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hapter C 5 Homework 4. An adventurous archaeologist crosses between t wo rock cliffs by slowly going hand over hand along a rope stretched between the cliffs. He s tops to rest at the middle of the rope. The rope will break if the tension in i t exceeds 2.50 x 104 N , and our hero's mass is 90.0 kg. (a) If the angle is 10.0, find the tension in the rope. Fx=0 T2cos- T1cos=0 T2cos= T1cos T2=T1 2Tsin=mg T=mg2sin =90.0 kg9.8 ms22sin10 =2540 nt (b) What is the smallest value the angle can have if the rope is not to break? 2Tsin=mg 22.50 104 Nsin=90.0 kg9.8 ms2 = 1.01 5. A picture frame hung against a wall is suspended by two wires attached to its upper corners. If the two wires make the same angle with the vertical, what must this angle be if t he tension in each wire is equal to 0.75 of the weight of the frame? (Ignore any friction between the wall and the picture frame.) T=0.75W 2Tcos=W 2(0.75 W)cos=W =48.2 7. Certain st reets in San Francisco make an angle of 17.5 with t he horizontal. What force parallel to the st reet surface is required to keep a loaded 1967 Corvette of mass 1390 kg from rolling down such a st reet? T=Wsin =1390 kg9.8 ms2sin17.5 =4096 nt 8. A large wrecking ball is held in place by two light steel cables. I f the mass m of the wrecking ball is 4090 kg, what are (a) the tension T B i n the cable that makes an angle of 40 with the vertical TBsin50=W TB=4090 kg9.8 ms2sin50 =52323 nt (b) the tension T A i n the horizontal cable? TA=TBcos50 =52323 ntcos50 =33633 nt Page | 2 9. Find the tension in each cord in the figure if the weight of the suspended object is w. (a) TC=TAsin30+ TBsin45 TAcos30=TBcos45 Since sin45= cos45: W=TAsin30+TAcos30 =TA(sin30+cos30) TA=0.732W TAcos30=TBcos45 (0.732W)cos30=TBcos45 TB=0.897W (b) TBcos45=TAsin30+ W TAcos30=TBsin45 Page | 3 Since sin45= cos45: W=TBcos45-TAsin30 =TAcos30-TAsin30 =TAcos30- sin30 TA=2.73W TAcos30=TBsin45 (2.73W)cos30=TBsin45 TB=3.35W 10. A 1130-kg car is held in place by a light cable on a very smooth (frictionless) ramp. The cable makes an angle of31.0 above the surface of the ramp, and the ramp itself rises at 25.0 above the horizontal. (a) Draw a free-body diagram for the car. (b) Find the tension in the cable. Fx=max 0=Tcos31- 1130 kg9.8 ms2sin25 T= 5460 N (c) How hard does the surface of the ramp push on the car? Page | 4 Fy=may 0= n+5460 Nsin31- 1130 kg9.8 ms2cos25 n=7224 N 11. A man pushes on a piano with mass 180 kg so that it slides at constant velocity down a ramp that is inclined at 11.0 above the horizontal floor. Neglect any friction acting on the piano. Calculate the magnitude of the force applied by the man if he pushes (a) parallel to the incline Fx=max 0=F-mgsin F=180 kg9.8 ms2sin11.0 =337 N (b) parallel to the floor Fy=may 0=ncos- W n=180 kg9.8 ms2cos11.0 = 1797 N Fx=max 0=F-nsin F=(1797 N)sin11.0 Page | 5 =343 N 14. Two blocks, each with weight w, are held in place on a frict ionless incline. In terms of w and t he angle of the incline, calculate the tension in (a) t he rope connect ing the blocks FxB=maxB 0=T1-wsin T1=wsin (b) t he rope that connects block A t o the wall. FxA=maxA 0= T2-wsin- T1 T2= 2wsin (c) Calculate the magnitude of the force that the incline exerts on each block. nA=nB=wcos (d) Interpret your answers for the cases = 0 and = 90. For = 0: T1=T2=0 For = 90: T1=wsin90=w T2=2wsin90=2w 18. Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal r opes. The pull is horizontal and of magnitude 125 N. Find (a) the acceleration of the system Fx=max 125 N=30.0 kg+20.0 kg+10.0 kgax ax=2.08 ms2 Page | 6 (b) t he tension in ropes A and B Rope A : P-TA=max TA=125 N-10.0 kg2.08 ms2 =104 N Rope B : P-TB=max TA=125 N-10.0 kg+20.0 kg2.08 ms2 =62.5 N 19. A 15.0-kg load of bricks hangs from one end of a rope that passes over a small, frictionless pulley. A 28.0-kg counterweight is suspended from the other end of the rope. The system is released from rest. (a) Draw two free-body diagrams, one for the load of bricks and one for the counterweight. Page | 7 (b) What is the magnitude of the upward acceleration of the load of bricks? T-mBg= mBay mCg -T =mCay mCg-mBg=mBay+ mCay 28.0 kg9.8 ms2- 15.0 kg9.8 ms2=15.0 kgay+ 28.0 kgay ay=2.96 ms2 (c) What is the tension in the rope while the load is moving? How does the tension compare to the weight of the load of bricks? To the weight of the counterweight? T-mBg= mBay T=15.0 kg9.8 ms2+ 15.0 kg2.96 ms2 =191 N Bricks: 191 N15.0 kg9.8 ms2=1.30 Counterweight: 191 N28.0 kg9.8 ms2=0.00475 Page | 8 20. A 8.00-kg block of ice, released from rest at the top of a 1.50-m-long fr ict ionless ramp, slides downhill, reaching a speed of 2.50 m /s a t the bottom. (a) What is the angle between the ramp and the horizontal? vf2=vi2+ 2ax 2.50 ms2=0 ms2+ 2a1.50 m a=2.08 ms2 Fx=max mgsin=max 8.00 kg9.8 ms2sin=8.00 kg2.08 ms2 =12.3 (b) What would be the speed of the ice at the bottom if the motion were opposed by a constant friction force of 10.0 N parallel to the surface of the ramp? mgsin- f=max Page | 9 8.00 kg9.8 ms2sin12.3- (10.0 N)=8.00 kgax ax= 0.833 ms2 vf2=vi2+ 2ax vf2=0 ms2+ 20.833 ms2(1.50 m) vf=1.58 ms 21. A light rope is attached to a block with mass 4.00 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 10.0 N. (a) Draw two free-body diagrams, one for the 4.00-kg block and one for the block with mass m. Page | 10 (b) What is the acceleration of either block? Fx=max T= ma 10.0 N=4.00 kga a=2.50 ms2 (c) Find the mass m of the hanging block. Fy=may mg-T=ma m9.8 ms2-10.0 N=m2.50 ms2 m=1.37 kg (d) How does the tension compare to the weight of the hanging block? W=mg =1.37 kg9.8 ms2 =13.4 N TW=10.0 N13.4 N=0.745 22. A transport plane takes off from a level landing field with two gliders in tow, one behind the other. The mass of each glider is 700 kg, and the total resistance (air drag plus friction Page | 11 with the runway) on each may be assumed constant and equal to 2500 N. The tension in the towrope between the transport plane and the first glider is not to exceed 12,000 N. (a) If a speed of 40 m/s is required for takeoff, what minimum length of runway is needed? Fx=max Tt- 2f=2max 12000 N-22500 N=2700 kgax ax=5.00 ms2 vf2=vi2+ 2ax 40 ms2=0 ms2+ 25.00 ms2x x=160 m (b) What is the tension in the towrope between the two gliders while they are accelerating for the takeoff? Tg- f=max Tg- 2500 N=700 kg5.00 ms2 Tg=6000 N 24. A 550-N physics student stands on a bathroom scale in an 850-kg (including the student) elevator that is supported by a cable. As the elevator starts moving, the scale reads 450 N. (a) Find the acceleration of the elevator (magnitude and direction). Fy=may n-w=may 450 N-550 N=550 N9.8 ms2ay ay= -1.78 ms2 downward (b) What