Vector mechanics for...ics 10th e solutions - beervectorism c13p001p201, Notas de estudo de Mecânica

Baixe Vector mechanics for...ics 10th e solutions - beervectorism c13p001p201 e outras Notas de estudo em PDF para Mecânica, somente na Docsity! CHAPTER 13 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 509 PROBLEM 13.2 A 1-lb stone is dropped down the “bottomless pit” at Carlsbad Caverns and strikes the ground with a speed of 95ft/s. Neglecting air resistance, determine (a) the kinetic energy of the stone as it strikes the ground and the height h from which it was dropped, (b) Solve Part a assuming that the same stone is dropped down a hole on the moon. (Acceleration of gravity on the moon = 5.31 ft/s2.) SOLUTION Mass of stone: 2 2 lb 1 lb 0.031056 lb s /ft 32.2 ft/s W m g = = = ⋅ Initial kinetic energy: 1 0 (rest)T = (a) Kinetic energy at ground strike: 2 22 2 1 1 (0.031056)(95) 140.14 ft lb 2 2 T mv= = = ⋅ 2 140.1 ft lbT = ⋅  Use work and energy: 1 1 2 2T U T→+ = where 1 2 2 2 1 0 2 U wh mgh mgh mv → = = + = 2 2 2 (95) 2 (2)(32.2) v h g = = 140.1 fth =  (b) On the moon: 25.31 ft/sg = T1 and T2 will be the same, hence 2 140.1 ft lbT = ⋅  2 2 2 (95) 2 (2)(5.31) v h g = = 850 fth =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 510 PROBLEM 13.3 A baseball player hits a 5.1-oz baseball with an initial velocity of 140 ft/s at an angle of 40° with the horizontal as shown. Determine (a) the kinetic energy of the ball immediately after it is hit, (b) the kinetic energy of the ball when it reaches its maximum height, (c) the maximum height above the ground reached by the ball. SOLUTION Mass of baseball: 1 lb (5.1 oz) 0.31875 lb 16 oz W  = =    2 2 0.31875 lb 0.009899 lb s /ft 32.2 ft/s W m g = = = ⋅ (a) Kinetic energy immediately after hit. 0 140 ft/sv v= = 2 21 1 1 (0.009899)(140) 2 2 T mv= = 1 97.0 ft lbT = ⋅  (b) Kinetic energy at maximum height: 0 cos 40 140cos 40 107.246 ft/sv v= ° = ° = 2 22 1 1 (0.009899)(107.246) 2 2 T mv= = 2 56.9 ft lbT = ⋅  Principle of work and energy: 1 1 2 2T U T→+ = 1 2 2 1 40.082 ft lbU T T→ = − = − ⋅ Work of weight: 1 2U Wd→ = − Maximum height above impact point. 2 1 40.082 ft lb 125.7 ft 0.31875 lb T T d W − − ⋅= = = − − 125.7 ft  (c) Maximum height above ground: 125.7 ft 2 fth = + 127.7 fth =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 511 PROBLEM 13.4 A 500-kg communications satellite is in a circular geosynchronous orbit and completes one revolution about the earth in 23 h and 56 min at an altitude of 35800 km above the surface of the earth. Knowing that the radius of the earth is 6370 km, determine the kinetic energy of the satellite. SOLUTION Radius of earth: 6370 kmR = Radius of orbit: 66370 35800 42170 km 42.170 10 mr R h= + = + = = × Time one revolution: 23 h 56 mint = + 3(23 h)(3600 s/h) (56 min)(60 s/min) 86.160 10 st = + = × Speed: 6 3 2 2 (42.170 10 ) 3075.2 m/s 86.160 10 r v t π π ×= = = × Kinetic energy: 2 1 2 T mv= 2 9 1 (500 kg)(3075.2 m/s) 2.3643 10 J 2 T = = × 2.36 GJT =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 514 PROBLEM 13.7 Determine the maximum theoretical speed that may be achieved over a distance of 110 m by a car starting from rest assuming there is no slipping. The coefficient of static friction between the tires and pavement is 0.75, and 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) front-wheel drive, (b) rear-wheel drive. SOLUTION Let W be the weight and m the mass. W mg= (a) Front wheel drive: 0.60 0.60 0.75s N W mg μ = = = Maximum friction force without slipping: 1 2 2 1 2 2 (0.75)(0.60 ) 0.45 0.45 1 0, 2 sF N W mg U Fd mgd T T mv μ → = = = = = = = Principle of work and energy: 1 1 2 2T U T→+ = 2 2 2 2 2 2 2 1 0 0.45 2 (2)(0.45 ) (2)(0.45)(9.81 m/s )(110 m) 971.19 m /s mgd mv v gd + = = = = 2 31.164 m/sv = 2 112.2 km/hv =  (b) Rear wheel drive: 0.40 0.40 0.75s N W mg μ = = = Maximum friction force without slipping: 1 2 2 1 2 2 (0.75)(0.40 ) 0.30 0.30 1 0, 2 sF N W mg U Fd mgd T T mv μ → = = = = = = = Principle of work and energy: 1 1 2 2T U T→+ = 22 2 2 2 2 2 1 0 0.30 2 (2)(0.30) (2)(0.30)(9.81 m/s )(110 m) 647.46 m /s mgd mv v gd + = = = = 2 25.445 m/sv = 2 91.6 km/hv =  Note: The car is treated as a particle in this problem. The weight distribution is assumed to be the same for static and dynamic conditions. Compare with sample Problem 16.1 where the vehicle is treated as a rigid body. