.
(a) If the 4.0 kg block does not slide off, then it must have the same top G
acceleration as the 12.0 kg block. That acceleration is caused by the force Ffr
of static friction between the two blocks. To find the minimum coefficient,
we use the maximum force of static friction. top

mtopgG

Ffr = mtopa = μFN = μmtop g → μ = a = 5.2 m s2 = 0.5306 ≈ 0.53 G G
g 9.80 m s2 FN −FN
top top
bottom Gtop
(b) If the coefficient of friction only has half the value, then the blocks will be FP
sliding with respect to one another, and so the friction will be kinetic. G
Ffr mbottomgG

bottom

μ = 1 ( 0.5306) = 0.2653 ; Ffr = mtopa = μFN = μmtop g →
2

top top

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134

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

( )a = μg = (0.2653) 9.80 m s2 = 2.6 m s2

(c) The bottom block is still accelerating to the right at 5.2 m s2 . Since the top block has a smaller

acceleration than that, it has a negative acceleration relative to the bottom block.
aG top rel = aG top rel + aG ground rel = aG top rel − aG bottom rel = 2.6 m s2 ˆi − 5.2 m s2 ˆi = −2.6 m s2 ˆi

bottom ground bottom ground ground

The top block has an acceleration of 2.6 m s2 to the left relative to the bottom block.

(d) No sliding:

Fx = FP − Ffr = m abottom bottom →

bottom bottom

net

FP = Ffr ( )+ m a = F + m a = m a + m a = m + m abottom bottom
fr bottom bottom top top bottom bottom top bottom

bottom top

( )= (16.0 kg) 5.2 m s2 = 83 N

This is the same as simply assuming that the external force is accelerating the total mass. The
internal friction need not be considered if the blocks are not moving relative to each other.

Sliding:

Fx = FP − Ffr = m abottom bottom →

bottom bottom

net

FP = Ffr + m a = F + m a = m a + m abottom bottom
fr bottom bottom top top bottom bottom

bottom top

= (4.0 kg) (2.6 m s2 ) + (12.0 kg) (5.2 m )s2 = 73 N

Again this can be interpreted as the external force providing the acceleration for each block.
The internal friction need not be considered.

33. To find the limiting value, we assume that the blocks are NOT slipping, G y
but that the force of static friction on the smaller block is at its FN
x
maximum value, so that Ffr = μFN. For the two-block system, there is G
Ffr
no friction on the system, and so F = ( M + m) a describes the

horizontal motion of the system. Thus the upper block has a vertical mθgG
acceleration of 0 and a horizontal acceleration of F . Write

(M + m)

Newton’s second law for the upper block, using the force diagram, and solve for the applied force F.
Note that the static friction force will be DOWN the plane, since the block is on the verge of sliding
UP the plane.

∑ Fy = FN cosθ − Ffr sinθ − mg = FN (cosθ − μ sinθ ) − mg = 0 → FN = (cosθ mg

− μ sinθ )

∑ Fx = FN sin θ + Ffr cosθ = FN (sinθ + μ cosθ ) = ma = m F m →
M+

F = FN (sinθ + μ cosθ ) M+ m = mg (sinθ + μ cosθ ) M+ m
m −μ m
(cosθ sin θ )

= (M + m) g ( sin θ + μ cosθ )
(cosθ − μ sinθ )

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Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

34. A free-body diagram for the car at one instant of time is shown. In the diagram, the G G
car is coming out of the paper at the reader, and the center of the circular path is to FN Ffr
the right of the car, in the plane of the paper. If the car has its maximum speed, it
would be on the verge of slipping, and the force of static friction would be at its mgG
maximum value. The vertical forces (gravity and normal force) are of the same
magnitude, because the car is not accelerating vertically. We assume that the force
of friction is the force causing the circular motion.

FR = Ffr → m v2 r = μs FN = μsmg →

( )v = μsrg = (0.65) (80.0 m) 9.80 m s2 = 22.57 m s ≈ 23 m s

Notice that the result is independent of the car’s mass .

35. (a) Find the centripetal acceleration from Eq. 5-1.

aR = v2 r = (1.30 m s)2 1.20 m = 1.408 m s2 ≈ 1.41m s2

(b) The net horizontal force is causing the centripetal motion, and so will be the centripetal force.

( )FR = maR = (22.5 kg) 1.408 m s2 = 31.68N ≈ 31.7 N

36. Find the centripetal acceleration from Eq. 5-1.

(525 m s)2 ⎛ 1g ⎞
⎜⎝ 9.80 m s2 ⎠⎟ =
4.80 ×103 m
( )aR = v2
r = = 57.42 m s2 5.86 g's

37. We assume the water is rotating in a vertical circle of radius r. When the bucket

is at the top of its motion, there would be two forces on the water (considering

the water as a single mass). The weight of the water would be directed down, FN mgG
and the normal force of the bottom of the bucket pushing on the water would
also be down. See the free-body diagram. If the water is moving in a circle,

then the net downward force would be a centripetal force.

∑ ( )F = FN + mg = ma = m v2 r → FN = m v2 r − g

The limiting condition of the water falling out of the bucket means that the water loses contact with

the bucket, and so the normal force becomes 0.

