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5 More Applications of Newton’s Laws

CHAPTER OUTLINE

5.1 Forces of Friction

5.2 Extending the Particle Uniform Circular Motion Model

5.3 Nonuniform Circular Motion

5.4 Motion in the Presence of Velocity-Dependent Resistive Forces

5.5 The Fundamental Forces of Nature

5.6 Context Connection: Drag Coefficients of Automobiles

* An asterisk indicates an item new to this edition.

ANSWERS TO OBJECTIVE QUESTIONS

*OQ5.1 Answer (d). The stopping distance will be the same if the mass of the truck is doubled. The normal force and the friction force both double, so the backward acceleration remains the same as without the load.

*OQ5.2 Answer (b). 200 N must be greater than the force of friction for the box’s acceleration to be forward.

*OQ5.3 Answers (a), (c), and (d). A free-body diagram shows the forces exerted on the object by other objects, and the net force is the sum of those forces.

*OQ5.4 (a) No. When v = 0, v2/r = 0.

(b) Yes. Its speed is changing because it is reversing direction.

*OQ5.5 Answer (d). All the other possibilities would make the total force on the crate be different from zero.

*OQ5.6 (a) Yes, point C. Total acceleration here is centripetal acceleration, straight up. (b)Yes, point A. The speed at A is zero where the bob is reversing direction. Total acceleration here is tangential acceleration, to the right and downward perpendicular to the cord. (c) No. (d) Yes,

Chapter 5 153

© 2013 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

point B. Total acceleration here is to the right and either downwards or upwards depending on whether the magnitude of the centripetal acceleration is smaller or larger than the magnitude of the tangential acceleration.

*OQ5.7 (a) Yes. Its path is an arc of a circle; the direction of its velocity is changing. (b) No. Its speed is not changing.

*OQ5.8 Answer (e). The stopping distance will decrease by a factor of four if the initial speed is cut in half.

*OQ5.9 (a) A > C = D > B = E = 0. At constant speed, centripetal acceleration is largest when radius is smallest. A straight path has infinite radius of curvature. (b) Velocity is north at A, west at B, and south at C. (c) Acceleration is west at A, nonexistent at B, east at C, to be radially inward.

*OQ5.10 Answer (c). When the truck accelerates forward, the crate has the natural tendency to remain at rest, so the truck tends to slip under the crate, leaving it behind. However, friction between the crate and the bed of the truck acts in such a manner as to oppose this relative motion between truck and crate. Thus, the friction force acting on the crate will be in the forward horizontal direction and tend to accelerate the crate forward. The crate will slide only when the coefficient of static friction is inadequate to prevent slipping.

*OQ5.11 (i) Answer (c). The iPod shifts backward relative to the student’s hand. The cord then pulls the iPod upward and forward, to make it gain speed horizontally forward along with the airplane. (ii) Answer (b). The angle stays constant while the plane has constant acceleration. This experiment is described in the book Science from your Airplane Window by Elizabeth Wood.

*OQ5.12 Answer (a). Her speed increases, until she reaches terminal speed.

*OQ5.13 Answer (b). The magnitude of acceleration decreases as the speed increases because the air resistance force increases, counterbalancing more and more of the gravitational force.

*OQ5.14 Answer (d). Formulas a, b, and e have the wrong units for speed. Formulas a and c would give an imaginary answer.

ANSWERS TO CONCEPTUAL QUESTIONS

*CQ5.1 If you have ever seen a car stuck on an icy road, with its wheels spinning wildly, you know the car has great difficulty moving forward until it “catches” on a rough patch. (a) Friction exerted by the road is the force making the car accelerate forward. Burning gasoline can



provide energy for the motion, but only external forces—forces exerted by objects outside—can accelerate the car. (b) If the car moves forward slowly as it speeds up, then its tires do not slip on the surface. The rubber contacting the road moves toward the rear of the car, and static friction opposes relative sliding motion by exerting a force on the rubber toward the front of the car. If the car is under control (and not skidding), the relative speed is zero along the lines where the rubber meets the road, and static friction acts rather than kinetic friction.

*CQ5.2 When the hands are shaken, there is a large acceleration of the surfaces of the hands. If the water drops were to stay on the hands, they must accelerate along with the hands. The only force that can provide this acceleration is the friction force between the water and the hands. (There are adhesive forces also, but let’s not worry about those.) The static friction force is not large enough to keep the water stationary with respect to the skin at this large acceleration. Therefore, the water breaks free and slides along the skin surface. Eventually, the water reaches the end of a finger and then slides off into the air. This is an example of Newton’s first law in action in that the drops continue in motion while the hand is stopped.

*CQ5.3 As you pull away from a stoplight, friction exerted by the ground on the tires of the car accelerates the car forward. As you begin running forward from rest, friction exerted by the floor on your shoes causes your acceleration.

*CQ5.4 (a) The object will move in a circle at a constant speed.

(b) The object will move in a straight line at a changing speed.

*CQ5.5 The speed changes. The tangential force component causes tangential acceleration.

*CQ5.6 (a) The larger drop has higher terminal speed. In the case of spheres, the text demonstrates that terminal speed is proportional to the square root of radius. (b) When moving with terminal speed, an object is in equilibrium and has zero acceleration.

*CQ5.7 If you slam on the brakes, your tires will skid on the road. The force of kinetic friction between the tires and the road is less than the maximum static friction force. Anti-lock brakes work by “pumping” the brakes (much more rapidly than you can) to minimize skidding of the tires on the road.

*CQ5.8 The drag force is proportional to the speed squared and to the effective area of the falling object. At terminal velocity, the drag and gravity forces are in balance. When the parachute opens, its effective area increases greatly, causing the drag force to increase greatly. Because the drag and gravity forces are no longer in balance, the greater drag

Chapter 5 155


force causes the speed to decrease, causing the drag force to decrease until it and the force of gravity are in balance again.

*CQ5.9 (a) Friction, either static or kinetic, exerted by the roadway where it meets the rubber tires accelerates the car forward and then maintains its speed by counterbalancing resistance forces. Most of the time static friction is at work. But even kinetic friction (racers starting) will still move the car forward, although not as efficiently. (b) The air around the propeller pushes forward on its blades. Evidence is that the propeller blade pushes the air toward the back of the plane. (c) The water pushes the blade of the oar toward the bow. Evidence is that the blade of the oar pushes the water toward the stern.

*CQ5.10 The water has inertia. The water tends to move along a straight line, but the bucket pulls it in and around in a circle.

*CQ5.11 This is the same principle as the centrifuge. All the material inside the cylinder tends to move along a straight-line path, but the walls of the cylinder exert an inward force to keep everything moving around in a circular path.

*CQ5.12 I would not accept that statement for two reasons. First, to be “beyond the pull of gravity,” one would have to be infinitely far away from all other matter. Second, astronauts in orbit are moving in a circular path. It is the gravitational pull of Earth on the astronauts that keeps them in orbit. In the space shuttle, just above the atmosphere, gravity is only slightly weaker than at the Earth’s surface. Gravity does its job most clearly on an orbiting spacecraft, because the craft feels no other forces and is in free fall.

*CQ5.13 Blood pressure cannot supply the force necessary both to balance the gravitational force and to provide the centripetal acceleration to keep blood flowing up to the pilot’s brain.

SOLUTIONS TO END-OF-CHAPTER PROBLEMS

Section 5.1 Forces of Friction P5.1 − f + mg sinθ = 0 and +n − mg cosθ = 0 with f = µs n yield

µs = tanθc = tan 36.0° = 0.727

µk = tanθc = tan 30.0° = 0.577

P5.2 If all the weight is on the rear wheels,

(a) F = ma : µsmg = ma or a = µsg



But

Δx =

at2

2=µs gt2

2

so µs =

2Δxgt2

µs =

2 0.250 mi( ) 1 609 m mi( )9.80 m s2( ) 4.43 s( )2 = 4.18

(b)

Time would increase, as the wheels would skid and only kineticfriction would act; or perhaps the car would flip over.

P5.3 For equilibrium: f = F and n = Fg. Also, f = µn, i.e.,

(a)

µ =fn=

FFg

µs =75.0 N

25.0 9.80( ) N= 0.306

(b) µk =

60.0 N25.0 9.80( ) N

= 0.245

P5.4

Fy∑ = may : +n − mg = 0

fs ≤µsn = µsmg

This maximum magnitude of static friction acts so long as the tires roll without skidding.