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 515 PROBLEM 13.8 Skid marks on a drag racetrack indicate that the rear (drive) wheels of a car slip for the first 20 m of the 400-m track. (a) Knowing that the coefficient of kinetic friction is 0.60, determine the speed of the car at the end of the first 20-m portion of the track if it starts from rest and the front wheels are just off the ground. (b) What is the maximum theoretical speed for the car at the finish line if, after skidding for 20 m, it is driven without the wheels slipping for the remainder of the race? Assume that while the car is rolling without slipping, 60 percent of the weight of the car is on the rear wheels and the coefficient of static friction is 0.75. Ignore air resistance and rolling resistance. SOLUTION (a) For the first 20 m, the normal force at the real wheels is equal to the weight of the car. Since the wheels are skidding, the friction force is k k kF N W mgμ μ μ= = = Principle of work and energy: 1 1 2 2T U T→+ = 2 2 2 2 2 2 2 2 2 1 0 2 1 0 2 2 (2)(0.6)(9.81 m/s )(20 m) 235.44 m /s k k Fd mv mgd mv v gd μ μ + = + = = = =   2 15.34 m/sv =  (b) Assume that for the remainder of the race, sliding is impending and N = 0.6 W (0.6 ) (0.75)(0.6 ) 0.45s sF N W mg mgμ μ= = = = Principle of work and energy: 2 2 3 3T U T→+ = 2 22 3 1 1 (0.45 ) 2 2 mv mg d mv′+ = 2 2 3 2 2 2 2 2 2 (2)(0.45) 235.44 m /s (2)(0.45)(9.81 m/s )(400 m 20 m) 3590.5 m /s v v gd ′= = + − = 3 59.9 m/sv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 516 PROBLEM 13.9 A package is projected up a 15° incline at A with an initial velocity of 8 m/s. Knowing that the coefficient of kinetic friction between the package and the incline is 0.12, determine (a) the maximum distance d that the package will move up the incline, (b) the velocity of the package as it returns to its original position. SOLUTION (a) Up the plane from A to B: 2 21 1 (8 m/s) 32 0 2 2 ( sin15 ) 0.12 N A A B A B k W W T mv T g g U W F d F Nμ− = = = = = − ° − = = 0 cos15 0 cos15F N W N WΣ = − ° = = ° (sin15 0.12cos15 ) (0.3747) : 32 (0.3743) 0 A B A A B B U W d Wd W T U T Wd g − − = − ° + ° = − + = − = 32 (9.81)(0.3747) d = 8.71 md =  (b) Down the plane from B to A: (F reverses direction) 2 2 1 0 8.71 m/s 2 ( sin15 ) (sin15 0.12cos15 )(8.70 m/s) 1.245 1 0 1.245 2 A A B B A B A B B A A A W T v T d g U W F d W U W W T U T W v g − − − = = = = ° − = ° − ° = + = + = 2 (2)(9.81)(1.245) 24.43 4.94 m/s A A v v = = = 4.94 m/sA =v 15°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 519 PROBLEM 13.12 Packages are thrown down an incline at A with a velocity of 1 m/s. The packages slide along the surface ABC to a conveyor belt which moves with a velocity of 2 m/s. Knowing that 7.5 md = and 0.25kμ = between the packages and all surfaces, determine (a) the speed of the package at C, (b) the distance a package will slide on the conveyor belt before it comes to rest relative to the belt. SOLUTION (a) On incline AB: cos 30 0.25 cos 30 sin 30 (sin 30 cos 30 ) AB AB k AB A B AB k N mg F N mg U mgd F d mgd μ μ → = ° = = ° = ° − = ° − ° On level surface BC: 7 m BC BC BC k B C k BC N mg x F mg U mg x μ μ→ = = = = − At A, 2 1 and 1 m/s 2A A A T mv v= = At C, 2 1 and 2 m/s 2C C C T mv v= = Assume that no energy is lost at the corner B. Work and energy. A A B B C CT U U T→ →+ + = 2 20 1 1 (sin 30 cos30 ) 2 2A k k BC mv mgd mg x mvμ μ+ ° − ° − = Solving for 2Cv , 2 2 2 2 (sin 30 cos30 ) 2 (1) (2)(9.81)(7.5)(sin 30 0.25cos30 ) (2)(0.25)(9.81)(7) C A k k BCv v gd g xμ μ= + ° − ° − = + ° − ° − 2 28.3811 m /s= 2.90 m/sCv =  (b) Box on belt: Let beltx be the distance moves by a package as it slides on the belt. 0y y x k k F ma N mg N mg F N mgμ μ Σ = − = = = = At the end of sliding, belt 2 m/sv v= = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 520 PROBLEM 13.12 (Continued) Principle of work and energy: 2 2 belt belt 2 2 belt belt 2 1 1 2 2 2 8.3811 (2) (2)(0.25)(9.81) C k C k mv mg x mv v v x g μ μ − = − = −= belt 0.893 mx =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 521 PROBLEM 13.13 Boxes are transported by a conveyor belt with a velocity v0 to a fixed incline at A where they slide and eventually fall off at B. Knowing that 0.40,kμ = determine the velocity of the conveyor belt if the boxes leave the incline at B with a velocity of 8 ft/s. SOLUTION Forces when box is on AB. 0: cos15 0 cos15 yF N W N W Σ = − ° = = ° Box is sliding on AB. cos15f k kF N Wμ μ= = ° Distance 20 ftAB d= = Work of gravity force: ( ) sin15A B gU Wd− = ° Work of friction force: cos15f kF d Wdμ− = − ° Total work (sin15 sin15 )A B kU Wd μ→ = ° − ° Kinetic energy: 2 0 2 1 