Physics 4A Chapters 7 and 8: Newton’s Third Law and Dynamics II

Conceptual Questions and Example Problems from Chapters 7 and 8

Conceptual Question 7.2 How does a sprinter sprint? What is the force on a sprinter as she accelerates? Where does that force come from?

7.2. The sprinter pushes backward on the ground, which pushes back (forward) on her. This is the only horizontal force on the sprinter, so she accelerates forward.

Conceptual Question 7.5 A mosquito collides head-on with a car traveling at 60 mph. Is the force of the mosquito on the car larger than, smaller than, or equal to the force of the car on the mosquito? Explain.

7.5. Newton’s third law tells us that the force of the mosquito on the car has the same magnitude as the force of the car on the mosquito. Conceptual Question 7.6 A mosquito collides head-on with a car traveling at 60 mph. Is the magnitude of the mosquito’s acceleration larger than, smaller than, or equal to the magnitude of the car’s acceleration?

7.6. The mosquito has a much smaller mass than the car, so the magnitude of the interaction force between the car and mosquito, although equal on each, causes the mosquito to have a much larger acceleration. In fact, the acceleration is usually fatal to the mosquito. Conceptual Question 7.15 In case a in the figure below, block A is accelerated across a frictionless table by a hanging 10 N weight (1.02 kg). In case b, block A is accelerated across a frictionless table by a steady 10 N tension in the string. The string is massless, and the pulley is massless and frictionless. Is A’s acceleration in case b greater than, less than, or equal to its acceleration in case a? Explain.

7.15. Block A’s acceleration is greater in case b. In case a, the hanging 10 N must accelerate both the mass of A and its own mass, leading to a smaller acceleration than case b, where the entire 10 N force accelerates the mass of block A.

Case a Case b

A 10 N

A 10 N

10 N ( )10 N

( )

M M a

aM M

= +

=+

A

A

10 N10 N

M a

aM

=

=

Conceptual Question 8.5 The figure below shows two balls of equal mass moving in vertical circles. Is the tension in string A greater than, less than, or equal to the tension in string B if the balls travel over the top of the circle with equal speed?

8.5. The difference in the tension between A and B is due to the centripetal force 2

net( )rmvF

r= . Since the velocity v is

the same for both, the greater radius for B means that the tension in case A is greater than for case B. Conceptual Question 8.6 Ramon and Sally are observing a toy car speed up as it goes around a circular track. Ramon says, “The car’s speeding up, so there must be a net force parallel to the track.” “I don’t think so,” replies Sally. “It’s moving in a circle, and that requires centripetal acceleration. The net force has to point to the center of the circle.” Do you agree with Ramon, Sally, or neither? Explain.

8.6. Neither Sally nor Raymond is completely correct. Both have partially correct descriptions, but are missing key points. In order to speed up, there must be a nonzero acceleration parallel to the track. In order to move in a circle, there must be a nonzero centripetal acceleration. Since both of these are required, the net force points somewhere between the forward direction (parallel to the track) and the center of the circle. Problem 7.14 The sled dog in the figure below drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. If the tension in rope 1 is 150 N, what is the tension in rope 2?

7.14. Model: Sled A, sled B, and the dog (D) are treated like particles in the model. The rope is considered to be massless. Visualize:

Solve: The acceleration constraint is A B( ) ( )x x xa a a= = . Newton’s second law applied to sled A gives

on A A G A A G A A

on A 1 on A A A

( ) ( ) 0 N ( )

( )y

x x

F n F n F m g

F T f m a

∑ = − = ⇒ = =

∑ = − =

Using A k A,f nµ= the x-equation yields 2 2

1 on A k A A 150 N (0 10)(100 kg)(9 8 m/s ) (100 kg) 0 52 m/sx x xT n m a a aµ− = ⇒ − . . = ⇒ = .