∑Fx = max → − fs = ma

The maximum acceleration is

a = −µs g

The initial and final conditions are: xi = 0, vi = 50.0 mi/h = 22.4 m/s, vf = 0. Then,

v f2 = vi

2 + 2a(x f − xi )→ vi2 = 2µs gx f

(a) x f =

vi2

2µs g

x f =

22.4 m s( )2

2 0.100( ) 9.80 m s2( ) = 256 m

ANS. FIG. P5.3

Chapter 5 157


(b) x f =

vi2

2µs g

x f =

22.4 m s( )2

2 0.600( ) 9.80 m s2( ) = 42.7 m

P5.5 Maximum static friction provides the force that produces maximum acceleration, resulting in a minimum time interval to accelerate through ∆x = 3.00 m.

We know that the maximum force of static friction is fs = µsn. If the shoe is on a horizontal surface, friction acts in the horizontal direction. Assuming that the vertical normal force is maximal, equal in magnitude to the force of gravity on the person, we have n = Fg = mg; therefore, the maximum static friction force is

fs = µsn = µsmg

Applying Newton’s second law:

∑Fx = max

fs = ma

µs mg = ma → a = µs g

Find the time interval ∆t = t to accelerate from rest through ∆x = 3.00 m using xf = xi + vxit + ½axt

2:

Δx =

12

ax(Δt)2 → Δt =2Δxax

=2Δxµs g

(a) For µs = 0.500, ∆t = 1.11 s

(b) For µs = 0.800, ∆t = 0.875 s

P5.6 The free-body diagrams for this problem are shown below.

22.0° 22.0°

+y +y

+x +xf

F = 45.8 lb22.0°Fg = 170 lb

F2 F1

ntipn Fgground lb= =2 85 0.

Free-Body Diagram Free-Body Diagram of Person of Crutch Tip



From the free body diagram for the person,

Fx∑ = F1 sin 22.0°( ) − F2 sin 22.0°( ) = 0

which gives F1 = F2 = F . Then, Fy∑ = 2F cos22.0° + 85.0 lbs − 170 lbs = 0 yields F = 45.8 lb.

(a) Now consider the free-body diagram of a crutch tip.

Fx∑ = f − 45.8 lb( )sin 22.0° = 0 or f = 17.2 lb.

Fy∑ = ntip − 45.8 lb( )cos22.0° = 0

which gives ntip = 42.5 lb.

For minimum coefficient of friction, the crutch tip will be on the verge of slipping, so f = fs( )max = µsntip and

µs =

fntip

= 17.2 lb42.5 lb

= 0.404

(b) As found above, the compression force in each crutch is

F1 = F2 = F = 45.8 lb

P5.7 Newton’s second law for the 5.00-kg mass gives

T – fk = (5.00 kg)a

Similarly, for the 9.00-kg mass,

(9.00 kg)g – T = (9.00 kg)a

Adding these two equations gives:

9.00 kg( ) 9.80 m/s2( ) − 0.200 5.00 kg( ) 9.80 m/s2( ) = 14.0 kg( )a

Which yields a = 5.60 m/s2. Plugging this into the first equation above gives

T = 5.00 kg( ) 5.60 m/s2( ) + 0.200 5.00 kg( ) 9.80 m/s2( ) = 37.8 N

ANS. FIG. P5.7

Chapter 5 159


P5.8 If the load is on the point of sliding forward on the bed of the slowing truck, static friction acts backward on the load with its maximum value, to give it the same acceleration as the truck:

ΣFx = max : − f = mloadax

ΣFy = may : n − mload g = 0

Solving for the normal force and substituting into the x equation gives:

−µsmload g = mloadax or ax = −µs g

We can then use

vxf

2 = vxi2 + 2ax x f − xi( )

Which becomes

0 = vxi2 + 2 −µs g( ) x f − 0( )

(a) x f =

vxi2

2µs g= 12 m s( )2

2 0.5( ) 9.8 m s2( ) = 14.7 m

(b) From the expression x f =

vxi2

2µs g,

neither mass affects the answer P5.9 m = 3.00 kg, θ = 30.0°, x = 2.00 m, t = 1.50 s

(a) At constant acceleration,

x f = vit +

12 at2

Solving,

a =

2 x f – vit( )t2 =

2(2.00 m – 0)(1.50 s)2 = 1.78 m/s2

From the acceleration, we can calculate the friction force, answer (c), next.

(c) Take the positive x axis down parallel to the incline, in the direction of the acceleration. We apply Newton’s second law:

Fx = mg sin θ − f = ma∑

Solving, f = m(g sin θ − a )

r

rr

ANS. FIG. P5.8

ANS. FIG. P5.9



Substituting,

f = (3.00 kg)[(9.80 m/s2)sin 30.0° − 1.78 m/s2] = 9.37 N

(b) Applying Newton’s law in the y direction (perpendicular to the incline), we have no burrowing-in or taking-off motion. Then the y component of acceleration is zero:

Fy = n – mg cos θ = 0∑

Thus n = mg cos θ

Because f = µkn

we have µk =

fmg cosθ =

9.37 N(3.00 kg) 9.80 m/s2( )cos 30.0

= 0.368

(d) v f = vi + at so v f = 0 + (1.78 m/s2 )(1.50 s) = 2.67 m/s

P5.10 (a) See the free-body diagram of the suitcase in the figure on the right.

(b) msuitcase = 20.0 kg, F = 35.0 N

Fx∑ = max : −20.0 N + F cosθ = 0Fy∑ = may : +n + F sinθ − Fg = 0

F cosθ = 20.0 N

cosθ =20.0 N35.0 N

= 0.571

θ = 55.2°

(c) With Fg = (20.0 kg)(9.80 m/s2),

n = Fg − F sinθ = 196 − 35.0 0.821( )[ ] N

n = 167 N

P5.11 We must consider separately the rock when it is in contact with the roof and when it has gone over the top into free fall. In the first case, we take x and y as parallel and perpendicular to the surface of the roof:

Fy∑ = may : +n − mg cosθ = 0n = mg cosθ

then friction is fk = µkn = µkmg cosθ

ANS. FIG. P5.10(a)

ANS. FIG. P5.11

Chapter 5 161


Fx∑ = max : − fk − mg sinθ = max

ax = −µk g cosθ − g sinθ = −0.400cos37.0° − sin 37.0°( )(9.80 m s2 )

= −9.03 m s2

The rock goes ballistic with speed given by

vxf2 = vxi

2 + 2ax x f − xi( ) = 15 m s( )2 + 2 −9.03 m s2( ) 10 m − 0( )= 44.4 m2 s2

vxf = 6.67 m s

For the free fall, we take x and y horizontal and vertical:

vyf2 = vyi

2 + 2ay y f − yi( )y f =

vyf2 − vyi

2

2ay

=0 − 6.67 m s sin 37°( )2

2 −9.8 m s2( ) = 0.822 m above the top of the roof

Then ytot = 10.0 m sin 37.0° + 0.822 m = 6.84 m

P5.12 (a) To find the maximum possible value of P, imagine impending upward motion as case 1. Setting Fx∑ = 0:

Pcos50.0° − n = 0

with fs , max = µsn:

fs , max = µsPcos50.0°

= 0.250 0.643( )P = 0.161P

Setting Fy∑ = 0:

Psin 50.0° − 0.161P

− 3.00 kg( ) 9.80 m/s2( ) = 0

Pmax = 48.6 N

To find the minimum possible value of P, consider impending downward motion. As in case 1,

fs , max = 0.161P

Setting Fy∑ = 0:

ANS. FIG. P5.12



Psin 50.0° + 0.161P − 3.00 kg( ) 9.80 m/s2( ) = 0

Pmin = 31.7 N

(b)

If P > 48.6 N, the block slides up the wall. If P < 31.7 N, the blockslides down the wall.

(c) We repeat the calculation as in part (a) with the new angle.

Consider impending upward motion as case 1. Setting

Fx∑ = 0: Pcos13° − n = 0fs , max = µsn: fs , max = µsPcos13°

= 0.250 0.974( )P = 0.244P

Setting

Fy∑ = 0: Psin 13° − 0.244P − 3.00 kg( ) 9.80 m/s2( ) = 0

Pmax = −1 580 N

The push cannot really be negative. However large or small it is, it cannot produce upward motion. To find the minimum possible value of P, consider impending downward motion. As in case 1,

fs , max = 0.244P

Setting

Fy∑ = 0: Psin 13° + 0.244P − 3.00 kg( ) 9.80 m/s2( ) = 0

Pmin = 62.7 N

P ≥ 62.7 N. The block cannot slide up the wall. If P < 62.7 N, theblock slides down the wall.

