CHAPTER 6 USING NEWTON’S LAWS

ActivPhysics can help with these problems:

All Activities in Section 2 “Forces and Motion”

and Section 4 “Circular Motion.”

Section 6-1: Using Newton’s Second Law

Problem

1. Two forces, both in the x-y plane, act on a 1.5-kgmass, which accelerates at 7.3 m/s

2in a direction

30◦ counter-clockwise from the x-axis. Oneforce has magnitude 6.8 N and points in the+x direction. Find the other force.

Solution

Newton’s second law for this mass says Fnet =F1 +

F2 =ma, where we assume no other significant forcesare acting. Since the acceleration and the first forceare given, one can solve for the second, F2 =ma−

F1 = (1.5 kg)(7.3 m/s2)(̂ı cos 30◦ + ̂ sin 30◦) −

(6.8 N)̂ı = (2.68̂ı+ 5.48̂) N. This has magnitude6.10 N and direction 63.9◦ CCW from the x-axis.

Problem

2. Two forces act on a 3.1-kg mass, which undergoesacceleration a= 0.91̂ı− 0.27̂ m/s

2. If one of the

forces is F1 =−1.2̂ı− 2.5̂ N, what is the otherforce?

Solution

As in the previous problem, F2 =ma−F1 =

(3.1 kg)(0.91̂ı− 0.27̂)(m/s2) − (−1.2̂ı− 2.5̂) N =

(4.02̂ı+ 1.66̂)N.

Problem

3. A 3700-kg barge is being pulled along a canal bytwo mules, as shown in Fig. 6-59. The tension ineach tow rope is 1100 N, and the ropes make25◦ angles with the forward direction. What forcedoes the water exert on the barge (a) if it moveswith constant velocity and (b) if it accelerates

forward at 0.16 m/s2?

Solution

The horizontal forces on the barge are the twotensions and the resistance of the water, as shown onFigure 6-59. The net force is in the x direction, so

2T cos 25◦ − Fres = max, since T1 = T2 = T.(a) If ax = 0, Fres = 2(1100 N) cos 25◦ = 1.99 kN.

(b) If ax = 0.16 m/s2, Fres = 1.99 kN − (3700 kg)×

(0.16 m/s2) = 1.40 kN.

25°

25°

Fres T1

T2

x

figure 6-59 Problem 3 Solution.

Problem

4. At what angle should you tilt an air table tosimulate motion on the moon’s surface, whereg = 1.6 m/s

2?

Solution

The acceleration down an incline is a|| = g sin θ (seeExample 6-1). To replicate the moon’s surface gravity,the angle of tilt should be θ = sin−1(1.62/9.81) =9.51◦ (see Appendix E).

Problem

5. A block of mass m slides with acceleration a downa frictionless slope that makes an angle θ to thehorizontal; the only forces acting on it are the forceof gravity Fg and the normal force N of the slope.Show that the magnitude of the normal force isgiven by N = m

√

g2 − a2.

Solution

Choose the x-axis down the slope (parallel to theacceleration) and the y-axis parallel to the normal.Then ax = a, ay = 0, Nx = 0, Ny = N, Fgx =Fg cos (90◦ − θ) = mg sin θ, and Fgy = −mg cos θ.Newton’s second law, N+Fg =ma, in componentsgives mg sin θ = ma, and N − mg cos θ = 0.Eliminate θ (using sin2 θ + cos2 θ = 1) to find (a/g)2 +

(N/mg)2 = 1, or N = m√

g2 − a2.

60 CHAPTER 6

Problem 5 Solution.

Problem

6. A skier starts from rest at the top of a 24◦ slope1.3 km long. Neglecting friction, how long does ittake to reach the bottom?

Solution

The acceleration down a frictionless incline is a =g sin θ (see Example 6-1), and the distance traveleddown the incline, starting from rest (v0 = 0) at the top(x0 = 0), is x = 1

2at2. Therefore, t =

√

2x/g sin θ =√

2(1.3 km)/(9.8 m/s2) sin 24◦ = 25.5 s.

Problem

7. A block is launched up a frictionless ramp thatmakes an angle of 35◦ to the horizontal. If theblock’s initial speed is 2.2 m/s, how far up theramp does it slide?

Solution

The acceleration up the ramp is −g sin 35◦, so theblock goes a distance in this direction calculated fromthe equation v2

0x − 2g sin 35◦(x − x0) = 0. Thus,

x − x0 = (2.2 m/s)2/2(9.8 m/s2) sin 35◦ = 43.1 cm.

Problem

8. At the start of a race, a 70-kg swimmer pushes offthe starting block with a force of 950 N directed at15◦ below the horizontal. (a) What is theswimmer’s horizontal acceleration? (b) If theswimmer is in contact with the starting block for0.29 s, what is the horizontal component of hisvelocity when he hits the water?

Solution

(a) If we assume that the reaction force of the startingblock and gravity are the only significant forces actingon the swimmer, the horizontal acceleration is justFx/m, or ax = (950 N) cos 15◦/70 kg = 13.1 m/s2.

(b) vx = axt = (13.1 m/s2)(0.29 s) = 3.80 m/s.

Problem 8 Solution.

Problem

9. A 15-kg monkey hangs from the middle of amassless rope as shown in Fig. 6-60. What is thetension in the rope? Compare with the monkey’sweight.

Solution

The sum of the forces at the center of the rope (shownon Fig. 6-60) is zero (if the monkey is at rest), T1 +

T2 +W= 0. The x component of this equationrequires that the tension is the same on both sides:T1 cos 8◦ + T2 cos 172◦ = 0, or T1 = T2. The ycomponent gives 2T sin 8◦ = W, or T = W/2 sin 8◦ =

3.59W = 3.59(15 kg)(9.8 m/s2) = 528 N.

8° 8° T2 T1

x

y

W

figure 6-60 Problem 9 Solution.

Problem

10. A tow truck is connected to a 1400-kg car by acable that makes a 25◦ angle to the horizontal, asshown in Fig. 6-61. If the truck accelerates at0.57 m/s2, what is the magnitude of the cabletension? Neglect friction and the mass of thecable.

Solution

The only force on the car with a horizontal component(in the direction of the acceleration) is the tension.Therefore, T cos θ = ma, or T = (1400 kg)×

CHAPTER 6 61

(0.57 m/s2)/cos 25◦ = 880 N. (The force of static

friction acts between the tires and the road withmagnitude sufficient to keep the wheels turning, but isassumed to be negligible.)

25° a

T

negligible

figure 6-61 Problem 10 Solution.

Problem

11. A 10-kg mass is suspended at rest by two stringsattached to walls, as shown in Fig. 6-62. Find thetension forces in the two strings.

Solution

The force diagram is superimposed on Fig. 6-62. Sincethe mass is at rest, the sum of the forces is zero,T1 +T2 +W= 0, which is true for the x andy components separately, T1 cos 45◦ − T2 = 0, andT1 sin 45◦ − W = 0. Solving for the magnitudes of thetensions, and substituting 98 N for the weight, we findT1 =

√2 98 N = 139 N, and T2 = T1/

√2 = 98 N.

45°

10 kg

W

yT1

T2

x

figure 6-62 Problem 11 Solution.

Problem

12. A 1100-kg car goes off the road and plunges downa 23◦ embankment, coming to rest against a tree.The contact between tree and car is such that theforce exerted on the car by the tree is purelyhorizontal, as suggested in Fig. 6-63. Find themagnitude of that force once the car is fullystopped.

23°

F

figure 6-63 Problem 12.

Solution

If the slope of the embankment exerts only a normalforce on the car (no friction), the situation is the sameas in Example 6-4, Fg+N+Fh = 0. Then Fh =

mg tan θ = (1100 kg)(9.8 m/s2) tan 23◦ = 4.58 kN.

Problem

13. A camper hangs a 26-kg pack between two trees,using two separate pieces of rope of differentlengths, as shown in Fig. 6-64. What is thetension in each rope?

figure 6-64 Problem 13.

62 CHAPTER 6

Solution

The sum of the forces acting on the pack (gravity andthe tension along each rope) is zero, since it is at rest,Fg +T1 +T2 = 0. In a coordinate system with x-axishorizontal to the right and y-axis vertical upward, thex and y components of the net force are −T1 cos 71◦ +T2 cos 28◦ = 0, and −26×9.8 N + T1 sin 71◦ +T2 sin 28◦ = 0 (see Example 6-3). Solving for T1 andT2, one finds T1 = (26×9.8 N)(sin 71◦ +tan28◦ cos 71◦) = 228 N and T2 = (26×9.8 N)÷(sin 28◦ + tan 71◦ cos 28◦) = 84.0 N.

Problem

14. A construction worker is lifting a 92-kg bundle ofplywood onto an upper floor, using thearrangement shown in Fig. 6-65. What force mustthe worker apply to lift the bundle at constantspeed? Assume the pulley is massless andfrictionless.

T1 T2

mg

figure 6-65 Problem 14.

