8.1. Visualize:

Solve: Figure (i) shows a weightlifter (WL) holding a heavy barbell (BB) across his shoulders. He is standing ona rough surface (S) that is a part of the earth (E). We distinguish between the surface (S), which exerts a contactforce, and the earth, which exerts the long range force of gravity. Figure (ii) shows the barbell, the weightlifter, andthe earth separated from one another. This separation helps to indicate the forces acting on each object. Figures (iii)and (iv) are free body diagrams for the barbell and the weightlifter, respectively.

Altogether there are four interactions. There is the interaction between the barbell and the weightlifter, theweightlifter and the surface of the earth (contact force), the barbell and the earth, and the weightlifter and the earth.

The interaction between the barbell and the weightlifter leads to two forces: rF WL on BB and

rF BB on WL. These are an

action/reaction pair. The interaction between the weightlifter and the surface of the earth is a contact interactionleading to

rn S on WL and

rn WL on S. These two forces also constitute an action/reaction pair. The weight force

rw E on BB

has its action/reaction force rw BB on E, and the weight force

rw E on WL has its action/reaction force

rw WL on E.

8.2. Visualize:

Solve: Figure (i) shows a softball player (P) with a ball (B) in her hand that has come forward beside her headand is just about ready to release the ball. We have represented the earth (E) and its rough surface (S). Contactforces exist between the surface and the softball player. Figure (ii) shows the ball, the player, and the earthseparated. This separation helps to indicate the various forces on each object. We have altogether 6 interactions.The action/reaction pairs of all the forces have been indicated by dotted lines. Figure (iii) shows the free-bodydiagrams for the ball and the softball player.

8.3. Visualize:

Solve: Figure (i) shows the head-on collision of a bowling ball (BB) and a soccer ball (SB) on a hard surface (S)of the earth (E). There are 3 types of interactions in the present case: between the bowling ball and the soccer ball,between the earth and the balls (long-range force of gravity), and between the balls and the surface (contact force).Figure (ii) shows the bowling ball, the soccer ball, and the surface separated from one another. The force on the

bowling ball due to the soccer ball rF SB on BB and the force on the soccer ball due to the bowling ball

rF BB on SB make

an action/reaction pair. The weights rw E on SB and

rw SB on E are an action/reaction pair, and so are the weights

rw E on BB

and rw BB on E. The contact forces

rn S on BB and

rn S on SB have action/reaction forces in

rn BB on S and

rn SB on S, respectively.

All the action/reaction pairs are indicated by dotted lines. Figure (iii) shows the free-body diagrams for the bowlingball and the soccer ball.

8.4. Visualize:

Solve: Figure (i) shows the mountain climber (C), the rope (R), the bag of supplies (B), the surface of themountain (S), and the earth (E). To indicate the forces on each of the various objects, we have separated C, R, B, S,and E. In the diagram,

rn denotes the normal (contact) forces,

rf denotes the frictional (contact) forces, and

rw

denotes the weights. All action/reaction forces have also been identified in this figure with dotted lines. Figure (ii)shows the free-body diagrams on the bag, the rope, and the climber.

8.5. Visualize:

Solve: The top figure shows the pulley (P), block A, block B, the surface S of the incline, the rope (R), and theearth (E). In indicating the various forces, we have denoted the normal (contact) forces by

rn , the kinetic frictional

(contact) forces by rf , and the weights by

rw . All the action/reaction forces have been identified with dotted lines.

The bottom figure shows the free body diagrams of each block, the rope, and the pulley.

8.6. (a) Visualize: The upper magnet is labeled U, the lower magnet L. Each magnet exerts a long-rangemagnetic force on the other. Each magnet and the table exert a contact force (normal force) on each other. In addition,the table experiences a normal force due to the surface.

(b) Solve: Each object is in static equilibrium with (Fnet)y = 0. Start with the lower magnet. Because FU on L = 3wL =6.0 N, equilibrium requires nT on L = 4.0 N. For the upper magnet, FL on U = FU on L = 6.0 N because these are anaction/reaction pair. Equilibrium for the upper magnet requires nT on U = 8.0 N. For the table, action/reaction pairsare nL on T = nT on L = 4.0 N and nU on T = nT on U = 8.0. The table’s weight is wT = 20 N, leaving nS on T = 24.0 N in orderfor the table to be in equilibrium. Summarizing,

Upper magnet Table Lower magnet

wU = 2.0 N wT = 20.0 N wL = 2.0 N nT on U = 8.0 N nU on T = 8.0 N nT on L = 4.0 N FL on U = 6.0 N nL on T = 4.0 N FU on L = 6.0 N

n S on T = 24.0 N

Assess: The result nS on T = 24.0 N makes sense. The combined weight of the table and two magnets is 24.0 N.Because the table is in equilibrium, the upward normal force of the surface has to exactly balance the total weightof the table and magnets.

8.7. Model: We will model the astronaut and the chair as particles. The astronaut and the chair will be denotedby A and C, respectively, and they are separate systems. The launch pad is a part of the environment.Visualize:

Solve: (a) Newton’s second law for the astronaut is

F n w m ayon A C on A A A A N( ) = − = =∑ 0 ⇒ = =n w m gC on A A A

By Newton’s third law, the astronaut’s force on the chair is

n n m gA on C C on A A280 kg 9.8 m/s 784 N= = = ( )( ) =

(b) Newton’s second law for the astronaut is:

F n w m ayon A C on A A A A( ) = − =∑ ⇒ = + = +( )n w m a m g aC on A A A A A A

By Newton’s third law, the astronaut’s force on the chair is

n n m g aA on C C on A A A2 280 kg 9.8 m/s 10 m/s 1580 N= = +( ) = ( ) +( ) =

Assess: This is a reasonable value because the astronaut’s acceleration is more than g.

8.8. Model: The car and the truck will be modeled as particles and denoted by the symbols C and T, respectively.The ground will be denoted by the symbol G.Visualize:

Solve: (a) The x-component of Newton’s second law for the car is

F F F m axon C G on C T on C C C( ) = − =∑

The x-component of Newton’s second law for the truck is

F F m axon T C on T T T( ) = =∑

Using aC = aT = a and FT on C = FC on T, we get

F Fm

aC on G C on TC

−( )

=1 F

maC on T

T

( )

=1

Combining these two equations,

F Fm

FmC on G C on T

CC on T

T

−( )

= ( )

1 1 ⇒ +

= ( )

Fm m

FmC on T

C TC on G

C

1 1 1

⇒ = ( ) +

F Fm

m mC on T C on GT

C T

= ( )+

4500 N2000 kg

1000 kg 2000 kg= 3000 N

(b) Due to Newton’s third law, FT on C = 3000 N.

8.9. Model: The blocks are to be modeled as particles and denoted as 1, 2, and 3. The surface is frictionlessand along with the earth it is a part of the environment. The three blocks are our three systems of interest.Visualize:

The force applied on block 1 is FA on 1 = 12 N. The acceleration for all the blocks is the same and is denoted by a.Solve: Newton’s second law for the three blocks along the x-direction is

F F F m axon 1 A on 1 2 on 1( ) = − =∑ 1 F F F m a

xon 2 1 on 2 3 on 2( ) = − =∑ 2 F F m axon 3 2 on 3( ) = =∑ 3

Adding these three equations and using Newton’s third law (F2 on 1 = F1 on 2 and F3 on 2 = F2 on 3), we get

F m m m aA on 1 = + +( )1 2 3 ⇒ ( ) = + +( )12 N 1 kg 2 kg 3 kg a ⇒ =a 2 m/s2

Using this value of a, the force equation on block 3 gives

F m a2 on 323 kg 2 m/s 6 N= = ( )( ) =3

Substituting into the force equation on block 1,

12 2 N 1 kg m/s2 on 12− = ( )( )F ⇒ =F2 on 1 10 N

Assess: Because all three blocks are pushed forward by a force of 12 N, the value of 10 N for the force that the2 kg block exerts on the 1 kg block is reasonable.