P5.13 (a) See the figure on the right.

(b) For Block #1,

T − µm1g = m1a

For Block #2,

68.0 N −T − µm2 g = m2a

Adding these equations gives,

68.0 N − µk m1 + m2( )g = m1 + m2( )a

ANS. FIG. P5.13

Chapter 5 163


ANS. FIG. P5.14(a)

or

a =

68.0 Nm1 + m2( ) − µg = 1.29 m/s2

(c) T = µk m1g + m1a = m1(µg + a) = (12.0 kg)[(0.100)(9.80 m/s2)

+ (1.29 m/s2)]

T = 27.2 N

P5.14 (a) The free-body diagrams for each object appear on the right.

(b) Let a represent the positive magnitude of the acceleration −aj of m1, of the

acceleration −ai of m2, and of the acceleration +aj of m3. Call T12 the tension in the left rope and T23 the tension in the cord on the right.

For m1, Fy∑ = may

+T12 − m1g = −m1a

For m2, Fx∑ = max

−T12 + µkn +T23 = −m2a

and Fy∑ = may , giving n − m2 g = 0

For m3, Fy∑ = may , giving T23 − m3g = +m3a

we have three simultaneous equations:

−T12 + 39.2 N = 4.00 kg( )a+T12 − 0.350 9.80 N( ) −T23 = 1.00 kg( )a

+T23 − 19.6 N = 2.00 kg( )a

Add them up (this cancels out the tensions):

+39.2 N − 3.43 N − 19.6 N = 7.00 kg( )a

a = 2.31 m s2 , down for m1 , left for m2 , and up for m3

(c) Now −T12 + 39.2 N = 4.00 kg( ) 2.31 m s2( )

T12 = 30.0 N



and T23 − 19.6 N = 2.00 kg( ) 2.31 m s2( )

T23 = 24.2 N

(d) If the tabletop were smooth, friction disappears (µk = 0), and so the acceleration would become larger. For a larger acceleration, according to the equations above, the tensions change:

T12 = m1g − m1a → T12 decreases

T23 = m3g + m3a → T23 increases

*P5.15 Find the acceleration of the car, which is the same as the acceleration of the book because the book does not slide.

For the car: vi = 72.0 km/h = 20.0 m/s, vf = 0, ∆x = (xf – xi) = 30.0 m. Using vf

2 = vi2 + 2a(xf – xi), we find the acceleration of the car:

a = –6.67 m/s2

Now, find the maximum acceleration that friction can provide. Because the book does not slide, static friction provides the force that slows down the book. We have the coefficient of static friction, µs = 0.550, and we know fs ≤ µsn. The book is on a horizontal seat, so friction acts in the horizontal direction, and the vertical normal force that the seat exerts on the book is equal in magnitude to the force of gravity on the book: n = Fg = mg. For maximum acceleration, the static friction force will be a maximum, so fs = µsn = µsmg. Applying Newton’s second law, we find the acceleration that friction can provide for the book:

∑Fx = max

–fs = ma

–µsmg = ma → a = –µs g = –(0.550)(9.80 m/s2) = –5.39 m/s2, which is too small for the stated conditions.

The situation is impossible because maximum static friction cannot provide the acceleration necessary to keep the book stationary on the seat.

Chapter 5 165


Section 5.2 Extending the Particle Uniform Circular Motion

Model

P5.16 (a)

F =mv2

r=

9.11× 10−31 kg( ) 2.20 × 106 m s( )2

0.530 × 10−10 m

= 8.32 × 10−8 N inward

(b) a =

v2

r=

2.20 × 106 m s( )2

0.530 × 10−10 m= 9.13 × 1022 m s2 inward

P5.17 m = 3.00 kg, r = 0.800 m. The string will break if the tension exceeds the weight corresponding to 25.0 kg, so

Tmax = Mg = (25.0 kg)(9.80 m/s2) = 245 N

When the 3.00-kg mass rotates in a horizontal circle, the tension causes the centripetal acceleration,

so T =

mv2

r

Then

v2 =rTm

=0.800 m( )T

3.00 kg≤

0.800 m( )Tmax

3.00 kg

=0.800 m( ) 245 N( )

3.00 kg= 65.3 m2/s2

and 0 ≤ v ≤ 65.3 m/s

or

0 ≤ v ≤ 8.08 m s

P5.18 Solve for the tensions.

Fg = mg = 4 kg( ) 9.8 m s2( ) = 39.2 N

sinθ =1.5 m2 m

θ = 48.6°

r = 2 m( )cos 48.6° = 1.32 m

ANS. FIG. P5.17

ANS. FIG. P5.18



Fx∑ = max =mv2

r

Ta cos 48.6° +Tb cos 48.6° =4 kg( ) 3 m s( )2

1.32 m

Ta +Tb =27.27 Ncos 48.6°

= 41.2 N

[1]

Fy∑ = may : Ta sin 48.6° −Tb sin 48.6° − 39.2 N = 0

Ta −Tb =39.2 N

sin 48.6°= 52.3 N

[2]

To solve simultaneously, we add the equations in Ta and Tb:

(Ta + Tb) + (Ta – Tb) = 41.2 N + 52.3 N

Ta =

93.8 N2

= 46.9 N

This means that Tb = 41.2 N – Ta = –5.7 N, which we may interpret as meaning the lower string pushes rather than pulls!

The situation is impossible because the speed of the object is toosmall, requiring that the lower string act like a rod and push ratherthan like a string and pull.

To answer the What if?, we go back to equation [2] above and substitute mg for the weight of the object. Then,

Fy∑ = may : Ta sin 48.6° −Tb sin 48.6° − mg = 0

Ta −Tb =(4.00 kg)gsin 48.6°

= 5.33g

We then add this equation to equation [2] to obtain

(Ta + Tb) + (Ta – Tb) = 41.2 N + 5.33g

or Ta = 20.6 N + 2.67g and Tb = 41.2 N – Ta = 41.2 N – 20.6 N – 2.67g

For this situation to be possible, Tb must be > 0, or g < 7.72 m/s2. This is certainly the case on the surface of the Moon and on Mars.

Chapter 5 167


P5.19 n = mg since ay = 0

The force causing the centripetal acceleration is the friction force f directed toward the center of the curve.

From Newton’s second law, f = mac =

mv2

r.

But the friction condition is f ≤ µsn ,

i.e.,

mv2

r≤ µsmg

Then

v ≤ µsrg = 0.600 35.0 m( ) 9.80 m/s2( ) → v ≤ 14.3 m/s

P5.20 (a) Fy∑ = may : mgmoon down =

mv2

r down

v = gmoonr = 1.52 m s2( ) 1.7 × 106 m + 100 × 103 m( )v = 1.65 × 103 m s

(b) v =

2πrT

, T =

2π 1.8 × 106 m( )1.65 × 103 m s

= 6.84 × 103 s = 1.90 h

P5.21 T cos5.00° = mg = 80.0 kg( ) 9.80 m s2( )

(a) T = 787 N:

T = 68.6 N( ) i + 784 N( ) j

(b) T sin 5.00° = mac : ac = 0.857 m s2 toward

the center of the circle.

The length of the wire is unnecessary information. We could, on the other hand, use it to find the radius of the circle, the speed of the bob, and the period of the motion.

ANS. FIG. P5.19

ANS. FIG. P5.21



ANS. FIG. P5.23

Section 5.3 Nonuniform Circular Motion

P5.22 (a) ac =

v2

r

r =

v2

ac

=13.0 m s( )2

2 9.80 m s2( ) = 8.62 m

(b) Let n be the force exerted by the rail.

Newton’s second law gives

Mg + n =

Mv2

r

n = M

v2

r− g

⎛⎝⎜

⎞⎠⎟= M 2g − g( ) = Mg, downward

(c) ac =

v2

r, or

ac =

13.0 m s( )2

20.0 m= 8.45 m s2

(d) If the force exerted by the rail is n1,

then n1 + Mg =

Mv2

r= Mac

n1 = M ac − g( ) which is < 0 since ac = 8.45 m/s2

Thus, the normal force would have to point away from thecenter of the curve. Unless they have belts, the riders will fallfrom the cars.