Solution

The tension in each rope pulls upward on the pulley,while the weight of the plywood pulls downward.Since the mass of the pulley is negligible, its weightcan be neglected, and the tension in each rope is thesame (this will be evident after Chapter 12). Theacceleration of the pulley, whose speed is constant, iszero; therefore the vertical component of Newton’ssecond law gives T1 + T2 − mg = 2T − mg = 0, orT = 1

2mg = 1

2(92 kg)(9.8 m/s

2) = 451 N.

Section 6-2: Multiple Objects

Problem

15. Your 12-kg baby sister is hanging on the bottomof the tablecloth with all her weight. In themiddle of the table, 60 cm from each edge, is a6.8-kg roast turkey. (a) What is the accelerationof the turkey? (b) From the time she startspulling, how long do you have to intervene beforethe turkey goes over the edge of the table?

Solution

The vertical motion of your baby sister and thehorizontal motion of the turkey are analogous to theclimber and rock in Example 6-5. If we assume thatboth have accelerations of the same magnitude as thetablecloth, which has negligible mass, no friction withthe table, etc., then (a) a = arx = mcg/(mc + mr) =

(12 kg)(9.8 m/s2)/(12 kg + 6.8 kg) = 6.26 m/s2 is theturkey’s horizontal acceleration, and (b) t =

√

2x/a =√

2(60 cm)/(6.26 m/s2) = 0.438 s is the time you have

to save the turkey from going over the edge. (Theassumptions relevant to Example 6-5 might besomewhat over-restrictive in this situation.)

Problem

16. Find expressions for the acceleration of the blocksin Fig. 6-66, where the string is fastened securelyto the ceiling. Neglect friction and assume thatthe masses of pulley and string are negligible.

Solution

The equations of motion (components parallel to theaccelerations) for the two masses (m1 includes theattached pulley) are T = m2a2 and m1g − 2T =m1a1. (The assumptions stated ensure that the tensionhas the same magnitude at all points in the string, asshown.) If the length of the string is fixed, when m1

moves down a distance d, m2 moves to the right adistance 2d, so a2 = 2a1. Thus T = m2(2a1), m1a1 =m1g − 2(2m2a1), and a1 = m1g/(m1 + 4m2).

Problem

17. If the left-hand slope in Fig. 6-56 makes a60◦ angle with the horizontal, and the right-handslope makes a 20◦ angle, how should the massescompare if the objects are not to slide along thefrictionless slopes?

Solution

The free-body force diagrams for the left- andright-hand masses are shown in the sketch, wherethere is only a normal contact force since each slope is

CHAPTER 6 63

m1

m2

a2

N

m2g

m1g

TT T

a1

figure 6-66 Problem 16 Solution.

frictionless, and we indicate separate parallel andperpendicular x-y-axes. If the masses don’t slide, thenet force on each must be zero, or Tℓ −mℓg sin 60◦ = 0,and mrg sin 20◦ − Tr = 0 (we only need the parallelcomponents in this problem). If the masses of thestring and pulley are negligible and there is no friction,then Tℓ = Tr. Adding the force equations, we findmrg sin 20◦ − mℓg sin 60◦ = 0, or the mass ratio mustbe mr/mℓ = sin 60◦/ sin 20◦ = 2.53 for no motion.

Problem 17 Solution.

Problem

18. In a setup like that shown in Fig. 6-37, but withdifferent masses, a 4.34-kg block starts from reston the left edge of a frictionless tabletop 1.25 mwide. It accelerates to the right, and reaches theright edge in 2.84 s. If the mass of the blockhanging from the left side is 3.56 kg, what is themass hanging from the right side?

figure 6-37 Problem 18.

Solution

The acceleration of the 4.34-kg block slidinghorizontally to the right across the frictionlesstabletop is a = 2(1.25 m)/(2.84 s)2 = 0.310 m/s2

(from Equation 2-10 and the given conditions). This isalso the magnitude of the accelerations of the othertwo blocks. Since we are not interested in the tensionin either string, we may use a shortcut to find theunknown mass. The net force on all three masses inthe direction of motion is the difference in the right-and left-hand weights, Fnet = (m − 3.56 kg)g, whichequals the total mass times the acceleration,

64 CHAPTER 6

(m + 3.56 kg + 4.34 kg)a. Solving for m, we find

m =(3.56 kg + 4.34 kg)(0.310 m/s2)+(3.56 kg)(9.8 m/s2)

(9.8 m/s2 − 0.310 m/s2)

= 3.93 kg

(The shortcut can be justified by adding thecomponent of the equations of motion of the threeblocks in the direction of motion: Tleft − (3.56 kg)g =(3.56 kg)a, Tright − Tleft = (4.34 kg)a, andmg − Tright = ma. In this setup, the ropes aremassless, and the pulleys are massless and frictionless,so the tension in each rope is constant.)

Problem

19. Suppose the angles shown in Fig. 6-56 are 60◦

and 20◦. If the left-hand mass is 2.1 kg, whatshould be the right-hand mass in order that(a) it accelerates downslope at 0.64 m/s

2;

(b) it accelerates upslope at 0.76 m/s2?

Solution

With reference to the solution to Problem 17, theparallel component of the equations of motionfor the masses are Tℓ − mℓg sin 60◦ = mℓaℓ andmrg sin 20◦ − Tr = mrar. The accelerations andtensions are equal, respectively, provided the stringdoesn’t stretch and the other assumptions inProblem 17 hold. Then mrar + mℓaℓ = (mr + mℓ)a =mrg sin 20◦ − Tr + Tℓ − mℓg sin 60◦ = mrg sin 20◦ −mℓg sin 60◦, or mr = mℓ(g sin 60◦ + a)/(g sin 20◦ − a).(a) A downslope right-hand acceleration is positive forthe coordinate systems we have chosen, so substitutinga = 0.64 m/s

2and mℓ = 2.1 kg, we find mr = 7.07 kg.

(b) If a = −0.76 m/s2, then mr = 3.95 kg.

Problem

20. Two unfortunate climbers, roped together, aresliding freely down an icy mountainside. Theupper climber (mass 75 kg) is on a slope at 12◦ tothe horizontal, but the lower climber (mass 63 kg)has gone over the edge to a steeper slope at 38◦.(a) Assuming frictionless ice and a massless rope,what is the acceleration of the pair? (b) Theupper climber manages to stop the slide with anice ax. Once the climbers have come to a completestop, what force must the ax exert against theice?

Solution

(a) If we assume that the climbers move together as aunit, with the same magnitude of downslopeacceleration a, then the net force acting on them is thesum of the downslope components of gravity on each,

Fnet = (75 kg)g sin 12◦ + (63 kg)g sin 38◦ = 533 N =(75 kg + 63 kg)a. Therefore, a = 533 N/138 kg =

3.86 m/s2. (b) After they have stopped, the force of

the ice ax against the ice must balance the downslopecomponents of gravity calculated in part (a).

Problem

21. In a florist’s display, hanging plants of mass3.85 kg and 9.28 kg are suspended from anessentially massless wire, as shown in Fig. 6-67.Find the tension in each section of the wire.

Solution

Let the tensions in each section of wire be denoted byT1, T2, and T3 as shown in the figure. The horizontaland vertical components of the net force on thejunction of the wire with each plant are equal to zero,since the system is stationary. Thus:

T1 sin 54.0◦ − T2 sin 13.9◦ − (3.85 × 9.8) N = 0

−T1 cos 54.0◦ + T2 cos 13.9◦ = 0

T2 sin 13.9◦ − T3 sin 68.0◦ − (9.28 × 9.8) N = 0

−T2 cos 13.9◦ + T3 cos 68.0◦ = 0

68.0° 54.0°

13.9°

(3.85 kg)g

(9.28 kg)g

T2

T1 T3

figure 6-67 Problem 21 Solution.

One can solve any three of these equations for theunknown tensions, perhaps using the fourth equationas a check (if you do, remember not to round off). Forexample, T1 = (3.85 × 9.8) N/(sin 54.0◦ − cos 54.0◦×tan 13.9◦) = 56.9 N, T2 = T1 cos 54.0◦/ cos 13.9◦ =34.4 N, and T3 = T2 cos 13.9◦/ cos 68.0◦ = 89.2 N.(Note that the given angles and weights are notindependent of one another.)

Problem

22. A rectangular block of mass m1 rests on awedge-shaped block of mass m2, as shown inFig. 6-68. All contact surfaces are frictionless.Find an expression for the magnitude of thehorizontal force F that must be applied to the

CHAPTER 6 65

wedge in order that the rectangular block not slidealong the wedge.

m2

m1F

θ

figure 6-68 Problem 22.

Solution

The forces acting on the wedge and block are shownseparately in free-body diagrams for clarity. Thecontact forces N1 (between wedge and block) and N2

(between wedge and horizontal surface) are purelynormal because the surfaces are frictionless. For norelative motion between the wedge and block, bothmust have the same horizontal acceleration, a. Thehorizontal and vertical components of Newton’s secondlaw for the block, and the horizontal component ofNewton’s second law for the wedge (all that’s neededin this problem) give N1 sin θ = m1a, N1 cos θ −m1g = 0, F − N1 sin θ = m2a. Eliminating a and N1,we find F = (m1 + m2)a = (m1 + m2)g tan θ.

