This preview shows page 1. Sign up to view the full content.
Unformatted text preview: 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 61: Using Newton's Second Law Problem
1. Two forces, both in the xy plane, act on a 1.5kg mass, which accelerates at 7.3m/i in a direction 30° counterclockwise from the xaxis. One force has magnitude 6.8 N and points in the +x direction. Find the other force. 2Tcos 25°  Fres = max, since T1 = T2 = T. (a) If ax = 0, Fres = 2(1100 N) cos 25° = 1.99 kN. 2 (b) If ax = 0.16 m/s , Fres = 1.99 kN  (3700 kg)x 2 (0.16 m/s ) = 1.40 kN. . Solution
Newton's second law for this mass says F net = F 1 F2 = ma, where we assume no other significant forces are acting. Since the acceleration and the first force are given, one can solve for the second, F2 = ma2 F1 = (1.5 kg)(7.3 m/s )(icos30° + jsin300)(6.8 N)i = (2.68i 5.48j) N. This has magnitude 6.10 N and direction 63.9° CCW from the xaxis. + FIGURE 659 Problem 3 Solution. Problem
4. At what angle should you tilt an air table to simulate motion on the moon's surface, where 9 = 1.6 m/s2? . + Problem
2. Two forces act on a 3.1kg mass, which undergoes acceleration a=0.9li  0.27j m/i. If one of the forces is F1 = 1.2i  2.5j N, what is the other force? Solution
The acceleration down an incline is all = 9 sin (} (see Example 61). To replicate the moon's surface gravity, the angle of tilt should be (}= sin1 (1.62/9.81) = 9.51° (see Appendix E). Solution
(3.1 kg)(0.91i  0.27j)(m/s (4.02i+1.66j)N. . As in the previous problem F2 = ma  F 1 = 2 Problem )  (1.2i  2.5j) N = Problem
3. A 3700kg barge is being pulled along a canal by two mules, as shown in Fig. 659. The tension in each tow rope is 1100 N, and the ropes make 25° angles with the forward direction. What force does the water exert on the barge (a) if it moves with constant velocity and (b) if it accelerates forward at 0.16 m/s2? 5. A block of mass m slides with acceleration a down a frictionless slope that makes an angle (}to the horizontal; the only forces acting on it are the force of gravity F 9 and the normal force N of the slope. Show that the ma~nitude of the normal force is gi\ren by N = mvg2  a2• Solution
Choose the xaxis down the slope (parallel to the acceleration) and the yaxis 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 components gives 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. Solution
The horizontal forces on the barge are the two tensions and the resistance of the water, as shown on Figure 659. The net force is in the x direction, so + CHAPTER 6 87 k=M'1
Problem 5 Solution. ~t .~ ~~r )(
9501J ....
Problem 8 Solution. Problem
9. A 15kg monkey hangs from the middle of a massless rope as shown in Fig. 660. What is the tension in the rope? Compare with the monkey's weight. Problem
6. A skier starts from rest at the top of a 24° slope 1.3 km long; Neglecting friction, how long does it take to reach the bottom? Solution
The sum of the forces at the center of the rope (shown on Fig. 660) is zero (if the monkey is at rest), T1 + T2 W O.The x component of this equation requires that the tension is the same on both sides: T1 cos 8° + T2 cos 172° = 0, or T1 = T2. The y component gives 2Tsin8° = W, or T = W/2 sin8° = 3.59W = 3.59(15 kg)(9.8 m/s2) = 528 N. Solution
The acceleration down a frictionless incline is a = gsin6 (see Example 61), and the distance traveled down 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/s
2 ) + = sin 24° = 25.5 s. Problem
7. A block is launched up a frictionless ramp that makes an angle of 35° to the horizontal. If the block's initial speed is 2.2 mis, how far up the ramp does it slide?
T2 'tI
I T( SO .x
w SO Solution
The acceleration up the ramp is g sin 35°, so the block goes a distance in this direction calculated from the equation 2gsin35°(x  xo) = O. Thus, 2 x  xo = (2.2 m/s)2/2(9.8 m/s ) sin 35° = 43.1 cm. v5x Problem
8. At the start of a race, a 70kg swimmer pushes off the starting block with a force of 950 N directed at 15° below the horizontal. (a) What is the swimmer's horizontal acceleration? (b) If the swimmer is in contact with the starting block for 0.29 s, what is the horizontal component of his velocity when he hits the water?
FIGURE 660 Problem 9 Solution. Problem
10. A tow truck is connected to a 1400kg car by a \ cable that makes a 25° angle to the horizontal, as shown in Fig. 661. If the truck accelerates at 0.57 m/s2, what is the magnitude of the cable tension? Neglect friction and the mass of the cable. Solution
(a) If we assume that the reaction force of the starting block and gravity are the only significant forces acting on the swimmer, the horizontal acceleration is just Fx/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. Solution
The only force on the car with a horizontal component (in the direction of the acceleration) is the tension. Therefore, T cos 6 = rna, or T = (1400 kg) x (0.57 m/s2)/cos25° = 880 N. (The force of static 88 CHAPTER 6 friction acts between the tires and the road with magnitude sufficient to keep the wheels turning, but is assumed to be negligible.) FIGURE 661 Problem 10 Solution. Problem
11. A 100kg mass is suspended at rest by two strings attached to walls, as shown in Fig. 662. Find the tension forces in the two strings. FIGURE 663 Problem 12. Problem
13. A camper hangs a 26kg pack between two trees, using two separate pieces of rope of different lengths, as shown in Fig. 664. What is the tension in each rope? Solution
The force diagram is superimposed on Fig. 662. Since the mass is at rest, the sum of the forces is zero, T} +T2 + W=O, which is true for the x and y components separately, T} cos 45°  T2 = 0, and T} sin 45°  W = O. Solving for the magnitudes of the tensions, and substituting 98 N for the weight, we find T} = ../2 98 N = 139 N, and T2 = T}/V2 = 98 N. FIGURE 664 Problem FIGURE 662 Problem 13. 11 Solution. Solution Problem
12. A 1I00kg 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 the force exerted on the car by the tree is purely horizontal, as suggested in Fig. 663. Find the magnitude of that force once the car is fully stopped. The sum of the forces acting on the pack (gravity and the tension along each rope) is zero, since it is at rest, F 9 + T} + T2 O. In a coordinate system with xaxis horizontal to the right and yaxis vertical upward, the x 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 63). Solving for T1 and T2, 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. = Solution
If the slope of the embankment exerts only a normal force on the car (no friction), the situation is the same as in Example 64, Fg+ N + Fh = O. Then Fh = mg tan 0 = (1100 kg)(9.8 m/s2) tan 23° = 4.58 kN. Problem.