2 1 2 A B B W T v g W T v g = = Principle of work and energy: A A B BT U T→+ = 2 20 1 1 (sin15 cos15 ) 2 2k B W W v Wd v g g μ+ ° − ° = 2 2 0 2 2 2 2 (sin15 cos15 ) (8) (2)(32.2)(20)[sin15 (0.40)(cos15 )] 228.29 ft /s B kv v gd μ= − ° − ° = − ° − ° = 0 15.11 ft/s=v 15°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 524 PROBLEM 13.16 A trailer truck enters a 2 percent uphill grade traveling at 72 km/h and reaches a speed of 108 km/h in 300 m. The cab has a mass of 1800 kg and the trailer 5400 kg. Determine (a) the average force at the wheels of the cab, (b) the average force in the coupling between the cab and the trailer. SOLUTION Initial speed: 1 72 km/h 20 m/sv = = Final speed: 2 108 km/h 30 m/sv = = Vertical rise: (0.02)(300) 6.00 mh = = Distance traveled: 300 md = (a) Traction force. Use cab and trailer as a free body. 1800 5400 7200 kgm = + = 3(7200)(9.81) 70.632 10 NW mg= = = × Work and energy: 1 1 2 2T U T→+ = 2 2 1 2 1 1 2 2t mv Wh F d mv− + = 2 2 2 3 21 1 1 1 1 1 1 (7200)(30) (70.632 10 )(6.00) (7200)(20) 2 300 2 2 tF mv Wh mvd    = + − = + × −      37.4126 10 N= × 7.41 kNtF =  (b) Coupling force .cF Use the trailer alone as a free body. 35400 kg (5400)(9.81) 52.974 10 Nm W mg= = = = × Assume that the tangential force at the trailer wheels is zero. Work and energy: 1 1 2 2T U T→+ = 2 2 1 2 1 1 2 2c mv Wh F d mv− + = The plus sign before cF means that we have assumed that the coupling is in tension. 2 2 2 3 22 1 1 1 1 1 1 1 (5400)(30) (52.974 10 )(6.00) (5400)(20) 2 2 300 2 2 cF mv Wh mvd    = + − = + × −      35.5595 10 N= × 5.56 kN (tension)cF =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 525 PROBLEM 13.17 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars B and C, causing them to slide on the track, but are not applied on the wheels of car A. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling. SOLUTION 0.35 (0.35)(100 kips) 35 kips (0.35)(80 kips) 28 kips k B C F F μ = = = = = 1 30 mi/h 44 ft/sv = = 2 20 0v T= = (a) Entire train: 1 1 2 2T U T−+ = 2 2 1 (80 kips 100 kips 80 kips) (44 ft/s) (28 kips 35 kips) 0 2 32.2 ft/s x + + − + =   124.07 ftx =  124.1 ftx =  (b) Force in each coupling: Recall that 124.07 ftx = Car A: Assume ABF to be in tension 1 1 2 2 21 80 kips (44) (124.07 ft) 0 2 32.2 19.38 kips AB AB T V T F F −+ = − = = + 19.38 kips (tension)ABF =  Car C: 1 1 2 2T U T−+ = 2 1 80 kips (44) ( 28 kips)(124.07 ft) 0 2 32.2 28 kips 19.38 kips BC BC F F + − = − = − 8.62 kipsBCF = + 8.62 kips (tension)BCF =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 526 PROBLEM 13.18 The subway train shown is traveling at a speed of 30 mi/h when the brakes are fully applied on the wheels of cars A, causing it to slide on the track, but are not applied on the wheels of cars A or B. Knowing that the coefficient of kinetic friction is 0.35 between the wheels and the track, determine (a) the distance required to bring the train to a stop, (b) the force in each coupling. SOLUTION (a) Entire train: 1 (0.35)(80 kips) 28 kips 30 mi/h 44 ft/s A AF N v μ= = = = = ← 2 20 0v T= = 1 1 2 2 2 2 1 (80 kips 100 kips 80 kips) (44 ft/s) (28 kips) 0 2 32.2 ft/s T V T x −+ = + + − = 279.1 ftx = 279 ftx =  (b) Force in each coupling: Car A: Assume ABF to be in tension 1 1 2 2T V T−+ = 2 2 1 80 kips (44 ft/s) (28 kips )(279.1 ft) 0 2 32.2 ft/s ABF− + = 28 kips 8.62 kipsABF+ = + 19.38 kipsABF = − 19.38 kips (compression)ABF =  Car C: 1 1 2 2T V T−+ = 2 2 1 80 kips (44 ft/s) (279.1 ft) 0 2 32.2 ft/s BCF+ = 8.617 kipsBCF = − 8.62 kips (compression)BCF =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 529 PROBLEM 13.20 The system shown is at rest when a constant 30 lb force is applied to collar B. (a) If the force acts through the entire motion, determine the speed of collar B as it strikes the support at C. (b) After what distance d should the 30 lb force be removed if the collar is to reach support C with zero velocity? SOLUTION Let F be the cable tension and vB be the velocity of collar B when it strikes the support. Consider the collar B. Its movement is horizontal so only horizontal forces acting on B do work. Let d be the distance through which the 30 lb applied force moves. 