Newton’s second law applied to sled B gives

on B B G B B G B B

on B 2 1 on B B B

( ) ( ) 0 N ( )

( )y

x x

F n F n F m g

F T T f m a

∑ = − = ⇒ = =

∑ = − − =

1 on BT and 1 on AT act as if they are an action/reaction pair, so 1 on B 150 NT = . Using B k Bf nµ= = 2(0 10)(80 kg)(9 8 m/s ) 78 4 N,. . = . we find

22 2150 N 78 4 N (80 kg)(0 52 m/s ) 270 NT T− − . = . ⇒ =

Thus the tension 22 2 7 10 NT = . × .

Problem 7.16 The figure to the right shows two 1.0 kg blocks connected by a rope. A second rope hangs beneath the lower block. Both ropes have a mass of 250 g. The entire assembly is accelerated upward at 3.0 m/s2 by force F. (a) What is F? (b) What is the tension at the top end of rope 1? (c) What is the tension at the bottom end of rope 1? (d) What is the tension at the top end of rope 2? 7.16. 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 2 5 kgM = . . This combined system is

accelerating upward at 23 0 m/sa = . under the influence of a force F and the gravitational force ˆMg j− . Newton’s second law applied to the combined system gives

2 2net( ) ( ) (2.5 kg)(3.0 m/s 9.8 m/s ) 32 NyF F Mg Ma F M a g= − = ⇒ = + = + =

(b) The ropes are not massless. We must consider both the blocks and the ropes as systems. The force F acts only on block A because it does not contact the other objects. We can proceed to apply the y-component of Newton’s second law to each system, starting at the top. Each object accelerates upward at 23 0 m/sa = . . For block A,

net on A A 1 on A A 1 on A A( ) ( ) 19 NyF F m g T m a T F m a g= − − = ⇒ = − + =

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

net on 1 A on 1 1 B on 1 1( )yF T m g T m a= − − =

where A on 1T

and 1 on AT

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

and

B on 1T

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

B on 1 A on 1 1( ) 16 NT T m a g= − + =

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

1 on B B on 1T T= and 2 on B B on 2T T=

Newton’s second law applied to block B is

net on B 1 on B B 2 on B B 2 on B 1 on B B( ) ( ) 3 2 NyF T m g T m a T T m a g= − − = ⇒ = − + = .

Problem 7.20 Two blocks are attached to opposite ends of a massless rope that goes over a massless, frictionless, stationary pulley. One of the blocks, with a mass of 6.0 kg, accelerates downward at ¾ g. What is the mass of the other block?

7.20. Model: The blocks are particles, the rope is massless, and the pulley is massless and frictionless. Visualize: Because “the tension in a massless string remains constant as it passes over a massless, frictionless pulley” the tension will be the same everywhere in the rope. There is an acceleration constraint: 1 2 .a a a− = =

Solve: First find the tension in the rope above the 6.0 kg mass. Then use that tension to find the mass of the other block.

( ) ( )3 11 1 1 1 1 4 4( ) (6.0 kg)F T m g m a T m g a m g g gΣ = − = − ⇒ = − = − =

( ) ( ) ( )143 61

2 2 2 2 2 24 4 3 74

(6.0 kg)(6.0 kg) kg 0.86 kg

gF T m g m a g m g m g m

g gΣ = − = ⇒ − = ⇒ = = =

+

Since the tension is the same everywhere along the rope the woman must pull with a force of 250 N to keep a constant speed. Assess: This can be checked by considering the two blocks as one compound object with a net force of 1 2( )m m g− and a total mass of 1 2.m m+

Problem 7.36 The block of mass M in the figure to the right slides on a frictionless surface. Find an expression for the tension in the string.