14. A construction worker is lifting a 92kg bundle of plywood onto an upper floor, using the arrangement shown in Fig. 665. What force must CHAPTER 6 89 the worker apply to lift the bundle at constant speed? Assume the pulley is massless and frictionless. to save the turkey from going over the edge. (The assumptions relevant to Example 65 might be somewhat overrestrictive in this situation.) Problem
16. Find expressions for the acceleration of the blocks in Fig. 666, where the string is fastened securely to the ceiling. Neglect friction and assume that the masses of pulley and string are negligible. Solution
The equations of motion (components parallel to the accelerations) for the two masses (ml includes the attached pulley) are T = m2a2 and mIg  2T = mIal' (The assumptions stated ensure that the tension has the same magnitude at all points in the string, as shown.) If the length of the string is fixed, when ml moves down a distance d, m2 moves to the right a distance 2d, so a2 = 2al' Thus T= m2(2al), mIal = mIg  2(2m2al), and al = mIg/(ml + 4m2)'
N '"'' 0'. ;: FIGURE 665 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 weight can be neglected, and the tension in each rope is the same (this will be evident after Chapter 12). The acceleration of the pulley, whose speed is constant, is zeroj therefore the vertical component of Newton's second law gives TI + T2  m~ = 2T  mg = 0, or T= = H92 kg)(9.8 m/s ) = 451 N. !mg Section 62: Problem Multiple Objects
FIGURE 666 Problem 16 Solution. 15. Your 12kg baby sister is hanging on the bottom of the tablecloth with all her weight. In the middle of the table, 60 cm from each edge, is a 6.8kg roast turkey. (a) What is the acceleration of the turkey? (b) From the time she starts pulling, how long do you have to intervene before the turkey goes over the edge of the table? Problem
17. If the lefthand slope in Fig. 656 makes a 60° angle with the horizontal, and the righthand slope makes a 20° angle, how should the masses compare if the objects are not to slide along the frictionless slopes?
\ Solution
The vertical motion of your baby sister and the horizontal motion of the turkey are analogous to the climber and rock in Example 65. If we assume that both have accelerations of the same magnitude as the tablecloth, which has negligible mass, no friction with the table, etc., then (a) a = arx = meg/(me +mr) = 2 (12 kg)(9.8 m/s2)/(12 kg + 6.8 kg) = 6.26 m/s is the turkey's horizontal acceleration, and (b) t = J2x / a = 2 2(60 cm)/(6.26 m/s ) = 0.438 s is the time you have Solution
The freebody force diagrams for the left and righthand masses are shown in the sketch, where there is only a normal contact force since each slope is frictionless, and we indicate separate parallel and perpendicular xyaxes. If the masses don't slide, the net force on each must be zero, or Tl  mig sin 60° = 0, and mrgsin20°  Tr = 0 (we only need the parallel components in this problem). If the masses of the string and pulley are negligible and there is no friction, V . ":' "'. .t.~en Ti == Tr. Adding the force equations, we find mrg sin 20°  meg sin 60° = 0, or the mass ratio must be mr/me == sin 60°/ sin 20° = 2.53 for no motion. (m + 3.56 kg + 4.34 kg)a.
= Solving for m, we find
m/s2)+(3.56 m/s2)
kg)(9.8 m (3.56 kg + 4.34 kg)(O.310 m/s2) (9.8 m/s2  0.310 = 3.93 kg (The shortcut can be justified by adding the component of the equations of motion of the three blocks in the direction of motion: l1ert  (3.56 kg)g = (3.56 kg)a, Tright l1ert = (4.34 kg)a, and mg  Tright = mao In this setup, the ropes are massless, and the pulleys are massless and frictionless, so the tension in each rope is constant.) Problem
Problem 17 Solution. 19. Suppose the angles shown in Fig. 656 are 60° and 20°. If the lefthand mass is 2.1 kg, what should be the righthand mass in order that (a) it accelerates downslope at 0.64 m/s2; (b) it accelerates upslope at 0.76 m/s2? Problem
18. In a setup like that shown in Fig. 637, but with different masses, ,a 4.34kg block starts from rest on the left edge of a frictionless tabletop 1.25 m wide. It accelerates to the right, and reaches the right edge in 2.84 s. If the mass of the block hanging from the left side is 3.56 kg, what is the mass hanging from the right side? Solution
With reference to the solution to Problem 17, the parallel component of the equations of motion for the masses are Te  meg sin 60° = meae and mrg sin 20°  Tr = mrar. The accelerations and tensions are equal, respectively, provided the string doesn't stretch and the other assumptions in Problem 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 righthand acceleration is positive for the coordinate systems we have chosen, so substituting a = 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. Problem
FIGURE 637 Problem 18. Solution
The acceleration of the 4.34kg block sliding horizontally to the right across the frictionless tabletop is a = 2(1.25 m)/(2.84 S)2 = 0.310 m/s2 (from Equation 210 and the given conditions). This is also the magnitude of the accelerations of the other two blocks. Since we are not interested in the tension in either string, we may use a shortcut to find the unknown mass. The net force on all three masses in the direction of motion is the difference in the rightand lefthand weights, Fnet = (m  3.56 kg)g, which equals the total mass times the acceleration, 20. Two unfortunate climbers, roped together, are sliding freely down an icy mountainside. The upper 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) The upper climber manages to stop the slide with an lee ax. Once the climbers have come to a complete stop, what force must the ax exert against the ice? Solution
(a) If we assume that the 6limbers move together as a unit, with the same magnitude of downslope acceleration a, then the net force acting on them is the Sum of the downslope components of gravity on each, CHAPTER 6 91 + (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 downslope components of gravity calculated in part (a).
Fnet = (75 kg)g sin 12° wedge in order that the rectangular block not slide along the wedge. Problem
21. In a florist's display, hanging plants of mass 3.85 kg and 9.28 kg are suspended from an essentially massless wire, as shown in Fig. 667. Find the tension in each section of the wire.
FIGURE 668 Problem 22. Solution Solution
Let the tensions in each section of wire be denoted by T1, T2, and T3 as shown in the figure. The horizontal and vertical components ofthe net force on the junction 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 x 9.8) N = 0  TI cos 54.0° + T2 cos 13.9° = 0 T2 sin 13.9°  T3 sin 68.0°  (9.28 x 9.8) N = 0 T2cos13.9° +Tgcos68.0° = 0 The forces acting on the wedge and block are shown separately in freebody diagrams for clarity. The contact forces N1 (between wedge and block) and N2 (betweenwedge and horizontal surface) are purely normal because the surfaces are frictionless. For no relative motion between the wedge and block, both must have the same horizontal acceleration, a. The horizontal and vertical components of Newton's second law for the block, and the horizontal component of Newton's second law for the wedge (all that's needed in 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. F
(9.28kg)g Problem 22 Solution.