is the acceleration if the scale reads 670 N? 670 N-550 N=550 N9.8 ms2ay ay=2.14 ms2 (c) If the scale reads zero, should the student worry? Explain. Being weightless would mean the elevator was in free fall, which would be Page | 12 bad news for the student. (d) What is the tension in the cable in parts (a) and (c)? Part a: T-mTotg=mTotay T-850 kg9.8 ms2=850 kg-1.78 ms2 T=6815 N Part c: T-850 kg9.8 ms2=850 kg2.14 ms2 T=10148 N 28. In a laboratory experiment on frict ion, a 135-N block rest ing on a rough horizontal table is p ulled by a horizontal wire. The pull gradually increases unt il the block begins to move and cont inues to increase thereafter. Figure shows a graph of the fr ict ion force on this b lock as a funct ion of the pull. (a) Ident ify the regions of the graph where stat ic and k inet ic fr ict ion occur. The frict ion is stat ic for P = 0 t o P = 75.0 N . The friction is kinetic for P > 75.0 N . (b) Find the coefficients of static and kinetic fr iction between the block and the t able. s=fsmaxn =75.0 N135 N =0.556 =0.370 k=fkn =50.0 N135 N (c) Why does the graph slant upward in the first part but then level out? W hen the block is moving the friction is kinetic and has the constant value fk= kn i ndependent of P. This is why the graph is horizontal for P > 75.0. When the block is at rest, fs=P since this prevents relative motion. This is w hy the graph for P < 75.0 N has slope +1. Page | 13 (d) What would the graph look like if a 135-N brick were placed on the box, and w hat would be the coefficients of fr ict ion be in t hat case? Max fs and fk would double. The values of f on the vertical axis would double but the shape of the graph would be unchanged. The coefficients of friction a re independent of the normal force. 29. A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant speed of 3.50 m/s. The coefficient of kinetic friction between the box and the surface is 0.20. (a) What horizontal force must the worker apply to maintain the motion? fk= kn =0.2011.2 kg9.80 ms2 =22.0 nt (b) If the force calculated in part (a) is removed, how far does the box slide before coming to rest? Page | 14 Fx=max -fk=max -kn=max vf2=vi2+ 2ax ax=-kg = -0.209.80 ms2 = -1.96 ms2 0 ms2=3.50 ms2+ 2-1.96 ms2x x=3.13 m 30. A box of bananas weighing 40.0 N rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.40, and the coefficient of kinetic friction is 0.20. (a) If no horizontal force is applied to the box and the box is at rest, how large is the friction force exerted on the box? I f there is no applied force, no friction force is needed to keep the box at rest. (b) What is the magnitude of the friction force if a monkey applies a horizontal force of 6.0 N to the box and the box is initially at rest? fsmax=sn =0.40(40.0 N) =16.0 N fs=6.0 N (c) What minimum horizontal force must the monkey apply to start the box in motion? The monkey must apply a force equal to fsmax, 16.0 N (d) What minimum horizontal force must the monkey apply to keep the box moving at constant velocity once it has been started? Page | 15 fk= kn =0.20(40.0 N) =8.0 N (e) If the monkey applies a horizontal force of 18.0 N, what is the magnitude of the friction force and what is the box's acceleration? F-fk=ma a=18.0 N-8.0 N(40.0 N)/(9.80 m/s2 ) =2.45ms2 33. You are lowering two boxes, one on top of the other, down the ramp shown in