In a teardrop-shaped loop, the radius of curvature r decreases,causing the centripetal acceleration to increase. The speedwould decrease as the car rises (because of gravity), but theoverall effect is that the required centripetal force increases,meaning the normal force increases − there is less danger ifnot wearing a seatbelt.

*P5.23 (a) v = 30 km h( ) 1 h

3 600 s⎛⎝⎜

⎞⎠⎟

1 000 m1 km

⎛⎝⎜

⎞⎠⎟ = 8.33 m s

ANS. FIG. P5.22

Chapter 5 169


ANS. FIG. P5.25

Fy∑ = may : + n − mg = −mv2

r

n = m g −v2

r⎛⎝⎜

⎞⎠⎟

= 1 800 kg( ) 9.80 m/s2 −8.33 m/s( )2

20.4 m⎡

⎣⎢

⎤

⎦⎥

= 1.15 × 104 N up

(b) At the maximum speed, the weight of the car is just enough to

provide the centripetal force, so n = 0. Then mg =

mv2

r.

v = gr = 9.80 m/s2( ) 20.4 m( ) = 14.1 m/s = 50.9 km/h

P5.24 (a)

Fy∑ = may =mv2

R

mg − n =

mv2

R

n = mg −

mv2

R

(b) When n = 0 mg =

mv2

R

Then, v = gR

A more gently curved bump, with larger radius, allows the car to have a higher speed without leaving the road. This speed is proportional to the square root of the radius.

P5.25 Let the tension at the lowest point be T. From Newton’s second law, F∑ = ma and

T − mg = mac =mv2

r

T = m g +v2

r⎛⎝⎜

⎞⎠⎟

T = 85.0 kg( ) 9.80 m s2 +8.00 m s( )2

10.0 m

⎡

⎣⎢⎢

⎤

⎦⎥⎥

= 1.38 kN > 1 000 N

He doesn’t make it across the river because the vine breaks.



ANS. FIG. P5.27

P5.26 (a) The external forces acting on the water are

the gravitational force

and the contact force exerted on the water by the pail .

(b) The contact force exerted by the pail is the most important in

causing the water to move in a circle. If the gravitational force acted alone, the water would follow the parabolic path of a projectile.

(c) When the pail is inverted at the top of the circular path, it cannot hold the water up to prevent it from falling out. If the water is not to spill, the pail must be moving fast enough that the required centripetal force is at least as large as the gravitational force. That is, we must have

m

v2

r≥ mg or

v ≥ rg = 1.00m( ) 9.80m s2( ) = 3.13 m s

(d) If the pail were to suddenly disappear when it is at the top of the circle and moving at 3.13 m/s, the water would follow the

parabolic path of a projectile launched with initial velocity

components of vxi = 3.13 m/s, vyi = 0.

*P5.27 M = 40.0 kg, R = 3.00 m, T = 350 N

(a)

F∑ = 2T − Mg =Mv2

R

v2 =

RM

⎛⎝⎜

⎞⎠⎟ 2T − Mg( )

v2 =3.00 m40.0 kg

⎛⎝⎜

⎞⎠⎟

700 N − 40.0 kg( ) 9.80 m/s2( )⎡⎣ ⎤⎦= 23.1 m2/s2( )

v = 4.81 m s

(b) The normal force from the seat on the child accelerates the child in the same way that the total tension in the chain accelerates the child-seat system. Therefore, n = 2T = 700 N .

Chapter 5 171


*P5.28 See the forces acting on seat (child) in ANS. FIG. P5.27.

(a)

F∑ = 2T − Mg =Mv2

R

v2 = 2T − Mg( ) R

M⎛⎝⎜

⎞⎠⎟

v = 2T − Mg( ) R

M⎛⎝⎜

⎞⎠⎟

(b) n − Mg = F =

Mv2

R

n = Mg +

Mv2

R

Section 5.4 Motion in the Presence of Velocity-Dependent Resistive Forces

P5.29 (a) The speed v varies with time according to Equation 5.6,

v =

mgb 1 – e–bt/m( ) = vT 1 – e–t/τ( )

where vT = mg/b is the terminal speed. Hence,

b =

mgvT

=3.00 × 10−3 kg( ) 9.80 m s2( )

2.00 × 10−2 m s= 1.47 N ⋅ s m

(b) To find the time interval for v to reach 0.632vT, we substitute v = 0.632vT into Equation 5.6, giving

0.632vT = vT (1 − e−bt/m) or 0.368 = e−(1.47t/0.003 00)

Solve for t by taking the natural logarithm of each side of the equation:

ln(0.368) = –

1.47 t3.00 × 10–3 or –1 = –

1.47 t3.00 × 10–3

or t = −

mb

⎛⎝⎜

⎞⎠⎟ ln 0.368( ) = 2.04 × 10−3 s

(c) At terminal velocity, R = vTb = mg = 2.94 × 10−2 N



P5.30 (a) ρ =

mV

, A = 0.020 1 m2, R =

12ρairADvT

2 = mg

m = ρbeadV = 0.830 g cm3 4

3π 8.00 cm( )3⎡

⎣⎢⎤⎦⎥= 1.78 kg

Assuming a drag coefficient of D = 0.500 for this spherical object, and taking the density of air at 20°C from the endpapers, we have

vT =

2 1.78 kg( ) 9.80 m s2( )0.500 1.20 kg m3( ) 0.020 1 m2( ) = 53.8 m s

(b) vf2 = vi

2 + 2gh = 0 + 2gh: h =

v f2

2g=

53.8 m s( )2

2 9.80 m s2( ) = 148 m

P5.31 (a) a = g – Bv

When v = vT, a = 0 and g = BvT → B =

gvT

The Styrofoam falls 1.50 m at constant speed vT in 5.00 s.

Thus, vT =

hΔt

=1.50m5.00s

= 0.300m s

Then B =

9.80 m/s2

0.300 m/s= 32.7 s−1

(b) At t = 0, v = 0, and a = g = 9.80 m s2 down

(c) When v = 0.150 m/s,

a = g – Bv

= 9.80 m/s2 – (32.7 s–1) (0.150 m/s) = 4.90 m/s2 down

P5.32 F∑ = ma

−kmv2 = mdvdt

−kdt =dvv2

−k dt0

t

∫ = v−2 dvvi

v

∫

Chapter 5 173


−k t − 0( ) = v−1

−1 vi

v

= −1v+

1vi

1v=

1vi

+ kt =1+ vikt

vi

v =vi

1+ vikt

P5.33 (a) We must fit the equation v = vie−ct to the two data points:

At t = 0, v = 10.0 m/s, so v = vie−ct becomes

10.0 m/s = vi e0 = (vi)(1)

which gives vi = 10.0 m/s

At t = 20.0 s, v = 5.00 m/s so the equation becomes

5.00 m/s = (10.0 m/s)e−c(20.0 s)

giving 0.500 = e−c(20.0 s)

or −20.0c = ln

12

⎛⎝⎜

⎞⎠⎟ → c = −

ln 12( )

20.0= 3.47 × 10−2 s−1

(b) At t = 40.0 s

v = 10.0 m s( )e−40.0c = 10.0 m s( ) 0.250( ) = 2.50 m s

(c) The acceleration is the rate of change of the velocity:

a =dvdt =

ddt vie

–ct = vi e–ct( )(–c) = – c vie–ct( )

= –cv

Thus, the acceleration is a negative constant times the speed.

P5.34 v =

mgb

⎛⎝⎜

⎞⎠⎟ 1− exp

−btm

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

where exp(x) = ex is the exponential function.

