Physics Problems

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 LawProblem1. Two forces, both in the x-y plane, act on a 1.5-kgmass, which accelerates at 7.3m/i in a direction30° counter-clockwise from the x-axis. Oneforce has magnitude 6.8 N and points in the+x direction. Find the other force.

SolutionNewton's second law for this mass says F net = F 1+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)(icos30° + jsin300)-(6.8 N)i = (2.68i +5.48j) N. This has magnitude6.10 N and direction 63.9° CCW from the x-axis.

Problem2. Two forces act on a 3.1-kg mass, which undergoesacceleration a=0.9li - 0.27j m/i. If one of theforces is F1 = -1.2i - 2.5j N, what is the otherforce?

SolutionAs in the previous problem2 F2 = ma - F 1 =(3.1 kg)(0.91i - 0.27j)(m/s ) - (-1.2i - 2.5j) N =(4.02i+1.66j)N. .

Problem3. 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 acceleratesforward at 0.16 m/s2?

SolutionThe 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

2Tcos 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)x(0.16 m/s2) = 1.40 kN. .

FIGURE 6-59 Problem 3 Solution.

Problem4. At what angle should you tilt an air table to

simulate motion on the moon's surface, where9 = 1.6 m/s2? .

SolutionThe acceleration down an incline is all = 9 sin (} (seeExample 6-1). To replicate the moon's surface gravity,the angle of tilt should be (}= sin-1 (1.62/9.81) =9.51° (see Appendix E).

Problem5. A block of mass m slides with acceleration a down

a frictionless slope that makes an angle (}to thehorizontal; the only forces acting on it are the forceof gravity F 9 and the normal force N of the slope.Show that the ma~nitude of the normal force isgi\ren by N =mvg2 - a2•

SolutionChoose the x-axis down the slope (parallel to theacceleration) and the y-axis parallel to the normal.Then ax ~. a, ay = 0, Nx =O,Ny = N, Fgx =Fg cos (90° - (})= mg sin (}, and Fgy = -mg cos (}.Newton's second law, N+ F 9 =ma, in componentsgives mg sin (}= ma, and N - mg cos (}= O.Eliminate (} (using sin 2

(} + cos2 (} = 1) to find (a/ 9)2 +(N/mg)2 = 1, or N = mJg2 - a2.

87

Problem 8 Solution.

CHAPTER 6

~t--k=--- )(.~-~~-rM'1 9501J ....


w

Problem10. A tow truck is connected to a 1400-kg car by a\cable 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.

SolutionThe only force on the car with a horizontal component(in the direction of the acceleration) is the tension.Therefore, T cos6 = rna, or T = (1400 kg)x(0.57 m/s2)/cos25° = 880 N. (The force of static

Problem9. 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.

SolutionThe 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=O.The x component of this equationrequires that the tension is the same on both sides:T1 cos8° + T2 cos 172°= 0, or T1 = T2. The ycomponent gives 2Tsin8° = W, or T = W/2 sin8° =3.59W = 3.59(15 kg)(9.8 m/s2) = 528 N.

'tT2 I T(SO I SO---.x

Problem 5 Solution.

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) cos15°/70 kg = 13.1 m/s2.(b) Vx = axt = (13.1 m/s2)(0.29 s) = 3.80 m/s.

SolutionThe acceleration up the ramp is -g sin 35°, so theblock goes a distance in this direction calculated fromthe equation v5x -2gsin35°(x - xo) =O.Thus,x - xo = (2.2 m/s)2/2(9.8 m/s2) sin 35° = 43.1 cm.

Problem8. 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?

Problem6. 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?

SolutionThe acceleration down a frictionless incline is a =gsin6 (see Example 6-1), and the distance traveleddown the incline, starting from rest (vo= 0 at the top(xo = 0), is x = !at2. Therefore, t = 2x/gsin6 =J2(1.3 km)/(9.8 m/s2) sin 24° = 25.5 s.

Problem7. A block is launched up a frictionless ramp thatmakes an angle of 35° to the horizontal. If theblock's initial speed is 2.2 mis, how far up theramp does it slide?

88 CHAPTER 6

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


Problem11. A 100kg mass is suspended at rest by two strings

attached to walls, as shown in Fig. 6-62. Find thetension forces in the two strings.

SolutionThe force diagram is superimposed on Fig. 6-62. Sincethe mass is at rest, the sum of the forces is zero,T} +T2 + W=O, which is true for the x andy components separately, T} cos 45° - T2 = 0, andT} sin 45° - W = O. Solving for the magnitudes of thetensions, and substituting 98 N for the weight, we findT} = ../2 98 N = 139 N, and T2 = T}/V2 = 98 N.


Problem12. A 1I00-kg car goes off the road and plunges down

a 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.

SolutionIf 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=O. Then Fh =mg tan 0 = (1100 kg)(9.8 m/s2) tan 23° = 4.58 kN.

FIGURE 6-63 Problem 12.

Problem13. 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?


SolutionThe sum of the forces acting on the pack (gravity andthe tension along each rope) is zero, since it is at rest,F 9 + T} + T2 = O. In a coordinate system with x-axishorizontal to the right and y-axis vertical upward, thex and y components of the net force are - T} cos 71° +T2cos28° = 0, and -26x9.8 N +T}sin71° +T2 sin 28° = 0 (see Example 6-3). Solving for T1 andT2, one finds T} = (26x9.8 N)(sin 71° +tan 28° cos 71°) = 228N and T2 = (26x9.8 N)+(sin 28°+ tan 71° cos 28°) = 84.0 N.