Figure 5.53 by pulling on a rope parallel to the surface of the ramp. Both boxes move together at a constant speed of 15.0 cm/s. The coefficient of kinetic friction between the ramp and the lower box is 0.444, and the coefficient of static friction between the two boxes is 0.800. (a) What force do you need to exert to accomplish this? = tan-12.50 m4.75 m=27.8 fk= knTot =0.444(694 N) =308 N fk+ T-mTotgsin=mTotax 308 N+ T-32.0 kg+48.0 kg9.8 ms2sin27.8= 0 T=57.1 N up the ramp Page | 16 (b) What are the magnitude and direction of the friction force on the upper box? fs=mgsin =32.0 kg9.8 ms2sin27.8 =146 N 34. Stopping Distance. (a) If the coefficient of kinetic friction between tires and dry pavement is 0.80, what is the shortest distance in which you can stop an automobile by locking the brakes when traveling at 28.7 m/s Fx=max -kmg=max ax= -kg vf2=vi2+ 2ax vf2=vi2+ 2-kgx 0 ms2=28.7 ms2+ 2-0.809.8 ms2x x=52.5 m Page | 17 (b) On wet pavement the coefficient of k inet ic fr ict ion may be only 0.25. How fast should you drive on wet pavement in order to be able to stop in the same distance as in part (a)? 0 ms2=vi2+ 2-0.259.8 ms252.5 m vi=16.0 ms 35. A clean brass washer slides along a horizontal clean steel surface unt il it stops. Using the values from Table 5.1, how many t imes farther would it slide wit h the same init ial speed if t he washer were Te f lon-coated? 0.440.04=11 t imes farther 36. Consider the system shown in Fig. 5.54. B lock A weighs 45.0 N and block B weighs 25.0 N. Once block B is set into downward motion, it descends at a constant speed. (a) Calculate the coefficient of kinetic friction between block A and the tabletop. mBg-T=0 N 25.0 N-T=0 T=25.0 N Page | 18 mAg-nA=0 N 45.0 N-nA=0 N nA=45.0 N T-fk=0 N 25.0 N-fk=0 N fk=25.0 N fk= knA 25.0 N=k(45.0 N) k=0.556 (b) A cat, also of weight 45.0 N, falls asleep on top of block A. If block B is now set i nto downward motion, what is its acceleration (magnitude and direction)? mAg-nA=0 N (45.0 N+45.0 N)-nA=0 N nA=90.0 N fk= knA =(0.556)(90.0 N) =50.0 N T-fk=mAax mBg-T=mBay -----------------------mBg-fk=mA+ mBay 25.0 N- 50.0 N=90.0 N9.8 ms2+25.0 N9.8 ms2ay ay= -2.13 ms2 (upward; the block is slowing down) Page | 19 37. Two crates connected by a rope lie on a horizontal surface. Crate A has mass m A and crate B has mass m B . T he coefficient of k inetic friction between each crate and the surface is k. The crates are pulled to the r ight a t constant velocity by a horizontal force F. I n terms of m A, m B , and k, calculate (a) the magnitude of the force F C rate A: Fy=0 nA=mAg Crate B: Fy=0 nB=mBg Fx=0 F=T+fkB = kmAg + kmBg = kgmA+ mB (b) the tension in the rope connecting the blocks. T=kmAg Fx=0 T=fkA= knA= kmAg Page | 20 Page | 21 41. Block A (mass 2.25 kg) rests on a tabletop. I t is connected by a horizontal cord passing over a light, frictionless pulley to a hanging block B ( mass 1.30 kg). The coefficient of k inetic friction between block A and the tabletop is 0.450. After the blocks are released f rom rest, find (a) the speed of each block after moving 3.00 cm B lock on table: fk= kmAg Fx=max T-fk=mAa T- kmAg=mAa Hanging block: Fy=may mBg-T= mBa T= mBg- mBa mBg- mBa - kmAg=mAa 1.30 kg9.8 ms2- 1.30 kga-0.4502.25 kg9.8 ms2=2.25 kga a=0.794 ms2 vf2=vi2+ vf2=0 2ax ms2+ 20.794 ms2(0.003 m) vf=0.0690 ms (b) the tension in the cord. T= mBg- mBa =1.30 kg9.8 ms2- 1.30 kg0.794 ms2 =11.7 N Fy=0 n=mAg 42. A 25.0-kg box of textbooks rests on a loading ramp that makes an angle w ith the horizontal. The coefficient of kinetic friction is 0.25, and the coefficient of static friction is 0.35. (a) As the angle i s increased, find the minimum angle at which the box starts to slip. Fx=0 N=mgcos Fy=0 fs=mgsin Page | 22 When the box begins to slip, fs>sN: 25.0 kg9.8 ms2sin> 25.0 kg9.8 ms2(0.35)cos tan>0.35 >19.29 (b) At this angle, find the accelerat ion once the box bas begun to move. Fx=max mgsin- kmgcos=max 9.8 ms2sin19.29- 0.259.8 ms2cos19.29=ax ax=0.925 ms2 (c) At this angle, how fast will the box be moving after it has slid 5.0 m along the loading ramp? vf2=vi2+ 2ax vf2=0 ms2+ 20.925 ms2(5.0 m) vf=3.04 ms 43. A large crate with mass m r ests on a horizontal f loor. The coefficients of friction between t he crate and the f loor are s and k. A woman pushes downward at an angle below the horizontal on the crate with a force F. (a) What magnitude of force F is required to keep the crate moving at constant velocity? Fy=0 n=mg+Fsin fk= k(mg+Fsin) Fx=0 Fcos- fk=0 Fcos- kmg-kFsin=0 F=kmgcos-ksin (b) If s is greater than some critical value, the woman cannot start the crate moving no matter how hard she pushes. Calculate this critical value of s F=smgcos-ssin Page | 23 s=FcosFsin+mg 44. A box with mass m is dragged across a level f loor having a coefficient of kinetic friction k by a rope that is pulled upward at an angle above the horizontal with a force of m agnitude F. (a) In terms of m, k, , and g, obtain an expression for the magnitude of force r equired to move the box with constant speed. Fy=0 n+Fsin=mg n= mg-Fsin fk= kmg-Fsin Fx=0 Fcos=fk Fcos= kmg-kFsin F=kmgcos+ksin (b) Knowing that you are studying physics, a CPR instructor asks you how much force it would take to slide a 90-kg patient across a f loor at constant speed by pulling on him at an angle of 25 above the horizontal. By dragging some weights wrapped in an old pair of pants down the hall with a spring balance, you find that k = 0.35. Use the result of part (a) to answer the instructor's question. F=kmgcos+ksin =0.3590 kg9.8 ms2cos25+(0.35)sin25 =293 N Page | 24 45. Bloc ks A, B, and C are placed as in the figure to the r ight and connected by ropes of negligible m ass. Both A and B weigh 25.0 N each, and the coefficient of kinetic friction between each b lock and the surface is 0.35. Block C descends with constant velocity. (a) Draw two separate free-body diagrams showing the forces acting on A and on B. (b) Find the tension in the rope connecting blocks A and B. Fy=0 nA=wA Fx=0 T1=fA =kn =0.35(25.0 N) =8.75 N (c) What is the weight of block C? Fy=0 n2=wBcos Fx=0 T2=T1+fB+wBsin Page | 25 =8.75 N+0.3525.0 Ncos36.9+(25.0 N)sin36.9 =30.8 N wC=T2 =30.8 N (d) If t he rope connect ing A and B were cut, what would be the acceleration of C? Block B: Fx=max T2-fB-wBsin=mBa Add the two equations together: T2-fB-wBsin+wC-T2=mBa+mCa a=wC-fB-wBsinmB+mC =9.8ms230.8 N-0.3525.0 Ncos36.9-(25.0 N)sin36.925.0 N+30.8 N =1.54 ms2 49. A machine part consists of a thin 40.0-cm-Iong bar with Block C: Fy=may wC-T2=mCa small 1.15-kg masses fastened by screws to its ends. The screws can support a maximum force of 75.0 N without pulling out. This bar rotates about an axis perpendicular to it at i ts center. (a) As the