( ) ( )FN = m v2 r − g
→ m v2 r−g =0 → v =critical rg
critical

From this, we see that yes , it is possible to whirl the bucket of water fast enough. The minimum

speed is rg .

38. The centripetal acceleration of a rotating object is given by aR = v2 r .

( ) ( ) ( ) ( )v = aRr = 1.25 ×105 g r = 1.25 ×105 9.80 m s2 8.00 ×10−2 m = 3.13×102 m s .

⎛ 1 rev ⎞ ⎝⎛⎜ 60 s ⎞⎠⎟
⎜⎝⎜ 2π 8.00 ×10−2m ⎟⎠⎟ 1 min
( ) ( )3.13 ×102 m s = 3.74 ×104 rpm

39. For an unbanked curve, the centripetal force to move the car in a circular path must G G
be provided by the static frictional force. Also, since the roadway is level, the FN Ffr
normal force on the car is equal to its weight. Assume the static frictional force is
at its maximum value, and use the force relationships to calculate the radius of the mgG

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136

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

curve. See the free-body diagram, which assumes the center of the curve is to the right in the
diagram.

FR = Ffr → m v2 r = μs FN = μsmg →

⎡⎢⎣(30 km h ) ⎛ 1m s ⎞⎤2
⎝⎜ 3.6 km h ⎟⎠⎦⎥
( )r = v2μsg = = 28 m ≈ 30 m
(0.7) 9.80 m s2

40. At the top of a circle, a free-body diagram for the passengers would be as

shown, assuming the passengers are upside down. Then the car’s normal

force would be pushing DOWN on the passengers, as shown in the diagram.

We assume no safety devices are present. Choose the positive direction to G mgG
be down, and write Newton’s second law for the passengers. FN

∑ ( )F = FN + mg = ma = m v2 r → FN = m v2 r − g

We see from this expression that for a high speed, the normal force is positive, meaning the

passengers are in contact with the car. But as the speed decreases, the normal force also decreases. If

the normal force becomes 0, the passengers are no longer in contact with the car – they are in free

fall. The limiting condition is as follows.

( )v2
min
r−g =0 → vmin = rg = 9.80 m s2 (7.6 m) = 8.6 m s

41. A free-body diagram for the car is shown. Write Newton’s second law for the car G mgG
in the vertical direction, assuming that up is positive. The normal force is twice FN
the weight.

∑ F = FN − mg = ma → 2mg − mg = m v2 r →

( )v = rg = (95 m) 9.80 m s2 = 30.51m s ≈ 31m s

G
42. In the free-body diagram, the car is coming out of the paper at the reader, and the FN G
center of the circular path is to the right of the car, in the plane of the paper. The Ffr
vertical forces (gravity and normal force) are of the same magnitude, because the
car is not accelerating vertically. We assume that the force of friction is the force mgG
causing the circular motion. If the car has its maximum speed, it would be on the
verge of slipping, and the force of static friction would be at its maximum value.

v2 ⎡⎣⎢(95 km hr ) ⎛ 1m s ⎞⎤2
rg ⎝⎜ 3.6 km hr ⎠⎟⎦⎥
m v2 r = μs FN = μsmg
( )FR = Ffr→ → μs = = = 0.84
(85 m) 9.80 m s2

Notice that the result is independent of the car’s mass.

43. The orbit radius will be the sum of the Earth’s radius plus the 400 km orbit height. The orbital
period is about 90 minutes. Find the centripetal acceleration from these data.

r = 6380 km + 400 km = 6780 km = 6.78 ×106 m T = 90 min ⎛ 60 sec ⎞ = 5400 sec
⎝⎜ 1 min ⎠⎟

( ) ( )aR
= 4π 2r = 4π 2 6.78 ×106 m = 9.18 m s2 ⎛ 1g s2 ⎞ = 0.937 ≈ 0.9 g's
T2 ⎝⎜ 9.80 m ⎟⎠
(5400 sec)2

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137

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

Notice how close this is to g, because the shuttle is not very far above the surface of the Earth,
relative to the radius of the Earth.

44. (a) At the bottom of the motion, a free-body diagram of the bucket would be as G
shown. Since the bucket is moving in a circle, there must be a net force on it FT
towards the center of the circle, and a centripetal acceleration. Write mgG
Newton’s second law for the bucket, with up as the positive direction.

∑ FR = FT − mg = ma = m v2 r →

( )v = s2
r ( FT − mg ) = (1.10 m) ⎣⎡25.0 N − (2.00 kg) 9.80 m ⎤⎦ = 1.723 ≈ 1.7 m s

m 2.00 kg

(b) A free-body diagram of the bucket at the top of the motion is shown. Since the FT mgG
bucket is moving in a circle, there must be a net force on it towards the center
of the circle, and a centripetal acceleration. Write Newton’s second law for the
bucket, with down as the positive direction.

∑ FR = FT + mg = ma = m v2 r → v= r ( FT + mg )

If the tension is to be zero, then m

r (0 + mg )

m
( )v =
= rg = (1.10 m) 9.80 m s2 = 3.28 m s

The bucket must move faster than 3.28 m/s in order for the rope not to go slack.