FIGURE 667 Problem 21 Solution. One can solve any thr~ of these equations for the unknown tensions, perhaps using the fourth equation as a check (if you do, remember not to round off). For example, T1 = (3.85 x 9.8) N/(sin54.0° cos54.00x tan 13.9°) = 56.9 N, T2 = T1 cos54.0°/ cos 13.9° = 34.4 N, and Tg = T2 cos 13.9° / cos68.0° = 89.2 N. (Note that the given angles and weights are not independent of one another.) Section 63: Problem Circular Motion 23. 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's orbital speed is 2.18x106 mis, what is the magnitude of the force between the electron and the proton? Problem
22. A rectangular block of mass ml rests on a wedgeshaped block of mass m2, as shown in Fig. 668. All contact surfaces are frictionless. Find an expression for the magnitude of the horizontal force F that must be applied to the Solution
For a particle in uniform circular motion, the net force equals the mass times the centripetal acceleration, F = mv2/r = (9.11xlO31 kg)(2.18x106 m/s)2+ (5.29x1011 m) = 8.18x10B N. 92 CHAPTER 6 period. (We'wrote a Greek letter, "tau," for the period because we used "tee" for the tension.) Problem
24. 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. (See Appendix E for relevant data.) Solution
We are asked to estimate the net force on the moon, which, according toNewton's second law, is the product of its mass and its acceleration. Since the moon describes approximately uniform circular motion about the Earth, F = m(v2/r) = mr(21r/T)2, where r and T are the radius and period of the orbit. Thus, F = (7.35x 1022kg)(3.85x 108 m)(21r/27.3x 86,400 s)2 = 2.01x102o N = m(2.73xlO3 m/s2). We displayed the numerical value of the moon's (centripetal) acceleration b~ause, comparing it to 9.8 m/s2 (the gravitational acceleration of an apple at the Earth's surface), Newton is said to have arrived at his famous inverse square law of gravitation. The distances of the moon and apple from the center of the Earth are about 60RE and RE, and (2.73x 103 /9.8) ~ (1/60)2. (See also Problem 16, Chapter 9.) FIGURE 669 Problem 26 Solution. Problem
27. A 94og rock is whirled in a horizontal circle at the end of a 1.3mIong string. (a) If the breaking strength of the string is 120 N, what is the maximum allowable speed of the ro.ck? (b) At this maximum speed, what angle does the string make with the horizontal? Solution
The situation is the same as described in Example 66. The horizontal component of the tension is the centripetal force, Tcose = mv2/r = mv2/lcos(), and the vertical component balances the weight, T sin () = mg. (b) At the maximum speed, the tension in the string is at its breakingstrength, Tmax = 120 Nj therefore the minimum angle the string makes with the horizontal is given by sin()min= mg/Tmax, or Omin= sin1(0.940x9.8 N/120 N) = 4.40°. a At these values ofT andO, the speed is Vrnax = Tmaxlcos2emin/m = J(120 N)(1.3 m)(cos4.400)2/(0.940 kg) = 12.8 m/s. Problem
25. Show that the force needed to keep a mass m in a circular path of radius r with period T is 4tr2mr1T2. Solution
For an object of mass m in uniform circular motion, the net force has magnitude mv2/r (Equation 61). The period ofthe motion (time for one revolution) is T = 21rrlv, so the centripetal force can also be written as m(21rr /T)2 /r = mr(21r /T)2 = 41r2mr /T2 (see Equation 418). Problem
28. If the rock of the previous problem is whirled in a vertical circle, what is the minimum speed needed at the top of the circle in order that the string remain taut? Problem
26. A mass ml uridergoes circular motion of radius R on a horizontal frictionless table, connected by a massless string through a hole in the table to a second mass m2 (Fig. 669). If m2 is stationary, find (a) the tension in the string and (b) the pericid of the circular motion. Solu~ion
At the top of the circle, the tension, gravity, and the centripetal acceleration are all vertically downward (as in Example 68). Then T + mg = mv2/l. If the string remains taut, T ~ 0, or v ~ ../9l = /(9.8 m/s2)(1.3 m) = 3.57 m/s. Solution
(a) Newton's second law applied to the stationary mass 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 = 21rymlR/m2g the is Problem
29. A subway train rounds an unbanked curve at 67 km/h. A passenger hanging onto a strap CHAPTER 6 93 notices that an adjacent unused strap makes an angle of 15° to the vertical. What is the radius of the turn? Solution
(a) As shown in Example 68, 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 the difference between the normal force of the seat and the force exerted by the seatbelt, i.e., N = 308 N + Fbelt. The seatbelt, firmly adjusted, perhaps adds a few pounds (lIb = 4.45 N), providing a feeling of security. (c) The seatbelt is required in case of accidents or rapid tangential decelerations; it is not needed to contribute to the centripetal force. olution
he 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 the mass times the horizontal centripetal acceleration. The freebody diagram for the strap is the same as Fig. 618, except that the angle from the vertical is now given, as shown. Thus, T cosO = mg, and Tsin8 = mv2/r. Dividing these equations to eliminate , T , and solving for the radius of the turn, one finds r = 2 v2/gtan8 = (67 m/3.6 s)2/(9.8tan 15°m/s ) = 132 m. Problem
32. A 45kg skater rounds a 5.0mradius turn at 6.3 m/s. (a) What are the horizontal and vertical components of the force the ice exerts on her skate blades? (b) At what angle can she lean without falling over? Solution sfrap FIGURE 618 Problem 29 Solution. (a) If the ice is level, the contact force exerted on the skater has vertic8.J. omponent (normal force) equal to c the 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) In Chapter 14 it will be shown that stability requires that the center of gravity of the skater should be along the line of action of the contact force. She should lean at 0= tan1(fs/N) = 39.0°, relative to the vertical. Problem
30. An Olympic hammer thrower whirls a 7.3kg hammer on the end of a 12ocm chain. If the chain makes a 10° angle with the horizontal, what is the speed of the hammer? Solution
From Example 66, v = Jglcos20/sinO where we used l = 1.2 m and 8 = 10°. = 8.10 mis,
Prpblem Problem 32 Solution. Problem
31. Riders on the "Great American Revolution" looptheIoop roller coaster of Example 68 wear seatbelts as the roller coaster negotiates its 6.3mradius loop with a speed of 9.7 m/s. At the top of the loop, what are the magnitude and direction of the force exerted on a 6okg rider (a) by the rollercoaster seat and (b) by the seatbelt? (c) What would happen if the rider unbuckled at this point? 33. An indoor running track is squareshaped with rounded corners; each corner has a radius of 6.5 m on its inside edge. The track includes six 1.omwide lanes. What should be the banking angles on (a) the innermost and (b) the outermost lanes if the design speed of the track is 24 km/h? Solution
The banking angle is 0 = tan1(v2 /gr) (see Example 67). A competitive runner rounds a turn on the inside 94 CHAPTER 6 edge of his or her lane. (a) () = tan1 [(24 mj3.6 s)2 : \ (9.8 m/s2) (6.5 m)] = 34.9°. (b) The radius of the inside edge of the outermost lane is 6.5 m + 5(1 m) = 11.5 m, so (J = 21.5°. (This type of banking is shown in Fig. 658.) Problem
34. A jetliner flying horizontally at 850 km/h banks at 32° to make a turn. What is the radius of the turn? Solution
The forces on an airplane making a horizontal circular turn are analogous to those on a car negotiating a banked curve (see Fig. 621), where the main .aerodynamic force on the airplane is normal to the wing surfaces. Thus, Faerosin(J = mv2/r, and Faero 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. .•. " (Ibwllf'd ce.f1ffr of curve. ) FIGURE 670 Problem 35. in the sixth step of Example 66, we find mg tan 18° = mv2/r, or v = y'gr tan 18° = V(9.8 m/s2)(180 m) tan 18 = 23.9 m/s = 86.2 km/h.