Problem.14. A construction worker is lifting a 92-kg bundle of

plywood onto an upper floor, using thearrangement shown in Fig. 6-65. What force must

'"''0'.;:

the worker apply to lift the bundle at constantspeed? Assume the pulley is massless andfrictionless.


SolutionThe 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, iszeroj therefore the vertical component of Newton'ssecond law gives TI +T2 - m~ = 2T - mg = 0, orT = !mg = H92 kg)(9.8 m/s ) = 451 N.

Section 6-2: Multiple ObjectsProblem15. Your 12-kg baby sister is hanging on the bottom

of 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?

SolutionThe 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 = meg/(me +mr) =(12 kg)(9.8 m/s2)/(12 kg + 6.8 kg) = 6.26 m/s2 is theturkey's horizontal acceleration, and (b) t = J2x / a =

V2(60 cm)/(6.26 m/s2) = 0.438 s is the time you have

CHAPTER 6 89

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

Problem16. Find expressions for the acceleration of the blocks

in Fig. 6-66, where the string is fastened securelyto the ceiling. Neglect friction and assume thatthe masses of pulley and string are negligible.

SolutionThe equations of motion (components parallel to theaccelerations) for the two masses (ml includes theattached pulley) are T = m2a2 and mIg - 2T =mIal' (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 mlmoves down a distance d, m2 moves to the right adistance 2d, so a2 = 2al' Thus T= m2(2al), mIal =mIg - 2(2m2al), and al = mIg/(ml + 4m2)'

N


Problem17. If the left-hand slope in Fig. 6-56 makes a

60° 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?

\

SolutionThe 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 isfrictionless, and we indicate separate parallel andperpendicular x-y-axes. If the masses don't slide, thenet force on each must be zero, or Tl - mig sin 60° = 0,and mrgsin20° - Tr = 0 (we only need the parallelcomponents in this problem). If the masses of thestring and pulley are negligible and there is no friction,

m = (3.56 kg + 4.34 kg)(O.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: l1ert - (3.56 kg)g =(3.56 kg)a, Tright -l1ert = (4.34 kg)a, andmg - Tright =mao In this setup, the ropes aremassless, and the pulleys are massless and frictionless,so the tension in each rope is constant.)

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

Problem19. 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/s2;

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

SolutionWith reference to the solution to Problem 17, theparallel component of the equations of motionfor the masses are Te - meg sin 60° =meae 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 +meae = (m~ +me)a =mrg sin 20° - Tr +Te - meg sin 60° = mrg sin 20" -meg sin 60°, or mr = me(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/s2 and me = 2.1 kg, we find mr = 7.07 kg.(b) If a = -0.76 m/s2

, then mr = 3.95 kg.

Problem20. Two unfortunate climbers, roped together, are

sliding freely down an icy mountainside. Theupper climber (mass 75 kg) is on a slope at 12° to.the horizontal, but the lower climber (mass 63 kg)has gone over the edge to a steeper slope at 38°.(a) Assumingfrictionless ice and a massless rope,what is the acceleration of the pair? (b) Theupper climber manages to stop the slide with anlee ax. Once the climbers have come to a completestop, what force must the ax exert against theice?

Solution(a) If we assume that the 6limbers 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,

Problem 17 Solution.

Problem18. In a setup like that shown in Fig. 6-37, but with

different 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?


. ":' "'.

.t.~enTi == Tr. Adding the force equations, we findmrg sin20° - meg sin 60° = 0, or the mass ratio mustbe mr/me == sin 60°/ sin 20° = 2.53 for no motion.

SolutionThe 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,

F

Section 6-3: Circular MotionProblem23. A simplistic model for the hydrogen atom pictures

its single electron in a circular orbit of radius\ 0.0529 nm about the fixed proton. If the electron'sorbital speed is 2.18x106 mis, what is themagnitude of the force between the electron andthe proton?



CHAPTER 6 91

SolutionFor a particle in uniform circular motion, the net forceequals the mass times the centripetal acceleration,F = mv2/r = (9.11xlO-31 kg)(2.18x106 m/s)2+(5.29x10-11 m) = 8.18x10-B N.

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

SolutionThe 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 sinO = mIa, N1 cosO -mIg = 0, F - N1 sin 0 = m2a. Eliminating a and N1,

we find F = (ml + m2)a = (ml + m2)g tan O.

Problem22. A rectangular block of mass ml rests on a

wedge-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

Fnet = (75 kg)g sin 12° + (63 kg)g sin 38° =.533 N =(75 kg + 63 kg)a. Therefore, a = 533N/138 kg =3.86m/s2• (b) After they have stopped, the force ofthe ice ax against the ice must balance the downslopecomponents of gravity calculated in part (a).

Problem21. In a florist's display, hanging plants of mass

3.85 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.

(9.28kg)g


One can solve any thr~ 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 x 9.8) N/(sin54.0° -cos54.00xtan 13.9°) = 56.9 N, T2 = T1cos54.0°/ cos 13.9° =34.4N, and Tg= T2 cos13.9° / cos68.0° = 89.2 N.(Note that the given angles and weights are notindependent of one another.)