45. The free-body diagram for passengers at the top of a Ferris wheel is as shown. G mgG
FN is the normal force of the seat pushing up on the passenger. The sum of the FN
forces on the passenger is producing the centripetal motion, and so must be a
centripetal force. Call the downward direction positive, and write Newton’s
second law for the passenger.

∑ FR = mg − FN = ma = m v2 r

Since the passenger is to feel “weightless,” they must lose contact with their seat, and so the normal
force will be 0. The diameter is 22 m, so the radius is 11 m.

( )mg = m v2 r → v = gr = 9.80 m s2 (11m) = 10.38 m s

(10.38 m s) ⎛ 1 rev ⎞ ⎛ 60 s ⎞ = 9.0 rpm
⎜ ⎟ ⎜⎝ 1 min ⎟⎠
⎝ 2π (11m) ⎠

46. To describe the motion in a circle, two independent quantities are needed. The radius of the circle
and the speed of the object are independent of each other, so we choose those two quantities. The

radius has dimensions of [L] and the speed has dimensions of [L T]. These two dimensions need

to be combined to get dimensions of ⎣⎡L T2 ⎦⎤. The speed must be squared, which gives ⎡⎣L2 T2 ⎤⎦ ,

and then dividing by the radius gives ⎣⎡L T2 ⎤⎦. So aR = v2 r is a possible form for the centripetal

acceleration. Note that we are unable to get numerical factors, like π or 1 , from dimensional
2

analysis.

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138

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

47. (a) See the free-body diagram for the pilot in the jet at the bottom of the G
FN
loop. We have aR = v2 r = 6g .
mgG
⎡⎢⎣(1200 km h) ⎛ 1m s 2
v2 ⎝⎜ 3.6 km h
( )v2 r = 6.0g → r = 6.0g = ⎞⎤ = 1900 m
⎟⎠⎥⎦

6.0 9.80 m s2

(b) The net force must be centripetal, to make the pilot go in a circle. Write Newton’s second
law for the vertical direction, with up as positive. The normal force is the apparent weight.

∑ FR = FN − mg = m v2 r

The centripetal acceleration is to be v2 r = 6.0g.

( )FN = mg + m v2 r = 7mg = 7 (78 kg) 9.80 m s2 = 5350 N = 5400 N

(c) See the free-body diagram for the pilot at the top of the loop. Notice that

∑the normal force is down, because the pilot is upside down. Write Newton’s G mgG
FN
second law in the vertical direction, with down as positive.
FR = FN + mg = m v2 r = 6mg → FN = 5mg = 3800 N

48. To experience a gravity-type force, objects must be on the inside of the outer G
wall of the tube, so that there can be a centripetal force to move the objects in FN
a circle. See the free-body diagram for an object on the inside of the outer
wall, and a portion of the tube. The normal force of contact between the
object and the wall must be maintaining the circular motion. Write
Newton’s second law for the radial direction.

∑ FR = FN = ma = m v2 r

If this is to have the same effect as Earth gravity, then we must also have that

FN = mg. Equate the two expressions for normal force and solve for the speed.

( )FN = m v2 r = mg → v = gr = 9.80 m s2 (550 m) = 73.42 m s

(73.42 m s) ⎛ 2π 1 rev m) ⎞ ⎛⎝⎜ 86, 400 s ⎞⎠⎟ = 1836 rev d ≈ 1.8 ×103 rev d
⎜ ⎟ 1d
⎝ (550 ⎠

49. The radius of either skater’s motion is 0.80 m, and the period is 2.5 sec. Thus their speed is given by

v = 2π r T = 2π (0.80 m) = 2.0 m s. Since each skater is moving in a circle, the net radial force on

2.5 s

each one is given by Eq. 5-3.

FR = mv2 r = ( 60.0 kg ) ( 2.0 m s)2 = 3.0 ×102 N .

0.80 m

50. A free-body diagram for the ball is shown. The tension in the G
θ FT
suspending cord must not only hold the ball up, but also provide the mgG
centripetal force needed to make the ball move in a circle. Write
Newton’s second law for the vertical direction, noting that the ball is

not accelerating vertically.

∑ Fy = FT sin θ − mg = 0 → FT = mg
sin θ

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139

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

The force moving the ball in a circle is the horizontal portion of the tension. Write Newton’s second
law for that radial motion.

∑F = F cos θ = ma = m v2 r
R T R

Substitute the expression for the tension from the first equation into the second equation, and solve

for the angle. Also substitute in the fact that for a rotating object, v = 2π r T . Finally we recognize

that if the string is of length l , then the radius of the circle is r = l cosθ.

FT cosθ = mg cosθ = mv 2 = 4π 2mr = 4π 2ml cosθ →
sin θ r T2 T2

( )sinθ
= gT 2 → θ = sin −1 gT 2 = sin−1 9.80 m s2 (0.500 s)2 = 5.94°
4π 2l 4π 2l 4π 2 (0.600 m)

( )The s2
tension is then given by FT = mg = (0.150 kg) 9.80 m = 14.2 N
sin θ
sin 5.94°

51. The force of static friction is causing the circular motion – it is the centripetal G G
force. The coin slides off when the static frictional force is not large enough to FN Ffr
move the coin in a circle. The maximum static frictional force is the coefficient
of static friction times the normal force, and the normal force is equal to the mgG
weight of the coin as seen in the free-body diagram, since there is no vertical
acceleration. In the free-body diagram, the coin is coming out of the paper and
the center of the circle is to the right of the coin, in the plane of the paper.