Problem 22 Solution.

Section 6-3: Circular Motion

Problem

23. A simplistic model for the hydrogen atom picturesits single electron in a circular orbit of radius0.0529 nm about the fixed proton. If the electron’sorbital speed is 2.18×106 m/s, what is themagnitude of the force between the electron andthe proton?

Solution

For a particle in uniform circular motion, the net forceequals the mass times the centripetal acceleration,F = mv2/r = (9.11×10−31 kg)(2.18×106 m/s)2÷(5.29×10−11 m) = 8.18×10−8 N.

Problem

24. Suppose the moon were held in its orbit not bygravity but by the tension in a massless cable.Estimate the magnitude of the cable tension. (SeeAppendix E for relevant data.)

Solution

We are asked to estimate the net force on the moon,which, according to Newton’s second law, is theproduct of its mass and its acceleration. Since themoon describes approximately uniform circular motionabout the Earth, F = m(v2/r) = mr(2π/T )2, where rand T are the radius and period of the orbit. Thus,F = (7.35×1022 kg)(3.85×108 m)(2π/27.3×86,400 s)2 = 2.01×1020 N = m(2.73×10−3 m/s

2). We

displayed the numerical value of the moon’s(centripetal) acceleration because, comparing it to

9.8 m/s2

(the gravitational acceleration of an apple atthe Earth’s surface), Newton is said to have arrived athis famous inverse square law of gravitation. Thedistances of the moon and apple from the center of theEarth are about 60RE and RE , and(2.73×10−3/9.8) ≈ (1/60)2. (See also Problem 16,Chapter 9.)

Problem

25. Show that the force needed to keep a mass m in acircular path of radius r with period T is4π2mr/T 2.

Solution

For an object of mass m in uniform circular motion,the net force has magnitude mv2/r (Equation 6-1).The period of the motion (time for one revolution) isT = 2πr/v, so the centripetal force can also be writtenas m(2πr/T )2/r = mr(2π/T )2 = 4π2mr/T 2 (seeEquation 4-18).

Problem

26. A mass m1 undergoes circular motion of radius Ron a horizontal frictionless table, connected by amassless string through a hole in the table to asecond mass m2 (Fig. 6-69). If m2 is stationary,find (a) the tension in the string and (b) theperiod of the circular motion.

Solution

(a) Newton’s second law applied to the stationarymass yields T − m2g = 0, so the tension is T = m2g.(b) This is also the magnitude of the net (horizontal)force on the mass in uniform circular motion, so,with the aid of the result of Problem 25,m2g = m1R(2π/τ)2, hence τ = 2π

√

m1R/m2g is the

66 CHAPTER 6

period. (We wrote a Greek letter, “tau,” for theperiod because we used “tee” for the tension.)

m2

m2g

R m1

m1g

T

T

N

figure 6-69 Problem 26 Solution.

Problem

27. A 940-g rock is whirled in a horizontal circle atthe end of a 1.3-m-long string. (a) If the breakingstrength of the string is 120 N, what is themaximum allowable speed of the rock? (b) At thismaximum speed, what angle does the string makewith the horizontal?

Solution

The situation is the same as described in Example 6-6.The horizontal component of the tension is thecentripetal force, T cos θ = mv2/r = mv2/ℓ cos θ, andthe vertical component balances the weight, T sin θ =mg. (b) At the maximum speed, the tension in thestring is at its breaking strength, Tmax = 120 N;therefore the minimum angle the string makes with thehorizontal is given by sin θmin = mg/Tmax, or θmin =sin−1(0.940×9.8 N/120 N) = 4.40◦. (a) At these valuesof T and θ, the speed is vmax =

√

Tmaxℓ cos2 θmin/m =√

(120 N)(1.3 m)(cos 4.40◦)2/(0.940 kg) = 12.8 m/s.

Problem

28. If the rock of the previous problem is whirled in avertical circle, what is the minimum speed neededat the top of the circle in order that the stringremain taut?

Solution

At the top of the circle, the tension, gravity, and thecentripetal acceleration are all vertically downward (asin Example 6-8). Then T + mg = mv2/ℓ. If the stringremains taut, T ≥ 0, or v ≥

√gℓ =

√

(9.8 m/s2)(1.3 m) = 3.57 m/s.

Problem

29. A subway train rounds an unbanked curve at67 km/h. A passenger hanging onto a strapnotices that an adjacent unused strap makes anangle of 15◦ to the vertical. What is the radius ofthe turn?

Solution

The net force on the unused strap is the vector sum ofthe tension in the strap (acting along its length at 15◦

to the vertical) and its weight. This must equal themass times the horizontal centripetal acceleration.The free-body diagram for the strap is the same asFig. 6-18, except that the angle from the vertical isnow given, as shown. Thus, T cos θ = mg, andT sin θ = mv2/r. Dividing these equations to eliminateT , and solving for the radius of the turn, one finds r =v2/g tan θ = (67 m/3.6 s)2/(9.8 tan 15◦m/s

2) = 132 m.

figure 6-18 Problem 29 Solution.

Problem

30. An Olympic hammer thrower whirls a 7.3-kghammer on the end of a 120-cm chain. If the chainmakes a 10◦ angle with the horizontal, what is thespeed of the hammer?

Solution

From Example 6-6, v =√

gℓ cos2 θ/ sin θ = 8.10 m/s,where we used ℓ = 1.2 m and θ = 10◦.

Problem

31. Riders on the “Great American Revolution”loop-the-loop roller coaster of Example 6-8 wearseatbelts as the roller coaster negotiates its6.3-m-radius loop with a speed of 9.7 m/s. At thetop of the loop, what are the magnitude anddirection of the force exerted on a 60-kg rider(a) by the roller-coaster seat and (b) by the

CHAPTER 6 67

seatbelt? (c) What would happen if the riderunbuckled at this point?

Solution

(a) As shown in Example 6-8, at the top of the loop,

N + mg = mv2/r, so N = (60 kg)[−9.8 m/s2 +(9.7 m/s)2/6.3 m] = 308 N. (b) Actually, 308 N is thedifference between the normal force of the seat and theforce exerted by the seatbelt, i.e., N = 308 N + Fbelt.The seatbelt, firmly adjusted, perhaps adds a fewpounds (1 lb = 4.45 N), providing a feeling of security.(c) The seatbelt is required in case of accidents orrapid tangential decelerations; it is not needed tocontribute to the centripetal force.

Problem

32. A 45-kg skater rounds a 5.0-m-radius turn at6.3 m/s. (a) What are the horizontal and verticalcomponents of the force the ice exerts on her skateblades? (b) At what angle can she lean withoutfalling over?

Solution

(a) If the ice is level, the contact force exerted on theskater has vertical component (normal force) equal tothe weight, and horizontal component (static friction)in the direction of the centripetal acceleration. Thus,N = mg = (45 kg)(9.8 m/s

2) = 441 N, and fs =

mv2/r = (45 kg)(6.3 m/s)2/(5 m) = 357 N. (b) InChapter 14 it will be shown that stability requires thatthe center of gravity of the skater should be along theline of action of the contact force. She should lean atθ = tan−1(fs/N) = 39.0◦, relative to the vertical.

Problem

33. An indoor running track is square-shaped withrounded corners; each corner has a radius of 6.5 mon its inside edge. The track includes six1.0-m-wide lanes. What should be the bankingangles on (a) the innermost and (b) the outermostlanes if the design speed of the track is 24 km/h?

Solution

The banking angle is θ = tan−1(v2/gr) (see Example6-7). A competitive runner rounds a turn on the insideedge of his or her lane. (a) θ = tan−1[(24 m/3.6 s)2 ÷(9.8 m/s2)(6.5 m)] = 34.9◦. (b) The radius of theinside edge of the outermost lane is 6.5 m + 5(1 m) =11.5 m, so θ = 21.5◦. (This type of banking is shownin Fig. 6-58.)

Problem 32 Solution.

Problem

34. A jetliner flying horizontally at 850 km/h banks at32◦ to make a turn. What is the radius of theturn?

Solution

The forces on an airplane making a horizontal circularturn are analogous to those on a car negotiating abanked curve (see Fig. 6-21), where the mainaerodynamic force on the airplane is normal to thewing surfaces. Thus, Faero sin θ = mv2/r, andFaero cos θ = mg, so tan θ = v2/gr. In this case,

r = (850 m/3.6 s)2/(9.8 m/s2) tan 32◦ = 9.10 km.

Problem 34 Solution.

Problem

35. You’re a passenger in a car rounding a turn withradius 180 m. You take your keys from yourpocket and dangle them from the end of your

68 CHAPTER 6

keychain. They make an 18◦ angle with thevertical, as shown in Fig. 6-70. What is the car’sspeed?

figure 6-70 Problem 35.