8.10. Model: The astronaut and the satellite, the two systems of our interest, will be treated as particles.Visualize:

Solve: The astronaut and the satellite accelerate in opposite directions for 0.5 s. The force on the satellite and theforce on the astronaut are an action/reaction pair, so both are 100 N. Newton’s second law for the satellite along thex-direction is

F F m axon S A on S S S( ) = =∑ ⇒ = =

−( ) = −aF

mSA on S

S

2100 N

640 kg0.156 m/s

Newton’s second law for the astronaut along the x-direction is

F F m axon A S on A A A( ) = =∑ ⇒ = = = =a

F

m

F

mAS on A

A

A on S

A

2100 N

80 kg1.25 m/s

Let us first calculate the positions and velocities of the astronaut and the satellite at t1 = 0.5s under the accelerationsaA and aS:

x x v t t a t t1A 0A 0A A= + −( ) + −( )1 012 1 0

2 = + + ( ) −( )0 0 12

2m m 1.25 m/s 0.5 s 0 s2 = 0.156 m

x x v t t a t t1S 0S 0S S= + −( ) + −( )1 012 1 0

2 = + + −( ) −( ) = −0 0 0 156 0 0212

2m m m/s 0.5 s 0 s m2. .

v v a t t1A 0A A= + −( )1 0 = + ( ) −( )0 1 25 0m/s m/s 0.5 s s2. = 0.625 m/s

v v a t t1S 0S S= + −( )1 0 = + −( ) −( ) = −0 0 156 0 078m/s m/s 0.5 s 0 s m/s2. .

With x1A and x1S as initial positions, v1A and v1S as initial velocities, and zero accelerations, we can now obtain thenew positions at t t2 1−( ) = 59.5 s:

x x v t t2A 1A 1A= + −( )2 1 = + ( )( )0.156 m 0.625 m/s 59.5 s = 37.34 m

x x v t t2S 1S 1S= + −( )2 1 = − + −( )( ) = −0 0 078 4 66. . .02 m m/s 59.5 s m

Thus the astronaut and the satellite are x x2A 2S 37.34 m 4.66 m 42.0 m− = ( ) − −( ) = apart.

8.11. Model: The man (M) and the block (B) are interacting with each other through a rope. We will assume thepulley to be frictionless. This assumption implies that the tension in the rope is the same on both sides of the pulley.The two systems are the man and the block.Visualize:

Solve: Clearly the entire system remains in equilibrium since mB > mM. The block would move downward but it isalready on the ground. From the free-body diagrams, we can write down Newton’s second law in the verticaldirection as

F T wyon M R on M M N( ) = − =∑ 0 ⇒ = = ( )( ) =T wR on M M

260 kg 9.8 m/s 588 N

Since the tension is the same on both sides, T T TB on R M on R 588 N= = = .

8.12. Model: The carp (C) and the trout (T) are two systems that will be represented through the particlemodel. The fishing rod line (R) is assumed to be massless.Visualize:

Solve: Jimmy’s pull T2 is larger than the total weight of the fish, so they accelerate upward. They are tied together, soeach fish has the same acceleration a. Newton’s second law along the y-direction for the carp and the trout is

F T T w m ayon C C C( ) = − − =∑ 2 1 F T w m a

yon T T T( ) = − =∑ 1

Adding these two equations gives

aT w w

m m= − −

+( ) =− ( )( ) − ( )( )

+2 C T

C T

2 260 N 1.5 kg 9.8 m/s 3 kg 9.8 m/s

1.5 kg 3.0 kg= 3.533 m/s2

Substituting this value of acceleration back into the force equation for the trout, we find that

T m a g1 = +( ) = ( ) +( )T2 23 kg 3.533 m/s 9.8 m/s = 40 N

w m gT T23 kg 9.8 m/s 29.4 N= = ( )( ) = w m gC C

21.5 kg 9.8 m/s 14.7 N= = ( )( ) =

Thus, T T w w2 1> > >T C .

8.13. Model: The block of ice (I) is a particle and so is the rope (R) because it is not massless. We musttherefore consider both the block of ice and the rope as systems.Visualize:

Solve: The force rFext acts only on the rope. Since the rope and the ice block move together, they have the same

acceleration. Also because the rope has mass, Fext on the front end of the rope is not the same as FI on R that acts onthe rear end of the rope.

Newton’s second law along the x-axis for the ice block and the rope is

F F m axon I R on I I

210 kg 2.0 m/s 20 N( ) = = = ( )( ) =∑F F F m a

xon R ext I on R R( ) = − =∑ ⇒ − =F F m aext R on I R

⇒ = + = + ( )( ) =F F m aext R on I R220 N 0.5 kg 2.0 m/s 21 N

8.14. Model: The hanging block and the rail car are separate systems.Visualize:

Solve: The mass of the rope is very small in comparison to the 2000 kg block, so we will assume a massless rope.

In that case, the forces rT1 and

r′T1 act as if they are an action/reaction pair. The hanging block is in static

equilibrium, with rFnet N= 0 , so ′ = =T m g1 block 19,600 N . The rail car with the pulley is also in static equilibrium:

T T T2 3 1 0+ − = N

Notice how the tension force in the cable pulls both the top and bottom of the pulley to the right. Now,T T1 1 19 600= ′ = , N by Newton’s third law. Also, the cable tension is T T T2 3= = . Thus, T T= ′ =1

2 1 9800 N .

8.15. Model: The two hanging blocks, which can be modeled as particles, form two systems. The other twosystems are the two knots where rope 1 meets with rope 2 and rope 2 meets with rope 3. All the four systems are instatic equilibrium. The ropes are assumed to be massless.Visualize:

Solve: (a) We will consider both the two hanging blocks and the two knots. The blocks are in static equilibrium

with rFnet N= 0 . Note that there are three action/reaction pairs. For Block 1 and Block 2,

rFnet N= 0 and we have

′ = =T w m g4 1 1 ′ = =T w m g5 2 2

Then, by Newton’s third law:T T m g4 4 1= ′ = T T m g5 5 2= ′ =

The knots are also in equilibrium. Newton’s law applied to the left knot is

F T Txnet N( ) = − =2 1 1 0cosθ F T T T m g

ynet N( ) = − = − =1 1 4 1 1 1 0sin sinθ θ

The y-equation gives T m g1 1 1= sin .θ Substitute this into the x-equation to find

Tm g m g

21 1

1

1

1

= =cos

sin tan

θθ θ

Newton’s law applied to the right knot is

F T Txnet N( ) = − ′ =3 3 2 0cosθ F T T T m g

ynet N( ) = − = − =3 3 5 3 3 2 0sin sinθ θ

These can be combined just like the equations for the left knot to give

′ = =Tm g m g

22 3

3

2

3

cos

sin tan

θθ θ

But the forces rT2 and

r′T2 are an action/reaction pair, so T T2 2= ′ . Therefore,

m g m g m

m1

1

2

33

2

11tan tan

tan tanθ θ

θ θ= ⇒ = ⇒ = °( ) = °−θ31 2 20 36tan tan

We can now use the y-equation for the right knot to find T m g3 2 3= =sinθ 66.6 N .