The rotational speed must be changed into a linear speed.

v = ⎛ 35.0 rev ⎞ ⎛ 1 min ⎞ ⎛ 2π (0.120 m) ⎞ = 0.4398 m s
⎝⎜ min ⎠⎟ ⎝⎜ 60 s ⎠⎟ ⎜⎝
1 rev ⎠⎟

m v2 r = μs FN = μsmg v2 (0.4398 m s)2 = 0.164
rg (0.120 m) 9.80 m
( )FR = Ffr → → μs = = s2

52. For the car to stay on the road, the normal force must be greater Final G Initial
than 0. See the free-body diagram, write the net radial force, and Road FN Road
solve for the radius. mgGθ θ

FR = mg cosθ − FN = mv 2 → r = mg mv 2 − FN
r cosθ

For the car to be on the verge of leaving the road, the normal force

would be 0, and so rcritical = mv 2 = v2 . This expression
mg cosθ g cosθ

gets larger as the angle increases, and so we must evaluate at the
largest angle to find a radius that is good for all angles in the range.

v2 ⎢⎡⎣95 km h ⎛ 1m s h ⎞⎤2
g cosθmax ⎝⎜ 3.6 km ⎟⎠⎥⎦
( )rcritical = = = 77 m
maximum
9.80 m s2 cos 22°

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140

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

53. (a) A free-body diagram of the car at the instant it is on the top of the hill is G mgG
shown. Since the car is moving in a circular path, there must be a net FN
centripetal force downward. Write Newton’s second law for the car, with
down as the positive direction.

∑ FR = mg − FN = ma = m v2 r →

( )FN = m ⎛ s2 (12.0 m s)2 ⎞
g − v2 r = ( 975 kg) ⎜ 9.80 m − ⎟= 7960 N
88.0 m ⎠
⎝

(b) The free-body diagram for the passengers would be the same as the one for the car, leading to

the same equation for the normal force on the passengers.

( )FN = m ⎛ s2 (12.0 m s)2 ⎞
g − v2 r = ( 72.0 kg) ⎜ 9.80 m − ⎟= 588 N
88.0 m ⎠
⎝

Notice that this is significantly less than the 700-N weight of the passenger. Thus the passenger

will feel “light” as they drive over the hill.

(c) For the normal force to be zero, we must have the following.

( ) ( )FN = m g − v2 r = 0 → g = v2 r → v = gr = 9.80 m s2 (88.0 m) = 29.4 m s

54. If the masses are in line and both have the same frequency of G G
rotation, then they will always stay in line. Consider a free- FNB G G FNA G
body diagram for both masses, from a side view, at the FTB FTB FTA
instant that they are to the left of the post. Note that the same
tension that pulls inward on mass 2 pulls outward on mass 1, mB mA
by Newton’s third law. Also notice that since there is no
vertical acceleration, the normal force on each mass is equal mBgG mAgG

to its weight. Write Newton’s second law for the horizontal

direction for both masses, noting that they are in uniform circular motion.

∑ FRA = FTA − FTB = mAaA = mA vA2 rA ∑ FRB = FTB = mBaB = mB vB2 rB

The speeds can be expressed in terms of the frequency as follows: v = ⎛⎝⎜ f rev ⎞ ⎛ 2π r ⎠⎞⎟ = 2π rf .
sec ⎠⎟ ⎝⎜ 1 rev

( )FTB = mB vB2 rB = mB 2π rB f 2 rB = 4π 2mBrB f 2

( ) ( )FTA = FTB + mA vA2 rA = 4π mBrB f 2 + mA 2π rA f 2 rA = 4π 2 f 2 mArA + mBrB

55. A free-body diagram of Tarzan at the bottom of his swing is shown. The upward G
FT
tension force is created by his pulling down on the vine. Write Newton’s second law
in the vertical direction. Since he is moving in a circle, his acceleration will be mgG
centripetal, and points upward when he is at the bottom.

∑ F = FT − mg = ma = m v2 r → v= ( FT − mg ) r

m

The maximum speed will be obtained with the maximum tension.
( )G
( ( ))vmax =
FT max − mg r = 1350 N − (78 kg) 9.80 m s2 5.2 m
m = 6.2 m s
78 kg

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141

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

56. The fact that the pilot can withstand 9.0 g’s without blacking out, along with the
speed of the aircraft, will determine the radius of the circle that he must fly as he
pulls out of the dive. To just avoid crashing into the sea, he must begin to form
that circle (pull out of the dive) at a height equal to the radius of that circle.

v2 (310 m s)2 = 1.1×103 m
9.0g
9.0 9.80 m s2
( )aR = v2 r = 9.0g → r = =

57. (a) We are given that x = (2.0 m) cos (3.0 rad s t) and y = (2.0 m) sin (3.0 rad s t) . Square both

components and add them together.

x2 + y2 = [(2.0 m) cos (3.0 rad s t)]2 + [(2.0 m) sin (3.0 rad s t)]2

= (2.0 m)2 ⎣⎡cos2 (3.0 rad s t) + sin2 (3.0 rad s t)⎤⎦ = (2.0 m)2

This is the equation of a circle, x2 + y2 = r2 , with a radius of 2.0 m.

