Chapter 4 Problems 1

1

CHAPTER 4 FORCES AND NEWTON'S LAWS OF MOTION Homework assignments: 1,2,5,6,11,14,18,20,26,31,34,36,39,46,50,55,63,67,70,76,81 1. REASONING AND SOLUTION According to Newton’s second law, the acceleration is

a = ΣF/m. Since the pilot and the plane have the same acceleration, we can write

( ) PILOTPILOTPILOT PLANE PLANE

orF F FF mm m m

Σ Σ Σ⎛ ⎞ ⎛ ⎞ ⎛ ⎞= Σ =⎜ ⎟ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠

Therefore, we find

( ) ( )4

4PILOT

3.7 10 N78 kg 93 N3.1 10 kg

F⎛ ⎞×

Σ = =⎜ ⎟×⎝ ⎠

______________________________________________________________________________ 2. REASONING Newton’s second law of motion gives the relationship between the net force

ΣF and the acceleration a that it causes for an object of mass m. The net force is the vector sum of all the external forces that act on the object. Here the external forces are the drive force, the force due to the wind, and the resistive force of the water.

SOLUTION We choose the direction of the drive force (due west) as the positive direction.

Solving Newton’s second law ( )F maΣ = for the acceleration gives

24100 N 800 N 1200 N 0.31 m/s6800 kg

Fam

Σ + − −= = = +

The positive sign for the acceleration indicates that its direction is due west .

t =v – v0

ΣF / m=

m v – v0( )ΣF

=5.0 kg( ) 4.0 ×10 3 m/s( )– 0 m/s( )[ ]

4.9 × 105 N= 4.1 ×10 –2 s

______________________________________________________________________________ 5. SSM REASONING The net force acting on the ball can be calculated using Newton's

second law. Before we can use Newton's second law, however, we must use Equation 2.9 from the equations of kinematics to determine the acceleration of the ball.

SOLUTION According to Equation 2.9, the acceleration of the ball is given by

a =v2 − v0

2

2x Thus, the magnitude of the net force on the ball is given by

2 2 2 20 (45 m/s) ? 0 m/s)(0.058 kg) 130 N

2 2(0.44 m)v v

F ma mx

⎛ ⎞ ⎡ ⎤−∑ = = = =⎜ ⎟ ⎢ ⎥

⎣ ⎦⎝ ⎠

FORCES AND NEWTON'S LAWS OF MOTION

2

_________________________________________________________________________ 6. REASONING AND SOLUTION The acceleration required is

a = v2 - v

02

2x = - (15.0 m/s)

2

2(50.0 m) = - 2.25 m/s2

Newton's second law then gives the magnitude of the net force as

F = ma = (1580 kg)(2.25 m/s2) = 3560 N ____________________________________________________________________________________________ 11. REASONING Newton’s second law gives the acceleration as a = (ΣF)/m. Since we seek

only the horizontal acceleration, it is the x component of this equation that we will use; ax = (ΣFx)/m. For completeness, however, the free-body diagram will include the vertical forces also (the normal force FN and the weight W).

SOLUTION The free body diagram is

shown at the right, where

F1 = 59.0 N F2 = 33.0 N θ = 70.0°

When F1 is replaced by its x and y components, we obtain the free body diagram in the second drawing.

Choosing right to be the positive direction, we

have

( ) ( )

1 2

2

cos

59.0 N cos 70.0 33.0 N1.83 m/s

7.00 kg

xx

x

F F Fa

m m

a

θΣ −= =

° −= = −

The minus sign indicates that the horizontal acceleration points to the left .

14. REASONING Newton’s second law, ΣF = ma, states that a net force of ΣF must act on an

object of mass m in order to impart an acceleration a to the object. In the impact shock test the box is subjected to a large deceleration and, hence, a correspondingly large net force.

+y

+x

FN

F2

F1W

θ

+y

+x

FN

F2 F1cos θ

WF1sin θ

Chapter 4 Problems 3

3

To determine the net force we will determine the deceleration in a kinematics calculation and use it in Newton’s second law.

SOLUTION According to Newton’s second law, the net force is ΣF = ma, where the

acceleration can be determined with the aid of Equation 2.4 (v = v0 + at). According to this equation

0v va

t−

=

Substituting this result for the acceleration into the second law gives

0v vF ma m

t−⎛ ⎞

Σ = = ⎜ ⎟⎝ ⎠

Since the initial velocity (v0 = +220 m/s), final velocity (v = 0 m/s), and the duration of the collision (t = 6.5 × 10−3 s) are known, we find

( ) 603

0 m/s 220 m/s41 kg 1.39 10 N6.5 10 s

v vF m

t −

−⎛ ⎞ −⎛ ⎞Σ = = = − ×⎜ ⎟ ⎜ ⎟×⎝ ⎠⎝ ⎠

The minus sign indicates that the net force points opposite to the direction in which the box

is thrown, which has been assumed to be the positive direction. The magnitude of the net force is 1.39 × 106 N , which is over three hundred thousand pounds.