8.16. Model: The block (B) and the steel cable (C), the two systems of interest to us, are treated like particles.The motion of these systems is governed by the constant-acceleration kinematic equations.Visualize:

Solve: Using v v a x xx x x12

02

1 02= + −( ) ,

4.0 m/s m /s 2.0 m2 2( ) = + ( )20 2ax ⇒ ax = 4.0 m/s2

From the free-body diagram on the block:

F F m ax xon B C on B B( ) = =∑ ⇒ = ( )( ) =FC on B

220 kg 4.0 m/s 80 N

Also, according to Newton’s third law FB on C = FC on B = 80 N. Newton’s second law on the cable is:

F F F m ax xon C ext B on C C( ) = − =∑ ⇒ − = ( )100 N 80 N 4.0 m/sC

2m ⇒ =mC 5 kg

8.17. Model: The block (B) and the steel cable (C), the two systems of our interest, are considered particles,and their motion is determined by the constant-acceleration kinematic equations.Visualize:

Solve: Using v v a t tx x x1 0 1 0= + −( ) ,

4.0 m/s m/s 2.0 s 0 s= + −( )0 ax ⇒ =ax 2.0 m/s2

Newton’s second law along the x-direction for the block is

F F m ax xon B C on B B( ) = =∑ = ( )( ) =20 kg 2.0 m/s 40 N2

Fext acts on the right end of the cable and FB on C acts on the left end. According to Newton’s third law, FB on C = FC on B =40 N. The difference in tension between the two ends of the cable is thus

Fext − FB on C = 100 N – 40 N = 60 N

8.18. Model: The block (B) and the steel cable (C), the two systems of our interest, are treated as particles,and their motion is determined by constant-acceleration kinematic equations. We ignore the hanging shape of thecable.Visualize:

Solve: Using x x v t t a t tx x1 0 0 1 012 1 0

2= + −( ) + −( ) ,

4.0 m 0 m m 2.0 s 0 s= + + −( )0 12

2ax ⇒ =ax 2.0 m/s2

Newton’s second law along the x-direction for the block is

F F m ax xonB C on B B

220 kg 2.0 m/s 40 N( ) = = = ( )( ) =∑Therefore, the change in tension (T) in the cable from one end to the other is F Fext B on C 100 N 40 N 60 N.− = − =The tension in the cable as a function of x is

T x x( ) = + ( )40 N

60 N

1 m= +( )40 60x N

with x in meters. x = 0 m is where the cable attaches to the box and x = 1.0 m is at the right end of the cable.

8.19. Visualize: Consider a segment of the rope of length y, starting from the bottom of the rope. The weightof this segment of rope is a downward force. It is balanced by the tension force at height y.

Solve: The mass m of this segment of rope is the same fraction of the total mass M = 2.2 kg as length y is afraction of the total length L = 3.0 m. That is, m/M = y/L, from which we can write the mass of the rope segment

mM

Ly=

This segment of rope is in static equilibrium, so the tension force pulling up on it is

T w mgMg

Ly y y= = = = =( . )

..

2 2

3 07 19

kg)(9.8 m/s

m N

2

where y is in m. The tension increases linearly from 0 N at the bottom (y = 0 m) to 21.6 N at the top (y = 3 m). Thisis shown in the graph.

8.20. Model: Sled A, sled B, and the dog (D) are treated like particles in the model of kinetic friction.Visualize:

Solve: The acceleration constraint is a a ax x xA B( ) = ( ) = . Newton’s second law on sled A is

rF n w

yon A A A N( ) = − =∑ 0 ⇒ =n wA A

rF T f m a

xxon A 1 on A A A( ) = − =∑

Using fA = µknA, the x-equation yields

T n m ax1 on A k A A− =µ ⇒ − ( )( )( ) = ( )150 N 100 kg 9.8 m/s 100 kg20 1. ax ⇒ =ax 0.52 m/s2

On sled B:rF n w

yon B B B N( ) = − =∑ 0 ⇒ =n wB B

rF T T f m a

xxon B 2 1 on B B B( ) = − − =∑

T1 on B and T1 on A act as if they are an action/reaction pair, so T1 on B N.= 150 Using f nB k B 80 kg= = ( )( )µ 0 10.

9.8 m/s 78.4 N2( ) = , we get

T22150 N 78.4 N 80 kg 0.52 m/s− − = ( )( ) ⇒ =T2 270 N

Thus the tension T2 = 270 N.

8.21. Model: The coffee mug (M) is our system of interest, and it will be treated as a particle. The model offriction and the constant-acceleration kinematic equations will also be used.Visualize:

Solve: The mug and the car have the same velocity. If the mug does not slip, their accelerations will also be the same.Using v v a x xx x x1

202

1 02= + −( ) , we get

0 22

m /s 20 m/s 50 m2 2 = ( ) + ( )ax ⇒ = −ax 4.0 m / s2

The static force needed to stop the mug is

F f max xnet s

20.5 kg 4.0 m/s( ) = − = = ( ) −( ) = −2.0 N ⇒ =fs .0 N2

The maximum force of static friction is

f n w mgs max s s s= = =µ µ µ = ( )( )( )0 50. 0.50 kg 9.8 m/s2 = 2.45 N

Since fs < fs max, the mug does not slide.

8.22. Visualize:

The car and the ground are denoted by C and G, respectively.Solve: (a) As the drive wheels turn they push backward against the ground. This is a static friction force

rFC on G

because the wheels don’t slip against the ground. By Newton’s third law, the ground exerts a reaction force rFG on C .

This reaction force is opposite in direction to rFC on G , hence, is in the forward direction. This is the force that

accelerates the car.

(b) The car has an internal source of energy – fuel – that allows it to turn the wheels and exert the force rFC on G .

This is an active push by the car, generating the force rFG on C in response. Houses do not have an internal source of

energy that allows them to push sideways against the ground.(c) The car presses down against the ground at both the drive wheels (assumed to be the front wheels F, although thatis not critical) and the nondrive wheels. For this car, two-thirds of the weight rests on the front wheels. Physically,

force rFG on C is a static friction force. The maximum acceleration of the car on the ground (or concrete surface) occurs

when the static friction reaches its maximum possible value.

F f n wG on C s s F s F= ( ) = =max

µ µ = ( )( )( )( ) =1 00 23. 1500 kg 9.8 m/s 9800 N2

⇒ = = =aF

mmaxG on C 29800 N

1500 kg6.53 m/s

8.23. Model: The starship and the shuttlecraft will be denoted as M and m, respectively, and both will betreated as particles. We will also use the constant-acceleration kinematic equations.Visualize:

Solve: (a) The tractor beam is some kind of long-range force rFM mon . Regardless of what kind of force it is, by

Newton’s third law there must be a reaction force rFm M on on the starship. As a result, both the shuttlecraft and the

starship move toward each other (rather than the starship remaining at rest as it pulls the shuttlecraft in). However,the very different masses of the two craft means that the distances they each move will also be very different. Thepictorial representation shows that they meet at time t1 when xM1 = xm1. There’s only one force on each craft, soNewton’s second law is very simple. Furthermore, because the forces are an action/reaction pair,

FM on m = Fm on M = Ftractor beam = 4 × 104 N

The accelerations of the two craft are

aF

MMm M= = ×

×=on

4

624 10 N

2 10 kg0.020 m/s a

F

mmM m= = − ×

×= −

ron

4

424 10 N

2 10 kg2.0 m/s

Acceleration am is negative because the force and acceleration vectors point in the negative x-direction. Theaccelerations are very different even though the forces are the same. Now we have a constant-acceleration problemin kinematics. At a later time t1 the positions of the crafts are

x x v t t a t t a tM M M M M1 0 0= + −( ) + −( ) =1 012 1 0

2 12 1

2

x x v t t a t t x a tm m m m m m1 0 0 0= + −( ) + −( ) = +1 012 1 0

2 12 1

2

The craft meet when xM1 = xm1, so

12 1

2 12 1

2a t x a tM m m= +0 ⇒ =−

=+

= ( ) =tx

a a

x

a am

M m

m

M m1

2 2 20 02

10,000 m

2.02 m/s99.5 s

Knowing t1, we can now find the starship’s position as it meets the shuttlecraft:

x a tM M1 99.0 m= =12 1

2

The starship moves 99.0 m as it pulls in the shuttlecraft from 10 km away.