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

T1 sin54.0° - T2 sin 13.9° - (3.85 x 9.8) N = 0- TI cos54.0° + T2 cos 13.9° = 0

T2 sin 13.9° - T3 sin68.0° - (9.28 x 9.8) N = 0-T2cos13.9° +Tgcos68.0° = 0

92 CHAPTER 6

Problem24. Suppose the moon were held in its orbit not by

gravity but by the tension in a massless cable.Estimate the magnitude of the cable tension. (SeeAppendix E for relevant data.)

SolutionWe are asked to estimate the net force on the moon,which, according toNewton's second law, is theproduct of its mass and its acceleration. Sincethemoon describes approximately uniform circular motionabout the Earth, F = m(v2/r) = mr(21r/T)2, where rand T are the radius and period of the orbit. Thus,F = (7.35x1022kg)(3.85x 108 m)(21r /27.3x86,400s)2= 2.01x102o N = m(2.73xlO-3 m/s2). Wedisplayed the numerical value of the moon's(centripetal) acceleration b~ause, comparing it to9.8 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(2.73x 10-3 /9.8) ~ (1/60)2. (See also Problem 16,Chapter 9.)

Problem25. Show that the force needed to keep a mass m in a

circular path of radius r with period T is4tr2mr1T2.

SolutionFor an object of mass m in uniform circular motion,the net force has magnitude mv2/r (Equation 6-1).The period ofthe motion (time for one revolution) isT = 21rrlv, so the centripetal force can also be writtenas m(21rr /T)2 /r = mr(21r /T)2 = 41r2mr /T2 (seeEquation 4-18).

Problem26. A mass ml uridergoes circular motion of radius R

on 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) thepericidof 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(21r/r)2, hence r = 21ry-m-l--R-/-m-2-gis the

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


Problem27. A 94o-g rock is whirled in a horizontal circle at

the end of a 1.3-m-Iongstring. (a) If the breakingstrength of the string is 120N, what is themaximum allowable speed of the ro.ck? (b) At thismaximum speed, what angle does the string makewith the horizontal?

SolutionThe situation is the same as described in Example 6-6.The horizontal component of the tension is thecentripetal force, Tcose = mv2/r = mv2/lcos(), andthe vertical component balances the weight, T sin ()=mg. (b) At the maximum speed, the tension in thestring is at its breaking-strength, Tmax = 120 Njtherefore the minimum angle the string makes with thehorizontal is given by sin()min=mg/Tmax, or Omin=sin-1(0.940x9.8 N/120 N) = 4.40°. a At these valuesofT andO, the speed is Vrnax = Tmaxlcos2emin/m =J(120 N)(1.3 m)(cos4.400)2/(0.940 kg) = 12.8 m/s.

Problem28. If the rock of the previous problem is whirled in a

vertical circle, what is the minimum speed neededat the top of the circle in order that the stringremain taut?

Solu~ionAt 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 ~ ../9l =-/(9.8 m/s2)(1.3 m) = 3.57 m/s.

Problem29. A subway train rounds an unbanked curve at

67 km/h. A passenger hanging onto a strap


Solution(a) If the ice is level, the contact force exerted on theskater has vertic8.J.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/s2) = 441N, and Is =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 at0= tan-1(fs/N) = 39.0°, relative to the vertical.

Problem32. A 45-kg skater rounds a 5.0-m-radius turn at

6.3 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?

Prpblem33. An indoor running track is square-shaped with

rounded corners; each corner has a radius of 6.5 mon its inside edge. The track includes six1.o-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?


SolutionThe banking angle is 0 = tan-1(v2 /gr) (see Example6-7). A competitive runner rounds a turn on the inside

CHAPTER 6 93

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 (lIb = 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.

notices that an adjacent unused strap makes anangle of 15° to the vertical. What is the radius ofthe turn?

sfrap

Problem31. Riders on the "Great American Revolution"

loop-the-Ioop 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 6o-kg rider(a) by the roller-coaster seat and (b) by theseatbelt? (c) What would happen if the riderunbuckled at this point?

Problem30. An Olympic hammer thrower whirls a 7.3-kg

hammer on the end of a 12o-cmchain. If the chainmakes a 10° angle with the horizontal, what is thespeed of the hammer?

SolutionFrom Example 6-6, v = Jglcos20/sinO = 8.10 mis,where we used l = 1.2 m and 8 = 10°.

olutionhe net force on the unused strap is the vector sum of

, he 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 cosO= mg, andTsin8 = mv2/r. Dividing these equations to eliminate

, T and solving for the radius of the turn, one finds r =, 2v2/gtan8 = (67 m/3.6 s)2/(9.8tan 15°m/s ) = 132 m.

94 CHAPTER 6

edge of his or her lane. (a) ()= tan-1 [(24mj3.6 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) =11.5m, so (J = 21.5°. (This type of banking is shownin Fig. 6-58.)

Problem34. A jetliner flying horizontally at 850 km/h banks at

32° to make a turn. What is the radius of theturn?

SolutionThe forces on an airplane making a horizontal circularturn are analogous to those on a car negotiating abanked curve (see Fig. 6-21), where the main.aerodynamic force on the airplane is normal to thewing surfaces. Thus, Faerosin(J = mv2/r, andFaero cos (J = mg, so tan (J = v2j gr. In this case,r = (850 m/3.6 s)2/(9.8 m/s2) tan 32° = 9.10 km.