1 1 2 2 2 ( ) ( ) ( ) 1 18 0 30 (2 )(2) 2 32.2 B B B B T U T d F v →+ = + − = 230 4 0.27950 Bd F v− = (1) Now consider the weight A. When the collar moves 2 ft to the left, the weight moves 4 ft up, since the cable length is constant. Also, 2 .A Bv v= 1 1 2 2 2 2 ( ) ( ) ( ) 1 0 ( )(4) 2 1 6 4 (6)(4) (2 ) 2 32.2 A A B A A A B T U T W F W v g F v −+ = + − = − = 24 24 0.37267 BF v− = (2) Add Eqs. (1) and (2) to eliminate F. 230 24 0.65217 Bd v− = (3) (a) Case a: 2 ft, ?Bd v= = 2(30)(2) (24) 0.65217 Bv− = 2 2 255.2 ft /sBv = 7.43 ft/sBv =  (b) Case b: ?, 0.Bd v= = 30 24 0d − = 0.800 ftd =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 530 PROBLEM 13.21 Car B is towing car A at a constant speed of 10 m/s on an uphill grade when the brakes of car A are fully applied causing all four wheels to skid. The driver of car B does not change the throttle setting or change gears. The masses of the cars A and B are 1400 kg and 1200 kg, respectively, and the coefficient of kinetic friction is 0.8. Neglecting air resistance and rolling resistance, determine (a) the distance traveled by the cars before they come to a stop, (b) the tension in the cable. SOLUTION 0.8 AF N= Given: Car B tows car A at 10 m/s uphill. Car A brakes so 4 wheels skid. 0.8kμ = Car B continues in same gear and throttle setting. Find: (a) Distance d, traveled to stop (b) Tension in cable (a) 1F = traction force (from equilibrium) 1 (1400 )sin 5 (1200 )sin 5F g g= ° + ° 2600(9.81)sin 5= ° For system: A B+ 1 2 1[( 1400 sin 5 1200 sin 5 ) ]U F g g F d− = − ° − ° − 2 22 1 1 1 0 (2600)(10) 2 2A B T T m v+= − = − = − Since 1( 1400 sin 5 1200 sin 5 ) 0F g g− ° − ° = 0.8[1400(9.81)cos5 ] 130,000 N mFd d− = − ° = − ⋅ 11.88 md =  (b) Cable tension, T 1 2 2 1[ 0.8 ](11.88)AU T N T T− = − = − 21400( 0.8(1400)(9.81)cos5 )11.88 (10) 2 T − ° = − ( 10945) 5892T − = − 5.053 kN= 5.05 kNT =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 531 PROBLEM 13.22 The system shown is at rest when a constant 250-N force is applied to block A. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the horizontal surface, determine (a) the velocity of block B after block A has moved 2 m, (b) the tension in the cable. SOLUTION Constraint of cable: 3 constant 3 0 3 0 A B A B A B x y x y v v + = Δ + Δ = + = Let F be the tension in the cable. Block A: 130 kg, 250 N, ( ) 0A Am P T= = = 1 1 2 2 2 2 ( ) ( ) ( ) 1 0 ( )( ) 2 1 0 (250 )(2) (30)(3 ) 2 A A A A A A B T U T P F x m v F v →+ = + − Δ = + − = 2500 2 135 BF v− = (1) Block B: 25 kg, 245.25 NB B Bm W m g= = = 1 1 2 2 2 2 ( ) ( ) ( ) 1 0 (3 )( ) 2 2 1 (3 ) 245.25) (25) 3 2 B B B B B B B B T U T F W y m v F v →+ = + − −Δ =  − =    22 163.5 12.5 BF v− = (2) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 534 PROBLEM 13.23 (Continued) Block A: 1 1 1 2 2 30 kg, 250 N, ( ) 0. ( ) ( ) ( ) A A A A A m P T T U T→ = = = + = 2 2 1 0 ( )( ) 2 1 0 (250 58.86 )(2) (30)(3 ) 2 A A A A B P F F x m v F v + − − Δ = + − − = 2382.28 2 135 BF v− = (1) Block B: 25 kg, 245.25 NB B BM W m g= = = 1 1 2 2 2 2 ( ) ( ) ( ) 1 0 (3 )( ) 2 2 1 (3 245.25) (25) 3 2 B B B B B B B B T U T F W y m v F v →+ = + − −Δ =  − =    22 163.5 12.5 BF v− = (2) Add Eqs. (1) and (2) to eliminate F. 2 2 2 2 382.28 163.5 147.5 1.48325 m /s B B v v − = = (a) Velocity of B: 1.218 m/sB =v  (b) Tension in the cable: From Eq. (2), 2 163.5 (12.5)(1.48325)F − = 91.0 NF =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 535 PROBLEM 13.24 Two blocks A and B, of mass 4 kg and 5 kg, respectively, are connected by a cord which passes over pulleys as shown. A 3 kg collar C is placed on block A and the system is released from rest. After the blocks have moved 0.9 m, collar C is removed and blocks A and B continue to move. Determine the speed of block A just before it strikes the ground. SOLUTION Position to Position .  1 10 0v T= = At  before C is removed from the system 2 2 2 2 2 2 2 1 2 2 1 2 1 2 1 1 2 2 1 1 ( ) (12 kg) 6 2 2 ( ) (0.9 m) (4 3 5)( )(0.9 m) (2 kg)(9.81 m/s )(0.9 m) 17.658 J : A B C A C B T m m m v v v U m m m g U g U T U T − − − − = + + = = = + − = + − = = + = 2 22 20 17.658 6 2.943v v+ = = At Position , collar C is removed from the system. Position to Position .  