8.24. Model: The rock (R) and Bob (B) are two systems of our interest, and will be treated as particles. Wewill also use the constant-acceleration kinematic equations.Visualize:

Solve: (a) Bob exerts a forward force rF B on R on the rock to accelerate it forward. The rock’s acceleration is

calculated as follows:

v v a x av

x1R 0R 0R R1R 2 m/s

m450 m/s2 2

2 2

22

30

2 1= + ⇒ = = ( )

( )=∆

∆

The force is calculated from Newton’s second law:

F m aB on R R R2 kg 450 m/s = 225 N= = ( )( ).5

(b) Because Bob pushes on the rock, the rock pushes back on Bob with a force rF R on B. Forces

rF R on B and

rF B on R are

an action/reaction pair, so FR on B = FB on R = 225 N. The force causes Bob to accelerate backward with anacceleration equal to

aF

m

F

mx

B

net on B

B

R on B

B

2225 N

75 kg3.0 m/s=

( )= − = − = −

This is a rather large acceleration, but it lasts only until Bob releases the rock. We can determine the time intervalby returning to the kinematics of the rock:

v v a t a t tv

a1R 0R R R1R

R

0.0667 s= + = ⇒ = =∆ ∆ ∆

At the end of this interval, Bob’s velocity isv v a t a t1B 0B B B 0.200 m/s= + = = −∆ ∆

Thus his recoil speed is 0.200 m/s.

8.25. Model: Assume package A and package B are particles. Use the model of kinetic friction and theconstant-acceleration kinematic equations.Visualize:

Solve: Package B has a smaller coefficient of friction. It will try to overtake package A and push against it.Package A will push back on B. The acceleration constraint is a a a

x xA B( ) = ( ) = .Newton’s second law for each package is

F F w f m a

F m g m g m a

xon A B on A A kA A

B on A A kA A A

( ) = + − =

⇒ + − ( ) =

∑ sin

sin cos

θ

θ µ θ

F F f w m axon B A on B kB B B( ) = − − + =∑ sinθ

⇒ − − ( ) + =F m g m g m aA on B kB B B Bµ θ θcos sin

where we have used n m g n m gA A B B and = =cos cos .θ θ Adding the two force equations, and using F FA on B B on A=because they are an action/reaction pair, we get

a gm m g

m m= −

+( )( )+

sincos

θµ µ θkA A kB B

A B

= 1.817 m/s2

Finally, using x x v t t a t t1 0 1 012 1 0

2= + −( ) + −( )0x ,

2 m m m 1.817 m/s s2= + + ( ) −( )0 0 012 1

2t ⇒ t1 = 1.48 s

8.26. Model: The two ropes and the two blocks (A and B) will be treated as particles.Visualize:

Solve: (a) The two blocks and two ropes form a combined system of total mass M = 2.5 kg. This combined

system is accelerating upward at a = 3.0 m/s2 under the influence of a force F and the weight −Mgj . Newton’ssecond law applied to the combined system is

F F Mg Ma F M a gynet 32.0 N( ) = − = ⇒ = +( ) =

(b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts onlyon block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’ssecond law to each system, starting at the top. Each has an acceleration a = 3.0 m/s2. For block A:

F F m g T m aynet on A A 1 on A A( ) = − − = ⇒ = − +( ) =T F m a g1 on A A 19.2 N

(c) Applying Newton’s second law to rope 1:

F T m g T m aynet on 1 A on 1 B on 1( ) = − − =1 1

rT A on 1 and

rT 1 on A are an action/reaction pair. But, because the rope has mass, the two tension forces

rT A on 1 andr

T B on 1 are not the same. The tension at the lower end of rope 1, where it connects to B, is

T T m a gB on 1 A on 1 16.0 N= − +( ) =1

(d) We can continue to repeat this procedure, noting from Newton’s third law that

T T1 on B B on 1= and T T2 on B B on 2=

Newton’s second law applied to block B is

F T m g T m aynet on B 1 on B B 2 on B B( ) = − − = ⇒ = − +( ) =T T m a g2 on B 1 on B B 3.2 N

8.27. Model: The two blocks (1 and 2) are the systems of interest and will be treated as particles. The ropes areassumed to be massless, and the model of kinetic friction will be used.Visualize:

Solve: (a) The separate free body diagrams for the two blocks show that there are two action/reaction pairs.Notice how block 1 both pushes down on block 2 (force

r′n1 ) and exerts a retarding friction force

rf 2 top on the top

surface of block 2. Block 1 is in static equilibrium (a1 = 0 m/s2) but block 2 is accelerating. Newton’s second lawfor block 1 is

F f T T fxnet on 1 rope rope N( ) = − = ⇒ =1 10 F n m g n m g

ynet on 1 N( ) = − = ⇒ =1 1 1 10

Although block 1 is stationary, there is a kinetic force of friction because there is motion between block 1 and block 2.The friction model means f n m gk k1 1 1= =µ µ . Substitute this result into the x-equation to get the tension in the rope:

T f m grope k 3.92 N= = =1 1µ(b) Newton’s second law for block 2 is

a aF

m

T f f

mxx= =

( )=

− −net on 2 pull 2 top 2 bot

2 2

aF

m

n n m g

myy= =

( )= − ′ −

02

2 1 2

2

m/s2net on 2

Forces rn1 and

r′n1 are an action/reaction pair, so ′ = =n n m g1 1 1 . Substituting into the y-equation gives

n m m g2 1 2= +( ) . This is not surprising because the combined weight of both objects presses down on the surface.The kinetic friction on the bottom surface of block 2 is then

f n m m g2 bot k k= = +( )µ µ2 1 2

The forces rf1 and

rf2 top are an action/reaction pair, so f f m g2 bot k= =1 1µ . Inserting these friction results into the

x-equation gives

aF

m

T m g m m g

mx=

( )=

− − +( )=

net on 2 pull k k 22.16 m/s2

1 1 2

2

µ µ

8.28. Model: Blocks 1 and 2 are our systems of interest and will be treated as particles. Assume a frictionlessrope and massless pulley.Visualize:

Solve: The blocks accelerate with the same magnitude but in opposite directions. Thus the acceleration constraintis a2 = a = −a1, where a will have a positive value. There are two real action/reaction pairs. The two tension forceswill act as if they are action/reaction pairs because we are assuming a massless rope and a frictionless pulley. Makesure you understand why the friction forces point in the directions shown in the free-body diagrams, especially

force r′f1 exerted on block 2 by block 1. We have quite a few pieces of information to include. First, Newton’s

second law for blocks 1 and 2:rF f T n T m a m a

xnet on 1 k( ) = − = − = = −1 1 1 1 1 1 1µ F n m g n m g

ynet on 1 N( ) = − = ⇒ =1 1 1 10

F T f f T T f n T m a m axnet on 2 pull pull k( ) = − ′− − = − ′− − = =1 2 2 1 2 2 2 2 2µ

F n n m g n n m gynet on 2 N( ) = − ′ − = ⇒ = ′ +2 1 2 2 1 20

We’ve already used the kinetic friction model in both x-equations. Next, Newton’s third law:

′ = =n n m g1 1 1 ′ = = =f f n m g1 1 1 1µ µk k T T T1 2= =

Knowing ′n1 , we can now use the y-equation of block 2 to find n2. Substitute all these pieces into the two x-equations,and we end up with two equations in two unknowns:

µkm g T m a1 1− = − T T m g m m g m apull k k− − − +( ) =µ µ1 1 2 2

Subtract the first equation from the second to get

T m m g m m apull k− +( ) = +( )µ 3 1 2 1 2 ⇒ =− +( )

+=a

T m m g

m mpull k 21.77 m/s

µ 3 1 2

1 2

8.29. Model: The sled (S) and the box (B) will be treated in the particle model, and the model of friction will beused.Visualize:

In the sled’s physical representation nS is the normal (contact) force on the sled due to the snow. Similarly fkS is theforce of kinetic friction on the sled due to snow.Solve: Newton’s second law on the box in the y-direction is

n w nS on B B S on B2 N 10 kg 9.8 m/s 92.09 N− ° = ⇒ = ( )( ) ° =cos cos20 0 20

The static friction force rfS on B accelerates the box. The maximum acceleration occurs when static friction reaches

its maximum possible value.