7.36. Model: Blocks A and B make up the system of interest and will be treated as particles. There is no friction anywhere. Visualize:

Notice that the coordinate system of for block B is rotated so that the motion in the positive x-direction is consistent between the two free-body diagrams. Solve: The blocks are constrained to have the same magnitude acceleration. Applying Newton’s second law to block B gives

B( ) ( )y GF T F ma T mg ma∑ = − + = ⇒ − = −

Applying Newton’s second law in both the x- and y-directions to the block A gives

A( ) ( ) 0

( )y G

x

F n F n Mg

F T Ma T Ma

∑ = − = ⇒ =

∑ = = ⇒ =

Using the first equation to eliminate the acceleration a gives the tension:

( / ) mMgT Ma M g T m Tm M

= = − ⇒ =+

Assess: The result is positive, as it should be for our choice of coordinate system. Consider m = 0. In this case, T = 0, as expected. For ,m M>> the tension is independent of the mass of the hanging block because its acceleration will be g, as we can see by solving for the acceleration:

forT Mga g g g m Mm m M

= − + = − → >>+

Problem 7.41 The 1.0 kg physics book in the figure below is connected by a string to a 500 g coffee cup. The book is given a push up the slope and released with a speed of 3.0 m/s. The coefficients of friction are µs = 0.50 and µk = 0.20. (a) How far does the book slide? (b) At the highest point, does the book stick to the slope, or does it slide back down?

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

Solve: (a) Using 2 21 0 1 02 ( ),x xv v a x x= + − we find

2 2 2 2 21 10 m /s (3 0 m/s) 2 ( ) 4 5 m /sa x ax= . + ⇒ = − .

To find 1,x we must first find a. Newton’s second law applied to the book and the coffee cup gives 2

on B B G B B( ) ( ) cos(20 ) 0 N (1 0 kg)(9 8 m/s )cos(20 ) 9 21 NyF n F n∑ = − ° = ⇒ = . . ° = .

on B k G B B B on C G C C C( ) ( ) sin(20 ) ( ) ( )x yF T f F m a F T F m a∑ = − − − ° = ∑ = − =

The last two equations can be rewritten, using C B ,a a a= = as

k B B B C Csin(20 )T n m g m a T m g m aµ− − − ° = − =

Adding the two equations gives

C B C B k( ) [ sin(20 )] (9 21 N)a m m g m m µ+ = − + ° − . 2 2(1 5 kg) (9 8 m/s )[0 500 kg (1 0 kg)sin 20 ] (0 20)(9 21 N) 6 73 m/sa a. = − . . + . ° − . . ⇒ = − .

Using this value for a, we can now find 1x as follows: 2 2 2 2

1 24 5 m /s 4 5 m /s 0 67 m

6 73 m/sx

a− . − .

= = = .− .

(b) The maximum static friction force is s max s B( ) (0 50)(9 21 N) 4 60 Nf nµ= = . . = . . We’ll see if the force sf needed to keep the book in place is larger or smaller than s max( )f . When the cup is at rest, the string tension is CT m g= . Newton’s first law for the book is

on B s B s C B

s C B

( ) sin(20 ) sin(20 ) 0( sin 20 ) 8 25 N

xF f T w f m g m gf M M g

∑ = − − ° = − − ° =

= + ° = .

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

Problem 7.A (Problem from Exam 1 – Spring 2002) In the figure below, two blocks are connected over a frictionless, massless pulley. The weight of block A is 50 N and the coefficient of static friction between block A and the incline is 0.20. The angle θ of the incline is 30.0o. What is the maximum and minimum mass that block B can have such that both blocks remain stationary?

Problem 8.7 A 200 g block on a 50-cm-long string swings in a circle on a horizontal, frictionless table at 75 rpm. (a) What is the speed of the block? (b) What is the tension in the string?

8.7. Model: Treat the block as a particle attached to a massless string that is swinging in a circle on a frictionless table. Visualize:

Solve: (a) The angular velocity and speed are rev 2 rad75 471 2 rad/minmin 1 rev

πω = × = .1 min(0 50 m)(471 2 rad/min) 3 93 m/s60 stv rω= = . . × = .

The tangential velocity is 3.9 m/s. (b) The radial component of Newton’s second law is

2

rmvF T

r∑ = =

Thus 2(3 93 m/s)(0 20 kg) 6 2 N

0 50 mT .