18. REASONING Newton’s law of universal gravitation indicates that the gravitational force

that each uniform sphere exerts on the other has a magnitude that is inversely proportional to the square of the distance between the centers of the spheres. Therefore, the maximum gravitational force between two uniform spheres occurs when the centers of the spheres are as close together as possible, and this occurs when the surfaces of the spheres are touching. Then, the distance between the centers of the spheres is the sum of the two radii.

SOLUTION When the bowling ball and the billiard ball are touching, the distance between

their centers is r = rBowling + rBilliard. Using this expression in Newton’s law of universal gravitation gives

( )

( )( )( )( )

Bowling Billiard Bowling Billiard22

Bowling Billiard

11 2 29

2

6.67 10 N m / kg 7.2 kg 0.38 kg9.6 10 N

0.11 m 0.028 m

Gm m Gm mF

r r r

−−

= =+

× ⋅= = ×

+

FORCES AND NEWTON'S LAWS OF MOTION

4

20. REASONING AND SOLUTION The forces that act on the rock are shown at the right. Newton's second law (with the direction of motion as positive) is

ΣF = mg – R = ma

Solving for the acceleration a gives

( )( ) ( )22

45 kg 9.80 m/s ? 50 N– 4.2 m/s45 kg

mg Ram

= = =

mg

R

____________________________________________________________________________________________ 26. REASONING Newton’s law of gravitation shows how the weight W of an object of mass

m is related to the mass M and radius r of the planet on which the object is located: W = GMm/r2. In this expression G is the universal gravitational constant. Using the law of gravitation, we can express the weight of the object on each planet, set the two weights equal, and obtain the desired ratio.

SOLUTION According to Newton’s law of gravitation, we have

A B2 2

A B

Weight on planet A Weight on planet B

GM m GM mr r

=

The mass m of the object, being an intrinsic property, is the same on both planets and can be

eliminated algebraically from this equation. The universal gravitational constant can likewise be eliminated algebraically. As a result, we find that

2

A B A A2 2 2

A B B B

A A

B B

or

0.60 0.77

M M M rr r M r

r Mr M

= =

= = =

31. REASONING According to Equation 4.4, the weights of an object of mass m on the

surface of a planet (mass = M, radius = R ) and at a height H above the surface are

W =GMmR 2

On surface

and WH =GMm

R + H( )2

At height H above surface

The fact that W is one percent less than WH tells us that the WH/W = 0.9900, which is the

starting point for our solution.

Chapter 4 Problems 5

5

SOLUTION The ratio WH/W is

WH

W=

GMmR + H( )2

GMmR 2

=R2

R + H( )2 =1

1 + H / R( )2 = 0.9900

Solving for H/R gives

1 +HR

=1

0.9900 or

HR

= 0.0050

____________________________________________________________________________________________ 34. REASONING In each case the object is in equilibrium. According to Equation 4.9b,

ΣFy = 0, the net force acting in the y (vertical) direction must be zero. The net force is composed of the weight of the object(s) and the normal force exerted on them.

SOLUTION a. There are three vertical forces acting on the crate: an upward normal force +FN that the

floor exerts, the weight –m1g of the crate, and the weight –m2g of the person standing on the crate. Since the weights act downward, they are assigned negative numbers. Setting the sum of these forces equal to zero gives

N 1 2 ( ) ( ) 0

yF

F m g m gΣ

+ − + − =

The magnitude of the normal force is

FN = m1g + m2g = (35 kg + 65 kg)(9.80 m/s2) = 980 N b. There are only two vertical forces acting on the person: an upward normal force +FN that

the crate exerts and the weight –m2g of the person. Setting the sum of these forces equal to zero gives

N 2 ( ) 0

yF

F m gΣ

+ − =

The magnitude of the normal force is

FN = m2g = (65 kg)(9.80 m/s2) = 640 N ____________________________________________________________________________________________ 36. REASONING AND SOLUTION The apparent weight is

FN = mw(g + a) We need to find the acceleration, a. Let T represent the force applied by the hoisting cable.

Newton's second law applied to the elevator gives

FORCES AND NEWTON'S LAWS OF MOTION

6

T – (mw + me)g = (mw + me)a Solving for a gives

a = Tmw + me

- g = 9410 N60.0 kg + 815 kg - 9.80 m/s2 = 0.954 m/s2.