Problem35. You're a passenger in a car rounding a turn with

radius 180 tn. Youtake your keys from yourpocket and dangle them from the end of yourkeychain. They make an 18° angle with thevertical, as shown in Fig. 6-70. What is the car'sspeed?

SolutionYou 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:Tcos(90° -18°) = Tsin 18° = mac = mv2/r, andT cos18° - mg = O.Eliminating the tension, as

(-Ibwllf'd ce.f1ffr.•." of curve. )


in the sixth step of Example 6-6, we find mg tan 18° =mv2/r, or v = y'gr tan 18° =

V(9.8 m/s2)(180 m) tan 180 = 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.)

Problem36. A bucket of water is whirled in a vertical circle of

radius 85 em. What is the minimum speed thatwill keep the water from falling out?

SolutionAs 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 IT > g, orv> V(9.8 mjs2)(0.85 m) = 2.89 m/s.

Problem37. A 1200-kg car drives on the country road shown in

Fig, 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?



Problem40. You make a huge snowball with a mass of 33 kg. If

the 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 velocityand (b) with an acceleration of 0.84 m/l?

CHAPTER 6 95

SolutionIf the floor is level, the normal force on the cabinet isequal in magnitude to its weight, so the frictional forcehas magnitude!k = JLkN = JLkmg = (0.81)(73 kg)x(9.8 m/s2) = 579 N. The direction of sliding frictionopposes the motion.

Section 6-4: FrictionProblem39. Movers slide a.file cabinet along a floor, The mass

of the cabinet is 73 kg, and the coefficientofkinetic friction between cabinet and floor is 0.81.What is the frictional force on the cabinet?

Solution1Ihe 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 = JLkN = JLkmg. 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 = mao (a) Atconstant velocity, a = 0, and Fapp = fk = (0.16)x(33 kg)(9.8 m/s2) = 51.7 N. (b) Fapp = ma + fk =(33 kg)(0.84 m/s2) + 51.7 N = 79.5 N.

SolutionThe radial component of Newton's second law for theTSS (positive component toward the center of theEarth) is m!JTss - T =m(arhss. Since the TSS andthe shuttle have the same period, r (the tether wouldpull the TSS forward or backward until this was so)(21r/r)2 = (v/r)2 = ar/r is the same for both, or(ar/rhss = (ar/r)shuttle. If the shuttle's orbit ishardly affected by the much smaller TSS, then(ar )Shuttle = gShuttie' Therefore, the tension in thecable is

T = m[!JTSS - (arhssl = m [9TSS - rTSS (ar)shuttle]rshuttle

= (500 kg)(9.8 m/s2) (0..932 _ 0.926 x 6370+ 230)6370+ 250

= 43.1 N

--'---T

.-~20km

SolutionIf 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 F9 +N+f~=mao At 8: crest, N is vertically .upward, 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 ~ OJthus, v ~ .;gr =J(9.8 m/s2)(31 m) = 17.4 m/s = 62.7 km/h.

Problem38. The Tethered Satellite System (TSS) is a NASA

experiment consisting of a 500-kg satelliteconnected to the space shuttle by a 20-km-Iongcable of negligiblemass. 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 itstether (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?

96 CHAPTER 6

Problem41. Eight SO-kgrugby players climb on a 70-kg "scrum

machine," and their teammates proceed to pushthem with constant velocity across a field. If thecoefficientof kinetic friction between scrummachine and field is 0.78, with what force mustthey push?

SolutionAs 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(8xSO kg+ 70 kg)(9.8 m/s2) = 6.96 kNi thereforeFapp = Ik = JLkN = (0.7S)(6.96 kN) = 5.43 kN.

Problem42. 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?

SolutionThe force of friction is the only horizontal force acting,and the normal force is vertical and equal to thepuck's weight. Thus, !k = ma = -JLkmg, or a = -JLkg.We take the positive direction parallel to the initialvelocity, so that Equation 2-11 can be used for theacceleration, a = (0 - v6)/2(x - xo) = -(14 m/s)2 7-

2(56 m) = -1.75 m/i. Then JLk= -a/g =-(-1.75 m/s2)/(9.8 m/s2) = 0.18. (We express JLk totwo significant figures only since it is an empiricalconstant.)

Problem43. A child sleds down a 12° slope at constant speed.

What is the coefficientof friction between slopeand sled?

SolutionThe 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,mgsinlJ =!k = JLkN, andmgcoslJ = N, orJLk =tan 12° = 0.21 (as in Example 6-10).

Problem44. The handle of a 22-kg lawnmower makes a

35° angle with the horizontal. If the coefficient offriction between lawnmower and ground is 0.68,what magnitude of force is required to push the

mower at constant velocity? Assume the force isapplied in the direction of the handle. Comparewith the mower's weight.

SolutionAssuming 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 =O.The x andy components of thisequation are F cos35° - !k= 0, and N - F sin 35° -mg = O.Using Ik = JLkN = JLk(Fsin35° +mg), withiLk = 0.68, we find F = Jjkmg/(cos35° - JLksin 35°) =1.58mg = 342 N.

)(

- - --+~SOF


Problem45. Repeat Example 6-5, now assuming that the

coefficient of kinetic friction between rock and iceis 0.057.