f ns S S on B 92.09 N 46.05 Nmax .= = ( )( ) =µ 0 50

Newton’s second law along the x-direction thus gives the maximum acceleration

f w m aS on B B B− ° =sin20 ⇒ − ( )( ) ° = ( )46.05 N 10 kg 9.8 m/s 10 kg2 sin20 a ⇒ =a 1.25 m/s2

Newton’s second law for the sled along the y-direction isn n wS B on S S N− − ° =cos20 0

⇒ = + °n n m gS B on S S cos20 = ( ) + ( )( ) °92.09 N 20 kg 9.8 m/s2 cos20 = 276.27 N

Therefore, the force of friction on the sled by the snow is

fkS = (µk)nS = (0.12)(276.27 N) = 33.15 N

Newton’s second law along the x-direction is

Tpull – wSsin20° − fkS – fB on S = mSa

The friction force f fB on S S on B= because these are an action/reaction pair. We’re using the maximum acceleration,

so the maximum tension is

Tmax sin− ( )( ) ° − − = ( )( )20 kg 9.8 m/s 33.15 N 6.05 N 20 kg 1.25 m/s2 220 4

⇒ Tmax = 171.2 N

Assess: Tmax of 171.2 N corresponds to a mass of 171.2 N/9.8 m/s2 = 17.5 kg. For the masses and the angle in thisproblem, a maximum tension of 171.2 N is physically understandable.

8.30. Model: Masses m1 and m2 are considered particles. The string is assumed to be massless.Visualize:

Solve: The tension in the string causes the centripetal acceleration of the circular motion. If the hole is smooth, it

acts like a pulley. Thus tension forces rT1 and

rT2 act as if they were an action/reaction pair. Mass m1 is in circular

motion of radius r, so Newton’s second law for m1 is

F Tm v

rr = =∑ 11

2

Mass m2 is at rest, so the y-equation of Newton’s second law is

F T m g T m gy = − = ⇒ =∑ 2 2 2 20 N

Newton’s third law tells us that T T1 2= . Equating the two expressions for these quantities:

m v

rm g v

m rg

m1

2

22

1

= ⇒ =

8.31. Model: The boy (B) and the crate (C) are the two systems of our interest, and they will be treated in theparticle model. We will also use the static and kinetic friction models.Visualize:

Solve: The fact that the boy’s feet occasionally slip means that the maximum force of static friction must existbetween the boy’s feet and the sidewalk. That is, fsB = µsB nB. Also fkC = µkC nC.

Newton’s second law for the crate is

F n w n m gyon C C C C C N( ) = − = ⇒ =∑ 0 F F f

xon C B on C kC N( ) = − =∑ 0 ⇒ = = =F f n m gB on C kC kC C kC Cµ µ

Newton’s second law for the boy is

F n w n m gyon B B B B B N( ) = − = ⇒ =∑ 0 F f F

xon B sB C on B N( ) = − =∑ 0 ⇒ = = =F f n m gC on B sB sB B sB Bµ µrFC on B and

rFB on C are an action/reaction pair, so

F F m g m gC on B B on C sB B kC C= ⇒ =µ µ ⇒ = =( )( )

( )=m

mC

sB B

kC

50 kg200 kg

µµ

0 8

0 2

.

.

8.32. Model: The 3 kg and 4 kg blocks are to be treated as particles. The models of kinetic and static frictionand the constant-acceleration kinematic equations will be used.Visualize:

Solve: Minimum time will be achieved when static friction is at its maximum possible value. Newton’s secondlaw for the 4-kg block is

F n wyon 4 3 on 4 N( ) = − =∑ 4 0 ⇒ = = = ( )( ) =n w m g3 on 4

24.0 kg 9.8 m/s 39.2 N4 4

⇒ = = = ( )( ) =f f ns4 s max s 3 on 4 39.2 N 23.52 Nµ 0 60.

Newton’s second law for the 3-kg block is

F n n wyon 3 4 on 3 N( ) = − − =∑ 3 3 0 ⇒ n n w3 3= + = + ( )( ) =4 on 3

239.2 N 3.0 kg 9.8 m/s 68.6 N

Friction forces f and fs4 are an action/reaction pair. Thus

F f f m axon 3 s3 k3( ) = − =∑ 3 3 ⇒ − =f n m aks4 µ 3 3 3 ⇒ 23.52 N 68.6 N 3 kg− ( )( ) = ( )0 20 3. a ⇒ =a3 3.267 m/s2

Since block 3 does not slip, this is also the acceleration of block 4. The time is calculated as follows:

x x v t t a t tx1 0 0 1 012 1 0

2− + −( ) + −( ) ⇒ = + + ( ) −( )5.0 m m m 3.267 m/s s20 0 012 1

2t ⇒ t1 = 1.75 s

8.33. Model: The masses m and M are to be treated in the particle model. We will also assume a massless ropeand frictionless pulley, and use the constant-acceleration kinematic equations for m and M.Visualize:

Solve: Using y y v t t a t ty1 0 0 1 012 1 0

2= + −( ) + −( )M ,

−( ) = + + −( )1 m m m 6.0 s 0 s0 0 12

2aM ⇒ = −aM 0.0556 m/s2

Newton’s second law for m and M is:

F T w mam y m m mon R on( ) = − =∑ F T w MaM y M M Mon R on( ) = − =∑The acceleration constraint is a am M= − . Also, the tensions are an action/reaction pair, thus T Tm MR on R on = . Withthese, the second law equations are

T Mg Ma

T mg maM M

M M

R on

R on

− =− = −

Subtracting the second from the first gives

− + = +Mg mg Ma maM M ⇒ = +−

m M

g a

g aM

M

= ( ) −+

=1009 8 0 556

9 8 0 55698 9 kg kg

. .

. ..

Assess: Note that am = − aM = .0556 m/s2. For such a small acceleration, a mass of 98.9 kg for m compared to M =100 kg is understandable.

8.34. Model: Use the particle model for the block of mass M and the two massless pulleys. Additionally, the

rope is massless and the pulleys are frictionless. The block is kept in place by an applied forcerF .

Visualize:

Solve: Since there is no friction on the pulleys, T2 = T3 and T2 = T5. Newton’s second law for mass M is

T w T Mg1 10− = ⇒ = = ( )( ) = N 10.2 kg 9.8 m/s 100 N2

Newton’s second law for the small pulley is

T T T T TT

T F2 3 1 2 31

502

+ − = ⇒ = = = = = N 50 N

Newton’s second law for the large pulley isT T T T T T T T4 2 3 5 4 2 3 50− − − = ⇒ = + + = N 150 N

8.35. Model: Assume the particle model for m1, m2, and m3, and the model of kinetic friction. Assume the ropesto be massless, and the pulleys to be frictionless and massless.Visualize:

Solve: Newton’s second law for m1 is T1 – w1 = m1a1. Newton’s second law for m2 is

F n w nm yon2

2 N 2 kg 9.8 m/s 19.6 N( ) = − = ⇒ = ( )( ) =∑ 2 2 20

F T f T m am xon k22( ) = − − =∑ 2 1 2 2 ⇒ − − = ( )T n T a2 2 1 2µk 2 kg

Newton’s second law for m3 is

F T w m am yon 3( ) = − =∑ 2 3 3 3

Since m1, m2, and m3 move together, a1 = a2 = − a3 = a. The equations for the three masses thus become

T w m a a1 1 1− = = ( )1 kg T n T m a a2 2 1 2− − = = ( )µk 2 kg T w m a a2 3 3− = − = −( )3 kg

Subtracting the third equation from the sum of the first two equations yields:

− − + = ( )w n w a1 2 3µk 6 kg

⇒ −( )( ) − ( )( ) + ( )( ) = ( )1 kg 9.8 m/s 19.6 N 3 kg 9.8 m/s 6 kg2 20 300. a ⇒ =a 2.29 m/s2