(b) vG = (−6.0 m s) sin (3.0 rad s t) ˆi + (6.0 m s) cos (3.0 rad s t) ˆj

( ) ( )aG = −18 m s2 cos (3.0 rad s t) ˆi + −18 m s2 sin (3.0 rad s t) ˆj

(c) v= vx2 + v 2 = [(−6.0 m s) sin (3.0 rad s t)]2 + [(6.0 m s) cos (3.0 rad s t)]2 = 6.0 m s
y

⎡⎣ −18 m s2 cos (3.0 rad s t )⎤⎦2 + ⎡⎣ −18 m s2 sin (3.0 rad s t )⎤⎦2 = 18 m s2
( ) ( )a =ax2+ a 2 =
y

(d) v2 = (6.0 m s)2 = 18 m s2 = a
r
2.0 m

( ) ( )(e) aG = −18 m s2 cos (3.0 rad s t) ˆi + −18 m s2 sin (3.0 rad s t) ˆj
( ) ( )= −9.0 s2 ⎣⎡2.0 m cos (3.0 rad s t) ˆi + 2.0 m sin (3.0 rad s t) ˆj⎦⎤ = 9.0 s2 (−rG)

We see that the acceleration vector is directed oppositely of the position vector. Since the

position vector points outward from the center of the circle, the acceleration vector points

toward the center of the circle.

58. Since the curve is designed for 65 km/h, traveling at a higher speed with the same radius means that
more centripetal force will be required. That extra centripetal force will be supplied by a force of
static friction, downward along the incline. See the free-body diagram for the car on the incline.
Note that from Example 5-15 in the textbook, the no-friction banking angle is given by the
following.

v2 ⎢⎡⎣(65 km h ) ⎛ 1.0 m s ⎞⎤2
rg ⎝⎜ 3.6 km h ⎠⎟⎦⎥
tan −1 tan −1 21.4°
( )θ= = =
(85 m) 9.80 m s2

Write Newton’s second law in both the x and y directions. The car will have no acceleration in the y
direction, and centripetal acceleration in the x direction. We also assume that the car is on the verge

of skidding, so that the static frictional force has its maximum value of Ffr = μsFN. Solve each

equation for the normal force.

∑ Fy = FN cosθ − mg − Ffr sinθ = 0 → FN cosθ − μsFN sinθ = mg →

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142

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

FN = (cosθ mg sin θ )
− μs

∑ Fx = FN sinθ + Ffr cosθ = FR = mv2 r → FN sinθ + μs FN cosθ = mv2 r →

FN = mv2 r )

(sinθ + μs cosθ

Equate the two expressions for FN , and solve for the coefficient of friction. The speed of rounding

the curve is given by v = (95 km h ) ⎛ 1.0 m s ⎞ = 26.39 m s.
⎝⎜ 3.6 km h ⎟⎠

mg mv2 r

(cosθ − μs sinθ ) = (sinθ + μs cosθ ) →

⎛ v2 ⎞ ⎛ v2 ⎞ ⎛ (26.39 m s)2 ⎞
⎜⎝ r ⎟⎠ ⎝⎜ r ⎠⎟ ⎜ ⎟
⎝ 85 m
( )μs
cosθ − g sin θ − g tan θ − 9.80 m s2 tan 21.4°

= = = ⎠ = 0.33

⎛ g cosθ + v2 sin θ ⎞ ⎛ g + v2 tan θ ⎞ ⎛ 9.80 m s2 + (26.39 m s)2 tan 21.4° ⎞
⎝⎜ r ⎠⎟ ⎝⎜ r ⎠⎟ ⎜ ⎟
⎝ 85 m ⎠

59. Since the curve is designed for a speed of 85 km/h, traveling at that speed y θG
would mean no friction is needed to round the curve. From Example 5- x FN
15 in the textbook, the no-friction banking angle is given by

v2 ⎣⎢⎡(85 km h) ⎛ 1m s ⎞⎤2 mgG θ G
rg ⎝⎜ 3.6 km h ⎠⎟⎥⎦ Ffr
tan −1 tan −1 39.91°
( )θ= = =
(68 m) 9.80 m s2 θ

Driving at a higher speed with the same radius means that more centripetal force will be required
than is present by the normal force alone. That extra centripetal force will be supplied by a force of
static friction, downward along the incline, as shown in the first free-body diagram for the car on the
incline. Write Newton’s second law in both the x and y directions. The car will have no acceleration
in the y direction, and centripetal acceleration in the x direction. We also assume that the car is on

the verge of skidding, so that the static frictional force has its maximum value of Ffr = μs FN .