SolutionIf 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 :zrequation changes, Trx -JLkN = rnrarx' Now we need to use the rock'sy-equation to eliminate N, obtaining Trx - JLmrg =mrarz. Solving for arx as before, we find JLkmrg -mraey - meg = meacy, or arz = -aey =(me - JLkmr)g/(me + mr) = (70 kg - 0.057x940 kg)x(9.8 m/s2)/(1010 kg) = 0.159 m/s2. Now the climberhas more time, t = V2(51 m)/(0.159 m/s2) = 25.3 s,to pray for rescue.

Problem46. During an ice storm, the coefficients of friction

between car tires and road are reduced toJLk= 0.088 and JLs= 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, 0 = tan-1J1.s= 7.97°. (b) From Example 6-10, a = gsinO -J1.kgcosO = (9.8 m/s2) (sin7.9'r' - 0.088cos7.97°) =50.5cm/s2. (Using a little trigonometry, we could havewritten a = g(J1.s- I-Lk)/VI + J1.~.)

Problem47. A bat crashes into the vertical front of an

accelerating 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?

SolutionSince N is parallel to the acceleration, butperpendicular to gravity and friction, N = ma, andfs =mg ~ l-LaN = f..tsma. Therefore, in order toremain in place, a ~ g/f..ts = (9.8 m/s2)/0.86 =11.4m/s2.


Problem48. In a factory, boxes drop vertically onto a conveyor

belt moving horizontally at 1.7 m/s. If thecoefficientof kinetic friction is 0.46, how long doesit take each box to come to rest with respect tothe belt?

SolutionKinetic friction accelerates each box up to the speed ofthe belt: !k = f..tkN = f..tkmg = rna (if we suppose thebelt to be horizontal). This takes time t = via =V/f..tkg = (1.7 m/s)/(0.46)(9.8 m/s2) = 0.377 s.

Problem49. The coefficientof static friction between steel train

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

CHAPTER 6 97

SolutionWhen stopping on a level track, the maximumacceleration due to friction is a = -f..tsg, as explainedin Example 6-12. The minimum stopping distancefrom an initial speed of (140/3.6) m/s is .1x =vV( -2a) = (38.9 m/s)2/(2xO.58x9.8 m/s2) =133m. With split-second timing, an accident could beaverted.

Problem50. If you neglect to fasten your seatbelt, and if the

coefficient 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 a6O-km/h car to rest in a distance of 1.6 m.

SolutionIf the seat is horizontal (parallel to the acceleration)then fs = ma ~ I-LsN = J.Lsmg, or a 5"J.Lsg. Themaximum deceleration is therefore (0.42)(9.8 m/s2) =4.12 m/s2. For the accident described, the magnitudeof the deceleration is vU2(x - xo) = (60 m/3.6 s)2-;-

. 22(1.6 m) = 86.8 m/s , or about twenty-one timesgreater. (Friction is no substitute for a seatbelt!)

Problem51. A bug crawls outward from the center of a

compact 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?

SolutionAssume 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(27r/T)2 5, J.LsN = J.Lsmg, orr ~ f..tsg(T /27r)2= (1.2)(9.8 m/s2)(60 s/27r(200))2 =2.68 em. Note that the period of revolution is 60 sdivided by the number of revolutions per minute.

\

Problem52. A 31D-gpaperback book rests on a 1.2-kg

textbook. 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?

98 CHAPTER 6

SolutionThe direction of motion and the orientation of thesurface of contact are not specified; assumeboth are horizontal. Then, for the acceleration,Is = ma ~ JLsN = JLsmg (where m is the mass of thepaperback), or JLs~ a/g. From the given data,JLs~ (0.96 m/s/0.42 s)/(9.8 m/s2) = 0.23. During thedeceleration, I::'ax = psmg < ma' (the magnitude ofthe paperback's acceleration is smaller than that ofthe textbook, because the paper~ack slides off), soPs < (0.96 m/s/0.33 s)/(9.8 m/s ) = 0.30.

Problem53. A 2.5-kg block and a 3.1-kg block slide down a

30° incline as shown in Fig. 6-73. The coefficien~~of 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.


SolutionThe forces on the blocks are as shown. (SinceJLk2> JLkl, there will be a contact force of magnitudeF. such that the acceleration of both blocks down thee, .

~f l'iFe /

a/ N2,/...... '"I ....../

/3a fk2/M.' I/3t!


. incline is a.) The x and y components of Newton'ssecond law for each block are

mIg sin 30° - PklNI - Fe= mIa, Nkl - mIg cos 30° = 0

m2g sin 30° - Pk2N2 + Fe= m2a, N2 - m2g cos 30° = O.

(a) To solve for a, add the x equations and use valuesof N from the y equations: a = [(ml +m2)g sin 30° -(mlJLkl +m2Pk2)g cos 300]/(ml +m2), ora = 1.63 m/s2, when the given m's and Pk'S aresubstituted. (b) To solve for Fe, divide each xequation by the corresponding m, and subtract:Fe = (JLk2- Pkl)mlm2g cos 30° /(ml +m2) = 3.29 N.

Problem54. Children sled down a 41-m-Iong hill inclined at

25°. At the bottom the slope levels out. If thecoefficient of friction is 0.12, how far do thechildren slide on the level?