Solution

You and your keys are rounding the curve along withthe car, so all have the same centripetal acceleration,ac = v2/r, which we assume is horizontal and directedtoward the center of the curve. The forces on the keysare the tension along the direction of the keychain andgravity downward, as shown on Fig. 6-70, so thesituation is just like Example 6-6, where the horizontalcomponent of the tension supplies the centripetalacceleration, and the vertical component balances theweight of the keys. Then the horizontal and verticalcomponents of Newton’s second law for the keys are:T cos(90◦ − 18◦) = T sin 18◦ = mac = mv2/r, andT cos 18◦ − mg = 0. Eliminating the tension, asin the sixth step of Example 6-6, we find mg tan 18◦ =mv2/r, or v =

√gr tan 18◦ =

√

(9.8 m/s2)(180 m) tan 18◦ = 23.9 m/s = 86.2 km/h.

(This problem could also be approached using anaccelerated coordinate system at rest relative to thecar, and introducing a fictitious force, −mac, calledthe centrifugal force, to account for the acceleratedmotion. The beginning student is advised to stick withinertial coordinate systems, however, in which there isless chance for confusion.)

Problem

36. A bucket of water is whirled in a vertical circle ofradius 85 cm. What is the minimum speed thatwill keep the water from falling out?

Solution

As long as the magnitude of the acceleration at thetop of the circle is greater than g, the water willremain in the bucket. (Then, the bottom of the bucketexerts a normal force on the water so the two are incontact; see Example 6-8). Thus, v2/r > g, or

v >

√

(9.8 m/s2)(0.85 m) = 2.89 m/s.

Problem

37. A 1200-kg car drives on the country road shown inFig. 6-71. The radius of curvature of the crestsand dips is 31 m. What is the maximum speed atwhich the car can maintain road contact at thecrests?

N

Fg

fs

figure 6-71 Problem 37.

Solution

If air resistance is ignored, the forces on the car aregravity and the contact force of the road, which isrepresented by the sum of the normal force(perpendicularly away from the road) and frictionbetween the tires and the road (parallel to the road inthe direction of motion). Newton’s second law for thecar is Fg +N+fs =ma. At a crest, N is verticallyupward, fs is horizontal, and the vertical component ofa is the radial acceleration −v2/r (downward in thiscase). The vertical component of Newton’s second lawis then −mg + N = −mv2/r. As long as the car is incontact with the road, N ≥ 0; thus, v ≤ √

gr =√

(9.8 m/s2)(31 m) = 17.4 m/s = 62.7 km/h.

Problem

38. The Tethered Satellite System (TSS) is a NASAexperiment consisting of a 500-kg satelliteconnected to the space shuttle by a 20-km-longcable of negligible mass. Suppose the shuttle is ina 250-km-high circular orbit, where theacceleration of gravity is 0.926 times its value atEarth’s surface. The TSS hangs vertically on its

CHAPTER 6 69

tether (Fig. 6-72), and at its 230-km altitude theacceleration of gravity is 0.932 times its surfacevalue. What is the tension in the cable?

20 km

mgTSS

T

figure 6-72 Problem 38 Solution.

Solution

The radial component of Newton’s second law for theTSS (positive component toward the center of theEarth) is mgTSS − T = m(ar)TSS. Since the TSS andthe shuttle have the same period, τ (the tether wouldpull the TSS forward or backward until this was so)(2π/τ)2 = (v/r)2 = ar/r is the same for both, or(ar/r)TSS = (ar/r)shuttle. If the shuttle’s orbit ishardly affected by the much smaller TSS, then(ar)Shuttle = gShuttle. Therefore, the tension in thecable is

T = m[gTSS − (ar)TSS] = m

[

gTSS − rTSS

rshuttle

(ar)shuttle

]

= (500 kg)(9.8 m/s2)

(

0.932 − 0.926× 6370 + 230

6370 + 250

)

= 43.1 N

Section 6-4: Friction

Problem

39. Movers slide a file cabinet along a floor. The massof the cabinet is 73 kg, and the coefficient ofkinetic friction between cabinet and floor is 0.81.What is the frictional force on the cabinet?

Solution

If the floor is level, the normal force on the cabinet isequal in magnitude to its weight, so the frictional forcehas magnitude fk = µkN = µkmg = (0.81)(73 kg)×(9.8 m/s

2) = 579 N. The direction of sliding friction

opposes the motion.

Problem

40. You make a huge snowball with a mass of 33 kg. Ifthe coefficient of friction between the ball and anice-covered pond is 0.16, with what force must youpush the ball to move it (a) at constant velocity

and (b) with an acceleration of 0.84 m/s2 ?

Solution

The pond has a level frozen surface, so the normalforce on the snowball is equal in magnitude to itsweight. The frictional force opposing the motion hasmagnitude fk = µkN = µkmg. The horizontal forcesacting are friction and the applied force, so thehorizontal component of Newton’s second law (positivein the direction of motion) is Fapp − fk = ma. (a) Atconstant velocity, a = 0, and Fapp = fk = (0.16)×(33 kg)(9.8 m/s

2) = 51.7 N. (b) Fapp = ma + fk =

(33 kg)(0.84 m/s2) + 51.7 N = 79.5 N.

Problem

41. Eight 80-kg rugby players climb on a 70-kg “scrummachine,” and their teammates proceed to pushthem with constant velocity across a field. If thecoefficient of kinetic friction between scrummachine and field is 0.78, with what force mustthey push?

Solution

As in Problem 40(a), a horizontal applied force musthave magnitude equal to the frictional force in orderto push an object at constant velocity along a levelsurface. The total weight on the scrum machine is(8×80 kg + 70 kg)(9.8 m/s

2) = 6.96 kN; therefore

Fapp = fk = µkN = (0.78)(6.96 kN) = 5.43 kN.

Problem

42. A hockey puck is given an initial speed of 14 m/s.If it comes to rest in 56 m, what is the coefficientof kinetic friction?

Solution

The force of friction is the only horizontal force acting,and the normal force is vertical and equal to thepuck’s weight. Thus, fk = ma = −µkmg, or a = −µkg.We take the positive direction parallel to the initialvelocity, so that Equation 2-11 can be used for theacceleration, a = (0 − v2

0)/2(x − x0) = −(14 m/s)2 ÷2(56 m) = −1.75 m/s

2. Then µk = −a/g =

−(−1.75 m/s2)/(9.8 m/s2) = 0.18. (We express µk totwo significant figures only since it is an empiricalconstant.)

70 CHAPTER 6

Problem

43. A child sleds down a 12◦ slope at constant speed.What is the coefficient of friction between slopeand sled?

Solution

The frictional force must balance the downslopecomponent of gravity on the sled to produce aconstant speed. The normal force on the sled mustbalance the perpendicular component of gravity if justgravity and the contact force are acting. Thus,mg sin θ = fk = µkN , and mg cos θ = N, or µk =tan12◦ = 0.21 (as in Example 6-10).

Problem

44. The handle of a 22-kg lawnmower makes a35◦ angle with the horizontal. If the coefficient offriction between lawnmower and ground is 0.68,what magnitude of force is required to push themower at constant velocity? Assume the force isapplied in the direction of the handle. Comparewith the mower’s weight.

Solution

Assuming the ground is also horizontal, we may depictthe forces on the lawnmower as shown. At constantvelocity (constant speed in a straight line) a= 0, andF+fk +N+mg= 0. The x and y components of thisequation are F cos 35◦ − fk = 0, and N − F sin 35◦ −mg = 0. Using fk = µkN = µk(F sin 35◦ + mg), withµk = 0.68, we find F = µkmg/(cos 35◦ − µk sin 35◦) =1.58mg = 342 N.

Problem 44 Solution.

Problem

45. Repeat Example 6-5, now assuming that thecoefficient of kinetic friction between rock and iceis 0.057.

Solution

If there is friction between the rock and the ice, wemust modify the rock’s equation of motion, Tr +

Fgr +N+fk = mar. Since the ice surface ishorizontal, only the rock’s x-equation changes, Trx −µkN = mrarx. Now we need to use the rock’sy-equation to eliminate N, obtaining Trx − µmrg =mrarx. Solving for arx as before, we find µkmrg −mracy − mcg = mcacy, or arx = −acy =(mc − µkmr)g/(mc + mr) = (70 kg − 0.057×940 kg)×(9.8 m/s2)/(1010 kg) = 0.159 m/s2. Now the climber

has more time, t =

√

2(51 m)/(0.159 m/s2) = 25.3 s,to pray for rescue.

Problem

46. During an ice storm, the coefficients of frictionbetween car tires and road are reduced toµk = 0.088 and µs = 0.14. (a) What is themaximum slope on which a car can be parkedwithout sliding? (b) On a slope just steeper thanthis maximum, with what acceleration will a carslide down the slope?