0 Problem 34 Solution. (This problem could also be approached using an accelerated coordinate system at rest relative to the car, and introducing a fictitious force, mac, called the centrifugal force, to account for the accelerated motion. The beginning student is advised to stick with inertial coordinate systems, however, in which there is less chance for confusion.) Problem
35. You're a passenger in a car rounding a turn with radius 180 tn. You take your keys from your pocket and dangle them from the end of your keychain. They make an 18° angle with the vertical, as shown in Fig. 670. What is the car's speed? Problem
36. A bucket of water is whirled in a vertical circle of radius 85 em. What is the minimum speed that will keep the water from falling out? Solution
As long as the magnitude of the acceleration at the top of the circle is greater than g, the water will remain in the bucket. (Then, the bottom of the bucket exerts a normal force on the water so the two are in contact; see Example 68). Thus, v2 IT > g, or Solution
You and your keys are rounding the curve along with the car, so all have the same centripetal acceleration, ac = v2/r, which we assume is horizontal and directed toward the center of the curve. The forces on the keys are the tension along the direction of the keychain and gravity downward, as shown on Fig. 670, so the situation is just like Example 66, where the horizontal component of the tension supplies the centripetal acceleration, and the vertical component balances the weight of the keys. Then the horizontal and vertical components of Newton's second law for the keys are: Tcos(90° 18°) = Tsin 18° = mac = mv2/r, and T cos 18°  mg = O.Eliminating the tension, as v> V(9.8 mjs2)(0.85 m) = 2.89 m/s. Problem
37. A 1200kg car drives on the country road shown in Fig, 671. The radius of curvature of the crests and dips is 31 m. What is the maximum speed at which the car can maintain road contact at the crests? CHAPTER 6 95 Solution
The radial component of Newton's second law for the TSS (positive component toward the center of the Earth) is m!JTss  T = m(arhss. Since the TSS and the shuttle have the same period, r (the tether would pull 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 is hardly affected by the much smaller TSS, then (ar )Shuttle = gShuttie' Therefore, the tension in the cable is T = m[!JTSS  (arhssl
= m [9TSS rTSS (ar)shuttle] rshuttle FIGURE 671 Problem 37. Solution
If air resistance is ignored, the forces on the car are gravity and the contact force of the road, which is represented by the sum of the normal force (perpendicularly away from the road) and friction between the tires and the road (parallel to the road in the direction of motion). Newton's second law for the car is F 9 + N + f~= mao At 8: crest, N is vertically . upward, fs is horizontal, and the vertical component of a is the radial acceleration v2 / r (downward in this case). The vertical component of Newton's second law is then mg + N = mv2/r. As long as the car is in contact with the road, N ~ OJthus, v ~ .;gr = J(9.8 m/s2)(31 m) = (500 kg)(9.8 m/s2) (0..932 _ 0.926 x 6370 + 230) 6370 + 250 = 43.1 N = 17.4 m/s = 62.7 km/h. Section 64: Problem Friction Problem
38. The Tethered Satellite System (TSS) is a NASA experiment consisting of a 500kg satellite connected to the space shuttle by a 20kmIong cable of negligible mass. Suppose the shuttle is in a 250kmhigh circular orbit, where the acceleration of gravity is 0.926 times its value at Earth's surface. The TSS hangs vertically on its tether (Fig. 672), and at its 230km altitude the acceleration of gravity is 0.932 times its surface value. What is the tension in the cable? 39. Movers slide a.file cabinet along a floor, The mass of the cabinet is 73 kg, and the coefficient of kinetic 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 is equal in magnitude to its weight, so the frictional force has magnitude!k = JLkN = JLkmg = (0.81)(73 kg)x (9.8 m/s2) = 579 N. The direction of sliding friction opposes the motion. Problem
40. You make a huge snowball with a mass of 33 kg. If the coefficient of friction between the ball and an icecovered pond is 0.16, with what force must you push the ball to move it (a) at constant velocity and (b) with an acceleration of 0.84 m/l? .~
20km T '  Solution
1Ihe pond has a level frozen surface, so the normal force on the snowball is equal in magnitude to its weight. The frictional force opposing the motion has magnitude fk = JLkN = JLkmg. The horizontal forces acting are friction and the applied force, so the horizontal component. of Newton's second law (positive in the direction of motion) is Fapp  fk = mao (a) At constant 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. FIGURE 672 Problem 38 Solution. 96 CHAPTER 6 Problem
41. Eight SOkgrugby players climb on a 70kg "scrum machine," and their teammates proceed to push them with constant velocity across a field. If the coefficientof kinetic friction between scrum machine and field is 0.78, with what force must they push? mower at constant velocity? Assume the force is applied in the direction of the handle. Compare with the mower's weight. Solution
Assuming the ground is also horizontal, we may depict the forces on the lawnmower as shown. At constant velocity (constant speed in a straight line) a = 0, and F + fk + N + mg = O.The x andy components of this equation are F cos 35°  !k= 0, and N  F sin 35° mg = O. Using Ik = JLkN = JLk(Fsin35° + mg), with iLk = 0.68, we find F = Jjkmg/(cos35°  JLksin 35°) = 1.58mg = 342 N. Solution
As in Problem 40(a), a horizontal applied force must have magnitude equal to the frictional force in order to push an object at constant velocity along a level surface. The total weight on the scrum machine is (8xSO kg + 70 kg)(9.8 m/s2) = 6.96 kNi therefore Fapp = Ik = JLkN = (0.7S)(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 coefficient of kinetic friction?