SolutionThe acceleration down the incline is a = g(sinO -PkCOSO) = (9.8 m/s2) (sin 25° - 0.12cos25°) =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 - Ik/m = -Pkmg/m = -Pk9, sothe distance traveled before stoppin~ is x =-v2/2(-Pkg) = as/Pkg = (3.08 m/s )(41 m)/(0.12)x(9.8 m/s2) = 107 m.

Problem55. In a typical front-wheel-drive car, 70% of the car's

.weight rides on the front wheels. If the coefficientof friction between tires and road is 0.61, what isthe maximum acceleration of the car?

SolutionOn a level road, the maximum acceleration from staticfriction between the tires and the road isamax = PsN fm (see Example 6-12). In this case, thenorm'al force on the front tires (the ones producing thefrictional force which accelerates the car) is 70% ofmg whereas the whole mass must be accelerated.

, 2 . 2Thus, amax = (0.61)(0.70)(9.8 m/s ) = 4.18 m/s .

Problein56. Repeat the previous problem for a rear-wheel-

drive car with the same portion of its weight overthe front wheels.

..

'I

Problem60. A block of mass m is being pulled at constant

speed v down a slope that makes an angle 8 withthe horizontal. The pulling force is appliedthrough a horizontal rope, as shown in Fig. 6-75.If the coefficient of kinetic friction is J.Lk,find anexpression for the rope tension.


CHAPTER 6 99


SolutionThe trunk remains at rest if the sum of the forces on it(x and y components) is 0: Fa cos500 - fs = 0, N -mg - Fa sin 500 = O. Since fs = Fa cos500 5 ItsN =J.Ls(mg + Fa sin 500), the condition for equilibrium canbe written Fa (cos500 - J.Lssin 500) :5 J.Lsmg, or(cos500 /lts) - sin 500 :5 (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 ofFa. Thus, the equilibrium condition will always besatisfied if sin 500 > cos500/ Its, or Its > cot 500=0.84.

Problem59. You try to push a heavy trunk, exerting a force at

an angle of 500 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.

corresponding times is l, so U)2 = (t/t')2 =a' /a = 1 - Itk cot 8, or Itk = 8 tan 350/9 = 0.62.


SolutionThe time it takes to slide down, starting from rest. atthe top, depends on the acceleration in the direction ofthe slope, t = .j2!::i..x/a, where D.x is the length of theslide (see Equation 2-10). Without friction, a = gsinO,and with friction, a' = 9 sin 0 - J.Lkgcos0 =a(l - J-Lk cot 0) (see Example 6-10). The ratio of the

Problem58. A slide inclined at 350 takes bathers into a

swimming 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?

fi ~~OO/3rng O.7W1fj-

Problem57. A police officerinvestigating an accident estimates

.from 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 ofkineti~ friction is 0.71, what was the initial speedof the moving car?

SolutionOn a level road, the acceleration of a skidding car is-J.Lkg (see Example 6-12). From the kinematicsof the reconstructed accident, v2 = v5 -2ltk9(X - xo), from which we calculate that

Vo = )(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 mifh. (Add speeding tothe traffic citation!)

SolutionThe frictional force on the rear wheels accelerates thecar the frictional force on the front wheels, assumedne~ligible,just makes them turn. If the car is on levelground Nr = 0.3mg, so Fs = ma 5 Its(O.3mg), or.

, 2 2a ~ (0.61)(0.3)(9.8 m/s ) = 1.79m/s . (Putting moreweight over the drive axle gives better traction.)

100 CHAPTER 6

SolutionSince the speed is constant (a=O), the sum of theforces on the block is zero ,(Newton's second law).Taking components parallel and perpendicular to theincline, we have: T cos9+mg sin0 - fk = 0, andN +TsinO - mgcosO = O.Then TcosO +mgsinO=!k = f..tkN = f..tk(mgcosO - TsinO), andT = mg(f..tk cosO- sin O)/(cos 0 + f..tksinO). (Note thatT ~ 0, so the application of step 7 of the strategy boxin Section 6-1 is restricted by the range of0, 0 ::;0 ::;tan-1 f..tk')


Problem61. A block is shoved down a 22° slope with an initial

speed of 1.4 m/s. If it slides 34 em beforestopping, what is the coefficient of friction?

SolutionThe acceleration down the slope can be found fromkinematics (Equation 2-11) and from Newton's secondlaw (as in Example 6-10): a = gsinO - f..tkgcos0 =...;v5!2l. In this equation, the values oro, vo, and l aregiven; therefore we can solve for f..tk:

° (1.4 m/s)2f..tk= tan 22 + 2 = 0.72.

2(0.34 m)(9.8 m/s )cos22°

Problem62. If the block in the previous problem were shoved

up 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 = gsinO + f..t~gcosO. With valuesfrom Problem 61, a = 1O.2m/s (positive down slope).The distance traveled up the slope to where the blockstops is (from Equation 2-11) -v'6/2a = -(1.4 m/s)2 +2(10.2 m/s2) - 9.58 cm (negative up slope). (We did

not use a rounded-off value of f..tk in this calculation.)(b) Once the block has stopped, it will remain at restif f..ts~ tan 22° = 0.404 (see Example 6-14). Since f..ts isnormally greater than f..tk (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 ForcesProblem63. Find the drag force on a 7.4-cm-diameter baseball

moving through air (density 1.2 kg/m3) at 45 m/s.The drag coefficient is 0.50.