bar is tu rning at a constant rate on a horizontal frictionless surface, w hat is the Page | 26 m axim um speed the masses can have wit hout pulling out the screws? F=marad=mv2R 75.0 N=1.15 kgv20.200 m v=3.61 (b) Suppose the machine is redesigned so that t he bar t u rns at a constant rate in a vert ical circle. Will one of the screws be more likely to pullout when the mass is at the top of t he circle or at the bottom? Use a free-body diagram to see why. Top of the circle: F=marad-mg Bot tom of circle: F=marad+mg For any given velocity (and therefore, given accelerat ion), the value of F i s larger at the bottom of t he circle than at the top. (c) Using the result of part (b), what is the greatest speed the masses can have w ithout pulling a screw? F=marad+mg 75.0 N= 1.15 kgv20.200 m+ 1.15 kg9.8 ms2 v=3.33 ms Page | 27 50. A flat (unbanked) curve on a highway has a radius of 220.0 m. A car rounds the curve at a speed of 25.0 m/s. (a) What is the minimum coefficient of friction that will prevent sliding? Fy=0 n=mg Fx=max sn=mv2R smg=mv2R s=v2Rg = 25.0 ms2220.0 m9.8 ms2 =0.290 (b) Suppose the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in part (a). What should be the maximum speed of the car so it can round the curve safely? s=v2Rg 130.290=v2220.0 m9.8 ms2 v=14.4 ms 51. A 1125-kg car and a 2250-kg pickup truck approach a curve on the expressway that has a radius of 225 m. (a) At what angle should the highway engineer bank this curve so that vehicles traveling at Page | 28 65.0 mi/h can safely round it regardless of the condition of their tires? Should the heavy truck go slower than the lighter car? tan=v2gR =29.1 ms29.8 ms2225 m =21.0 (b) As the car and truck round the curve at 65.0 mi/h, find the normal force on each one due to the highway surface. n=mgcos =1125 kg9.8ms2cos21.0 =11806 N 52. The "Giant Swing" at a county fair consists of a vertical central shaft with a number of horizontal arms attached at its upper end. Each arm supports a seat suspended from a cable 5.00 m long, the upper end of the cable being fastened to the arm at a point 3.00 m from the central shaft. (a) Find the time of one revolution of the swing if the cable supporting a seat makes an angle of 30.0 with the vertical. Fy=0 Fcos=mg F=mgcos mgcos=mv2Rsin 9.8 ms2cos30=v23.00 m+5.00 msin30sin30 v=5.58 ms (b) Does the angle depend on the weight of the passenger for a given rate of revolution? No, because tan=v2gR, theta does not depend on the weight of the passenger for a Fx=max Fsin=mv2R F=mv2Rsin Page | 29 g iven rate of revolut ion. 54. A small but ton placed on a horizontal rotat ing platform with diameter 0.320 m will revolve with the platform when it is brought up to a speed of 40.0 rev/m in, provided the but ton is no more than 0.150 m from the axis. (a) What is the coefficient of stat ic fr ict ion between the but ton and the platform? Fy=0 n=mg Fx=max fs=mr2 smg=mr2 s=0.150 m40.0revmin2radrev1 min60 s29.8 ms2 =0.269 (b) How far from the axis can t he button be placed, wit hout slipping, if the platform rotates at 60.0 rev/m in? s=R2g 0.269= R60.0revmin2radrev1 min60 s29.8 ms2 R=0.0667 m 55. One problem for humans living in outer space is t hat they are apparent ly weight less. One way around t his problem is to design a space stat ion that spins about its center at a constant rate. This creates "art if icial