)(  ~SO + F
Problem 44 Solution. Solution
The force of friction is the only horizontal force acting, and the normal force is vertical and equal to the puck's weight. Thus, !k = ma = JLkmg, or a = JLkg. We take the positive direction parallel to the initial velocity, so that Equation 211 can be used for the acceleration, a = (0  v6)/2(x  xo) = (14 m/s)2 72(56 m) = 1.75 m/i. Then JLk = a/g = (1.75 m/s2)/(9.8 m/s2) = 0.18. (We express JLk to two significant figures only since it is an empirical constant.) Problem
45. Repeat Example 65, now assuming that the coefficient of kinetic friction between rock and ice is 0.057. Solution
If there is friction between the rock and the ice, we must modify the rock's equation of motion, Tr F gr + N + fk = mar. Since the ice surface is horizontal, only the rock's :zrequation changes, Trx JLkN = rnrarx' Now we need to use the rock's yequation 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 climber Problem
43. A child sleds down a 12° slope at constant speed. What is the coefficientof friction between slope and sled? + Solution
The frictional force must balance the downslope component of gravity on the sled to produce a constant speed. The normal force on the sled must balance the perpendicular component of gravity if just gravity and the contact force are acting. Thus, mgsinlJ =!k = JLkN, andmgcoslJ = N, orJLk = tan 12° = 0.21 (as in Example 610). has more time, t = V2(51 m)/(0.159 m/s2) = 25.3 s, to pray for rescue. Problem
46. During an ice storm, the coefficients of friction between car tires and road are reduced to JLk = 0.088 and JLs = 0.14. (a) What is the maximum slope on which a car can be parked without sliding? (b) On a slope just steeper than this maximum, with what acceleration will a car slide down the slope? Problem
44. The handle of a 22kg lawnmower makes a 35° angle with the horizontal. If the coefficient of friction between lawnmower and ground is 0.68, what magnitude of force is required to push the CHAPTER 6 97 Solution
(a) With reference to Example 614, 0 = tanJ1.s= 7.97°. (b) From Example 610, a = gsinO 2 J1.kgcosO = (9.8 m/s ) (sin 7.9'r'  0.088 cos 7.97°) =
1 Solution
When stopping on a level track, the maximum acceleration due to friction is a = f..tsg, as explained in Example 612. The minimum stopping distance from an initial speed of (140/3.6) m/s is .1x = vV( 2a) = (38.9 m/s)2/(2xO.58x9.8 m/s2) = 133 m. With splitsecond timing, an accident could be averted. 50.5 cm/s2. (Using a little trigonometry, we could have written a = g(J1.s ILk)/ + J1.~.) VI Problem
47. A bat crashes into the vertical front of an accelerating subway train. If the coefficient of friction between bat and train is 0.86, what is the minimum acceleration of the train that will allow the bat to remain in place? Problem
50. If you neglect to fasten your seatbelt, and if the coefficient of friction between you and your car seat is 0.42, what is the maximum deceleration for which you can remain in your seat? Compare with the deceleration in an accident that brings a 6Okm/h car to rest in a distance of 1.6 m. Solution
Since N is parallel to the acceleration, but perpendicular to gravity and friction, N = ma, and fs = mg ~ lLaN = f..tsma. Therefore, in order to remain in place, a ~ g/f..ts = (9.8 m/s2)/0.86 = 11.4 m/s2. Solution
If the seat is horizontal (parallel to the acceleration) then fs = ma ~ ILsN = J.Lsmg, or a 5"J.Lsg. The maximum deceleration is therefore (0.42)(9.8 m/s2) = 4.12 m/s2. For the accident described, the magnitude of the deceleration is vU2(x  xo) = (60 m/3.6 s)2;. 2 2(1.6 m) = 86.8 m/s , or about twentyone times greater. (Friction is no substitute for a seatbelt!) Problem
Problem 47 Solution. 51. A bug crawls outward from the center of a compact disc spinning at 200 revolutions per minute. The coefficient of static friction between the bug's sticky feet and the disc surface is 1.2. How far does the bug get from the center before slipping? Problem
48. In a factory, boxes drop vertically onto a conveyor belt moving horizontally at 1.7 m/s. If the coefficientof kinetic friction is 0.46, how long does it take each box to come to rest with respect to the belt? Solution
Assume that the disc is level. Then the frictional force produces 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, or r ~ 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 s divided by the number of revolutions per minute.
\ Solution
Kinetic friction accelerates each box up to the speed of the belt: !k = f..tkN = f..tkmg = rna (if we suppose the belt to be horizontal). This takes time t = via = 2 V/f..tkg = (1.7 m/s)/(0.46)(9.8 m/s ) = 0.377 s. Problem
49. The coefficientof static friction between steel train wheels and steel rails is 0.58. The engineer of a train moving at 140 km/h spots a stalled car on the tracks 150 m ahead. If he applies the brakes so that the wheels do not slip, will the train stop in time? Problem
52. A 31Dgpaperback book rests on a 1.2kg textbook. A force is applied to the textbook, and the two books accelerate together from rest to 96 cm/s in 0.42 s.The textbook is then brought to a stop in 0.33 s, during which time the paperback slides off. Within what range does the coefficient of static friction between the two books lie? 98 CHAPTER 6 . incline is a.) The x and y components of Newton's second law for each block are mIg sin Solution
The direction of motion and the orientation of the surface of contact are not specified; assume both are horizontal. Then, for the acceleration, Is = ma ~ JLsN = JLsmg (where m is the mass of the paperback), or JLs~ a/g. From the given data, 2 JLs~ (0.96 m/s/0.42 s)/(9.8 m/s ) = 0.23. During the deceleration, I::'ax = psmg < ma' (the magnitude of the paperback's acceleration is smaller than that of the textbook, because the paper~ack slides off), so Ps < (0.96 m/s/0.33 s)/(9.8 m/s ) = 0.30. 30°  PklNI  Fe = mIa, Nkl  mIg cos 30° = 0 m2g sin 30°  Pk2N2 + Fe
= O. = m2a, N2  m2g cos 30° Problem
53. A 2.5kg block and a 3.1kg block slide down a 30° incline as shown in Fig. 673. The coefficien~~ of kinetic friction between the 2.5kg block and the slope is 0.23; between the 3.1kg block and the slope it is 0.51. Determine the (a) acceleration of the pair and (b) the force the lighter block exerts on the heavier one. (a) To solve for a, add the x equations and use values of N from the y equations: a = [( + m2)g sin 30° ml (mlJLkl + m2Pk2)g cos 300]/(ml + m2), or 2 a = 1.63 m/s , when the given m's and Pk'S are substituted. (b) To solve for Fe, divide each x equation by the corresponding m, and subtract: Fe = (JLk2 Pkl)mlm2g cos 30° /(ml + m2) = 3.29 N. Problem
54. Children sled down a 41mIong hill inclined at 25°. At the bottom the slope levels out. If the coefficient of friction is 0.12, how far do the children slide on the level? Solution
The acceleration down the incline is a = g(sinO PkCOSO) = (9.8 m/s2) (sin 25°  0.12cos25°) = 3.08 m/s (see Example 610). The speed at the bottom is v2 = 2 as, where s = 41 m. On level ground, the deceleration is  Ik/m = Pkmg/m = Pk9, so the distance traveled before stoppin~ is x = v2/2(Pkg) = as/Pkg = (3.08 m/s )(41 m)/(0.12)x 2 (9.8 m/s ) = 107 m.