∑ Fy = FN cosθ − mg − Ffr sinθ = 0 → FN cosθ − μs FN sinθ = mg →

FN = ( cos θ mg sin θ )
− μs

∑ Fx = FN sin θ + Ffr cosθ = m v2 r → FN sin θ + μs FN cosθ = m v2 r →

FN = mv2 r )

(sinθ + μs cosθ

Equate the two expressions for the normal force, and solve for the speed.

mv2 r mg
(sinθ + μs cosθ ) = (cosθ − μs sinθ ) →

(sinθ + μs cosθ ) (sin 39.91° + 0.30 cos 39.91°)
(cosθ − μs sinθ ) (cos 39.91° − 0.30sin 39.91°)
( )v =
rg = (68 m) 9.80 m s2 = 32 m s

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143

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

Now for the slowest possible speed. Driving at a slower speed with G θG y
the same radius means that less centripetal force will be required than Ffr FN
that supplied by the normal force. That decline in centripetal force θ x
will be supplied by a force of static friction, upward along the incline,

as shown in the second free-body diagram for the car on the incline. mgG
Write Newton’s second law in both the x and y directions. The car

will have no acceleration in the y direction, and centripetal θ
acceleration in the x direction. We also assume that the car is on the

verge of skidding, so that the static frictional force has its maximum value of Ffr = μs FN .

∑ Fy = FN cosθ − mg + Ffr sin θ = 0 →

FN cosθ + μs FN sin θ = mg → FN = ( cos θ mg sin θ )
+ μs

∑ Fx = FN sin θ − Ffr cosθ = m v2 r → FN sin θ − μs FN cosθ = m v2 r →

FN = mv2 r )

(sinθ − μs cosθ

Equate the two expressions for the normal force, and solve for the speed.

mv2 r mg
(sinθ − μs cosθ ) = (cosθ + μs sinθ ) →

(sinθ − μs cosθ ) (sin 39.91° − 0.30 cos 39.91°)
(cosθ + μs sinθ ) (cos 39.91° + 0.30sin 39.91°)
( )v =
rg = (68 m) 9.80 m s2 = 17 m s

Thus the range is 17 m s ≤ v ≤ 32 m s , which is 61km h ≤ v ≤ 115 km h .

60. (a) The object has a uniformly increasing speed, which means the tangential acceleration is
constant, and so constant acceleration relationships can be used for the tangential motion. The
object is moving in a circle of radius 2.0 meters.

[ ]( ) ( )vtan + v0
Δxtan = ttan → vtan = 2Δxtan − v0 = 2 1 2π r =π 2.0 m = πm s
2 t 4 t 2.0 s
tan

(b) The initial location of the object is at 2.0 mˆj , and the final location is 2.0 mˆi.
rG rG0
2.0 mˆi − 2.0 mˆj
2.0 s
( )vG avg − s ˆi − ˆj
= t = = 1.0 m

(c) The velocity at the end of the 2.0 seconds is pointing in the −ˆj direction.
vG − vG 0
t − (π m s) ˆj

2.0 s
( )aGavg
= = = − π 2 m s2 ˆj

61. Apply uniform acceleration relationships to the tangential motion to find the tangential acceleration.

Use Eq. 2-12b.

[ ] ( )( ) ( )Δxtan 2 2π r
= v0 t+ a t1 2 → atan = 2Δxtan = 1 t2 = π 2.0 m = π 2 m s2
t2 4 2.0 s 2
2 tan
( )tan

The tangential acceleration is constant. The radial acceleration is found from arad = v2 = (atant )2 .
tan
r r

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144

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

(a) atan = (π 2) m ( )s2 , arad = atant 2 = ⎡⎣(π 2) m s2 (0 s)⎤⎦2 = 0
r
2.0 m

( )s2 , arad = atant 2 = ⎡⎣(π 2) m s2 (1.0 s)⎦⎤2
r
2.0 m
( )(b)
atan = (π 2) m = π 2 8 m s2

( )s2 , arad = atant 2 = ⎣⎡(π 2) m s2 (2.0 s)⎦⎤2
r
2.0 m
( )(c)
atan = (π 2) m = π 2 2 m s2

62. (a) The tangential acceleration is the time derivative of the speed.

( )atan
= dvtan d 3.6 + 1.5t2 = 3.0t → atan (3.0 s) = 3.0 (3.0) = 9.0 m s2
dt = dt

(b) The radial acceleration is given by Eq. 5-1.

( ) ( )arad 2
3.6 + 1.5t2 2 3.6 + 1.5(3.0)2
= v2 = r → arad (3.0 s) = = 13 m s2
tan 22 m
r

63. We show a top view of the particle in circular motion, traveling clockwise. aG tan aG
Because the particle is in circular motion, there must be a radially-inward
component of the acceleration. θ
(a) aR = a sinθ = v2 r → aG R

( )v = ar sinθ = 1.15 m s2 (3.80 m) sin 38.0o = 1.64 m s

(b) The particle’s speed change comes from the tangential acceleration,

which is given by atan = a cosθ . If the tangential acceleration is
constant, then using Eq. 2-12a,

vtan − v0 tan = atant →

( )vtan = v0 tan + atant = 1.64 m s + 1.15 m s2 (cos 38.0°) (2.00 s) = 3.45 m s

64. The tangential force is simply the mass times the tangential acceleration.

( )aT = b + ct2 → FT = maT = m b + ct2

To find the radial force, we need the tangential velocity, which is the anti-derivative of the tangential

acceleration. We evaluate the constant of integration so that v = v0 at t = 0.

aT = b + ct2 → vT = c + bt + 1 ct 3 → v (0) = c = v0 → vT = v0 + bt + 1 ct 3
3 3

=( )FR mvT2 = m v0 + bt + 1 ct 3 2
r r 3

65. The time constant τ must have dimensions of [T]. The units of m are [M]. Since the expression

bv is a force, we must have the dimensions of b as force units divided by speed units. So the

dimensions of b are as follows: Force units = [M] ⎣⎡L T2 ⎦⎤ = ⎡ M ⎤ . Thus to get dimensions of
speed units [L T] ⎢⎣ T ⎦⎥

[T] , we must have τ = m b .