gravity" at t he outside r im of the stat ion. (a) If t he diameter of t he space stat ion is 800 m, how many revolut ions per m inute a re needed for the "art if icial gravity" accelerat ion to be 9.80 m/s ? ac=v2R 9.8 ms2=v2(400 m) v=62.6 ms =vR Page | 30 2 =62.6 ms400 m =0.157rads 60 s1 min1 rev2 rad=1.49revmin (b) If t he space stat ion is a wait ing area for t ravelers going to Mars, it m ight be desirable to sim ulate the accelerat ion due to gravity on the Mart ian surface (3.70 m/s ). How m any revolutions per minute are needed in this case? 3.70 ms2=v2(400 m) v=38.5 ms 2 =38.5 ms400 m =0.0962rads 60 s1 min1 rev2 rad=0.918revmin 56. The Cosmoclock 21 Ferris wheel in Yokohama City, Japan, has a diameter of 100 m. I ts name comes from its 60 arms, each of which can function as a second hand (so that it m akes one revolution every 60.0 s). (a) Find the speed of the passengers when the Ferris wheel is rotating at this rate. =2T =260 s =0.105 rads v= r =0.105 rads50 m =5.24 ms (b) A passenger weighs 882 N at the weight-guessing booth on the ground. What is h is apparent weight at the highest and at the lowest point on the Ferris wheel? W=mg 882 N=m9.8 ms2 m=90 kg Page | 31 H ighest Point: mg-n=mr2 n=mg-mr2 =90 kg9.8 ms2-90 kg50 m0.105 rads2 =832.7 N Lowest Point: n-mg=mr2 n=mg+mr2 =90 kg9.8 ms2+90 kg50 m0.105 rads2 =931.3 N (c) What would be the time for one revolution if the passenger's apparent weight at the highest point were zero? n=mg-mr2 0 N=90 kg9.8 ms2-90 kg(50 m)2 =0.443 rads =2T 0.443 rads=2T T=14.2 s (d) What then would be the passenger's apparent weight at the lowest point? n=mg+mr2 =90 kg9.8 ms2+90 kg50 m0.443 rads2 =1764 N Page | 32 57. An airplane f lies in a loop (a circular path in a vert ical plane) of radius 150 m. The pilot's head always points toward t he center of the loop. The speed of t he airplane is not constant; t he airplane goes slowest at the top of the loop and fastest at the bottom. (a) At the top of the loop, the pilot feels weight less. What is the speed of the a irplane at this point? mg-(0 N)=mv2R g=v2R 9.8 ms2=v2150 m v=38.34 ms (b) At the bottom of the loop, the speed of t he airplane is 280 km/h. What is the apparent weight of t he pilot at this point? His t rue weight is 700 N. v=280kmhr1000 m1 km1 hr3600 s=77.78 ms m=700 N9.8 ms2=71.4 kg n=mg+mv2R =700 N+71.4 kg77.78 ms2150 m =3581 N 58. A 50.0-kg stunt pilot who has been diving her airplane vert ically pulls out of the dive by changing her course to a circle in a vert ical plane. (a) If t he plane's speed at the lowest point of the circle is 95.0 m /s, w hat is the m inimum radius of the circle for the acceleration at this point not to exceed 4.00g? arad=v2R 4.00 g 9.8 ms21 g=95.0 ms2R R=230 m (b) What is the apparent weight of the pilot at the lowest point of the pullout? n=mg+mv2R =50.0 kg9.8 ms2+50.0 kg95.0 ms2230 m =2450 N Page | 33 59. You t ie a cord to a pail of water, and you swing t he pail in a vert ical circle of radius 0.600 m . What m inim um speed must you give t he pail at t he highest point of the circle if no w ater is to spill from it? n=mg-mv2R 0 N=m9.8 ms2-v20.600 m v=2.42 ms 67. Block A weighs 1.20 N and block B weighs 3.60 N. The coefficient of kinetic friction between all surfaces is 0.300. Find the magnitude of the horizontal force F necessary to drag block B t o the left at constant speed (a) if A r ests on B and moves with it Fy=0 n=wA+wB Fx=0 F=fk F= k(wA+wB) F= k(wA+wB) =0.300(1.20 N+3.60 N) =1.44 N (b) if A is held at rest Block A: Fy=0 nA=wA=1.20 N Block B: Page | 34 nB=wA=1.20 N fkB= knB =(0.300)(1.20 N) =0.360 N F=fkB+fk =0.360 N+1.44 N =1.80 N 68. A window washer pushes his scrub b rush up a vert ical window at constant speed by applying a force F as shown in the figure. The brush weighs 12.0 N and t he coefficient of k inet ic fr ict ion is k=0.150. Calculate (a) t he magnitude of the force F Fx=0 n=Fcos53.1 fk= kn Page | 35 =(0.150)Fcos53.1 Fy=0 Fsin53.1=fk+w Fsin53.1-0.150Fcos53.1=12.0 N F=12.0 Nsin53.1-0.150cos53.1 =16.9 N (b) the normal force exerted by the window on the brush. n=Fcos53.1 =(16.9 N)cos53.1 =10.2 N 71. You are standing on a bathroom scale in an elevator in a tall building. Your mass is 72 kg. The elevator starts from rest and travels upward with a speed that varies with time according to vt=3.0ms2t+0.20ms3t2. When t = 4.0 s, what is the reading of the bathroom scale? at=v't=3.0ms2+0.40ms3t a4.0 s=3.0ms2+0.40ms3(4.0 s) =4.6 ms2 Fy=may fscale-w=ma fscale=72 kg9.8ms2+72 kg4.6ms2 =1037 N 72. You are designing an elevator for a hospital. The force exerted on a passenger by the floor of the elevator is not to exceed 1.60 times the passenger's weight. The elevator accelerates upward with constant acceleration for a distance of 3.0 m and then starts to slow down. What is the maximum speed of the elevator? Page | 36 n=1.60mg Fy=may n-mg=ma 1.60mg-mg=ma a=0.609.8 ms2 =5.88 ms2 vf2=vi2+2ax vf2=0 ms2+25.88 ms2(3.0 m) vf=5.94 ms 73. You are working for a shipping company. Your job is to stand at the bottom of a 8.0-m-Iong ramp that is inclined at 37 above the horizontal. You grab packages off a conveyor belt and propel them up the ramp. The coefficient of kinetic friction between the packages and the ramp is k=0.30. (a) What speed do you need to give a package at the bottom of the ramp so that it has zero speed at the top of the ramp? Fy=0 n=mgsin37 Fx=max -mgsin37-fk=ma -mgsin37-kmgcos37=ma a= -(9.8 ms2)sin37-0.309.8 ms2cos37 = -8.25 ms2 vf2=vi2+2ax 0 ms2=vi2+2-8.25 ms2(8.0 m) vi=11.5 ms Page | 37 (b) Your coworker is supposed to grab the packages as they arrive at the top of the ramp, but she misses one and it slides back down. What is its speed when it returns to you? Fx=max fk-mgsin37=ma kmgcos37-mgsin37=ma a=0.309.8 ms2cos37-(9.8 ms2)sin37 = -3.55 ms2 vf2=vi2+2ax vf2=0 ms2+2-3.55 ms2(-8.0 m) vi=7.54 ms 95. A curve with a 120-m radius on a level road is banked at the correct angle for a speed of 20 m/s. If an automobile rounds this curve at 30 m/s, what is the minimum coefficient of static friction needed between tires and road to prevent skidding? When v=20 ms: Fy=0 ncos=mg n=mgcos Fx=marad nsin=mv2R mgcossin=mv2R tan=v2Rg v2=Rgtan =tan-120 ms2120 m9.8 ms2 =18.8 When v=30 ms: Fy=0 ncos-snsin=mg Page | 38 n=mgcos-ssin Fx=marad nsin+sncos=mv2R nsin+sncos=30 ms20 ms2mRgtanR nsin+sncos=(2.25)mgtan n= (2.25)mgtansin+scos mgcos-ssin=(2.25)mgtansin+scos sin+scoscos-ssin=(2.25)tan sin18.8+scos18.8cos18.8-ssin18.8=(2.25)tan18.8 sin18.8+scos18.8=(2.25)tan18.8(cos18.8-ssin18.8) sin18.8+scos18.8=2.25sin18.8-2.25ssin18.8tan18.8 scos18.8+2.25ssin18.8tan18.8=1.25sin18.8 s=1.25sin18.8cos18.8+2.25sin18.8tan18.8 =0.337 (a) (b) Page | 39 (c) (d) (e) (f) Page | 40 ... View Full Document