2 FIGURE 673 Problem 53. Solution
The forces on the blocks are as shown. (Since JLk2> JLkl, there will be a contact force of magnitude F.e, such that the acceleration of both blocks down the . Problem
55. In a typical frontwheeldrive car, 70% of the car's .weight rides on the front wheels. If the coefficient of friction between tires and road is 0.61, what is the maximum acceleration of the car? ~f l'i
/ Solution
On a level road, the maximum acceleration from static friction between the tires and the road is amax = PsN fm (see Example 612). In this case, the norm'al force on the front tires (the ones producing the frictional force which accelerates the car) is 70% of mg whereas the whole mass must be accelerated. , 2 . 2 Thus, amax (0.61)(0.70)(9.8 m/s ) 4.18 m/s . Fe
/3a a/ /...... ...... I / '" N2, fk2 = = M.' /3t! I / Problein
56. Repeat the previous problem for a rearwheeldrive car with the same portion of its weight over the front wheels. .. Problem 53 Solution. CHAPTER 6 99 Solution
The frictional force on the rear wheels accelerates the car the frictional force on the front wheels, assumed ne~ligible,just makes them turn. If the car is on level ground, Nr = 0.3mg, so Fs = ma 5 Its(O.3mg), or. 2 2 a ~ (0.61)(0.3)(9.8 m/s ) = 1.79m/s . (Putting more weight over the drive axle gives better traction.) corresponding times is so U)2 = (t/t')2 = a' /a = 1  Itk cot 8, or Itk = 8 tan 35 /9 = 0.62.
0 l, Problem
59. You try to push a heavy trunk, exerting a force at an angle of 500 below the horizontal (Fig. 674). Show that, no matter how hard you try to push, it is impossible to budge the trunk if the coefficient of static friction exceeds 0.84. 'I fi
O/3rng ~~O O.7W1fjFIGURE Problem 56 Solution. 674 Problem 59. Problem
57. A police officerinvestigating an accident estimates .from the damage done that a moving car hit a stationary car at 25 km/h. If the moving car left skid marks 47 m long, and if the coefficient of kineti~ friction is 0.71, what was the initial speed of the moving car? 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 = O. Since fs = Fa cos 50 5 ItsN = J.Lsmg + Fa sin 50 ( the condition for equilibrium can be written Fa (cos 50 J.Ls 50 :5 J.Ls g, or sin m (cos50 /lts)  sin 50 :5 (mg/Fa). The righthand side is always positive (Fa and mg are magnitudes), but the lefthand side can be positive or negative. If it is negative, the trunk does not move, independent of Fa. Thus, the equilibrium condition will always be satisfied if sin 50 > cos 50 Its, or Its > cot 50 = 0.84.
0 0 0 0 ), 0 0 ) 0 0 0 0 0 / Solution
On a level road, the acceleration of a skidding car is J.Lkg (see Example 612). From the kinematics of the reconstructed accident, 2 = 2ltk9(X  xo), from which we calculate that v v5  Vo 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 to the traffic citation!) = )(25 Problem
58. A slide inclined at 35 takes bathers into a swimming pool. With water sprayed onto the slide to make it essentially frictionless, a bather spends only onethird as much time on the slide as when it is dry. What is the coefficient of friction on the dry slide?
0 Problem 59 Solution. Solution
The time it takes to slide down, starting from rest. at the top, depends on the acceleration in the direction of the slope, t = .j2!::i..x/a, where D.x is the length of the slide (see Equation 210). Without friction, a = gsinO, and with friction, a' = 9 sin 0  J.Lkgcos 0 = a(l  JLk cot 0) (see Example 610). The ratio of the Problem
60. A block of mass m is being pulled at constant speed v down a slope that makes an angle 8 with the horizontal. The pulling force is applied through a horizontal rope, as shown in Fig. 675. If the coefficient of kinetic friction is J.Lk, find an expression for the rope tension. 100 CHAPTER 6 Solution
Since the speed is constant (a=O), the sum of the forces on the block is zero ,(Newton's second law). Taking components parallel and perpendicular to the incline, we have: T cos 9 + mg sin 0  fk = 0, and N + TsinO  mgcosO = O. Then TcosO + mgsinO= !k = f..tkN = f..tk(mgcosO  TsinO), and T = mg(f..tk cosO sin O)/(cos 0 + f..tk sinO). (Note that T ~ 0, so the application of step 7 of the strategy box in Section 61 is restricted by the range of 0, 0 ::; 0 ::;tan1 f..tk') not use a roundedoff value of f..tk in this calculation.) (b) Once the block has stopped, it will remain at rest if f..ts ~ tan 22° = 0.404 (see Example 614). Since f..ts is normally greater than f..tk (which was 0.72 in Problem 61), the block does not slide back. (Notice how the direction of the frictional forces depends on the circumstances: it was up the incline in Problem 61, down the incline in Problem 62(a), and up the incline in Problem 62(b).) Section 65: Problem Drag Forces 63. Find the drag force on a 7.4cmdiameter baseball moving through air (density 1.2 kg/m3) at 45 m/s. The drag coefficient is 0.50. Solution
FIGURE 675 Problem 60 Solution. Problem
61. A block is shoved down a 22° slope with an initial speed of 1.4 m/s. If it slides 34 em before stopping, what is the coefficient of friction? The crosssectional area of the baseball (a sphere) is ill'd2, so Equation 64 and the other given quantities result 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. Problem
64. A football's terminal speed is 53 m/s when it's falling pointed end first and 43 m/s when it's falling with its long dimension horizontal. If the football is 28 em long at its longest, what is the diameter on its short dimension? Assume the drag coefficient C is the same in both orientations. Solution
The acceleration down the slope can be found from kinematics (Equation 211) and from Newton's second law (as in Example 610): a = gsinO  f..tkg cos 0 = ...;v5!2l. In this equation, the values oro, vo, and l are given; therefore we can solve for f..tk:
f..tk = tan 22 Solution
The terminal speed of the football is inversely proportional to the square root of its crosssectional area perpendicular to the direction of fall (see Example 615). If we assume that the drag coefficient ° + (1.4 m/s)2 2 2(0.34 m)(9.8 m/s )cos22° = 0.72. Problem
62. If the block in the previous problem were shoved up the slope with the same initial speed, (a) how far would it go? (b) Once it stopped, would it slide back down? Solution