SolutionThe cross-sectional area of the baseball (a sphere) isill'd2, so Equation 6-4 and the other given quantitiesresult in a drag force of magnitude FD = ~CpAv2 =~(0.50)(1.2 kg/m3)i1l'(7.4 cm)2(45 m/s}2 = 2.61 N.Problem64. A football's terminal speed is 53 m/s when it's

falling pointed end first and 43 m/s when it'sfalling with its long dimension horizontal. If thefootball is 28 em long at its longest, what is thediameter on its short dimension? Assume the dragcoefficient C is the same in both orientations.

SolutionThe 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 coefficient


i~',]

Problem69. A tether ball on a 1.7-m rope is struck so it goes

into circular motion in a horizontal plane, with the

Paired ProblemsProblem67. In Fig. 6-76, suppose ml = 5.0 kg and

m2 = 2.0 kg, and that the surface and pulley arefrictionless. Determine the magnitude anddirection of m2's acceleration.

FIGURE 6-76 Problems 67, 68.

CHAPTER6 101

SolutionSince we are not interested in the tension in the ropeconnecting the masses, this is a good opportunity totake 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), ml and m2move as a unit, with the same acceleration (inmagnj.tude) which we'll choose to be positive for m2upward and ml downslope. Gravity acts positivedownslope on ml (mig 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, ismIg sin 30° - m2g = (ml +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)x(9.8 m/s2) /(5 kg+ 2 kg) = 0.700Tn/s2

•

Problem68. Repeat the preceding problem, now taking

ml = 3.0 kg with m2 still 2.0 kg.

Sol~tionThe analysis in the solution to the previous problemgives a = (3 kg sin 30° - 2 kg)(9.8 mjs2)+(3 kg + 2 kg) = -0.98 m/s2

• The minus sign meansthat 1Jl2 goes downward and ml upslope, as specifiedpreviously.

SolutionThe expression for the terminal speed found inExample 6-15 can be used, Vt = V2 mg/CpA, wherethe mass of the drop is its volume, V, times thedensity of water, Pw.The volume and cross-sectionalarea of a spherical drop, in terms of the diameter, aret7l"d3 and lll'~, respectively, so their ratio is~d. Thenthe terminal speed becomes Vt = V4 gd (Pw/p)/3C =)4(9.8 m/s2)(10-3m)(1000/l.2)/3(0.50) = 4.67 mis,where we canceled identical units in the density ratio.

Problem65~Find the terminal speed of a l.O-mm-diameter

spherical raindrop in air. The densities of air andwater are 1.2 kg/m3and 1000 kg/m3

, respectively,and the drag coefficient is 0.50.

is the same for both orientations, then(Vt,IOng/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 !d, as shown. The areascan be found from geometry, in terms of l and d:Aend= 11rd2, and Along= R2(J -l(R - !d), whereR = (l2+ ~)/4d, and(J= 2 sin-l (l/2R). Thetranscendental equation for the ratio of these areascan be solved numerically on a PC. Our result forl = 28 em is d ~ 16.71 em.

Problem66. Will a golf ball of mass 45 g and diameter 4.3 em

reach terminal speed when dropped from a heightof 25 m? The drag coefficient is 0.35, and thedensity of air is 1.2kg/m 3

•

SolutionThe terminal speed for this golf ball, inair, is Vt = ../2 mg/CpA =

V2(45g)(9.8 m/s2)/(0.35)(l.2 kg/m3)1l'(4.3 cm/2)2 =38.0m/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 = V2 9 (25 m) =22.1 mis, 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 = vl(1- e-2gy/v;), as shown in many intermediatemechanics texts. This gives a speed of 20.4 m/s for thegolf ball dropped in this problem.)

102 CHAPTER 6

rope making a 15° angle to the horizontal. Whatis the ball's speed?

SolutionThe tetherball whirling in a horizontal circle isanalogous to the mass on a string in Example 6-6.From step 6, v = J gfcos2 ()/ sin B =

V (9.8mjs2)(1. 7 m) cos215°/ sin 15° = 7.75 m/s.

Problem70. An airplane goes into a turn 3.6 km in radius. If

the banking angle required is 28° from thehorizontal, what is the plane's speed?

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

Thus, v = Jgr tan 0 = V(9.8 m/s2)(3.6 km)tan28° =137m/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 e cos().)

Problem71. Starting from rest, a skier slides 100 m down a 28°

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

SolutionIf air resistance is ignored, the forces acting on theskier are analogous to those on the sled inExample 6-10, so the downslope acceleration isall = g(sin() - J.LkcosB). Starting from rest, thetime needed to coast a distance ~x downslopeis t = J2~x Iall' With no friction, t =

V2(100 m)/(9.8 mjs~) sin 28° =6.59 s. If J.Lk= 0.17, t' =V2(100 m)/(9.8 m/i)(sin 28° - 0.17cos28°) = 7.99 s,or about 1.40 s longer.

Problem72. At the end of a factory production line, boxes

start 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?

SolutionAs in the preceding problem, the time required to slidedown the incline is t = J2~xjall $ 3.3 s. Therefore,all = g(sinB - J.LkcosB) ~ 2~x/(3.3 s)2, or J.Lk$tan 30° - 2(5.4 m)j(3.3 s)2(9.8 m/s2) cos30° = 0.46.