Now the apparent weight is

FN = 60.0 kg(9.80 m/s2 + 0.954 m/s2) = 645 N ____________________________________________________________________________________________

39. SSM REASONING AND SOLUTION Four forces act on the sled. They are the pulling force P, the force of kinetic friction kf , the weight mg of the sled, and the normal force NF exerted on the sled by the surface on which it slides. The following figures show free-body diagrams for the sled. In the diagram on the right, the forces have been resolved into their x and y components.

fk

x

y

mg

FNP cos θ

P sin θ

θfk

x

y

mg

FNP

Since the sled is pulled at constant velocity, its acceleration is zero, and Newton's second

law in the direction of motion is (with right chosen as the positive direction)

kcos 0x xF P f maθ= − = =∑ From Equation 4.8, we know that k k Nf Fµ= , so that the above expression becomes k Ncos 0P Fθ µ− = (1) In the vertical direction, Nsin 0y yF P F mg maθ= + − = =∑ (2) Solving Equation (2) for the normal force, and substituting into Equation (1), we obtain

Chapter 4 Problems 7

7

( )kcos sin 0P mg Pθ µ θ− − =

Solving for kµ , the coefficient of kinetic friction, we find

k 2cos (80.0 N) cos 30.0 0.444

sin (20.0 kg) (9.80 m/s ) (80.0N) sin 30.0P

mg Pθµ

θ°

= = =− − °

____________________________________________________________________________________________ 46. REASONING At first glance there seems to be very little information given. However, it

is enough. In part a of the drawing the bucket is hanging stationary and, therefore, is in equilibrium. The forces acting on it are its weight and the two tension forces from the rope. There are two tension forces from the rope, because the rope is attached to the bucket handle at two places. These three forces must balance, which will allow us to determine the weight of the bucket. In part b of the drawing, the bucket is again in equilibrium, since it is traveling at a constant velocity and, therefore, has no acceleration. The forces acting on the bucket now are its weight and a single tension force from the rope, and they again must balance. In part b, there is only a single tension force, because the rope is attached to the bucket handle only at one place. This will allow us to determine the tension in part b, since the weight is known.

SOLUTION Let W be the weight of the bucket, and let T be the tension in the rope as the

bucket is being pulled up at a constant velocity. The free-body diagrams for the bucket in parts a and b of the drawing are as follows:

Since the bucket in part a is in equilibrium, the net force acting on it is zero. Taking upward

to be the positive direction, we have

92.0 N 92.0 N 0 or 184 NF W WΣ = + − = = Similarly, in part b we have

0 or 184 NF T W T WΣ = − = = =

Free-body diagram for part a

W

92.0 N 92.0 N

Free-body diagram for part b

W

T

FORCES AND NEWTON'S LAWS OF MOTION

8

50. REASONING The drawing shows the I-beam and the three forces that act on it, its weight W and the tension T in each of the cables. Since the I-beam is moving upward at a constant velocity, its acceleration is zero and it is in vertical equilibrium. According to Equation 4.9b, 0yF =∑ , the net force in the vertical (or y) direction must be zero. This relation will allow us to find the magnitude of the tension.

SOLUTION Taking up as the +y direction, Equation 4.9b

becomes

sin 70.0 sin 70.0 N 0

yF

T T 3

Σ

+ ° + ° − 8.00 ×10 =

Solving this equation for the tension gives 4260 NT = . ____________________________________________________________________________________________ 55. REASONING Since the boxes are at rest, they are in equilibrium. According to Equation

4.9b, the net force in the vertical, or y, direction is zero, ΣFy = 0. There are two unknowns in this problem, the normal force that the table exerts on box 1 and the tension in the rope that connects boxes 2 and 3. To determine these unknowns we will apply the relation ΣFy = 0 twice, once to the boxes on the left of the pulley and once to the box on the right.

SOLUTION There are four forces acting on the two boxes on the

left. The boxes are in equilibrium, so that the net force must be zero. Choosing the +y direction as being the upward direction, we have that

1 2 N 0

yF

W W F TΣ

− − + + = (1)

where W1 and W2 are the magnitudes of the weights of the boxes, FN is the magnitude of the normal force that the table exerts on box 1, and T is the magnitude of the tension in the rope. We know the weights. To find the unknown tension, note that the box 3 is also in equilibrium, so that the net force acting on it must be zero.

3 30 so that

yF

W T T WΣ

− + = =

Substituting this expression for T into Equation (1) and solving for the normal force gives

N 1 2 3 55 N + 35 N 28 N = 62 NF W W W= + − = − ____________________________________________________________________________________________

70.0° 70.0°

T T

W = –8.00 × 103 N

T

T

W1

W2

W3FN

1

2

3

Chapter 4 Problems 9

9

63. REASONING According to Newton’s second law, the acceleration has the same direction as the net force and a magnitude given by a = ΣF/m.