(a) Because the frictional force always opposes the relative motion of the surfaces in contact, Example 610 does not give the acceleration of a block shoved up an incline. Rather, fk is in the opposite direction to that shown in Fig. 633, and Newton's second law gives a = gsinO + f..t~gcosO. With values from Problem 61, a = 1O.2m/s (positive down slope). The distance traveled up the slope to where the block stops is (from Equation 211) v'6/2a = (1.4 m/s)2 + 2(10.2 m/s2)  9.58 cm (negative up slope). (We did Problem 64 Solution. CHAPTER6 is the same for both orientations, then (Vt,IOng/Vt,end)2(43/53)2 = Aend/Along.Aend (the = football's crosssectional area in a plane perpendicular to, and through, the midpoint of the long axis) is the area of a circle of diameter d. Along(the football's crosssectional area in a plane containing the long axis) is the area of two identical circular segments, of chord equal to the long dimension, l = 28 cm, and height (or saggita) equal to as shown. The areas can be found from geometry, in terms of l and d: Aend= 11rd2, and Along= R2(J l(R  !d), where R = (l2 + ~)/4d, and(J= 2 sinl (l/2R). The transcendental equation for the ratio of these areas can be solved numerically on a PC. Our result for l = 28 em is d ~ 16.71 em. 101 Paired Problems Problem
67. In Fig. 676, suppose ml = 5.0 kg and m2 = 2.0 kg, and that the surface and pulley are frictionless. Determine the magnitude and direction of m2's acceleration.
i~
',] !d, FIGURE Problem
65~Find the terminal speed of a l.Ommdiameter spherical raindrop in air. The densities of air and water are 1.2 kg/m3and 1000 kg/m3, respectively, and the drag coefficient is 0.50. 676 Problems 67, 68. Solution
Since we are not interested in the tension in the rope connecting the masses, this is a good opportunity to take advantage of the type of shortcut mentioned in the solution to Problem 18. (For a similar solution using equations of motion for each object, see Problem 17 or 19.) Being tied together by a rope (which is assumed to be unextensible), ml and m2 move as a unit, with the same acceleration (in magnj.tude) which we'll choose to be positive for m2 upward and ml downslope. Gravity acts positive downslope on ml (mig sin 30°) and negative downward on m2(m2g), so the net force on the system of masses, in the positive direction of motion, equal to the total mass times the acceleration, is mIg sin 30°  m2g = (ml + m2)a. (Here, we neglect the mass of the rope and pulley, which are also accelerated, and any frictional forces.) Thus, a = (5 kg sin 30°  2 kg)x (9.8 m/s2) /(5 kg + 2 kg) = 0.700 Tn/s2• Solution
The expression for the terminal speed found in Example 615 can be used, Vt = V2 mg/CpA, where the mass of the drop is its volume, V, times the density of water, Pw. The volume and crosssectional area of a spherical drop, in terms of the diameter, are t7l"d3 and lll'~, respectively, so their ratio is~d. Then the terminal speed becomes Vt = V4 gd (Pw/p)/3C = )4(9.8 m/s2)(103m)(1000/l.2)/3(0.50) = 4.67 mis, where we canceled identical units in the density ratio. Problem
66. Will a golf ball of mass 45 g and diameter 4.3 em reach terminal speed when dropped from a height of 25 m? The drag coefficient is 0.35, and the density of air is 1.2kg/m 3• Problem Solution
The terminal speed for this golf ball, in
air, is
Vt = ../2 mg/CpA 68. Repeat the preceding problem, now taking ml = 3.0 kg with m2 still 2.0 kg. = V2(45g)(9.8 m/s2)/(0.35)(l.2 kg/m3)1l'(4.3 cm/2)2 = 38.0m/s (see Example 615). Even in the absence of air resistance, a golf ball dropped from a height of 25 m would attain a speed of only v = V2 9 (25 m) = 22.1 mis, far short of the terminal speed. (The actual speed of an object, dropping through a distance y, subject to the drag force of Equation 64, is v2 = vl(1 e2gy/v;), as shown in many intermediate mechanics texts. This gives a speed of 20.4 m/s for the golf ball dropped in this problem.) Sol~tion
The analysis in the solution to the previous problem gives a = (3 kg sin 30°  2 kg)(9.8 mjs2)+ (3 kg + 2 kg) = 0.98 m/s2• The minus sign means that 1Jl2 goes downward and ml upslope, as specified previously. Problem
69. A tether ball on a 1.7m rope is struck so it goes into circular motion in a horizontal plane, with the 102 CHAPTER 6 rope making a 15° angle to the horizontal. What is the ball's speed? Problem
73. A car moving at 40 km/h negotiates a 130mradius banked turn designed for 60 km/h. (a) What coefficient of friction is needed to keep the car on the road? (b) To which side of the curve would it move if it hit an essentially frictionless icy patch? Solution
The tetherball whirling in a horizontal circle is analogous to the mass on a string in Example 66. From step 6, v = gfcos2 ()/ sin B = J V (9.8 mjs2)(1. 7 m) cos2 15°/ sin 15° = 7.75 m/s. Solution
The forces on a car (in a plane perpendicular to the velocity) rounding a banked curve at arbitrary speed are analyzed in detail in the solution to Problem 81 below. (a) It is shown there that to prevent skidding, J.Ls~ Iv2 /gR(l + v2v~/g2 R2), where R is the radius of the curve, and Vd is the design speed for the proper banking angle, tan()d = v~/gR. In this problem, Vd = (60/3.6) mis, v = (40/3.6) mis, and R = 130 m, so J.Ls ~ 0;12. (b) Since v < Vd, the car would slide down the bank of the curve in the absence of friction. Problem
70. An airplane goes into a turn 3.6 km in radius. If the banking angle required is 28° from the horizontal, what is the plane's speed? Solution
The airplane making a turn at the proper banking angle is analogous to the situation in Example 67. Thus, v = Jgr tan 0 = V(9.8 m/s2)(3.6 km)tan28° = 137 m/s = 493 km/h. (Note that the angles given in this and the previous problem are complementary, and that the radius of the circle in Problem 69 is e cos ().) v~1 Problem
74. A passenger sets a coffee cup on the seatback tray of an airplane fiying at 580 km/h. The plane goes into a 2.6kmradius turn, getting part of its turning force from its rudder and part from banking at 25° (Le., it's banking at a lower angle than required to give the full turning force). (a) What coefficient of friction is needed to keep the coffee cup on the tray? (b) If there were insufficient friction, which way would the cup slide? Problem
71. Starting from rest, a skier slides 100 m down a 28° slope. How much longer does the run take if the coefficientof kinetic friction is 0.17 instead of O? Solution
If air resistance is ignored, the forces acting on the skier are analogous to those on the sled in Example 610, so the downslope acceleration is all = g(sin()  J.Lkcos B). Starting from rest, the time needed to coast a distance ~x downslope is t = Iall' With no friction, t = Solution
This problem involves a coffee cup on a banked traytable moving with speed v in a horizontal circular arc of radius R. The forces and centripetal acceleration are analogous to those on a car rounding a banked curve, as in the previous problem. Here, the circular speed v =580 km/h = 161 m/s is greater than the design speed for a 25° banked turn, Vd = J gR tan 25° = J~(9.8m/s2)("";'2.6km) = tan250 109 m/&"so a coefficientof static friction 2  v~ gR (1 + V2V~/ g2 R2) = 0.37 is needed to keep the cup from sliding up the tray.