Solution

(a) With reference to Example 6-14, θ = tan−1

µs = 7.97◦. (b) From Example 6-10, a = g sin θ −µkg cos θ = (9.8 m/s2)(sin 7.97◦ − 0.088 cos7.97◦) =

50.5 cm/s2. (Using a little trigonometry, we could have

written a = g(µs− µk)/√

1 + µ2s.)

Problem

47. A bat crashes into the vertical front of anaccelerating subway train. If the coefficient offriction between bat and train is 0.86, what is theminimum acceleration of the train that will allowthe bat to remain in place?

Solution

Since N is parallel to the acceleration, butperpendicular to gravity and friction, N = ma, andfs = mg ≤ µsN = µsma. Therefore, in order toremain in place, a ≥ g/µs = (9.8 m/s

2)/0.86 =

11.4 m/s2.

Problem

48. In a factory, boxes drop vertically onto a conveyorbelt moving horizontally at 1.7 m/s. If thecoefficient of kinetic friction is 0.46, how long doesit take each box to come to rest with respect tothe belt?

CHAPTER 6 71

Problem 47 Solution.

Solution

Kinetic friction accelerates each box up to the speed ofthe belt: fk = µkN = µkmg = ma (if we suppose thebelt to be horizontal). This takes time t = v/a =

v/µkg = (1.7 m/s)/(0.46)(9.8 m/s2) = 0.377 s.

Problem

49. The coefficient of static friction between steel trainwheels and steel rails is 0.58. The engineer of atrain moving at 140 km/h spots a stalled car onthe tracks 150 m ahead. If he applies the brakes sothat the wheels do not slip, will the train stop intime?

Solution

When stopping on a level track, the maximumacceleration due to friction is a = −µsg, as explainedin Example 6-12. The minimum stopping distancefrom an initial speed of (140/3.6) m/s is ∆x =

v20/(−2a) = (38.9 m/s)2/(2×0.58×9.8 m/s

2) =

133 m. With split-second timing, an accident could beaverted.

Problem

50. If you neglect to fasten your seatbelt, and if thecoefficient of friction between you and your carseat is 0.42, what is the maximum deceleration forwhich you can remain in your seat? Compare withthe deceleration in an accident that brings a60-km/h car to rest in a distance of 1.6 m.

Solution

If the seat is horizontal (parallel to the acceleration)then fs = ma ≤ µsN = µsmg, or a ≤ µsg. Themaximum deceleration is therefore (0.42)(9.8 m/s

2) =

4.12 m/s2. For the accident described, the magnitude

of the deceleration is v20/2(x − x0) = (60 m/3.6 s)2÷

2(1.6 m) = 86.8 m/s2, or about twenty-one times

greater. (Friction is no substitute for a seatbelt!)

Problem

51. A bug crawls outward from the center of acompact disc spinning at 200 revolutions perminute. The coefficient of static friction betweenthe bug’s sticky feet and the disc surface is 1.2.How far does the bug get from the center beforeslipping?

Solution

Assume that the disc is level. Then the frictional forceproduces the (centripetal) acceleration of the bug,and the normal force equals its weight. Thus,fs = m(v2/r) = mr(2π/T )2 ≤ µsN = µsmg, or

r ≤ µsg(T/2π)2 = (1.2)(9.8 m/s2)(60 s/2π(200))2 =

2.68 cm. Note that the period of revolution is 60 sdivided by the number of revolutions per minute.

Problem

52. A 310-g paperback book rests on a 1.2-kgtextbook. A force is applied to the textbook, andthe two books accelerate together from rest to96 cm/s in 0.42 s. The textbook is then brought toa stop in 0.33 s, during which time the paperbackslides off. Within what range does the coefficientof static friction between the two books lie?

Solution

The direction of motion and the orientation of thesurface of contact are not specified; assumeboth are horizontal. Then, for the acceleration,fs = ma ≤ µsN = µsmg (where m is the mass of thepaperback), or µs ≥ a/g. From the given data,

µs ≥ (0.96 m/s/0.42 s)/(9.8 m/s2) = 0.23. During thedeceleration, fmax

s = µsmg < ma′ (the magnitude ofthe paperback’s acceleration is smaller than that ofthe textbook, because the paperback slides off), so

µs < (0.96 m/s/0.33 s)/(9.8 m/s2) = 0.30.

Problem

53. A 2.5-kg block and a 3.1-kg block slide down a30◦ incline as shown in Fig. 6-73. The coefficientof kinetic friction between the 2.5-kg block andthe slope is 0.23; between the 3.1-kg block and theslope it is 0.51. Determine the (a) acceleration ofthe pair and (b) the force the lighter block exertson the heavier one.

72 CHAPTER 6

2.5 kg3.1kg

30°

figure 6-73 Problem 53.

Problem 53 Solution.

Solution

The forces on the blocks are as shown. (Sinceµk2 > µk1, there will be a contact force of magnitudeFc, such that the acceleration of both blocks down theincline is a.) The x and y components of Newton’ssecond law for each block are

m1g sin 30◦ − µk1N1 − Fc

= m1a, Nk1 − m1g cos 30◦ = 0

m2g sin 30◦ − µk2N2 + Fc

= m2a, N2 − m2g cos 30◦ = 0.

(a) To solve for a, add the x equations and use valuesof N from the y equations: a = [(m1 + m2)g sin 30◦ −(m1µk1 + m2µk2)g cos 30◦]/(m1 + m2), or

a = 1.63 m/s2, when the given m’s and µk’s are

substituted. (b) To solve for Fc, divide each x

equation by the corresponding m, and subtract:Fc = (µk2 − µk1)m1m2g cos 30◦/(m1 + m2) = 3.29 N.

Problem

54. Children sled down a 41-m-long hill inclined at25◦. At the bottom the slope levels out. If thecoefficient of friction is 0.12, how far do thechildren slide on the level?

Solution

The acceleration down the incline is a = g(sin θ −µk cos θ) = (9.8 m/s2)(sin 25◦ − 0.12 cos25◦) =

3.08 m/s2

(see Example 6-10). The speed at thebottom is v2 = 2 as, where s = 41 m. On level ground,the deceleration is −fk/m = −µkmg/m = −µkg, sothe distance traveled before stopping is x =−v2/2(−µkg) = as/µkg = (3.08 m/s

2)(41 m)/(0.12)×

(9.8 m/s2) = 107 m.

Problem

55. In a typical front-wheel-drive car, 70% of the car’sweight rides on the front wheels. If the coefficientof friction between tires and road is 0.61, what isthe maximum acceleration of the car?

Solution

On a level road, the maximum acceleration from staticfriction between the tires and the road isamax = µsN/m (see Example 6-12). In this case, thenormal force on the front tires (the ones producing thefrictional force which accelerates the car) is 70% ofmg, whereas the whole mass must be accelerated.Thus, amax = (0.61)(0.70)(9.8 m/s

2) = 4.18 m/s

2.

Problem

56. Repeat the previous problem for a rear-wheel-drive car with the same portion of its weight overthe front wheels.

Solution

The frictional force on the rear wheels accelerates thecar, the frictional force on the front wheels, assumednegligible, just makes them turn. If the car is on levelground, Nr = 0.3mg, so Fs = ma ≤ µs(0.3mg), or

a ≤ (0.61)(0.3)(9.8 m/s2) = 1.79 m/s

2. (Putting more

weight over the drive axle gives better traction.)

Problem 56 Solution.

CHAPTER 6 73

Problem

57. A police officer investigating an accident estimatesfrom the damage done that a moving car hit astationary car at 25 km/h. If the moving car leftskid marks 47 m long, and if the coefficient ofkinetic friction is 0.71, what was the initial speedof the moving car?

Solution

On a level road, the acceleration of a skidding car is−µkg (see Example 6-12). From the kinematicsof the reconstructed accident, v2 = v2

0 −2µkg(x − x0), from which we calculate that

v0 =

√

(25 m/3.6 s)2 + 2(0.71)(9.8 m/s2)(47 m) =

26.5 m/s = 95.4 km/h = 59.3 mi/h. (Add speeding tothe traffic citation!)

Problem

58. A slide inclined at 35◦ takes bathers into aswimming pool. With water sprayed onto the slideto make it essentially frictionless, a bather spendsonly one-third as much time on the slide as whenit is dry. What is the coefficient of friction on thedry slide?

Solution

The time it takes to slide down, starting from rest atthe top, depends on the acceleration in the direction ofthe slope, t =

√

2∆x/a, where ∆x is the length of theslide (see Equation 2-10). Without friction, a = g sin θ,and with friction, a′ = g sin θ − µkg cos θ =a(1 − µk cot θ) (see Example 6-10). The ratio of the

corresponding times is 13, so

(

13

)2= (t/t′)2 =

a′/a = 1 − µk cot θ, or µk = 8 tan 35◦/9 = 0.62.

Problem

59. You try to push a heavy trunk, exerting a force atan angle of 50◦ below the horizontal (Fig. 6-74).Show that, no matter how hard you try to push, itis impossible to budge the trunk if the coefficientof static friction exceeds 0.84.