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145

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

66. (a) The terminal velocity is given by Eq. 5-9. This can be used to find the value of b.

( )( )vT
= mg → b = mg = 3 ×10−5 kg 9.80 m s2 = 3.27 ×10−5 kg s ≈ 3 ×10−5 kg s
b vT
(9m s)

(b) From Example 5-17, the time required for the velocity to reach 63% of terminal velocity is the

time constant, τ = m b.

τ = m = 3 ×10−5 kg s = 0.917 s ≈ 1s
b 3.27 ×10−5 kg

67. (a) We choose downward as the positive direction. Then the force of gravity is in the positive

direction, and the resistive force is upwards. We follow the analysis given in Example 5-17.

Fnet = mg − bv = ma → a = dv = g − b v = − b ⎜⎛⎝ v − mg ⎠⎟⎞ →
dt m m b

dv b dt v dv b t dt ln ⎣⎡⎢v mg ⎤v b t
∫ ∫v = − m → = − m 0 → − b ⎦⎥v0 = − m →
mg mg
− b v0 v − b

⎡ v − mg ⎤ b v − mg e− b t mg ⎛⎜1 − e− b t ⎞ v0e− b t
⎢ v0 − b ⎥ m b m b ⎝ m ⎟ m
ln ⎢ mg ⎥ = − t → = → v = ⎠ +
⎢⎣ b ⎦⎥ mg
v0 − b

Note that this motion has a terminal velocity of vterminal = mg b .

(b) We choose upwards as the positive direction. Then both the force of gravity and the resistive
force are in the negative direction.

Fnet = −mg − bv = ma → a = dv = −g − b v = − b ⎛⎜⎝ v + mg ⎠⎞⎟ →
dt m m b

b v dv t mg ⎤v b
∫ ∫dv m b ⎥⎦v0 m
mg = − dt → mg b dt → ln ⎡⎢⎣v + = − t →
b b =−m
v + v0 v + 0

⎡ v + mg ⎤ b v + mg e− b t mg ⎛ e− b t − 1⎟⎞ v0e− b t
⎢ v0 + b ⎥ m b m b ⎜ m ⎠ m
ln ⎢ mg ⎥ = − t → = → v= +
b mg
⎢⎣ ⎥⎦ v0 + b ⎝

After the object reaches its maximum height ⎡⎢trise = m ln ⎛⎝⎜1 + bv0 ⎞⎤ , at which point the speed
⎣ b mg ⎟⎠⎦⎥

will be 0, it will then start to fall. The equation from part (a) will then describe its falling

motion.

∑68. The net force on the falling object, taking downward as positive, will be F = mg − bv2 = ma.

(a) The terminal velocity occurs when the acceleration is 0.

mg − bv2 = ma → mg − bvT2 = 0 → vT = mg b

( )(b) s2
vT = mg → b = mg = (75kg) 9.80 m = 0.2 kg m
b vT2 (60 m s)2

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146

Chapter 5 Using Newton’s Laws: Friction, Circular Motion, Drag Forces

(c) The curve would be qualitatively like Fig. 5-27, because the speed would increase from 0 to the
terminal velocity, asymptotically. But this curve would be ABOVE the one in Fig. 5-27,
because the friction force increases more rapidly. For Fig. 5-27, if the speed doubles, the
friction force doubles. But in this case, if the speed doubles, the friction force would increase
by a factor of 4, bringing the friction force closer to the weight of the object in a shorter period
of time.

69. (a) See the free-body diagram for the coasting. Since the bicyclist has a G G
constant velocity, the net force on the bicycle must be 0. Use this to FD
find the value of the constant c. FN y

∑ Fx = mg sinθ − FD = mg sinθ − cv2 = 0 → x

( )c = θ θmgG
mg sinθ = (80.0kg) 9.80 m s2 sin 7.0° = 13.72 kg m
v2
⎢⎣⎡9.5 km h ⎛ 1m s ⎞⎤2
⎝⎜ 3.6 km h ⎠⎟⎥⎦

≈ 14 kg m

G G G y
(b) Now another force, FP, must be added down the plane to represent FD FN x

the additional force needed to descend at the higher speed. The G
velocity is still constant. See the new free-body diagram. FP

∑ Fx = mg sinθ + FP − FD = mg sinθ + FP − cv2 = 0 → θ θmgG

FP = cv2 − mg sinθ

m) ⎣⎢⎡25km ⎛ 1m s ⎞⎤2
⎜⎝ 3.6 km ⎟⎠⎦⎥
( )= (13.72 kg s2
h h − (80.0kg) 9.80 m sin 7.0° = 570 N

70. (a) The rolling drag force is given as FD1 ≈ 4.0 N. The air resistance drag force is proportional to

v2, and so FD2 = bv2. Use the data to find the proportionality constant, and then sum the two

drag forces to find the total drag force.