J.Ls~ J2~x 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°  0.17cos 28°) = 7.99 s, or about 1.40 s longer. Problem
72. At the end of a factory production line, boxes start from rest and slide down a 30° ramp 5.4 m long. If the slide is to take no more than 3.3 s, what is the maximum frictional coefficient that can be tolerated? Iv I/ Solution
As in the preceding problem, the time required to slide down 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) cos 30° = 0.46. Supplementary Problems Problem."
75. A space station is in the shape of a hollow ring, 450 m in diameter (Fig. 677). At how many revolutions per minute should it rotate in order to simulate Earth's gravitythat is, so that the CHAPTER 6 normal force on an astronaut at the outer edge would be the astronaut's weight on Earth? 103
:~ vertical component of the tension in the upper string must balance the weight of the ball, independent of v, hence Tu = mg/sinO = 12.4 N. 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. )1 , 450m mg FIGURE 677 Problem 75. _FIGURE 678 Problem 76 Solution. Solution
Standing on the outer edge of the space station, rotating with it, the astronaut experiences a normal force equal to the centripetal force, N = mac = m 41r2r /T2, where T is the period ofrotation (see Example 48). Since T is the time per revolution, the number of revolutions per unit time is liT (called the frequency of revolution). If the normal force is to duplicate Earth's gravity, ac = g, and liT = (1/27r) x
= (1/27rh/(9.8m/s )/(450 m/2) = (3.32 x 2 revIs)(60 sl min) = 1.99 rpm. 10 Problem
77. In the looptheIoop track of Fig. 625, show that the car leaves the track at an angle 4> given by cos 4> = v2/rg, where <p is the angle made by a vertical line through the center of the circular track and a line from the center to the point where the car leaves the track. J97T 2 Solution
 The angle 4> and the forces acting on the car are shown in the sketch. The radial component of the net force (towards the center of the track) equals the mass times the centripetal acceleration, N + mg cos 4> = mv2/r. (The tangential component is not of interest in this problem.) The car leaves the track when N = (mv2 /r)  mg cos 4> = 0 (no more contact) or cos 4> = v2 Igr. This implies that the car leaves the track at real angles for v2 < gr; otherwise, the car never leaves the track, as in Example 68. Problem
76. Figure 678 shows a 0.84kg 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 lower string to be taut and (b) the tension in each string if the ball's speed is 5.0 m/s. Solution
Consider the three forces acting on the ball, gravity and the tensions pulling along each string, as shown sketched on Fig. 678. The ball's acceleration is the centripetal acceleration; v2/r, directed along the lower string toward the axis of rotation, so the horizontal and vertical components of Newton's second law are Te + Tu cos(}= mv21r and Tu sin (}= mg. The angle between the tensions is given by the lengths of the strings, or () = cos1 (1.2/1.6) = 41.4°. (a) The lower string is taut provided Te ~ O.Eliminating T" from the above 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 Problem 77 Solution. 104 CHAPTER 6 Fg +Fa =mBe. (a) At the north pole, Be = 0, so the magnitudes of F 9 and Fa are equal, or Fa = mgj but at the equator, Be has a maximum magnitude, equal to the difference in the magnitudes of F 9 and Fa, or Fa = 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 106 m)/(9.81 m/s2) = 0.34%. Problem
78. An astronaut is training in a centrifuge that consists of a small chamber whirled around horizontally at the end of a 5.1mIong shaft. The astronaut places a notebook on the vertical wall of the chamber and it stays in place. If the coefficient of static friction is 0.62, what is the minimum rate at which the centrifuge must be revolving? Solution
The wall and gravity act on the notebook. If the latter doesn't fall, fs = mg ~ PaN = psmv2/r, or 2 v ~ gripB' In circular motion, the linear speed is related 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/PB' or w ~ vg/PBr = V(9.8 m/s2)/(0.62)(5.1 m) = 1.76 s1 = 16.8 rev/min. (Note: w rad/s, and 211' rad = 1 rev.) at pole Fg I I I / 1/ 8 I I = 211'/T has units f}<' g I I RE ~
I Problem 79 Solution. Problem 78 Solution. Problem
80. Driving in thick fog on a horizontal road, a driver spots a tractortrailer truck jackknifed across the road, as in Fig. 679. To avert a collision, the driver could brake to a stop or swerve in.a circular arc, as suggested in Fig. 679. Which offers the greater margin of safety? Assume that the same Problem
79. You stand on a spring scale at the north pole and again at the equator. (a) Which scale reading will be lower, and why? (b) By what percentage will the lower reading differ from the higher one? (Here you're neglecting variations in 9 due to geological factors.) Solution
When standing on the Earth's surface, you are rotating with the Earth about its axis through the poles, with a period of 1d. The radius of your circle of rotation (your perpendicular distance t~ the axis) is r = RECOSO, where REis the radius of the Earth (constant if geographical variations are neglected) and () is your lattitude. Your centripetal acceleration has magnitude ae = (211' /T)2 r and is directed toward the axis of rotation (see Example 48). We assume there are only two forces acting on you, gravity, F 9 (magnitude mg approximately constant, directed towards the center of the Earth), and the force exerted by the scale, Fa. Newton's second law requires that FIGURE 679 Problem 80. ...
View
Full
Document
 Spring '08
 WORMER
 Physics, Circular Motion, Force

Click to edit the document details