Solution

The trunk remains at rest if the sum of the forces on it(x and y components) is 0: Fa cos 50◦ − fs = 0, N −mg − Fa sin 50◦ = 0. Since fs = Fa cos 50◦ ≤ µsN =µs(mg + Fa sin 50◦), the condition for equilibrium canbe written Fa(cos 50◦ − µs sin 50◦) ≤ µsmg, or(cos 50◦/µs) − sin 50◦ ≤ (mg/Fa). The right-handside is always positive (Fa and mg are magnitudes),but the left-hand side can be positive or negative. If itis negative, the trunk does not move, independent of

50°

figure 6-74 Problem 59.

Fa. Thus, the equilibrium condition will always besatisfied if sin 50◦ > cos 50◦/µs, or µs > cot 50◦=0.84.

Problem 59 Solution.

Problem

60. A block of mass m is being pulled at constantspeed v down a slope that makes an angle θ withthe horizontal. The pulling force is appliedthrough a horizontal rope, as shown in Fig. 6-75.If the coefficient of kinetic friction is µk, find anexpression for the rope tension.

Solution

Since the speed is constant (a= 0), the sum of theforces on the block is zero (Newton’s second law).Taking components parallel and perpendicular to theincline, we have: T cos θ + mg sin θ − fk = 0, andN + T sin θ − mg cos θ = 0. Then T cos θ + mg sin θ =fk = µkN = µk(mg cos θ − T sin θ), andT = mg(µk cos θ − sin θ)/(cos θ + µk sin θ). (Note thatT ≥ 0, so the application of step 7 of the strategy boxin Section 6-1 is restricted by the range ofθ, 0 ≤ θ ≤ tan−1 µk.)

74 CHAPTER 6

θθ

N

Tfk

mg1

figure 6-75 Problem 60 Solution.

Problem

61. A block is shoved down a 22◦ slope with an initialspeed of 1.4 m/s. If it slides 34 cm beforestopping, what is the coefficient of friction?

Solution

The acceleration down the slope can be found fromkinematics (Equation 2-11) and from Newton’s secondlaw (as in Example 6-10): a = g sin θ − µkg cos θ =−v2

0/2ℓ. In this equation, the values of θ, v0, and ℓ aregiven; therefore we can solve for µk:

µk = tan 22◦ +(1.4 m/s)2

2(0.34 m)(9.8 m/s2) cos 22◦

= 0.72.

Problem

62. If the block in the previous problem were shovedup the slope with the same initial speed, (a) howfar would it go? (b) Once it stopped, would itslide back down?

Solution

(a) Because the frictional force always opposes therelative motion of the surfaces in contact,Example 6-10 does not give the acceleration of a blockshoved up an incline. Rather, fk is in the oppositedirection to that shown in Fig. 6-33, and Newton’ssecond law gives a = g sin θ + µkg cos θ. With valuesfrom Problem 61, a = 10.2 m/s

2(positive down slope).

The distance traveled up the slope to where the blockstops is (from Equation 2-11) −v2

0/2a = −(1.4 m/s)2 ÷2(10.2 m/s2) − 9.58 cm (negative up slope). (We didnot use a rounded-off value of µk in this calculation.)(b) Once the block has stopped, it will remain at restif µs ≥ tan 22◦ = 0.404 (see Example 6-14). Since µs isnormally greater than µk (which was 0.72 inProblem 61), the block does not slide back. (Noticehow the direction of the frictional forces depends onthe circumstances: it was up the incline inProblem 61, down the incline in Problem 62(a), andup the incline in Problem 62(b).)

Section 6-5: Drag Forces

Problem

63. Find the drag force on a 7.4-cm-diameter baseballmoving through air (density 1.2 kg/m3) at 45 m/s.The drag coefficient is 0.50.

Solution

The cross-sectional area of the baseball (a sphere) is14πd2, so Equation 6-4 and the other given quantities

result in a drag force of magnitude FD = 12CρAv2 =

12(0.50)(1.2 kg/m

3)14π(7.4 cm)2(45 m/s)2 = 2.61 N.

Problem

64. A football’s terminal speed is 53 m/s when it’sfalling pointed end first and 43 m/s when it’sfalling with its long dimension horizontal. If thefootball is 28 cm long at its longest, what is thediameter on its short dimension? Assume the dragcoefficient C is the same in both orientations.

Problem 64 Solution.

Solution

The terminal speed of the football is inverselyproportional to the square root of its cross-sectionalarea perpendicular to the direction of fall (seeExample 6-15). If we assume that the drag coefficientis the same for both orientations, then(vt,long/vt,end)

2 = (43/53)2 = Aend/Along. Aend (thefootball’s cross-sectional area in a plane perpendicularto, and through, the midpoint of the long axis) is thearea of a circle of diameter d. Along (the football’scross-sectional area in a plane containing the longaxis) is the area of two identical circular segments, ofchord equal to the long dimension, ℓ = 28 cm, andheight (or saggita) equal to 1

2d, as shown. The areas

can be found from geometry, in terms of ℓ and d:Aend = 1

4πd2, and Along = R2θ − ℓ(R − 1

2d), where

R = (ℓ2 + d2)/4d, andθ = 2 sin−1(ℓ/2R). The

CHAPTER 6 75

transcendental equation for the ratio of these areascan be solved numerically on a PC. Our result forℓ = 28 cm is d ≃ 16.71 cm.

Problem

65. Find the terminal speed of a 1.0-mm-diameterspherical raindrop in air. The densities of air andwater are 1.2 kg/m

3and 1000 kg/m

3, respectively,

and the drag coefficient is 0.50.

Solution

The expression for the terminal speed found inExample 6-15 can be used, vt =

√

2 mg/CρA, wherethe mass of the drop is its volume, V, times thedensity of water, ρw.The volume and cross-sectionalarea of a spherical drop, in terms of the diameter, are16πd3 and 1

4πd2, respectively, so their ratio is 2

3d. Then

the terminal speed becomes vt =√

4 gd (ρw/ρ)/3C =√

4(9.8 m/s2)(10−3m)(1000/1.2)/3(0.50) = 4.67 m/s,

where we canceled identical units in the density ratio.

Problem

66. Will a golf ball of mass 45 g and diameter 4.3 cmreach terminal speed when dropped from a heightof 25 m? The drag coefficient is 0.35, and thedensity of air is 1.2 kg/m

3.

Solution

The terminal speed for this golf ball, inair, is vt =

√

2 mg/CρA =√

2(45g)(9.8 m/s2)/(0.35)(1.2 kg/m

3)π(4.3 cm/2)2 =

38.0 m/s (see Example 6-15). Even in the absence ofair resistance, a golf ball dropped from a height of25 m would attain a speed of only v =

√

2 g (25 m) =22.1 m/s, far short of the terminal speed. (The actualspeed of an object, dropping through a distance y,subject to the drag force of Equation 6-4, isv2 = v2

t (1 − e−2gy/v2

t ), as shown in many intermediatemechanics texts. This gives a speed of 20.4 m/s for thegolf ball dropped in this problem.)

Paired Problems

Problem

67. In Fig. 6-76, suppose m1 = 5.0 kg andm2 = 2.0 kg, and that the surface and pulley arefrictionless. Determine the magnitude anddirection of m2’s acceleration.

Solution

Since we are not interested in the tension in the ropeconnecting the masses, this is a good opportunity to

30°

m1m2

figure 6-76 Problems 67, 68.

take advantage of the type of shortcut mentioned inthe solution to Problem 18. (For a similar solutionusing equations of motion for each object, seeProblem 17 or 19.) Being tied together by a rope(which is assumed to be unextensible), m1 and m2

move as a unit, with the same acceleration (inmagnitude) which we’ll choose to be positive for m2

upward and m1 downslope. Gravity acts positivedownslope on m1(m1g sin 30◦) and negative downwardon m2(−m2g), so the net force on the system ofmasses, in the positive direction of motion, equal tothe total mass times the acceleration, ism1g sin 30◦ − m2g = (m1 + m2)a. (Here, we neglectthe mass of the rope and pulley, which are alsoaccelerated, and any frictional forces.) Thus,a = (5 kg sin 30◦ − 2 kg)×(9.8 m/s2)/(5 kg + 2 kg) = 0.700 m/s2.

Problem

68. Repeat the preceding problem, now takingm1 = 3.0 kg with m2 still 2.0 kg.

Solution

The analysis in the solution to the previous problemgives a = (3 kg sin 30◦ − 2 kg)(9.8 m/s

2)÷

(3 kg + 2 kg) = −0.98 m/s2. The minus sign meansthat m2 goes downward and m1 upslope, as specifiedpreviously.

Problem

69. A tetherball on a 1.7-m rope is struck so it goesinto circular motion in a horizontal plane, with therope making a 15◦ angle to the horizontal. Whatis the ball’s speed?