At t → ∞ v → vT =

mgb

At t = 5.54 s 0.500vT = vT 1− exp

−b 5.54 s( )9.00 kg

⎛⎝⎜

⎞⎠⎟

⎡

⎣⎢

⎤

⎦⎥



exp−b 5.54 s( )

9.00 kg⎛⎝⎜

⎞⎠⎟= 0.500

−b 5.54 s( )9.00 kg

= ln 0.500 = −0.693

b =9.00 kg( ) 0.693( )

5.54 s= 1.13 kg s

(a) vT =

mgb

vT =

9.00 kg( ) 9.80 m s2( )1.13 kg s

= 78.3 m s

(b) 0.750vT = vT 1− exp

−1.13t9.00 s

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

exp

−1.13t9.00 s

⎛⎝⎜

⎞⎠⎟ = 0.250

t =

9.00 ln 0.250( )−1.13

s = 11.1 s

(c)

dxdt

=mgb

⎛⎝⎜

⎞⎠⎟ 1− exp −

btm

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

;

dxx0

x

∫ =mgb

⎛⎝⎜

⎞⎠⎟ 1− exp

−btm

⎛⎝⎜

⎞⎠⎟

⎡⎣⎢

⎤⎦⎥

dt0

t

∫

x − x0 =

mgtb

+m2 gb2

⎛⎝⎜

⎞⎠⎟

exp−btm

⎛⎝⎜

⎞⎠⎟

0

t

=mgt

b+

m2 gb2

⎛⎝⎜

⎞⎠⎟

exp−btm

⎛⎝⎜

⎞⎠⎟ − 1⎡

⎣⎢⎤⎦⎥

At t = 5.54 s,

x = 9.00 kg 9.80 m s2( ) 5.54 s1.13 kg s

+9.00 kg( )2 9.80 m s2( )

1.13 kg s( )2

⎛

⎝⎜⎜

⎞

⎠⎟⎟

exp −0.693( ) − 1[ ]

x = 434 m + 626 m −0.500( ) = 121 m

Chapter 5 175


Section 5.5 The Fundamental Forces of Nature P5.35 According to Newton’s law of universal gravitation, the force of

gravity on an object is given by

Fg =

Gm1m2

r2

Where G is the gravitational constant. The weight of an object on Earth

is given by mg, where g =

GME

r2 . Here, ME is the mass of the Earth. We

note that g, the acceleration of gravity due to the Earth, follows an inverse-square law. At the surface (at distance one Earth-radius RE from the center), it is 9.80 m/s2. At an altitude of 3.00RE above the surface (at distance 4.00RE from the center), the acceleration of gravity will be 4.002 = 16.0 times smaller:

a =GME

(4.00 RE )2 =GME

16.0 RE2 =

9.80 m s2

16.0

= 0.612 m/s2 toward the Earth

P5.36

F = keq1q2

r12( )2 = 8.99 × 109 N ⋅m2/C2( ) +40 C( ) −40 C( )2 000 m( )2

= −3.60 × 106 N (attractive) = 3.60 × 106 N downward

P5.37

F = Gm1m2

r2 =6.672 × 10−11 N · m2/kg2( ) 2.00 kg( ) 2.00 kg( )

0.30 m( )2

= 2.97 × 10−9 N

P5.38 For two 70–kg persons, modeled as spheres,

Fg =

Gm1m2

r2 =6.67 × 10−11 N ⋅m2/kg2( ) 70 kg( ) 70 kg( )

2 m( )2 ~ 10−7 N



Section 5.6 Context Connection: Drag Coefficients of Automobiles

P5.39 With 100 km/h = 27.8 m/s, the resistive force is

R =12

DρAv2 =12

0.250( ) 1.20 kg m3( ) 2.20 m2( ) 27.8 m s( )2 = 255 N

a = −Rm

= −255 N

1 200 kg= −0.212 m s2

P5.40 (a) The drag force on this car body at this speed is

R = 12

DρAv2 = 12

0.34( ) 1.2 kg/m3( ) 2.6 m2( ) 10 m/s( )2

= 53.0 N

Now Newton’s second law is Fx∑ = max :

+ fs − 53.0 N = 1 300 kg 3 m/s2( )fs = 3.95 × 103 N forward

(b) Newton’s second law changes to

+3 950 N − 12

0.2( ) 1.2 kg/m3( ) 2.6 m2( ) 10 m/s( )2

= 1 300 kg( )a

a = 3 950 N − 31.2 N1 300 kg

= 3.02 m/s2 forward

Streamlining to reduce drag makes little difference to the acceleration at this low speed.

(c) Fx∑ = max now reads

+3 950 N − 12

0.34( ) 1.2 kg/m3( ) 2.6 m2( )v2 = 0

v = 3 9500.530

kg ⋅m ⋅m

s2 ⋅kg= 86.3 m/s

(d) Still again, +3 950 N − 1

20.2( ) 1.2( ) 2.6( ) kg

mv2 = 0

v = 3 950

0.312m2

s2 = 113 m/s

According to this model, drag reduction significantly affects maximum speed.

Chapter 5 177


Additional Problems

P5.41 When the cloth is at a lower angle θ, the radial component of F∑ = ma reads

n + mg sinθ =

mv2

r

At θ = 68.0°, the normal force drops

to zero and g sin 68° =

v2

r:

v = rg sin 68° = 0.33 m( ) 9.8 m s2( )sin 68° = 1.73 m s

The rate of revolution is

angular speed

= 1.73 m s( ) 1 rev2πr

⎛⎝⎜

⎞⎠⎟

2πr2π 0.33 m( )

⎛⎝⎜

⎞⎠⎟

= 0.835 rev s = 50.1 rev min

P5.42 (a) The crate is in equilibrium, just before it starts to move. Let the normal force acting on it be n and the friction force, fs.

Resolving vertically: ∑Fy = may gives

n = Fg + Psin θ

Horizontally, ∑Fx = max gives

P cos θ = f

But,

fs ≤ µsn

i.e.,

Pcosθ ≤ µs Fg + Psinθ( )

or

P cosθ − µs sinθ( ) ≤ µsFg

ANS. FIG. P.41

ANS. FIG. P5.42



Divide by cos θ :

P 1− µs tanθ( ) ≤ µsFg secθ

Then

Pminimum =µsFg secθ

1− µs tanθ

(b)

To set the crate into motion, the x component (P cos θ) mustovercome friction fs = µsn:

P cos θ ≥ µsn = µs (Fg + P sin θ)

P cos θ − µs sin θ( ) ≥ µs Fg

For this condition to be satisfied, it must be true that

cos θ − µs sin θ( ) > 0→ µs tan θ < 1→ tanθ <1µs

.

If this condition is not met, no value of P can move the crate.

P5.43 Applying Newton’s second law to each object gives:

T1 = f1 + 2m g sinθ + a( ) [1]

T2 −T1 = f2 + m g sinθ + a( ) [2]

T2 = M g − a( ) [3]

(a), (b) Assuming that the system is in equilibrium (a = 0) and that the incline is frictionless, (f1 = f2 = 0), the equations reduce to

T1 = 2mg sinθ [1’]

T2 −T1 = mg sinθ [2’]

T2 = Mg [3’]

Substituting [1’] and [3’] into equation [2’] then gives

M = 3msinθ

so equation (3’) becomes

T2 = 3mg sinθ

(c), (d) M = 6msinθ (double the value found above), and f1 = f2 = 0. With these conditions present, the equations become

r rr

r

r

r

rrr

rr

r

ANS. FIG. P5.43

Chapter 5 179


ANS. FIG. P5.44

T1 = 2m g sinθ + a( ) , T2 −T1 = m g sinθ + a( ) and

T2 = 6msinθ g − a( ). Solved simultaneously, these yield

a = g sinθ1+ 2sinθ

,

T1 = 4mg sinθ 1+ sinθ1+ 2sinθ

⎛⎝⎜

⎞⎠⎟

and

T2 = 6mg sinθ 1+ sinθ1+ 2sinθ

⎛⎝⎜

⎞⎠⎟

(e) Equilibrium (a = 0) and impending motion up the incline, so

M = Mmax while f1 = 2µsmg cosθ and f2 = µsmg cosθ , both directed down the incline. Under these conditions, the equations become T1 = 2mg sinθ + µs cosθ( ) , T2 −T1 = mg sinθ + µs cosθ( ) ,

and T2 = Mmax g, which yield

Mmax = 3m sinθ + µs cosθ( ) .

(f) Equilibrium (a = 0) and impending motion down the incline, so

M = Mmin , while f1 = 2µsmg cosθ and f2 = µsmg cosθ , both directed up the incline. Under these conditions, the equations are

T1 = 2mg sinθ − µs cosθ( ) , T2 −T1 = mg sinθ − µs cosθ( ) , and

T2 = Mmin g, which yield

Mmin = 3m sinθ − µs cosθ( ) . When this

expression gives a negative value, it corresponds physically to a mass M hanging from a cord over a pulley at the bottom end of the incline.

(g)

T2,max −T2,min = Mmax g − Mmin g = 6µsmg cosθ

P5.44 (a) If the car is about to slip down the incline, f is directed up the incline.