22 2 1 9 ( ) kg (2.943) 13.244 J 2 2A B T m m v  ′ = + = =    2 23 3 3 1 9 ( )( ) 2 2A B T m m v v= + = 2 2 3 2 2 3 3 2 2 3 3 ( )( )(0.7 m) ( 1 kg)(9.81 m/s )(0.7 m) 6.867 J 13.244 6.867 4.5 1.417 A BU m m g T U T v v ′− − = − = − = − ′ + = − = = 3 1.190 m/sAv v= = 1.190 m/sAv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 536 PROBLEM 13.25 Four packages, each weighing 6 lb, are held in place by friction on a conveyor which is disengaged from its drive motor. When the system is released from rest, package 1 leaves the belt at A just as package 4 comes onto the inclined portion of the belt at B. Determine (a) the speed of package 2 as it leaves the belt at A, (b) the speed of package 3 as it leaves the belt at A. Neglect the mass of the belt and rollers. SOLUTION Slope angle: sin 6 ft 23.6 15 ft β β= = ° (a) Package falls off the belt and 2, 3, 4 move down 2 2 2 2 2 2 22 6 2 ft. 3 1 3 6 lb 3 0.2795 2 2 32.2 ft/s T mv v v =   = = =        1 2 (3)( )( ) (3)(6 lb)(2 ft) 36 lb ftU W R− = = = ⋅ 1 1 2 2 2 2 2 20 36 0.2795 128.8 T U T v v −+ = + = = 2 11.35 ft/s=v 23.6°  (b) Package 2 falls off the belt and its energy is lost to the system and 3 and 4 move down 2 ft.  2 2 2 2 2 2 2 2 3 3 3 32 1 6 lb (2) (128.8) 2 32 ft/s 24 lb ft 1 6 lb (2) ( ) 0.18634 2 32.2 ft/s T mv T T mv v v   ′ = =         ′ = ⋅   = = =          2 3 2 2 3 3 (2)( )(2) (2)(6 lb)(2 ft) 24 lb ftU W T U T − − = = = ⋅ + =   2 23 324 24 0.18634 257.6v v+ = =  3 16.05 ft/s=v 23.6°  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 539 PROBLEM 13.27 Solve Problem 13.26, assuming that the 2-kg block is attached to the spring. PROBLEM 13.26 A 3-kg block rests on top of a 2-kg block supported by, but not attached to, a spring of constant 40 N/m. The upper block is suddenly removed. Determine (a) the maximum speed reached by the 2-kg block, (b) the maximum height reached by the 2-kg block. SOLUTION Call blocks A and B. 2 kg, 3 kgA Bm m= = (a) Position 1: Block B has just been removed. Spring force: 1( )S A BF m m g kx= − + = − Spring stretch: 2 1 ( ) (5 kg)(9.81 m/s ) 1.22625 m 40 N/m A Bm m gx k + = − = − = − Let position 2 be a later position. Note that the spring remains attached to block A. Work of the force exerted by the spring: 2 1 2 1 1 2 2 2 2 1 2 2 2 2 2 2 ( ) 1 1 1 2 2 2 1 1 (40)( 1.22625) (40) 30.074 20 2 2 x e x x x U kxdx kx kx kx x x → = − = − = − = − − = −  Work of the gravitational force: 1 2 2 1 2 2 ( ) ( ) (2)(9.81)( 1.22625) 19.62 24.059 g AU m g x x x x → = − − = − + = − − Total work: 21 2 2 220 19.62 6.015U x x→ = − − + Kinetic energies: 1 2 2 2 2 2 2 2 0 1 1 (2) 2 2A T T m v v v = = = = Principle of work and energy: 1 1 2 2T U T→+ = 2 22 2 20 20 19.62 6.015x x v+ − + = Speed squared: 2 22 2 220 19.62 6.015v x x= − − + (1) At maximum speed, 2 2 0 dv dx = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 540 PROBLEM 13.27 (Continued) Differentiating Eq. (1) and setting equal to zero, 2 2 2 2 2 2 40 19.62 0 19.62 0.4905 m 40 dv v x dx x = − = − = = − = − Substituting into Eq. (1), 2 2 1 22 (20)( 0.4905) (19.62)( 0.4905) 6.015 10.827 m /sv = − − − − + = Maximum speed: 2 3.29 m/sv =  (b) Maximum height occurs when v2 = 0. Substituting into Eq. (1), 22 20 20 19.62 6.015x x= − − + Solving the quadratic equation 2 1.22625 m and 0.24525 mx = − Using the larger value, 2 0.24525 mx = Maximum height: 2 1 0.24525 1.22625h x x= − = + 1.472 mh =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 541 PROBLEM 13.28 An 8-lb collar C slides on a horizontal rod between springs A and B. If the collar is pushed to the right until spring B is compressed 2 in. and released, determine the distance through which the collar will travel assuming (a) no friction between the collar and the rod, (b) a coefficient of friction 0.35.kμ = SOLUTION (a) 144 lb/ft 216 lb/ft B A k k = = Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A. 2/12 0 0 2 2 0 0 144 lb/ft 2 216 lb/ft ft ( ) 2 12 2 A B y A B B A A B T T U k xdx k xdx U y − − = = = −     = −            2: 0 2 108 0 0.1361 ft 1.633 in. A A B BT U T y y −+ = + − = = = Total distance 2 16 (6 1.633)d = + − − 13.63 in.d =  (b) Assume that C does not reach the spring at B because of friction. 6 lb (0.35)(8 lb) 2.80 lb 0 f A D N W F T T = = = = = = 2/12 0 144 ( ) 2 2.80A D fU dx F y y− = × − = − 0 2 2.80 0 0.714 ft 8.57 in. A A D DT U T y y −+ = + − = = = The collar must travel 16 6 2 12 in.− + = before it engages the spring at B. Since 8.57 in.,y = it stops before engaging the spring at B. Total distance 8.57 in.d =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 544 PROBLEM 13.30 A 10-kg block is attached to spring A and connected to spring B by a cord and pulley. The block is held in the position shown with both springs unstretched when the support is removed and the block is released with no initial velocity. Knowing that the constant of each spring is 2 kN/m, determine (a) the velocity of the block after it has moved down 50 mm, (b) the maximum velocity achieved by the block. SOLUTION (a) W = weight of the block 10 (9.81) 98.1 N= = 1 2B A x x= 2 21 2 1 1 ( ) ( ) ( ) 2 2A A A B B U W x k x k x− = − − (Gravity) (Spring A) (Spring B) 21 2 1 (98.1 N)(0.05 m) (2000 N/m)(0.05 m) 2 U − = − 2 1 (2000 N/m)(0.025 m) 2 − 2 21 2 1 1 ( ) (10 kg) 2 2 U m v v− = = 2 1 4.905 2.5 0.625 (10) 2 v− − = 0.597 m/sv =  (b) Let distance moved down by the 10 kg blockx = 2 2 2 1 2 1 1 1 ( ) ( ) ( ) 2 2 2 2A B x U W x k x k m v−  = − − =    2 1 ( ) 0 ( ) (2 ) 2 8 B A d k m v W k x x dx   = = − −   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 545 PROBLEM 13.30 (Continued) 2000 0 98.1 2000 ( ) (2 ) 98.1 (2000 250) 8 x x x= − − = − + 0.0436 m (43.6 mm)x = For 2 1 0.0436, 4.2772 1.9010 0.4752 (10) 2 x U v= = − − = max 0.6166 m/sv = max 0.617 m/sv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 546 PROBLEM 13.31 A 5-kg collar A is at rest on top of, but not attached to, a spring with stiffness k1 = 400 N/m; when a constant 150-N force is applied to the cable. Knowing A has a speed of 1 m/s when the upper spring is compressed 75 mm, determine the spring stiffness k2. Ignore friction and the mass of the pulley. SOLUTION Use the method of work and energy applied to the collar A. 1 1 2 2T U T→+ = Since collar is initially at rest, 1 0.T = In position 2, where the upper spring is compressed 75 mm and 2 1.00 m/s,v = the kinetic energy is 2 22 2 1 1 (5 kg)(1.00 m/s) 2.5 J 2 2 T mv= = = As the collar is raised from level A to level B, the work of the weight force is 1 2( )gU mgh→ = − where 25 kg, 9.81 m/sm g= = and 450 mm 0.450 mh = = Thus, 1 2( ) (5)(9.81)(0.450) 22.0725 JgU → = − = − In position 1, the force exerted by the lower spring is equal to the weight of collar A. 1 (5 kg)(9.81 m/s) 49.05 NF mg= = − = − As the collar moves up a distance x1, the spring force is 1 1 2F F k x= − until the collar separates from the spring at 1 1 49.05 N 0.122625 m 122.625 mm 400 N/mf F x k = = = = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 549 PROBLEM 13.33 An uncontrolled automobile traveling at 65 mph strikes squarely a highway crash cushion of the type shown in which the automobile is brought to rest by successively crushing steel barrels. The magnitude F of the force required to crush the barrels is shown as a function of the distance x the automobile has moved into the cushion. Knowing that the weight of the automobile is 2250 lb and neglecting the effect of friction, determine (a) the distance the automobile will move into the cushion before it comes to rest, (b) the maximum deceleration of the automobile. SOLUTION (a) 65 mi/h 95.3 ft/s= Assume auto stops in 5 14 ft.d≤ ≤ 1 2 2 1 1 2 1 2 2 1 2 1 1 2 2 95.33 ft/s 1 1 2250 lb (95.3 ft/s) 2 2 32.2 ft/s 317,530 lb ft 317.63 k ft 0 0 (18 k)(5 ft) (27 k)( 5) 90 27 135 27 45 k ft 317.53 27 45 v T mv T v T U d d d T U T d − − =  =     = ⋅ = ⋅ = = = + − = + − = − ⋅ + = = − 13.43 ftd =  Assumption that 14 ftd ≤ is ok. (b) Maximum deceleration occurs when F is largest. For 13.43 ft, 27 k.d F= = Thus, DF ma= 2 2250 lb (27,000 lb) ( ) 32.2 ft/s Da  =     2386 ft/sDa =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 550 PROBLEM 13.34 Two types of energy-absorbing fenders designed to be used on a pier are statically loaded. The force-deflection curve for each type of fender is given in the graph. Determine the maximum deflection of each fender when a 90-ton ship moving at 1 mi/h strikes the fender and is brought to rest. SOLUTION Weight: 31 (90 ton)(2000 lb/ton) 180 10 lbW = = × Mass: 3 2180 10 5590 lb s /ft 32.2 W m g ×= = = ⋅ Speed: 1 5280 ft 1 mi/h 1.4667 ft/s 3600 s v = = = Kinetic energy: 2 21 1 2 1 1 (5590)(1.4667) 2 2 6012 ft lb 0 (rest) T mv T = = = ⋅ = Principle of work and energy: 1 1 2 2T U T→+ = 1 2 1 2 6012 0 6012 ft lb 72.15 kip in. U U → → + = = − ⋅ = − ⋅ The area under the force-deflection curve up to the maximum deflection is equal to 72.15 kip in.⋅ Fender A: From the force-deflection curve max max 60 5 kip/in. 12 F F kx k x = = = = 2 0 0 1 Area 2 x x fdx kx dx kx= = =  2 1 (5) 72.51 2 x = 2 228.86 in.x = 5.37 in.x =  Fender B: We divide area under curve B into trapezoids Partial area Total Area From 0x = to 2 in.:x = 1 (2 in.)(4 kips) 4 kip in. 2 = ⋅ 4 kip in.⋅ From 2 in.x = to 4 in.:x = 1 (2 in.)(4 10) 14 kip in. 2 + = ⋅ 18 kip in.⋅ From 4 in.x = to 6 in.:x = 1 (2 in.)(10 18) 28 kip in. 2 + = ⋅ 46 kip in.