SOLUTION Since the two forces are perpendicular, the magnitude of the net force is given

by the Pythagorean theorem as ΣF = 40.0 N( )2 + 60.0 N( )2 . Thus, according to Newton’s second law, the magnitude of the acceleration is

a =ΣFm

=40.0 N( )2 + 60.0 N( )2

4.00 kg= 18.0 m/s2

The direction of the acceleration vector is given by

θ = tan –1 60.0 N40.0 N

⎛ ⎝ ⎜ ⎞

⎠ ⎟ = 56.3° above the + x axis

______________________________________________________________________________ 67. REASONING AND SOLUTION

Newton's second law applied to object 1 (422 N) gives

T = m1a1

Similarly, for object 2 (185 N)

T – m2g = m2a2 If the string is not to break or go

slack, both objects must have accelerations of the same magnitude.

T

W2

Object 2

W1

T

N1

Object 1

Then a1 = a and a2 = –a. The above equations become T = m1a (1) T – m2g = – m2a (2) a. Substituting Equation (1) into Equation (2) and solving for a yields

a =m2 g

m1 + m2= 2.99 m/s2

b. Using this value in Equation (1) gives

T = m1 a = 129 N ____________________________________________________________________________________________ 70. REASONING AND SOLUTION

FORCES AND NEWTON'S LAWS OF MOTION

10

a. In the vertical direction ΣFy = may gives

T – mg = may so

T = may + mg = mg(1 + ay/g)

T = 822 N( ) 1 +1.10 m/s 2

9.80 m/s2

⎛

⎝ ⎜

⎞

⎠ ⎟ = 914 N

b. The acceleration of the man is zero if his velocity is constant. From part a

T = mg = 822 N

____________________________________________________________________________________________ 76. REASONING AND SOLUTION If the +x axis is taken to be parallel to and up the ramp,

then ΣFx = max gives T – fk – mg sin 30.0° = max

where fk = µkFN . Hence, T = max + µkFN + mg sin 30.0° (1)

Also, ΣFy = may gives FN – mg cos 30.0° = 0

since no acceleration occurs in this direction. Then

FN = mg cos 30.0° (2) Substitution of Equation (2) into Equation (1) yields

T = max + µkmg cos 30.0° + mg sin 30.0°

T = (205 kg)(0.800 m/s2) + (0.900)(205 kg)(9.80 m/s2)cos 30.0° + (205 kg)(9.80 m/s2)sin 30.0° = 2730 N ____________________________________________________________________________________________ 81. REASONING The tension in each coupling bar is responsible for accelerating the objects

behind it. The masses of the cars are m1, m2, and m3. We can use Newton’s second law to express the tension in each coupling bar, since friction is negligible:

( ) ( )1 2 3 2 3 3

Coupling bar CCoupling bar A Coupling bar B

A B CT m m m a T m m a T m a= + + = + =

In these expressions a = 0.12 m/s2 remains constant. Consequently, the tension in a given

bar will change only if the total mass of the objects accelerated by that bar changes as a

Chapter 4 Problems 11

11

result of the luggage transfer. Using ∆ (Greek capital delta) to denote a change in the usual fashion, we can express the changes in the above tensions as follows:

( ) ( ) ( )1 2 3 2 3 3

Coupling bar CCoupling bar A Coupling bar B

A B CT m m m a T m m a T m a⎡ ⎤ ⎡ ⎤∆ = ∆ + + ∆ = ∆ + ∆ = ∆⎣ ⎦ ⎣ ⎦

SOLUTION a. Moving luggage from car 2 to car 1 does not change the total mass m1 + m2 + m3, so

∆(m1 + m2 + m3) = 0 kg and 0 NAT∆ = . The transfer from car 2 to car 1 causes the total mass m2 + m3 to decrease by 39 kg, so

∆(m2 + m3) = –39 kg and

( ) ( )( )22 3 –39 kg 0.12 m/s 4.7 NBT m m a⎡ ⎤∆ = ∆ + = = −⎣ ⎦

The transfer from car 2 to car 1 does not change the mass m3, so ∆m3 = 0 kg and

0 NCT∆ = . b. Moving luggage from car 2 to car 3 does not change the total mass m1 + m2 + m3, so

∆(m1 + m2 + m3) = 0 kg and 0 NAT∆ = . The transfer from car 2 to car 3 does not change the total mass m2 + m3, so ∆(m2 + m3) = 0

kg and 0 NBT∆ = . The transfer from car 2 to car 3 causes the mass m3 to increase by 39 kg, so ∆m3 = +39 kg

and ( ) ( )( )2

3 39 kg 0.12 m/s 4.7 NCT m a∆ = ∆ = + = + ____________________________________________________________________________________________