Fy∑ = ncosθ + f sinθ − mg = 0

where f = µsn. Substituting,

n =

mgcosθ 1+ µs tanθ( )

and f =

µsmgcosθ 1+ µs tanθ( )

Then,

Fx∑ = nsinθ − f cosθ = mvmin

2

R yields



vmin =

Rg tanθ − µs( )1+ µs tanθ

When the car is about to slip up the incline, f is directed down the incline. Then,

Fy∑ = ncosθ − f sinθ − mg = 0 with f = µsn

This yields

n =

mgcosθ 1− µs tanθ( ) and

f =

µsmgcosθ 1− µs tanθ( )

In this case,

Fx∑ = nsinθ + f cosθ = mvmax

2

R, which gives

vmax =

Rg tanθ + µs( )1− µs tanθ

(b) If vmin =


= 0 , then µs = tanθ .

P5.45 For the right-hand block (m1), ΣF1 = m1a gives

−m1g sin 35.0° − fk ,1 +T = m1a

or

− 3.50 kg( ) 9.80 m/s2( )sin 35.0°

− µs 3.50 kg( ) 9.80 m/s2( )cos35.0° +T

= 3.50 kg( ) 1.50 m/s2( )

[1]

For the left-hand block (m2), ΣF2 = m2a gives

+m2 g sin 35.0° − fk ,2 −T = m2a

+ 8.00 kg( ) 9.80 m/s2( )sin 35.0° −

µs 8.00 kg( ) 9.80 m/s2( )cos35.0° −T = 8.00 kg( ) 1.50 m/s2( ) [2]

Solving equations [1] and [2] simultaneously gives

(a) µk = 0.0871

(b) T = 27.4 N

ANS. FIG. P5.45

Chapter 5 181


ANS. FIG. P5.46

ANS. FIG. P5.47

P5.46 (a) For the system to start to move when released, the force tending to move m2 down the incline, m2g sin θ, must exceed the maximum friction force which can retard the motion:

fmax = f1,max + f2,max = µs ,1n1 + µs ,2n2

fmax = µs ,1m1g + µs ,2m2 g cosθ

From the table of coefficients of friction in the text, we take µs,1 = 0.610 (aluminum on steel) and µs,2 = 0.530 (copper on steel). With

m1 = 2.00 kg, m2 = 6.00 kg, θ = 30.0°

the maximum friction force is found to be fmax = 38.9 N. This exceeds the force tending to cause the system to move,

m2 g sinθ = 6.00 kg 9.80 m s2( )sin 30° = 29.4 N . Hence,

the system will not start to move when released

(b) and (c) No answer because the blocks do not move.

(d) The friction forces increase in magnitude until the total friction force retarding the motion, f = f1 + f2, equals the force tending to set the system in motion. That is, until

f = m2 g sinθ = 29.4 N

P5.47 (a) As shown in the free-body diagram on the right, the mass at the end of the chain is in vertical equilibrium. Thus,

T cosθ = mg [1]

Horizontally, the mass is accelerating toward the center of a circle of radius r:

T sinθ = mar =

mv2

r [2]

Here, r is the sum of the radius of the circular platform R = D/2 = 4.00 m and 2.50 sinθ :



ANS. FIG. P5.48

r = 2.50sinθ + 4.00( ) mr = 2.50sin 28.0° + 4.00( ) m = 5.17 m

We solve for the tension T from [1]:

T cosθ = mg → T =

mgcosθ

and substitute into [2] to obtain

tanθ =

ar

g=

v2

gr

v

2 = gr tanθ = 9.80 m/s2( ) 5.17 m( ) tan 28.0°( )

v = 5.19 m/s

(b) The free-body diagram for the child is shown above.

(c) T =

mgcosθ

=40.0 kg( ) 9.80 m s2( )

cos28.0°= 444 N

P5.48 The force diagram is shown on the right. With motion impending,

n +T sinθ − mg = 0

f = µs mg −T sinθ( )

and

T cosθ − µsmg + µsT sinθ = 0

so

T =

µsmgcosθ + µs sinθ

To minimize T, we maximize cosθ + µs sinθ:

ddθ

cosθ + µs sinθ( ) = 0 = − sinθ + µs cosθ

Therefore, the angle where tension T is a minimum is

θ = tan−1 µs( ) = tan−1 0.350( ) = 19.3°

What is the tension at this angle? From above,

Chapter 5 183


T =

0.350 1.30 kg( ) 9.80 m/s2( )cos 19.3° + 0.350 sin 19.3°

= 4.21 N

The situation is impossible because at the angle of minimum tension,the tension exceeds 4.00 N.

P5.49 Standing on the inner surface of the rim, and moving with it, each person will feel a normal force exerted by the rim. This inward force causes the 3.00 m/s2 centripetal acceleration:

ac = v2/r so v = acr = 3.00 m s2( ) 60.0 m( ) = 13.4 m s

The period of rotation comes from v =

2πrT

:

T =

2πrv

=2π 60.0 m( )13.4 m s

= 28.1 s

so the frequency of rotation is

f =

1T=

128.1 s

=1

28.1 s60 s

1 min⎛⎝⎜

⎞⎠⎟ = 2.14 rev min .

P5.50 (a) The free-body diagrams are shown in the figure below.

rr

r

r

r

r

r r

ANS. FIG. P5.50(a)

f1 and n1 appear in both diagrams as action-reaction pairs.

(b) For the 5.00-kg mass, Newton’s second law in the y direction gives:

n1 = m1g = 5.00 kg( ) 9.80 m/s2( ) = 49.0 N

In the x direction,

f1 −T = 0

T = f1 = µmg = 0.200 5.00 kg( ) 9.80 m/s2( ) = 9.80 N

For the 10.0-kg mass, Newton’s second law in the x direction gives:



45.0 N − f1 − f2 = 10.0 kg( )a

In the y direction,

n2 − n1 − 98.0 N = 0

f2 = µn2 = µ n1 + 98.0 N( ) = 0.20 49.0 N + 98.0 N( ) = 29.4 N

45.0 N − 9.80 N − 29.4 N = 10.0 kg( )a

a = 0.580 m/s2

P5.51 There are three forces on the child, a vertical normal force, a horizontal force (combination of friction and a horizontal force from a seat belt), and gravity.

Fx : Fs = mv2 R∑

Fy : n − mg = 0→ n = mg∑

The magnitude of the net force is

Fnet = mv2 R( )2

+ mg( )2 with a direction of

θ = tan−1 mg

mv2 R⎡

⎣⎢

⎤

⎦⎥ = tan−1 gR

v2⎡⎣⎢

⎤⎦⎥

above the horizontal.

For m = 40.0 kg and R = 10.0 m:

Fnet =

40.0 kg( ) 3.00 m/s( )2

10.0 m

⎡

⎣⎢⎢

⎤

⎦⎥⎥

2

+ 40.0 kg( ) 9.80 m/s2( )⎡⎣ ⎤⎦2

Fnet = 394 N

direction:

θ = tan−1 9.80 m/s2( ) 10.0 m( )

3.00 m/s( )2

⎡

⎣⎢⎢

⎤

⎦⎥⎥

→ θ = 84.7°

P5.52 (a) While the car negotiates the curve, the accelerometer is at the angle θ .

Horizontally: T sinθ = mv2

r

Vertically: T cosθ = mg

where r is the radius of the curve, and v is the speed of the car.

ANS. FIG. P5.51

r

r

r

ANS. FIG. P5.52

Chapter 5 185


By division, tanθ = v2

rg

Then ac =

v2

r= g tanθ: ac = 9.80 m s2( )tan 15.0°

ac = 2.63 m/s2

(b) r = v2

ac

gives r = 23.0 m/s( )2

2.63 m/s2 = 201 m

(c) v2 = rg tanθ = 201 m( ) 9.80 m s2( )tan 9.00°

v = 17.7 m/s

*P5.53 (a) The free-body diagrams of the two blocks are:

ANS. FIG. P5.53(a)

Vertical forces sum to zero because the blocks move on a horizontal surface; therefore, ay = 0 for each block.

∑F1y = m1ay ∑F2y = m2ay

−m1g + n1 = 0→ n1 = m1g −m2 g + n2 = 0→ n2 = m2 g

Kinetic friction is: Kinetic friction is:

f1 = µ1n1 = µ1m1g f2 = µ2n2 = µ2m2 g

(b)

The net force on the system of the blocks would be equal to themagnitude of the force, F, minus the friction force on eachblock. The blocks will have the same acceleration.

(c)

The net force on the mass, m1 , would be equal to the force, F, minus the friction force on m1 and the force P 21 , as identi-fied in the free-body diagram.



(d)

The net force on the mass, m2 , would be equal to the force, P12 , minus the friction force on m2 , as identified in the free-body diagram.