⋅ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 551 PROBLEM 13.34 (Continued) We still need 72.15 46 26.15 kip in.UΔ = − = ⋅ Equation of straight line approximating curve B from 6 in.x = to 8 in.x = is 18 18 6 2 30 18 x F F x Δ −= = + Δ − 2 1 18 (6 ) 26.15 kip in. 2 ( ) 6 8.716 0 1.209 in. U x x x x x x Δ = Δ + Δ Δ = ⋅ Δ + Δ − = Δ = Thus: 6 in. 1.209 in. 7.209 in.x = + = 7.21 in.x =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 554 PROBLEM 13.37 Express the acceleration of gravity hg at an altitude h above the surface of the earth in terms of the acceleration of gravity 0g at the surface of the earth, the altitude h and the radius R of the earth. Determine the percent error if the weight that an object has on the surface of earth is used as its weight at an altitude of (a) 1 km, (b) 1000 km. SOLUTION ( ) 2 2 2 / ( ) 1 E E h h R GM m GM m R F mg h R = = + + At earth’s surface, (h = 0) 02 EGM m mg R = 2 02 2 1 GME E R h h R GM g g R  +    = = Thus, 0 2 1 6370 km h h R g g R   +    = = At altitude h, “true” weight h TF mg W= = Assumed weight 0 0W mg= ( ) ( ) ( ) 0 0 0 0 0 0 0 0 2 0 22 0 Error 1 1 1 11 T h h g h R h hh RR W W mg mg g g E W mg g g g g E g − − −= = = = − +  −= = =   + +   (a) h = 1 km: ( )216370 1 1100 100 1 P E  −= =   +   P = 0.0314%  (b) h = 1000 km: ( )210006370 1 1100 100 1 P E  −= =   +   P = 25.3%  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 555 PROBLEM 13.38 A golf ball struck on earth rises to a maximum height of 60 m and hits the ground 230 m away. How high will the same golf ball travel on the moon if the magnitude and direction of its velocity are the same as they were on earth immediately after the ball was hit? Assume that the ball is hit and lands at the same elevation in both cases and that the effect of the atmosphere on the earth is neglected, so that the trajectory in both cases is a parabola. The acceleration of gravity on the moon is 0.165 times that on earth. SOLUTION Solve for .mh At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the same in both cases. 2 2 1 2 1 2 1 2 1 1 2 2 Earth U Moon H e e m m T mv T mv U mg h mg h − − = = = − = − Earth 2 2 1 1 2 2e e H mv mg h mv− = Moon 2 2 1 1 2 2m m H mv mg h mv− = Subtracting 0 (60 m) 0.165 m e e e m m e m e m e h g g h g h h g g h g − + = =   =     364 mmh =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 556 PROBLEM 13.39 The sphere at A is given a downward velocity 0v of magnitude 5 m/s and swings in a vertical plane at the end of a rope of length 2 ml = attached to a support at O. Determine the angle θ at which the rope will break, knowing that it can withstand a maximum tension equal to twice the weight of the sphere. SOLUTION 2 2 1 0 1 2 2 1 2 1 1 (5) 2 2 12.5 m 1 2 ( )sin T mv m T T mv U mg l θ− = = = = = 21 1 2 2 1 12.5 2 sin 2 T U T m mg mvθ−+ = + = 225 4 sing vθ+ = (1) Newton’s law at . 2 2 2 2 sin 2 4 2 sin v v mg mg m m v g g θ θ − = = = −  (2) Substitute for 2v from Eq. (2) into Eq. (1) 25 4 sin 4 2 sin (4)(9.81) 25 sin 0.2419 (6)(9.81) g g gθ θ θ + = − −= = 14.00θ = °  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 559 PROBLEM 13.41 (Continued) (b) Cord rotates about 2 L A R  =    2 2 (cos 60 ) C L mv T mg− ° = (2)( ) 2 1 mg m g T = + 1 5 2 2 2 T mg mg  = + =    2.5T W=  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 560 PROBLEM 13.42 A roller coaster starts from rest at A, rolls down the track to B, describes a circular loop of 40-ft diameter, and moves up and down past Point E. Knowing that h = 60 ft and assuming no energy loss due to friction, determine (a) the force exerted by his seat on a 160-lb rider at B and D, (b) the minimum value of the radius of curvature at E if the roller coaster is not to leave the track at that point. SOLUTION Let yp be the vertical distance from Point A to any Point P on the track. Let position 1 be at A and position 2 be at P. Apply the principle of work and energy. 1 0T = 2 2 1 2 P T mv= 1 2 PU mgy→ = 2 1 1 2 2 2 1 : 0 2 2 P P P P T U T mgy mv v gy →+ = + = = Magnitude of normal acceleration at P: 2 2 ( ) P PP n P P v gy a ρ ρ = = (a) Rider at Point B. 60 ft 20 ft (2 )(60) 6 20 B B n y h r g a g ρ = = = = = = :F maΣ = (6 ) 7 7 (7)(160 lb) B B N mg m g N mg W − = = = = 1120 lbB =N  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 561 PROBLEM 13.42 (Continued) Rider at Point D. 2 20 ft 20 ft (2 )(20) 2 20 D D n y h r g a g ρ = − = = = = :F maΣ = (2 ) 160 lb D D N mg m g N mg W + = = = = 160 lbD =N  (b) Car at Point E. 40 ft 0 E E y h r N = − = = : 2 n E E F ma gy mg m ρ Σ = = ⋅ 2E Eyρ = 80.0 ftρ =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 564 PROBLEM 13.44 A small block slides at a speed v on a horizontal surface. Knowing that h = 0.9 m, determine the required speed of the block if it is to leave the cylindrical surface BCD when θ = 30°. SOLUTION