8.36. Model: Assume the particle model for the two blocks, and the model of kinetic and static friction.Visualize:

Solve: (a) If the mass m is too small, the hanging 2 kg mass will pull it up the slope. We want to find the smallestmass that will stick because of friction. The smallest mass will be the one for which the force of static friction is atits maximum possible value: f f ns = ( ) =s smax

µ . As long as the mass m is stuck, both blocks are at rest withrFnet N= 0 . Newton’s second law for the hanging mass M is

F T Mg T Mgy Mnet M N 19.6 N( ) = − = ⇒ = =0

For the smaller mass m,

F T f mgx mnet s N( ) = − − =sinθ 0 F n mg n mg

ynet( ) = − ⇒ =cos cosθ θ

For a massless string and frictionless pulley, forces rTm and

rTM act as if they are an action/reaction pair. Thus

T Tm M= . Mass m is a minimum when f n mgs s s( ) = =max

cosµ µ θ . Substituting these expressions into the x-equation,

T mg mgM − − =µ θ θs Ncos sin 0 ⇒ =+( ) =m

T

gM

s

1.83 kgµ θ θcos sin

(b) Because µk < µS, the 1.83 kg block will begin to slide up the ramp, and the 2 kg mass will begin to fall, if theblock is nudged ever so slightly. Now the net force and the acceleration are not zero. Notice how, in the pictorialrepresentation, we chose different coordinate systems for the two masses. This gives block M an acceleration withonly a y-component and block m an acceleration with only an x-component. The magnitudes of the accelerationsare the same because the blocks are tied together. But block M has a negative acceleration component ay (vector

ra

points down) whereas block m has a positive ax. Thus the acceleration constraint is (am)x = − (aM)y = a, where a willhave a positive value. Newton’s second law for block M is

F T Mg M a May M ynet( ) = − = ( ) = −

For block m we have

F T f mg T mg mg m a max m xnet k k( ) = − − = − − = ( ) =sin cos sinθ µ θ θ

In writing these equations, we used Newton’s third law to write T T Tm M= = . Also, we noticed that the y-equationand the friction model for block m don’t change, except for µs becoming µk, so we already know fk from part (a).Notice that the tension in the string is not the weight Mg. We have two equations in the two unknowns T and a:

T Mg Ma− = − T mg ma− +( ) =µ θ θk cos sin

Subtracting the second equation from the first to eliminate T,

− + +( ) = − − = − +Mg mg Ma ma M m aµ θ θk cos sin ( )

⇒ =− +( )

+=a

M m

M mg

µ θ θk 21.32 m/scos sin

8.37. Model: Assume the particle model for the two blocks.Visualize:

Solve: (a) The slope is frictionless, so the blocks stay in place only if held. Once m is released, the blocks willmove one way or the other. As long as m is held, the blocks are in static equilibrium with

rFnet N= 0 . Newton’s

second law for the hanging block M is

F T Mg T MgM y M Mnet on N 19.6 N( ) = − = ⇒ = =0

By Newton’s third law, T T TM m= = = 19 6. N is the tension in the string.(b) The free-body diagram shows box m after it is released. Whether it moves up or down the slope depends onwhether the acceleration a is positive or negative. The acceleration constraint is (am)x = a = −(aM)y. Newton’s secondlaw for each system gives

F T mg m a mam x m xnet on( ) = − ° = ( ) =sin35 F T Mg M a MaM y M ynet on( ) = − = ( ) = −

We have two equations in two unknowns. Subtract the second from the first to eliminate T:

− ° + = +( )mg Mg m M asin35 ⇒ = − °+

= −aM m

M mg

sin350.481 m/s2

Since a < 0 m/s2, the box accelerates down the slope.(c) It is now straightforward to compute T Mg Ma= − = 20.6 N . Notice how the tension is larger than when theblocks were motionless.

8.38. Model: Assume the particle model for the book (B) and the coffee cup (C), the models of kinetic andstatic friction, and the constant-acceleration kinematic equations.Visualize:

Solve: (a) Using v v a x xx x12

02

1 02= + −( ) ,

0 22

1 m /s 3.0 m/s2 2 = ( ) + ( )a x ⇒ = −ax1 4.5 m /s2 2

To find x1, we must first find a. Newton’s second law for the book and the coffee cup is

F n w nyon B B B B

2 N 1.0 kg 9.8 m/s 9.21 N( ) = − ° = ⇒ = ( )( ) ° =∑ cos cos20 0 20

F T f w m axon B k B B B( ) = − − − ° =∑ sin20 F T w m a

yon C C C C( ) = − =∑The last two equations can be rewritten, using aC = aB = a, as

− − − ° =T n m g m aµk B B Bsin20 T m g m a− =C C

Adding the two equations,

a m m g m mC B C B k 9.21 N+( ) = − + °( ) − ( )sin20 µ

⇒ ( ) = −( ) + ( ) °[ ] − ( )( )1.5 kg 9.8 m/s 0.5 kg 1.0 kg 0.20 9.21 N2a sin20 ⇒ = −a 6 m/s2.73

Using this value for a, we can now find x1 as follows:

xa1 6 73

0 669= − = −−

=4.5 m /s 4.5 m /s

m/s m

2 2 2 2

2..

(b) The maximum static friction force is f ns max s B N N.= = ( )( ) =µ 0 50 9 21 4 60. . . We’ll see if the force fs needed tokeep the book in place is larger or smaller than fs max . When the cup is at rest, the string tension is T m g= C .Newton’s first law for the book is

F f T w f m g m gxon B s B s C B( ) = − − ° = − − ° =∑ sin sin20 20 0

⇒ = + °( )f M M gs C B sin20 = 8.25 N

Because f fs s max ,> the book slides back down.

8.39. Model: Use the particle model for the cable car and the counterweight. Assume a massless cable.Visualize:

Solve: (a) Notice the separate coordinate systems for the cable car (system 1) and the counterweight (system 2).

Forces rT1 and

rT2 act as if they are an action/reaction pair. The braking force

rFB works with the cable tension

rT1 to

allow the cable car to descend at a constant speed. Constant speed means dynamic equilibrium, so rFnet = 0N for

both systems. Newton’s second law for the cable car is

F T F m gxnet on 1 B N( ) = + − =1 1 1 0sinθ F n m g

ynet on 1 N( ) = − =1 1 1 0cosθ

Newton’s second law for the counterweight is

F m g Txnet on 2 N( ) = − =2 2 2 0sinθ F n m g

ynet on 2 N( ) = − =2 2 2 0cosθ

From the x-equation for the counterweight, T m g2 2 2= sin .θ By Newton’s third law, T T1 2= . Thus the x-equationfor the cable car becomes

F m g T m g m gB 3770 N= − = − =1 1 1 1 1 2 2sin sin sinθ θ θ

(b) If the brakes fail, then FB N= 0 . The car will accelerate down the hill on one side while the counterweightaccelerates up the hill on the other side. Both will have negative accelerations because of the direction of theacceleration vectors. The constraint is a1x = a2x = a, where a will have a negative value. Using T T T1 2= = , the twox-equations are

F T m g m a m ax xnet on 1 1( ) = − = =1 1 1 1sinθ F m g T m a m a

x xnet on 2 2( ) = − = =2 2 2 2sinθ

Note that the y-equations aren’t needed in this problem. Add the two equations to eliminate T:

− + = +( )m g m g m m a1 1 2 1 2sin sinθ θ ⇒ = − −+

= −am m

m mg1 1 2 2

1 2

sin sinθ θ0.991 m/s2

Now we have a problem in kinematics. The speed at the bottom is calculated as follows:

v v a x x ax12

02

1 0 12 2= + −( ) = ⇒ = = −( ) −( ) =v ax1 12 2 0.991 m/s 400 m 28.2 m/s2

Assess: A speed of approximately 60 mph as the cable car travels a distance of 2000 m along a frictionless slopeof 30° is reasonable.