(e) The blocks are pushed to the right by force F, so kinetic friction

f

acts on each block to the left. Each block has the same horizontal acceleration, ax = a. Each block exerts an equal and opposite force on the other, so those forces have the same magnitude: P12 = P21 = P.

∑F1x = m1ax ∑F2x = m2ax

F – P –f1 = m1a P – f2 = m2a

F − P − µ1m1g = m1a

P − µ2m2 g = m2a

(f) Adding the above two equations of x components, we find

F − P − µ1m1g + P − µ2m2 g = m1a + m2aF − µ1m1g − µ2m2 g = (m1 + m2 )a →

a =F − µ1m1g − µ2m2 g

m1 + m2( )

(g) From the x component equation for block 2, we have

P − µ2m2 g = m2a → P = µ2m2 g + m2a

P =m2

m1 + m2

⎛⎝⎜

⎞⎠⎟

F + µ2 − µ1( )m1g⎡⎣ ⎤⎦

We see that when the coefficients of friction are equal, µ1 = µ2, the magnitude P is independent of friction.

*P5.54 The free-body diagram is shown on the right, where it is assumed that friction points up the incline, otherwise, the child would slide down the incline. The net force is directed left toward the center of the circular path in which the child travels. The radius of this path is R = d cos θ.

Three forces act on the child, a normal force, static friction, and gravity. The relations of their force components are:

Fx : fs cos θ − n sin θ = mv2/R∑ [1]

ANS. FIG. P5.54

Chapter 5 187


Fy : fs sin θ + n cos θ − mg = 0→∑fs sin θ + n cos θ = mg [2]

Solve for the static friction and normal force.

To solve for static friction, multiply equation [1] by cos θ and equation [2] by sin θ and add:

cos θ [fs cos θ – n sin θ] + sin θ [fs sin θ + n cos θ]

= cos θ (mv2/R) + sin θ (mg)

fs = (mg) sin θ + (mv2/R) cos θ

To solve for the normal force, multiply equation [1] by –sin θ and equation [2] by cos θ and add:

(–sin θ) [fs cos θ – n sin θ] + cos θ [fs sin θ + n cos θ]

= (–sin θ) (mv2/R) + cos θ (mg)

n = (mg) cos θ – (mv2/R) sin θ

In the above, we have used cos2 θ + sin2

θ = 1.

If the above equations are to be consistent, static friction and the normal force must satisfy the condition fs ≤ µsn; this means

(mg) sin θ + (mv2 R)cos θ ≤ µs[(mg) cos θ − (mv2 R)sin θ]→

v2(cos θ + µs sin θ) ≤ g R(µs cos θ − sin θ)

Using this result, and that R = d cos θ, we have the requirement that

v ≤

gd cosθ(µs cosθ − sinθ)(cosθ +µs sinθ)

If this condition cannot be met, if v is too large, the physical situation cannot exist.

The values given in the problem are d = 5.32 m, µs = 0.700, θ = 20.0°, and v = 3.75 m/s. Check whether the given value of v satisfies the above condition:

9.80 m/s2( ) 5.32 m( )cos20.0° − 0.700( )cos20.0° − sin 20.0°[ ]cos20.0° + 0.700 sin 20.0°( )

= 3.62 m/s



The situation is impossible because the speed of the child givenin the problem is too large: static friction could not keep the child in place on the incline.

P5.55 (a) First, draw a free-body diagram, (top figure) of the top block. Since ay = 0, n1 = 19.6 N, and

fk = µkn1 = 0.300 19.6 N( )= 5.88 N

From ΣFx = maT ,

10.0 N − 5.88 N = 2.00 kg( )aT

or aT = 2.06 m/s2 (for top block). Now draw a free-body diagram (middle figure) of the bottom block and observe that ΣFx = MaB gives

f = 5.88 N = 8.00 kg( )aB or

aB = 0.735 m/s2 (for the bottom block). In time t, the distance each block moves (starting from rest) is

dT = 1

2aTt2 and

dB = 1

2aBt2 . For the

top block to reach the right edge of the bottom block, (see bottom figure), it is necessary that dT = dB + L or

12

2.06 m/s2( )t2 = 12

0.735 m/s2( )t2 + 3.00 m

which gives t = 2.13 s .

(b) From above, dB = 1

20.735 m s2( ) 2.13 s( )2 = 1.67 m .

P5.56 (a) Since the object of mass m2 is in equilibrium, Fy∑ = T − m2 g = 0

or T = m2 g

(b) The tension in the string provides the required centripetal acceleration of the puck.

Thus, Fc = T = m2 g .

r

r

r

r

r

rr

ANS. FIG. P5.55

Chapter 5 189


ANS. FIG. P5.57

(c) From Fc =

m1v2

R,

we have

v =RFc

m1

=m2

m1

⎛⎝⎜

⎞⎠⎟

gR

(d) The puck will spiral inward, gaining speed as it does so. It gains

speed because the extra-large string tension pulls at an angle of less than 90° to the direction of the inward-spiraling velocity, producing forward tangential acceleration as well as inward radial acceleration of the puck.

(e) The puck will spiral outward, slowing down as it does so.

P5.57 The plane’s acceleration is toward the center of the circle of motion, so it is horizontal. The radius of the circle of motion is (60.0 m) cos 20.0° = 56.4 m and the acceleration is

ac =

v2

r =35 m s( ) 2

56.4 m = 21.7 m s2

We can also calculate the weight of the airplane:

Fg = mg = (0.750 kg)(9.80 m/s2)

= 7.35 N

We define our axes for convenience. In this case, two of the forces—one of them our force of interest—are directed along the 20.0° line. We define the x axis to be directed in the +

T direction, and the y axis to be

directed in the direction of lift. With these definitions, the x component of the centripetal acceleration is

acx = ac cos 20.0°

and ∑Fx = max yields T + Fg sin 20.0° = macx

Solving for T, T = macx − Fg sin 20.0°

Substituting, T = (0.750 kg)(21.7 m/s2) cos 20.0° − (7.35 N) sin 20.0°

Computing, T = (15.3 N) − (2.51 N) = 12.8 N

*P5.58 Find the acceleration of the block according to the kinematic equations. The book travels through a displacement of 1.00 m in a time interval of 0.483 s. Use the equation xf = xi + vxit + ½axt

2, where ∆x = xf – xi = 1.00 m,



∆t = t = 0.483 s, and vi = 0:

x f = xi + vxit + 1

2 axt2 → a =

2Δxt2 = 8.57 m/s2

Now, find the acceleration of the block caused by the forces. See the free-body diagram below. We take the positive y axis is perpendicular to the incline; the positive x axis is parallel and down the incline.

∑Fy = may

n − mg cos θ = 0→ n = mg cos θ

∑Fx = max

mg sin θ − fk = ma

where fk = µkn = µkmg cos θ

Substituting the express for kinetic friction into the x component equation gives

mg sin θ − µkmg cos θ = ma → a = g(sin θ − µk cos θ)

For µk = 0.300, a = 7.02 m/s2.

The situation is impossible because these forces on the book cannot produce the acceleration described.

P5.59 (a) We let R represent the radius of the hoop and T represent the period of its rotation. The bead moves in a circle with radius v = Rsinθ at a speed of

v =

2πrT

=2πRsinθ

T

The normal force has an inward radial component of n sinθ and an upward component of n cosθ .

Fy∑ = may : ncosθ − mg = 0

or

n =

mgcosθ

Then

Fx∑ = nsinθ = mv2

r becomes

ANS. FIG. P5.58

ANS. FIG. P5.59

Chapter 5 191


mgcosθ

⎛⎝⎜

⎞⎠⎟ sinθ =

mRsinθ

2πRsinθT

⎛⎝⎜

⎞⎠⎟

2

which reduces to

g sinθcosθ

=4π 2Rsinθ

T 2

This has two solutions: sinθ = 0 ⇒ θ = 0° [1]

and cosθ =

gT 2

4π 2R [2]

If R = 15.0 cm and T = 0.450 s, the second solution yields

cosθ =

9.80 m s2( ) 0.450 s( )2

4π 2 0.150 m( ) = 0.335 or θ = 70.4°

Thus, in this case, the bead can ride at two positions: θ = 70.4°

and θ = 0° .

(b) At this slower rotation, solution [2] above becomes

cosθ =

9.80 m s2( ) 0.850 s( )2

4π 2 0.150 m( ) = 1.20 , which is impossible.