Solution

The tetherball whirling in a horizontal circle isanalogous to the mass on a string in Example 6-6.From step 6, v =

√

gℓ cos2 θ/ sin θ =√

(9.8 m/s2)(1.7 m) cos2 15◦/ sin 15◦ = 7.75 m/s.

76 CHAPTER 6

Problem

70. An airplane goes into a turn 3.6 km in radius. Ifthe banking angle required is 28◦ from thehorizontal, what is the plane’s speed?

Solution

The airplane making a turn at the proper bankingangle is analogous to the situation in Example 6-7.

Thus, v =√

gr tan θ =

√

(9.8 m/s2)(3.6 km) tan 28◦ =

137 m/s = 493 km/h. (Note that the angles given inthis and the previous problem are complementary, andthat the radius of the circle in Problem 69 is ℓ cos θ.)

Problem

71. Starting from rest, a skier slides 100 m down a 28◦

slope. How much longer does the run take if thecoefficient of kinetic friction is 0.17 instead of 0?

Solution

If air resistance is ignored, the forces acting on theskier are analogous to those on the sled inExample 6-10, so the downslope acceleration isa‖ = g(sin θ − µk cos θ). Starting from rest, thetime needed to coast a distance ∆x downslopeis t =

√

2∆x/a‖. With no friction, t =√

2(100 m)/(9.8 m/s2) sin 28◦ =

6.59 s. If µk = 0.17, t′ =√

2(100 m)/(9.8 m/s2)(sin 28◦ − 0.17 cos 28◦) = 7.99 s,

or about 1.40 s longer.

Problem

72. At the end of a factory production line, boxesstart from rest and slide down a 30◦ ramp 5.4 mlong. If the slide is to take no more than 3.3 s,what is the maximum frictional coefficient thatcan be tolerated?

Solution

As in the preceding problem, the time required to slidedown the incline is t =

√

2∆x/a‖ ≤ 3.3 s. Therefore,a‖ = g(sin θ − µk cos θ) ≥ 2∆x/(3.3 s)2, or µk ≤tan 30◦ − 2(5.4 m)/(3.3 s)2(9.8 m/s

2) cos 30◦ = 0.46.

Problem

73. A car moving at 40 km/h negotiates a130-m-radius banked turn designed for 60 km/h.(a) What coefficient of friction is needed to keepthe car on the road? (b) To which side of thecurve would it move if it hit an essentiallyfrictionless icy patch?

Solution

The forces on a car (in a plane perpendicular to thevelocity) rounding a banked curve at arbitrary speedare analyzed in detail in the solution to Problem 81below. (a) It is shown there that to prevent skidding,µs ≥

∣

∣v2 − v2d

∣

∣ /gR(1 + v2v2d/g2R2), where R is the

radius of the curve, and vd is the design speed for theproper banking angle, tan θd = v2

d/gR. In this problem,vd = (60/3.6) m/s, v = (40/3.6) m/s, and R = 130 m,so µs ≥ 0.12. (b) Since v < vd, the car would slidedown the bank of the curve in the absence of friction.

Problem

74. A passenger sets a coffee cup on the seatback trayof an airplane flying at 580 km/h. The plane goesinto a 2.6-km-radius turn, getting part of itsturning force from its rudder and part frombanking at 25◦ (i.e., it’s banking at a lower anglethan required to give the full turning force).(a) What coefficient of friction is needed to keepthe coffee cup on the tray? (b) If there wereinsufficient friction, which way would the cupslide?

Solution

This problem involves a coffee cup on a bankedtraytable moving with speed v in a horizontal circulararc of radius R. The forces and centripetal accelerationare analogous to those on a car rounding a bankedcurve, as in the previous problem. Here, the circularspeed v = 580 km/h = 161 m/s is greater than thedesign speed for a 25◦ banked turn,

vd =√

gR tan 25◦ =

√

(9.8 m/s2)(2.6 km) tan 25◦ =

109 m/s, so a coefficient of static frictionµs ≥

∣

∣v2 − v2d

∣

∣ /gR (1 + v2v2d/g2R2) = 0.37 is needed

to keep the cup from sliding up the tray.

Supplementary Problems

Problem

75. A space station is in the shape of a hollow ring,450 m in diameter (Fig. 6-77). At how manyrevolutions per minute should it rotate in order tosimulate Earth’s gravity—that is, so that the

CHAPTER 6 77

normal force on an astronaut at the outer edgewould be the astronaut’s weight on Earth?

450 m

figure 6-77 Problem 75.

Solution

Standing on the outer edge of the space station,rotating with it, the astronaut experiences a normalforce equal to the centripetal force, N = mac =m 4π2r/T 2, where T is the period of rotation (seeExample 4-8). Since T is the time per revolution, thenumber of revolutions per unit time is 1/T (called thefrequency of revolution). If the normal force is toduplicate Earth’s gravity, ac = g, and 1/T = (1/2π) ×√

g/r = (1/2π)

√

(9.8 m/s2)/(450 m/2) = (3.32 ×

10−2 rev/s)(60 s/ min) = 1.99 rpm.

Problem

76. Figure 6-78 shows a 0.84-kg ball attached to avertical post by strings of length 1.2 m and 1.6 m.If the ball is set whirling in a horizontal circle, find(a) the minimum speed necessary for the lowerstring to be taut and (b) the tension in each stringif the ball’s speed is 5.0 m/s.

Solution

Consider the three forces acting on the ball, gravityand the tensions pulling along each string, as shownsketched on Fig. 6-78. The ball’s acceleration is thecentripetal acceleration, v2/r, directed along the lowerstring toward the axis of rotation, so the horizontaland vertical components of Newton’s second law areTℓ + Tu cos θ = mv2/r and Tu sin θ = mg. The anglebetween the tensions is given by the lengths of thestrings, or θ = cos−1(1.2/1.6) = 41.4◦. (a) The lowerstring is taut provided Tℓ ≥ 0. Eliminating Tu from theabove equations, we find Tℓ = mv2/r − mg cot θ ≥ 0,so this condition implies v ≥

√gr cot θ =

√

(9.8 m/s2)(1.2 m) cot 41.4◦ = 3.65 m/s. (b) The

vertical component of the tension in the upper string

must balance the weight of the ball, independent of v,hence Tu = mg/ sin θ = 12.4 N. When v = 5 m/s, Tℓ =(0.84 kg)(5 m/s)2(1.2 m)−1 − (12.4 N)(1.2/1.6) =8.17 N.

1.6 m

1.2 m Tℓ

Tu

mg

θ

figure 6-78 Problem 76 Solution.

Problem

77. In the loop-the-loop track of Fig. 6-25, show thatthe car leaves the track at an angle φ given bycosφ = v2/rg, where φ is the angle made by avertical line through the center of the circulartrack and a line from the center to the point wherethe car leaves the track.

Problem 77 Solution.

Solution

The angle φ and the forces acting on the car areshown in the sketch. The radial component of the netforce (towards the center of the track) equals the masstimes the centripetal acceleration, N + mg cosφ =

78 CHAPTER 6

mv2/r. (The tangential component is not of interest inthis problem.) The car leaves the track whenN = (mv2/r) − mg cosφ = 0 (no more contact) orcosφ = v2/gr. This implies that the car leaves thetrack at real angles for v2 < gr; otherwise, the carnever leaves the track, as in Example 6-8.

Problem

78. An astronaut is training in a centrifuge thatconsists of a small chamber whirled aroundhorizontally at the end of a 5.1-m-long shaft. Theastronaut places a notebook on the vertical wall ofthe chamber and it stays in place. If the coefficientof static friction is 0.62, what is the minimum rateat which the centrifuge must be revolving?

Solution

The wall and gravity act on the notebook. If the latterdoesn’t fall, fs = mg ≤ µsN = µsmv2/r, orv2 ≥ gr/µs. In circular motion, the linear speed isrelated to the rate of revolution (the angular speed,denoted by Greek letter “omega”) by v = 2πr/τ = ωr,where τ is the period. Thus, v2 = ω2r2 ≥ gr/µs, or

ω ≥√

g/µsr =

√

(9.8 m/s2)/(0.62)(5.1 m) =

1.76 s−1 = 16.8 rev/ min. (Note: ω = 2π/τ has unitsrad/s, and 2π rad = 1 rev.)

Problem 78 Solution.

Problem

79. You stand on a spring scale at the north pole andagain at the equator. (a) Which scale reading willbe lower, and why? (b) By what percentage willthe lower reading differ from the higher one?(Here you’re neglecting variations in g due togeological factors.)