8.40. Model: Use the particle model for the two blocks. Assume a massless rope, and massless, frictionlesspulleys.Visualize:

Note that for every meter block 1 moves forward, one meter is provided to block 2. So each rope on m2 has to belengthened by one-half meter. Thus the acceleration constraint is a a2

12 1= − .

Solve: Newton’s second law for block 1 is T = m1a1. Newton’s second law for block 2 is 2T – w 2 = m2a2.Combining these two equations gives

2 1 1 2 212 1m a m g m a( ) − = −( ) ⇒ +[ ] =a m m m g1 1 2 24 2 ⇒ =

+a

m g

m m12

1 2

2

4

where we have used a a212 1= − .

Assess: If m1 = 0 kg, then a2 = –g. This is what is expected for a freely falling object.

8.41. Model: Use the particle model for the two blocks. Assume a massless rope and massless, frictionless pulleys.Visualize:

For every one meter that the 1-kg block goes down, each rope on the 2-kg block will be shortened by one-halfmeter. Thus the acceleration constraint is a1 = –2a2.Solve: Newton’s second law for the two blocks is

2T = m2a2 T − w1 = m1a1

Since a a1 22= − , the above equations become

2T = m2a2 T − m1g = m1(− 2a2)

⇒ + ( ) =ma

m a m g22

1 2 122 ⇒ =

+=

( )( )+( )a

m g

m m21

2 1

2

4

2 1 kg 9.8 m/s

2 kg 4 kg

2

= 3.27 m/s2

Assess: If m1 = 0 kg, then a2 = 0 m/s2, which is expected.

8.42. Model: The painter and the chair are treated as a single system and represented as a particle. We assumethat the rope is massless and that the pulley is massless and frictionless.Visualize:

Solve: If the painter pulls down on the rope with force F, Newton’s third law requires the rope to pull up on thepainter with force F. This is just the tension in the rope. With our model of the rope and pulley, the same tensionforce F also pulls up on the painter’s chair. Newton’s second law for (painter + chair) is

2F w m m a− = +( )P C

⇒ = ( ) +( ) + +( )[ ] = +( ) +( )F m m a m m g m m a g12

12P C P C P C

= ( ) +( ) +( )12 70 kg 10 kg 0.20 m/s 9.8 m/s2 2 = 400 N

Assess: A force of 400 N, which is approximately one-half the total weight, is reasonable since the upwardacceleration is small.

8.43. Model: Use the particle model for the tightrope walker and the rope. The rope is assumed to be massless,so the tension in the rope is uniform.Visualize:

Solve: Newton’s second law for the tightrope walker is

F w ma F m a gR on W R on W2 270 kg 8.0 m/s 9.8 m/s 1246 N− = ⇒ = +( ) = ( ) +( ) =

Now, Newton’s second law for the rope gives

F T T Fyon R W on R N( ) = + − =∑ sin sinθ θ 0 ⇒ =

°=

°=

°=T

F FW on R R on W 3590 N2 10 2 10

1246

2 10sin sin sinWe used F FW on R R on W= because they are an action/reaction pair.

8.44. Model: Use the particle model for the wedge and the block.Visualize:

The block will not slip relative to the wedge if they both have the same acceleration a.Solve: The y-component of Newton’s second law for block m2 is

F n w nm g

yon 2 N( ) = − = ⇒ =∑ 2 2 220cos

cosθ

θCombining this equation with the x-component of Newton’s second law yields:

F n m a an

mg

xon 2( ) = = ⇒ = =∑ 2 22

2

sinsin

tanθ θ θ

Now, Newton’s second law for the wedge is

F F n m axon 1( ) = − =∑ 2 1sinθ

⇒ = + = +F m a n m a m a1 2 1 2sinθ = (m1 + m2) a = (m1 + m2) g tanθ

8.45. A 1.0 kg wood block is placed on top of a 2.0 kg wood block. A horizontal rope pulls the 2.0 kg blockacross a frictionless floor with a force of 21.0 N. Does the 1.0 kg block on top slide?Visualize:

Solve: The 1.0 block is accelerated by static friction. It moves smoothly with the lower block if fs < fs max. It slidesif the force that would be needed to keep it in place exceeds fs max. Begin by assuming the blocks move together withcommon acceleration a. Newton’s second is

Top block:

Bottom block:

on 1 s

on 2 pull s

( )

( )

F f m a

F T f m a

x

x

∑∑

= =

= − =1

2

Adding these two equations gives Tpull = (m1 + m2)a, or a = (21.0 N)/(1.0 kg + 3.0 kg) = 7.0 m/s2. The static frictionforce needed to accelerate the top block at 7.0 m/s2 is

fs = m1a = (1.0 kg)(7.0 m/s2) = 7.0 N

To find the maximum possible static friction force fs max = µsn1, the y-equation of Newton’s second law for the topblock shows that n1 = m1g. Thus

fs max = µsm1g = (0.50)(1.0 kg)(9.80 m/s2) = 4.9 N

Because 7.0 N > 4.9 N, static friction is not sufficient to accelerate the top block. It slides.

8.46. A 1.0 kg wood block is placed behind a 2.0 kg wood block on a horizontal table. The coefficients ofkinetic friction with the table are 0.3 for the 1.0 kg block and 0.5 for the 2.0 kg block. The 1.0 kg block is pushedforward, against the 2.0 block, and released with a speed of 2.0 m/s. How far do the blocks travel before stopping?Visualize:

Solve: The 2.0 kg block in front has a larger coefficient of friction. Thus the 1.0 kg block pushes against the rearof the 2.0 kg block and, in reaction, the 2.0 kg block pushes backward against the 1.0 kg block. There’s no verticalacceleration, so n1 = m1g and n2 = m2g, leading to f1 = µ1m1g and f2 = µ2m2g. Newton’s second law along the x-axis is

1 kg block:

2 kg block:

on 1 2 on 1 2 on 1

on 2 1 on 2 2 on 1

( )

( )

F F f F m g m a

F F f F m g m a

x

x

∑ = − − = − − =

= − = − =∑1 1 1 1

2 2 2 2

µ

µ

where we used a1 = a2 = a. Also, F1 on 2 = F2 on 1 because they are an action/reaction pair. Adding these two equationsgives

–( ) ( )

( . )( .. .

µ µµ µ

1 1 2 2 1 2

1 1 2 2

1 2

0 3 1 09 80 4 25

m m g m m a

am m

m mg

+ = +

= − ++

= − × = − kg) + (0.5)(2.0 kg)

1.0 kg + 2.0 kg m/s m/s2 2

We can now use constant-acceleration kinematics to find

v v a x x xv

ax xx

12

02

1 0 102

0 22

2 0

2 4 250 47= = + − ⇒ = − = −

−=( )

( .

( . ).

m/s)

m/s m

2

2

8.47. Model: Treat the ball of clay and the block as particles.Visualize:

Solve: (a) Forces F→

C on B and F→

B on C are an action/reaction pair, so FB on C = FC on B. Note that aB ≠ aC because theclay is decelerating while the block is accelerating. Newton’s second law in the x-direction is

Clay:

Block:

on C B on C C C

on B C on B B on C B B

( )

( )

F F m a

F F F m a

x

x

∑∑

= − =

= = =

Equating the two expressions for FB on C gives

am

maC

B

CB= −

Turning to kinematics, the velocity of each after ∆t is

( ) ( )

( ) ( )

v v a t

v v a t a tC C C

B B B B

1 0

1 0

= + ∆= + ∆ = ∆

But (vC)1 = (vB)1 because the clay and the block are moving together after ∆t has elapsed. Equating these twoexpressions gives (vC)0 + aC∆t = aB∆t, from which we find

a av

tC BC= −∆

( )0

We can now equate the two expressions for aC:

− = −∆

⇒ = ∆+

=+

=m

ma a

v

ta

v t

m mB

CB B

CB

C

B C

2 m/s) / (0.01 s)

g)/ (100 g) m/s

( ) ( ) /

/

(

(0 0

1

10

1 900100

Then aC = –9aB = –900 m/s2. With the acceleration now known, we can use either kinematic equation to find

(vC)1 = (vB)1 = (100 m/s2)(0.010 s) = 1.0 m/s

(b) FC on B = mBaB = (0.90 kg)(100 m/s2) = 90 N.(c) FB on C = mCaC = (0.10 kg)(900 m/s2) = 90 N.Assess: The two forces are of equal magnitude, as expected from Newton’s third law.