= . = ..

Problem 8.13 Mass m1 on the frictionless table of the figure to the right is connected by a string through a hole in the table to a hanging mass m2. With what speed must m1 rotate in a circle of radius r if m2 is to remain hanging at rest?

8.13. Model: Masses 1m and 2m 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 1T

and 2T

act as if they were an action/reaction pair. Mass 1m is in circular motion of radius r, so Newton’s second law for 1m is

21

1rm vF T

r∑ = =

Mass 2m is at rest, so the y-equation of Newton’s second law is

2 2 2 20 NyF T m g T m g∑ = − = ⇒ =

Newton’s third law tells us that 1 2.T T= Equating the two expressions for these quantities: Problem 8.20 A roller coaster car crosses the top of a circular loop-the-loop at twice the critical speed. What is the ratio of the normal force to the gravitational force?

8.20. Model: Model the roller coaster car as a particle at the top of a circular loop-the-loop undergoing uniform circular motion. Visualize:

Notice that the r-axis points downward, toward the center of the circle. Solve: The critical speed occurs when n goes to zero and GF

provides all the centripetal force pulling the car in the

vertical circle. At the critical speed 2c / ,mg mv r= therefore cv rg= . Since the car’s speed is twice the critical speed,

c2tv v= and the centripetal force is 2 2

cG

(4 ) (4 ) 4rmv m v m rgF n F mg

r r r∑ = + = = = =

Thus the normal force is 3n mg= . Consequently, G/ 3n F = . Problem 8.24 A 500 g ball swings in a vertical circle at the end of a 1.5-m-long string. When the ball is at the bottom of the circle, the tension in the string is 15 N. What is the speed of the ball at that point?

8.24. Model: Model the ball as a particle which is in a vertical circular motion. Visualize:

Solve: At the bottom of the circle,

2 22

G(0 500 kg)(15 N) (0 500 kg)(9 8 m/s ) 5 5 m/s

(1 5 m)rmv vF T F v

r.

∑ = − = ⇒ − . . = ⇒ = ..

Problem 8.29 An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diamter vertical circle. At the point where the plane is flying straight down, its speed is 55 m/s and it is speeding up at a rate of 12 m/s per second. (a) What is the magnitude of the net force on the plane? You can neglect air resistance. (b) What angle does the net force make with the horizontal? Let an angle above the horizontal be positive and an angle below the horizontal be negative.

8.29. Model: The plane is a particle undergoing nonuniform circular motion. Make the r-t plane vertical; then there will be on motion in the z-direction. Neglect air resistance. Visualize:

Solve: We are given the tangential acceleration and we find the centripetal acceleration from 2 / .ra v r= Use the second law to find the components of the force.

2 26

2 6

(55 m/s)(85000 kg) 1.978 10 N130 m

(85000 kg)(12 m/s ) 1.02 10 N

r

t t

vF mr

F ma

= = = ×

= = = ×

(a) Now find the magnitude of the net force. 2 2 6 2 6 2 6

net (1.978 10 N) (1.02 10 N) 2.2 10 Nr tF F F= + = × + × = × (b) Find the angle the net force makes with the horizontal.

61 1

61.02 10 Ntan tan 27

1.978 10 Nt

r

FF

θ − − − ×= = = − °

×

The angle is negative because it is below the horizontal. Assess: The plane’s velocity is down and increasing, so the net force must be below the horizontal. Problem 8.40 A concrete highway curve of radius 70 m is banked at a 15° angle. What is the maximum speed with which a 1500 kg rubber-tired car can take this curve without sliding? (µs = 1.0)

8.40. Model: We will use the particle model for the car, which is undergoing uniform circular motion on a banked highway, and the model of static friction. Visualize:

Note that we need to use the coefficient of static friction s ,µ which is 1.0 for rubber on concrete. Solve: Newton’s second law for the car is

2

s cos sinrmvF f n

rθ θ∑ = + = s Gcos sin 0zF n f Fθ θ∑ = − − = N

Maximum speed is when the static friction force reaches its maximum value s max s( )f nµ= . Then 2

s( cos15 sin15 ) mvnr

µ ° + ° =

s(cos15 sin15 )n mgµ° − ° =

Dividing these two equations and simplifying, we get 2

s s

s s

2

tan15 tan151 tan15 1 tan15

(1 0 0 268)(9 80 m/s )(70 m) 34 m/s(1 0 268)

v v grgr

µ µµ µ+ ° + °

= ⇒ =− ° − °

. + .= . =

− .