FD2 = bv2 → 1.0 N = b (2.2 m s)2 → b= 1.0 N = 0.2066 kg m

(2.2 m s)2

( )FD = FD1 + FD2 = 4.0 + 0.21v2 N

(b) See the free-body diagram for the coasting bicycle and rider. Take G G
the positive direction to be down the plane, parallel to the plane. FD FN y
The net force in that direction must be 0 for the bicycle to coast at a
constant speed. x

∑F = mg sinθ − FD = 0 → mg sinθ = FD → θ θmgG
x

( )θ
= sin−1 FD = sin−1 4.0 + 0.2066v2
mg mg

( ( ) )= sin−1
4.0N + (0.2066 kg m) (8.0 m s)2 = 1.3°
(78 kg) 9.80 m s2

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147

Physics for Scientists & Engineers with Modern Physics, 4th Edition Instructor Solutions Manual

71. From Example 5-17, we have that v = mg ⎛ 1 − e− b t ⎞⎟ . We use this expression to find the position
b ⎜ m ⎠

⎝

and acceleration expressions.

a = dv = mg ⎛ − e − b t ⎞ ⎜⎛⎝ − b ⎟⎠⎞ = ge − b t
dt b ⎜ m ⎟ m m
⎠
⎝

∫ ∫ ∫v dx x dx t v dt t mg ⎛ e − b t ⎞ dt
= dt → dx = vdt → 0 = 0 = 0 b ⎜ 1 − m ⎟ →

⎝ ⎠

x = ⎡ mg t + mg m e− b t ⎤t = mg t + m2g ⎛ e − b t − 1⎞⎟
⎣⎢ b b b m ⎥ b b2 ⎜ m ⎠
⎦0
⎝

72. We solve this problem by integrating the acceleration to find the velocity, and integrating the

velocity to find the position.

Fnet = −bv 1 = ma = m dv → dv = − b v1 → dv = − b dt →
2 dt 2 v1 m

dt m 2

∫ ∫vdv = − b t dt → 2v 1 − 2v 1 = − b t → v = ⎛ v1 − bt ⎞2
v1 m 0 2 2 m ⎜⎝ 2 2m ⎟⎠
v0 0 0
2

∫ ∫dx = ⎜⎝⎛ v1 − bt ⎞2 → dx = ⎛⎝⎜ v1 − bt ⎞2 dt → x dx = t ⎜⎝⎛ v1 − bt ⎞2 dt →
2 2m ⎠⎟ 2 2m ⎟⎠ 0 0 2 2m ⎠⎟
dt 0 0 0

x = − 2m ⎡⎛ v1 − bt 3 − v3 ⎤ = 2m ⎡ v3 − ⎛ v1 − bt 3 ⎤
3b ⎣⎢⎝⎜ 2 2m 2 ⎥ 3b ⎢ 2 ⎝⎜ 2 2m
0 ⎞ 0 0 0 ⎞
⎠⎟ ⎟⎠ ⎥
⎦ ⎣ ⎦

= 2m ⎛ v3 − ⎛ v3 − 3v0 bt + 3v 1 b2t 2 − b3t 3 ⎞⎞ = 2m ⎛ v3 − v3 + 3v0 bt − 3v 1 b2t 2 + b3t 3 ⎞
3b ⎜ 2 ⎜ 2 2m 2 4m2 8m3 ⎠⎟ ⎟ 3b ⎜ 2 2 2m 2 4m2 8m3 ⎟⎠
⎝ 0 ⎝ 0 0 ⎠ ⎝ 0 0 0

= ⎛ v0t − v b1 t2 + b2 t3 ⎞
⎝⎜⎜ 2 12m2 ⎠⎟⎟
0

2m

73. From problem 72, we have that v = ⎛ v1 − bt ⎞2 and x ⎛ v1 b t 2 b2 t 3 ⎞ . The maximum
⎝⎜ 2 2m ⎠⎟ = ⎝⎜⎜ v0t − 2 + 12m2 ⎠⎟⎟
0 0

2m

distance will occur at the time when the velocity is 0. From the equation for the velocity, we see that

happens at tmax = 2mv012 . Use this time in the expression for distance to find the maximum distance.
b

( )x t = tmax = v0 2mv012 − v b1 ⎛ 2mv012 ⎞2 + b2 ⎛ 2mv012 ⎞3 = 2mv023 − 2 mv 3 + 2mv023 = 2mv023
b 2 ⎜⎜⎝ b ⎠⎟⎟ 12m2 ⎜⎜⎝ b ⎟⎠⎟ b 2 3b 3b
0 0

2m b

74. The net force is the force of gravity downward, and the drag force upwards. Let the downward

direction be positive. Represent the value of 1.00 ×104 kg s by the symbol b, as in Eq. 5-6.

∑F = mg − Fd = mg − bv = ma = m dv → dv = g − b v →
dt dt m

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148