8.48. Model: Treat the basketball player (P) as a particle, and use the constant-acceleration kinematic equations.Visualize:

Solve: (a) While in the process of jumping, the basketball player is pressing down on the floor as he straightenshis legs. He exerts a force FP on F on the floor. The player experiences a weight force

rwP as well as a normal force

by the floor rn F on P. The only force that the floor experiences is the one exerted by the player

rF P on F.

(b) The player standing at rest exerts a force rF P on F on the floor. The normal force

rn F on P is the reaction force to

rF P on F.

But n FF on P P on F= , so rF net = 0 N. When the basketball player accelerates upwards by straightening his legs, his

speed has to increase from zero to v1y with which he leaves the floor. Thus, according to Newton’s second law,there must be a net upward force on him during this time. This can be true only if n wF on P P> . In other words, the

player presses on the floor with a force FP on F larger than his weight. The reaction force rn F on P then exceeds his

weight and accelerates him upward until his feet leave the floor.(c) The height of 80 cm = 0.80 m is sufficient to determine the speed v1y with which he leaves the floor. Once hisfeet are off the floor, he is simply in free fall, with a1 = − g. From kinematics,

v v a y yy y22

12

1 2 12= + −( ) ⇒ = + −( )( )0 212 m /s 0.80 m2 2 v gy

⇒ = ( ) =v y1 2 g 0.80 m 3.96 m/s

(d) The basketball player reached v1y = 3.96 m/s by accelerating from rest through a distance of 0.60 m. Assuminga0 to be constant during the jump, we find

v v a y y a yy y12

02

0 1 0 0 12 0 2= + −( ) = + −( ) m /s 0 m2 2 ⇒ = = ( )( )

=av

yy

012

1

2

2 2

3.96 m/s

0.60 m13.1 m/s2

(e) The scale reads the value of nF on P, the force exerted by the scale on the player. Before jumping,

n w n wF on P P F on P P2 N 100 kg 9.8 m/s 980 N− = ⇒ = = ( )( ) =0

While accelerating up,

n w ma n ma mgF on P P 0 F on P 0− = ⇒ = + = +

mg

a1

0

= ( ) +

=980 1

9 8

13 11710 N N

.

.

After leaving the scale, nF on P = 0 N because there is no contact with the scale.

8.49. Model: The hamster of mass m and the wedge with mass M will be treated as systems 1 and 2,respectively. They will be treated as particles.Visualize:

The scale is denoted by the letter s.Solve: (a) The reading of the scale is the magnitude of the force

rn2 that the scale exerts upward. There are two

action/reaction pairs. Initially the hamster of mass m is stuck in place and is in static equilibrium with rFnet = 0 N.

Because of the shape of the blocks, it is not clear whether the scale has to exert a horizontal friction force rf s on 2 to

prevent horizontal motion. We’ve included one just in case. Newton’s second law for the hamster is

F mg f f mgxnet on 1 2 on 1 2 on 1 N( ) = − = ⇒ =sin sinθ θ0

F n mgynet on 1 N( ) = − =1 0cosθ ⇒ =n mg1 cosθ

For the wedge, we see from Newton’s third law that ′ = =n n mg1 1 cosθ and that f f mg2 on 1 1 on 2= = sinθ . Using

these equations, Newton’s second law for the wedge is

F f f nxnet on 2 1 on 2 s on 2( ) = + − ′cos sinθ θ1 = + − =mg f mgsin cos cos sinθ θ θ θs on 2 N0 ⇒ =fs on 2 N0

F n n f Mgynet on 2 1 on 2( ) = − ′ − −2 1 cos sinθ θ = − − − =n mg mg Mg2

2 2 0cos sinθ θ N

⇒ = +( ) + = +( )n mg Mg M m g22 2cos sinθ θ = +( )( ) =0.8 kg 0.2 kg 9.8 m/s 9.8 N2

First we find that fs on 2 N= 0 , so no horizontal static friction is needed to prevent motion. More interesting, the

scale reading is M m g+( ) which is the total weight resting on the scale. This is the expected result.(b) Now suppose that the hamster is accelerating down the wedge. The total mass is still M + m, but is the readingstill (M + m)g? The frictional forces between the systems 1 and 2 have now vanished, and system 1 now has anacceleration. However, the acceleration is along the hamster’s x-axis, so a1y = 0 m/s2. The hamster’s y-equation is still

F n mg n mgynet on 1 N( ) = − = ⇒ =1 10cos cosθ θ

We still have ′ = =n n mg1 1 cosθ , so the y-equation for block 2 (with a2y = 0 m/s2) is

F n n Mg n mg Mgynet on 2 N( ) = − ′ − = − − =2 1 2

2 0cos cosθ θ

⇒ = + = +( ) =n mg Mg M m g22 2 8 99cos cos .θ θ N

Assess: The scale reads less than it did when the hamster was at rest. This makes sense if you consider the limit θ→ 90°, in which case cos θ → 0. If the face of the wedge is vertical, then the hamster is simply in free fall and canhave no effect on the scale (at least until impact!) So for θ = 90° we expect the scale to record Mg only, and that isindeed what the expression for n2 gives.

8.50. Model: The hanging masses m1, m2, and m3 are modeled as particles. Pulleys A and B are massless andfrictionless. The strings are massless.Visualize:

Solve: (a) The length of the string over pulley B is constant. Therefore,

y y y y L y y y LB B A B A B B−( ) + −( ) = ⇒ = − −3 32

The length of the string over pulley A is constant. Thus,

y y y y L y y yA A A A−( ) + −( ) = = − −2 1 1 22

⇒ − −( ) − − =2 2 3 1 2y y L y y LB B A ⇒ + + =2 3 2 1y y y constant

This constraint implies that

2 03 2 1dy

dt

dy

dt

dy

dt+ + = m/s = + +2 3 2 1v v vy y y

Also by differentiation, 2 03 2 1a a ay y y+ + = m/s2 .

(b) Newton’s second law for the masses m3, m2, m1, and pulley A is

TB − m3g = m3a3y TA – m2g = m2a2y TA – m1g = m1a1y

TB − 2TA = 0 N

The pulley equation is zero because the pulley is massless. These four equations plus the acceleration constraint arefive equations for the five unknowns (two tensions and three accelerations). To solve for TA, multiply the m3

equation by 2, substitute 2TB = 4TA, then divide each of the mass equations by the mass. This gives the threeequations

4 2 23 3

2 2

1 1

T m g a

T m g a

T m g a

y

y

y

A

A

A

/

/

/

− =

− =

− =

If these three equations are added, the right side adds to zero because of the acceleration constraint. Thus

( / / / )( / / / )

4 1 1 4 04

4 1 13 2 23 2 2

m m m T g Tg

m m m+ + − = ⇒ =

+ +A A

(c) Using numerical values, we find TA = 18.97 N. Then

a T m g

a T m g

a T m g

y

y

y

1 1

2 2

3 3

2 21

2 85

2 0 315

= − = −

= − =

= − = −

A2

A2

A2

m/s

m/s

m/s

/ .

/ .

/ .

(d) m3 = m1 + m2, so it looks at first like m3 should hang in equilibrium. For it to do so, tension TB would need toequal m3g. However, TB is not (m1 + m2)g because masses m1 and m2 are accelerating rather than hanging at rest.Consequently, tension TB is not able to balance the weight of m3.