Assess: The above value of 34 m/s 70 mph≈ is reasonable.

Problem 8.47 A conical pendulum is formed by attaching a ball of mass m to a string of length L, then allowing the ball to move in a horizontal circle of radius r. The figure below shows that the string traces out the surface of a cone, hence the name. Find an expression for the tension T in the string.

8.47. Use the particle model for the ball, which is undergoing uniform circular motion. Visualize: We are given L, r, and m, so our answers must be in terms of those variables. L is the hypotenuse of the right triangle. The ball moves in a horizontal circle of radius cos .r L θ= The acceleration and net force point toward the center of the circle, not along the string.

Solve: (a) Apply Newton’s second law in the -direction.z

cos 0coszmgF T mg Tθ

θ∑ = − = ⇒ =

From the right triangle 2 2cos / .L r Lθ = −

2 2cosmg mgLT

L rθ= =

−

(b) Apply Newton’s second law in the -direction.r 2 2 2sin ( sin ) .rF T m r m L T m Lθ ω ω θ ω∑ = = = ⇒ =

Set the two expressions for T equal to each other, cancel m and one L, and solve for .ω

22 2 2 2

mgL gm LL r L r

ω ω= ⇒ =− −

(c) Insert 1 0 m,L = . 0 20 mr = . and 0 50 kg.m = . 2

2 2 2 2

2

2 2 2 2

(0 50 kg)(9 8 m/s )(1 0 m) 5 0 N(1 0 m) (0 20 m)

9 8 m/s 1 rev 60 s3 163 rad/s 30 rpm2 rad 1 min(1 0 m) (0 20 m)

mgLTL r

g

L rω

π

. . .= = = .

− . − .

.= = = . =

− . − .

Assess: Notice that the mass canceled out of the equation for ,ω but not for T, so the 500 g was necessary information. Problem 8.56 A 100 g ball on a 60-cm-long string is swung in a vertical circle about a point 200 cm above the floor. The tension in the string when the ball is at the very bottom of the circle is 5.0 N. A very sharp knife is suddenly inserted, as shown in the figure below, to cut the string directly below the point of support. How far to the right of where the string was cut does the ball hit the floor?

8.56. Model: Model the ball as a particle swinging in a vertical circle, then as a projectile. Visualize:

Solve: Initially, the ball is moving in a circle. Once the string is cut, it becomes a projectile. The final circular-motion velocity is the initial velocity for the projectile. The free-body diagram for circular motion is shown at the bottom of the circle. Since G ,T F> there is a net force toward the center of the circle that causes the centripetal acceleration. The r-equation of Newton’s second law is

2

net G( )rmvF T F T mg

r= − = − =

2bottom

0 60 m( ) [5 0 N (0 10 kg)(9 8 m/s )] 4 91 m/s0 100 kg

rv T mgm

.⇒ = − = . − . . = .

.

As a projectile the ball starts at 0 1 4 my = . with 0ˆ 4 91 m/sv i= . .

The equation for the y-motion is 2 21 1

1 0 0 0 12 20 m ( )yy y v t g t y gt= = + ∆ − ∆ = −

This is easily solved to find that the ball hits the ground at time

01

2 0 535 sytg

= = .

During this time interval it travels a horizontal distance

1 0 0 1 (4 91 m/s)(0 535 s) 2 63 mxx x v t= + = . . = .

So the ball hits the floor 2.6 m to the right of the point where the string was cut.