Problem73. A car moving at 40 km/h negotiates a

130-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?

SolutionThe 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,J.Ls~ Iv2 - v~1/gR(l + v2v~/g2 R2), where R is theradius of the curve, and Vd is the design speed for theproper banking angle, tan()d = v~/gR. In this problem,Vd = (60/3.6) mis, v = (40/3.6) mis, and R = 130m,so J.Ls ~ 0;12. (b) Since v < Vd, the car would slidedown the bank of the curve in the absence of friction.

Problem74. A passenger sets a coffee cup on the seatback tray

of an airplane fiying 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° (Le., it's banking at a lower anglethan required to givethe 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?

SolutionThis problem involves a coffeecup 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 = J gR tan 25° = J~(9-.8-m-/s-2-)("";'2.-6-km-)t-a-n-25-0=109 m/&"so a coefficientof static frictionJ.Ls~ Iv2 - v~ I / gR (1+ V2V~/ g2 R2) = 0.37 is neededto keep the cup from sliding up the tray.

Supplementary ProblemsProblem."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

:~

)1,


Problem77. In the loop-the-Ioop track of Fig. 6-25, show that

the car leaves the track at an angle 4> given bycos 4> = v2/rg, where <p 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.

_FIGURE 6-78 Problem 76 Solution.

mg

vertical component of the tension in the upper stringmust balance the weight of the ball, independent of v,hence Tu = mg/sinO = 12.4N. When v = 5 mis, Te=(0.84 kg)(5 m/s)2(1.2 m)-l - (12.4 N)(1.2/1.6) =8.17 N.

CHAPTER 6 103

Solution- The angle 4> 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 cos 4> =mv2/r. (The tangential component is not of interest inthis problem.) The car leaves the track whenN = (mv2 /r) - mg cos 4> = 0 (no more contact) orcos 4> = v2 Igr. This implies that the car leaves thetrack at real angles for v2 < gr; otherwise, the carnever leaves the track, as in Example 6-8.

450m


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

Problem76. Figure 6-78 shows a 0.84-kg ball attached to. a

vertical 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.

SolutionStanding on the outer edge of the space station,rotating with it, the astronaut experiences a normalforce equal to the centripetal force, N = mac =m 41r2r /T2, where T is the period ofrotation (seeExample 4-8). Since T is the time per revolution, thenumber of revolutions per unit time is liT (called thefrequency of revolution). If the normal force is toduplicate Earth's gravity, ac = g, and liT = (1/27r) x

J97T = (1/27rh/(9.8m/s2)/(450 m/2) = (3.32 x10-2 revIs)(60 sl min) = 1.99 rpm.

SolutionConsider 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 areTe+Tu cos(}=mv21r and Tu sin (}= mg. The anglebetween the tensions is given by the lengths of thestrings, or ()= cos-1 (1.2/1.6) = 41.4°. (a) The lowerstring is taut provided Te ~ O.Eliminating T" from theabove equations, we find Te = mv21r - mg cot (}~ 0,so this condition implies v ~ Vgr cot ()=V(9.8 m/s2)(1.2 m)cot41.4° = 3.65 m/s. (b) The

104 CHAPTER 6

Problem78. An astronaut is training in a centrifuge that

consists of a small chamber whirled aroundhorizontally at the end of a 5.1-m-Iong 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?

SolutionThe wall and gravity act on the notebook. If the latterdoesn't fall, fs = mg ~ P-aN = p-smv2/r, orv2 ~ grip-B' In circular motion, the linear speed isrelated to the rate of revolution (the angular speed,denoted by Greek letter "omega") by v = 211'r/T = wr,where T is the period. Thus, v2 = w2r2 ~ gr/P-B' or

w ~ vg/P-Br = V(9.8 m/s2)/(0.62)(5.1 m) =1.76s-1 = 16.8 rev/min. (Note: w = 211'/T has unitsrad/s, and 211' rad = 1 rev.)


Problem79. You stand on a spring scale at the north pole and

again 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 9 due togeological factors.)

SolutionWhen 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 t~ the axis) isr = RECOSO, where REis the radius of the Earth(constant if geographical variations are neglected) and() is your lattitude. Your centripetal acceleration hasmagnitude ae = (211' /T)2 r and is directed toward theaxis of rotation (see Example 4-8). We assume thereare only two forces acting on you, gravity, F9

(magnitude mg approximately constant, directedtowards the center of the Earth), and the force exertedby the scale, Fa. Newton's second law requires that

Fg +Fa =mBe. (a) At the north pole, Be = 0, so themagnitudes of F9 and Fa are equal, or Fa = mgj butat the equator, Be has a maximum magnitude, equalto the difference in the magnitudes of F 9 and Fa, orFa = mg - m(211'/T)2RE. Therefore Fa (your ."weight") is lower at the equator th8.n at the pole.(b) The fractional difference of these two values is(Fa,pole - Fa,eq.)/Fa,pole = (211'/T)2RE/g =(21r /86,400 s)2(6.37x 106m)/(9.81 m/s2) = 0.34%.

at pole

FgIII /1/ 8f----I }<'I g

IIRE

~I


Problem80. Driving in thick fog on a horizontal road, a driver

spots 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 thegreater margin of safety? Assume that the same


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