In this case, the bead can ride only at the bottom of the loop,

θ = 0° .

(c) There is only one solution for (b) because the period is too large.

(d)

The equation that the angle must satisfy has two solutions

whenever 4π 2R > gT 2 but only the solution 0° otherwise.The loop’s rotation must be faster than a certain thresholdvalue in order for the bead to move away from the lowestposition. Zero is always a solution for the angle.

(e)

From the derivation of the solution in (a), there are never morethan two solutions.



ANS. FIG. P5.60

P5.60 (a) The wall’s normal force pushes inward:

∑Finward = mainward

becomes n =

mv2

R =mR

2πRT

⎛⎝⎜

⎞⎠⎟

2

=4π 2Rm

T 2

The friction and weight balance:

∑Fupward = maupward becomes

+f – mg = 0

so with the person just ready to start sliding down, fs = μsn = mg

Substituting,

µsn = µs

4π 2RmT 2 = mg

Solving,

T 2 =

4π 2Rµs

g

gives T =

4π 2Rµs

g

(b)

The gravitational and friction forces remain constant. (Static friction adjusts to support the weight.) The normalforce increases. The person remains in motion with the wall.

(c)

The gravitational force remains constant. The normal and friction forces decrease. The person slides relative to the wall and downward into the pit.

P5.61 At terminal velocity, the accelerating force of gravity is balanced by friction drag:

mg = arv + br2v2

Chapter 5 193


(a) With r = 10.0 µm, mg = 3.10 × 10−9( )v + 0.870 × 10−10( )v2

For water, m = ρV = 1 000 kg m3 4

3π 10−5 m( )3⎡

⎣⎢⎤⎦⎥

4.11× 10−11 = 3.10 × 10−9( )v + 0.870 × 10−10( )v2

Assuming v is small, ignore the second term on the right hand side:

v = 0.013 2 m/s

(b) With r = 100 µm, mg = 3.10 × 10−8( )v + 0.870 × 10−8( )v2

Here we cannot ignore the second term because the coefficients are of nearly equal magnitude.

4.11× 10−8 = 3.10 × 10−8( )v + 0.870 × 10−8( )v2

v =−3.10 ± 3.10( )2 + 4 0.870( ) 4.11( )

2 0.870( ) = 1.03 m s

(c) With r = 1.00 mm, mg = 3.10 × 10−7( )v + 0.870 × 10−6( )v2

Assuming v > 1 m/s, and ignoring the first term:

4.11× 10−5 = 0.870 × 10−6( )v2

v = 6.87 m s



P5.62 (a)

t(s) d (m)

1.00 4.88

2.00 18.9

3.00 42.1

4.00 43.8

5.00 112

6.00 154

7.00 199

8.00 246

9.00 296

10.0 347

11.0 399

12.0 452

13.0 505

14.0 558

15.0 611

16.0 664

17.0 717

18.0 770

19.0 823

20.0 876

Chapter 5 195


ANS. FIG. P5.63

(b)

(c) A straight line fits the points from t = 11.0 s to 20.0 s quite precisely. Its slope is the terminal speed.

vT = slope =

876 m − 399 m20.0 s − 11.0 s

= 53.0 m s

P5.63 (a) The gravitational force exerted by the planet on the person is

mg = (75.0 kg)(9.80 m/s2)

= 735 N down

Let n represent the force exerted on the person by a scale, which is an upward

force whose size is her “apparent weight.” The true weight is mg down. For the person at the equator, summing up forces on the object in the direction towards the Earth’s center gives ∑F = ma:

mg − n = mac

where ac = v2/RE = 0.033 7 m/s2

is the centripetal acceleration directed toward the center of the Earth.

Thus, we can solve part (c) before part (b) by noting that



n = m(g − ac) < mg

(c) or mg = n + mac > n.

The gravitational force is greater. The normal force is smaller,just as one experiences at the top of a moving ferris wheel.

(b) If m = 75.0 kg and g = 9.80 m/s2, at the equator we have

n = m(g − ac) = (75.0 kg)(9.800 m/s2 – 0.033 7 m/s2) = 732 N

P5.64 v = vi − kx implies the acceleration is

a = dv

dt= 0 − k

dxdt

= −kv

Then the total force is

F∑ = ma = m −kv( )

The resistive force is opposite to the velocity:

F∑ = −km

v

Chapter 5 197


ANSWERS TO EVEN-NUMBERED PROBLEMS P5.2 (a) 4.18; (b) Time would increase, as the wheels would skid and only

kinetic friction would act; or perhaps the car would flip over.

P5.4 (a) 256 m; (b) 42.7 m

P5.6 (a) 0.404; (b) 45.8 lb

P5.8 (a) 14.7 m; (b) neither mass is necessary

P5.10 (a) See ANS. FIG. P5.10; (b) θ = 55.2° ; (c) n = 167 N

P5.12 (a) 48.6 N, 31. 7 N; (b) If P > 48.6 N, the block slides up the wall. If P < 31.7 N, the block slides down the wall; (c) 62.7 N, P > 62.7 N, the block cannot slide up the wall. If P < 62.7 N, the block slides down the wall

P5.14 (a) See ANS. FIG. P5.14; (b) See P5.14 for complete list of accelerations and directions; (c) 2.31 m/s2, down for m1, left for m2, and up for m3; (d) T12 = 30.0 N and T23 = 24.2 N; (e) T12 decreases and T23 increases

P5.16 (a) 8.32 × 10–8 N inward; (b) 9.13 × 1022 m/s2 inward

P5.18 The situation is impossible because the speed of the object is too small, requiring that the lower string act like a rod and push rather than like a string and pull.

P5.20 (a) 1.65 × 103 m/s; (b) 6.84 × 103 s

P5.22 (a) 8.62 m; (b) Mg, downward; (c) 8.45 m/s2; (d) The normal force would have to point away from the center of the curve. Unless they have belts, the riders will fall from the cars. In a teardrop-shaped loop, the radius of curvature r decreases, causing the centripetal acceleration to increase. The speed would decrease as the car rises (because of gravity), but the overall effect is that the required centripetal force increases, meaning the normal force increases – there is less danger if not wearing a seatbelt.

P5.24 (a) mg −

mv2

R; (b) gr

P5.26 (a) the gravitational force and the contact force exerted on the water by the pail; (b) contact force exerted by the pail; (c) 3.13 m/s; (d) the water would follow the parabolic path of a projectile

P5.28 (a)

2T − Mg( ) RM

⎛⎝⎜

⎞⎠⎟

; (b) Mg +

Mv2

R

P5.30 (a) 53.8 m/s; (b) 148 m



P5.32

vi

1+ vikt

P5.34 (a) 78.3 m/s; (b) 11.1 s; (c) 121 m

P5.36 3.60 × 106 N downward

P5.38 ~10–7 N toward you

P5.40 (a) 3.95 × 103 N forward; (b) 3.02 m/s2 forward; (c) 86.3 m/s; (d) 113 m/s

P5.42 (a)

µsFg secθ1− µs tanθ

; (b) tanθ <

1µs

P5.44 (a)


, Rg tanθ + µs( )

1− µs tanθ; (b) tanθ

P5.46 (a) the system will not start to move when released; (b and c) no answer; (d) f = m2 g sinθ = 29.4 N

P5.48 The situation is impossible because at the angle of minimum tension, the tension exceeds 4.00 N

P5.50 (a) See P5.50 for complete solution; (b) 9.80 N, 0.580 m/s2

P5.52 (a) 2.63 m/s2; (b) 201 m; (c) 17.7 m/s

P5.54 The situation is impossible because the speed of the child given in the problem is too large: static friction could not keep the child in place on the incline

P5.56 (a) m2g; (b) m2g; (c)

m2

m1

⎛⎝⎜

⎞⎠⎟

gR ; (d) The puck will spiral inward, gaining

speed as it does so; (e) The puck will spiral outward, slowing down as it does so

P5.58 The situation is impossible because these forces on the book cannot produce the acceleration described.

P5.60 (a)

4π 2Rµs

g ; (b) The gravitational and friction forces remain constant

(Static friction adjusts to support the weight). Normal force increases. The person remains in motion with the wall; (c) The gravitational force remains constant. The normal force and friction forces decrease. The person slides relative to the wall and downward into the pit.

P5.62 (a) See table in P5.62 (a); (b) See graph in P5.62 (b); (c) 53.0 m/s

P5.64 F∑ = −km

v

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