Solution

When standing on the Earth’s surface, you arerotating with the Earth about its axis through thepoles, with a period of 1d. The radius of your circle ofrotation (your perpendicular distance to the axis) isr = RE cos θ, where RE is the radius of the Earth(constant if geographical variations are neglected) andθ is your lattitude. Your centripetal acceleration hasmagnitude ac = (2π/T )

2r and is directed toward the

axis of rotation (see Example 4-8). We assume thereare only two forces acting on you, gravity, Fg

(magnitude mg approximately constant, directedtowards the center of the Earth), and the force exertedby the scale, Fs. Newton’s second law requires thatFg +Fs =mac. (a) At the north pole, ac = 0, so themagnitudes of Fg and Fs are equal, or Fs = mg; butat the equator, ac has a maximum magnitude, equalto the difference in the magnitudes of Fg and Fs, orFs = mg − m(2π/T )2RE . Therefore Fs (your“weight”) is lower at the equator than at the pole.(b) The fractional difference of these two values is(Fs,pole − Fs,eq.)/Fs,pole = (2π/T )2RE/g =

(2π/86, 400 s)2(6.37×106 m)/(9.81 m/s2) = 0.34%.

Fs

Fs

Fs

Fg

RE

Fg

Fg

at pole

θat equator

r

Problem 79 Solution.

Problem

80. Driving in thick fog on a horizontal road, a driverspots a tractor-trailer truck jackknifed across theroad, as in Fig. 6-79. To avert a collision, thedriver could brake to a stop or swerve in a circulararc, as suggested in Fig. 6-79. Which offers the

CHAPTER 6 79

figure 6-79 Problem 80.

greater margin of safety? Assume that the samecoefficient of static friction is operative in bothcases, and that the car maintains constant speed ifit swerves.

Solution

On a level road, the maximum force of friction isµsN = µsmg, so the maximum acceleration is µsg (inmagnitude). Applied in a straight line (ax ≤ µsg), thestopping distance from an initial speed v is0 = v2 − 2axx, or x ≥ v2/2µsg. On a swerving circularpath at constant speed (ar ≤ µsg), the turning radiusis ar = v2/r, or r ≥ v2/µsg. Evidently, rmin = 2xmin,so braking is safer than swerving.

Problem

81. A highway turn of radius R is banked for a designspeed vd. If a car enters the turn at speed v =vd + ∆v, where ∆v can be positive or negative,show that the minimum coefficient of staticfriction needed to prevent slipping is

µs =|∆v|gR

(2vd + ∆v)

[1 + (vdv/gR)2].

Solution

The equation of motion for a car rounding a bankedturn is N+mg+fs =mar, where ar = v2/R is theradical acceleration (assumed horizontal and constantin magnitude) and the forces are as shown. Note thatthe frictional force changes direction for v greater or

Problem 81 Solution.

less than the design speed. Taking components paralleland perpendicular to the road, we find N − mg cos θ =m(v2/R) sin θ, mg sin θ ± fs = (v2/R) cos θ, where theupper sign is for v > vd, and the lower for v < vd. (Wechose these components because the solution for Nand fs is direct.) This argument applies if the car doesnot skid (otherwise a 6= ar) so fs ≤ µsN. Therefore:

µs ≥ fs

N=

∓g sin θ ± v2 cos θ/R

g cos θ + v2 sin θ/R=

∓g tan θ ± v2/R

g + v2 tan θ/R

=∓v2

d ± v2

gR(1 + v2v2d/g2R2)

since v2d/gR = tan θ. If we set ∆v = v − vd, the

condition on µs becomes:

µs ≥ ± ∆v(2vd + ∆v)

gR(1 + v2dv2/g2R2)

=|∆v| (2vd + ∆v)

gR(1 + v2dv2/g2R2)

.

(Note that |∆v| is +∆v for v > vd and −∆v forv < vd.) The expression for the minimum coefficient offriction is not particularly simple, but for v = 0 (car atrest) it reduces to |−vd| (2vd − vd)/gR = v2

d/gR =tan θ, as in Example 6-14.

Problem

82. Suppose the coefficient of friction between a blockand a horizontal surface is proportional to theblock’s speed: µ = µ1v/v1, where µ1 and v1 areconstants. If the block is given an initial speed v0,show that it comes to rest in a distance x =v0v1/µ1g.

Solution

Take the x-axis in the direction of the initial velocity,with origin at the initial position (x(0) = 0,vx(0) = v0). The equation of motion (x component) ism(dvx/dt) = −µN = −(µ1/v1)vxmg, since N = mg ona horizontal surface. The solution of this differentialequation for vx, which satisfies the initial conditions,is: vx(t) = (dx/dt) = v0 exp[−(µ1g/v1)t]). The solutionfor x is: x(t) = (v0v1/µ1g)(1 − exp[−(µ1g/v1)t]). Theblock comes to rest in the limit t → ∞, sincevx(∞) = 0. Thus x(∞) = v0v1/µ1g.

80 CHAPTER 6

Problem

83. A block is projected up an incline making an angleθ with the horizontal. It returns to its initialposition with half its initial speed. Show that thecoefficient of kinetic friction is µk = 3

5tan θ.

Problem 83 Solution.

Solution

Going up the incline, the block’s acceleration (positivedown the incline) is g(sin θ + µk cos θ) = aup,whereas going down, the acceleration isg(sin θ − µk cos θ) = adown. If the block slides up adistance ℓ, its initial speed upward was

√

2aupℓ,whereas, sliding down the same distance, it returns tothe bottom with speed

√2adownℓ. Given that the

latter speed is half the former, 4adown = aup =4g(sin θ − µk cos θ) = g(sin θ + µk cos θ). Therefore3 sin θ = 5µk cos θ, or µk = 3

5tan θ.

Problem

84. The victim of a political kidnapping is forced intoa north-facing car and then blindfolded. The carpulls into traffic and, from the sound of thesurrounding traffic, the victim knows that the caris moving at about the legal speed limit of85 km/h. The car then turns to the right; thevictim estimates that the force the seat exerts onhim is one-fifth of his weight. The victimexperiences this force for 28 s. At the end of thattime, in what direction can the victim concludethat he is heading?

Solution

In a circular turn at constant speed on level ground,the force felt by the victim (friction exerted by theseat) is mv2/r = 1

5mg, as given. Therefore, the radius

of the turn is r = 5v2/g. The arclength of the turn iss = rθ = vt, so the angle (in radians) is θ = vt/r =vt/(5v2/g) = gt/5v. The velocity turns through thesame angle (measured clockwise from north), so thevictim’s final heading is

θ =(9.8 m/s

2)(28 s)

5(85 m/3.6 s)= 2.32 rad = 133◦.

(This can be expressed as 43◦ S of E, or approximatelysoutheast.)

Problem 84 Solution.

Problem

85. A 2.1-kg mass is connected to a spring of springconstant k = 150 N/m and unstretched length18 cm. The pair are mounted on a frictionless airtable, with the free end of the spring attached to africtionless pivot. The mass is set into circularmotion at 1.4 m/s. Find the radius of its path.

Solution

Since the airtable is frictionless, the only horizontalforce acting on the mass is the spring force, ofmagnitude k(ℓ − ℓ0) and in the direction of thecentripetal acceleration v2/ℓ. Here, the radius of thecircle is ℓ, the length of the spring, while ℓ0 is theunstretched length. Therefore, k(ℓ − ℓ0) = mv2/ℓ.This is a quadratic equation forℓ, ℓ2 − ℓ0ℓ − mv2/k = 0, with positive solutionℓ = 1

2[ℓ0 +

√

ℓ20 + 4 mv2/k] = 1

2[0.18 m +

√

(0.18 m)2 + 4(2.1 kg)(1.4 m/s)2/(150 N/m)] =27.9 cm.

Problem

86. This problem is for those with unusually advancedmath skills. Set up Newton’s law for an objectfalling from rest subject to the drag force ofEquation 6-4. Use a coordinate system with they-axis vertically downward, and write theacceleration as ay = dvy/dt. Your Newton’s lawthen becomes a differential equation. Integrate itto show that the speed as a function of time is

given by v(t) =√

2gmCAρ tanh

(

√

CAgρ2m t

)

, where

tanh is the hyperbolic tangent functiontanh(x) = (ex − e−x)/(ex + e−x).

Solution

For an object falling downward (in the y-direction)under the influence of gravity and the quadratic dragforce of Equation 6-4, Newton’s second law is

CHAPTER 6 81

m dv/dt = mg − 12CρAv2. (The drag force is really

− 12CρAv |v| , but v is always positive for a falling

object in the coordinate system chosen.) Thisequation can be separated and simplified (using vt

from Example 6-15) to yield (g/vt)dt = vtdv/(v2t − v2),

which can be integrated using partial fractions orintegral tables. The result, expressed in terms oflogarithms or inverse hyperbolic functions, depends onwhether vt > v or vt < v, which is determined by theinitial speed, v0. Here, we consider an object droppedfrom rest, so v0 = 0, and we obtain

∫ t

0

gdt

vt=

gt

vt=

1

2

∫ v

0

(

1

vt + v+

1

vt − v

)

dv

=1

2log

(

vt + v

vt − v

)

= tanh−1 v

vt.

Solving for v as a function of t is quicker withhyperbolic functions, but the result in terms ofexponentials is the same:

v

vt= tanh

(

gt

vt

)

=e2gt/vt − 1

e2gt/vt + 1.