At Point C where the block leaves the surface BCD the contact force is reduced to zero. Apply Newton’s second law at Point C. n-direction: 2 cos Cn mv N mg ma h θ− = − = − With N = 0, we get 2 coscv gh θ= Apply the work-energy principle to the block sliding over the path BC. Let position 1 correspond to Point B and position 2 to C. 21 1 2 B T mv= 22 1 1 cos 2 2C T mv mgh θ= = 1 2 weight change in vertical distance (1 cos ) U mgh θ → = × = − 2 1 1 2 2 2 1 1 : (1 cos ) cos 2 2 cos 2 (1 cos ) (3cos 2) B B T U T mv mg mgh v gh gh gh θ θ θ θ θ →+ = + − = = − − = − Data: 2 2 2 2 9.81 m/s , 0.9 m, 30 . (9.81)(0.9)(3cos30 2) 5.2804 m /sB g h v θ= = = ° = ° − = 2.30 m/sBv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 565 PROBLEM 13.45 A small block slides at a speed 8v = ft/s on a horizontal surface at a height 3h = ft above the ground. Determine (a) the angleθ at which it will leave the cylindrical surface BCD, (b) the distance x at which it will hit the ground. Neglect friction and air resistance. SOLUTION Block leaves surface at C when the normal force 0.N = 2 cos cos n C mg ma v g h θ θ = = 2 cosCv gh gyθ= = (1) Work-energy principle. (a) 2 2 1 1 (8) 32 2 2B T mv m m= = = 21 ( ) ( ) 2C C B C C B B C C T mv U W h g mg h y T U T − − = = − = − + = 2 1 32 ( ) 2 C m mg h y mv+ − = Use Eq. (1) 1 32 ( ) 2C C g h y gy+ − = (2) ( )32 3 2 3 32 2 (32 ) (32 (32.2)(3)) (32.2) C C C gh gy gh y g y + = += += 2.6625 ftCy = (3) 2.6625 cos cos 0.8875 3 C C y y h h θ θ= = = = 27.4θ = °  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 566 PROBLEM 13.45 (Continued) (b) From (1) and (3) (32.2)(2.6625) 9.259 ft/s C C C v gy v v = = = At C: ( ) cos (9.259)(cos 27.4 ) 8.220 ft/sC x Cv v θ= = ° = ( ) sin (9.259)(sin 27.4 ) 4.261 ft/sC y Cv v θ= − = − ° = 2 2 1 ( ) 2.6625 4.261 16.1 2C C y y y v t gt t t= + − = − − At E: 20: 0.2647 0.1654 0Ey t t= + − = 0.2953 st = At E: (sin ) ( ) (3)(sin 27.4 ) (8.220)(0.2953)C xx h v tθ= + = ° + 1.381 2.427 3.808 ftx = + = 3.81 ftx =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 569 PROBLEM 13.48 The velocity of the lift of Problem 13.47 increases uniformly from zero to its maximum value at mid-height 7.5 s and then decreases uniformly to zero in 7.5 s. Knowing that the peak power output of the hydraulic pump is 6 kW when the velocity is maximum, determine the maximum life force provided by the pump. PROBLEM 13.47 It takes 15 s to raise a 1200-kg car and the supporting 300-kg hydraulic car-lift platform to a height of 2.8 m. Determine (a) the average output power delivered by the hydraulic pump to lift the system, (b) the average power electric required, knowing that the overall conversion efficiency from electric to mechanical power for the system is 82 percent. SOLUTION Newton’s law ( ) (1200 300) 1500 C LMg M M g g Mg g = + = + = 1500 1500F F g aΣ = − = (1) Since motion is uniformly accelerated, a = constant Thus, from (1), F is constant and peak power occurs when the velocity is a maximum at 7.5 s. max 7.5 s v a = max max (6000 W) ( )( ) (6000)/ P F v v F = = = Thus, (6000)/(7.5)( )a F= (2) Substitute (2) into (1) 1500 (1500)(6000)/(7.5)( )F g F− = 2 2 (1500 kg)(6000 N m/s) (1500 kg)(9.81 m/s ) 0 (7.5 s) F F ⋅− − = 2 614,715 1.2 10 0 14,800 N F F F − − × = = 14.8 kNF =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 570 PROBLEM 13.49 (a) A 120-lb woman rides a 15-lb bicycle up a 3-percent slope at a constant speed of 5 ft/s. How much power must be developed by the woman? (b) A 180-lb man on an 18-lb bicycle starts down the same slope and maintains a constant speed of 20 ft/s by braking. How much power is dissipated by the brakes? Ignore air resistance and rolling resistance. SOLUTION 3 tan 1.718 100 θ θ= = ° (a) 15 120B WW W W= + = + 135 lbW = ( sin )( )WP W vθ= ⋅ =W v (135)(sin 1.718 )(5)WP = ° (a) 20.24 ft lb/sWP = ⋅ 20.2 ft lb/sWP = ⋅  (b) 18 180B mW W W= + = + 198 lbW = Brakes must dissipate the power generated by the bike and the man going down the slope at 20 ft/s. ( sin )( )BP W vθ= ⋅ =W v (198)(sin 1.718)(20)BP = 118.7 ft lb/sBP = ⋅      (b) PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 571 PROBLEM 13.50 A power specification formula is to be derived for electric motors which drive conveyor belts moving solid material at different rates to different heights and distances. Denoting the efficiency of the motors by η and neglecting the power needed to drive the belt itself, derive a formula (a) in the SI system of units for the power P in kW, in terms of the mass flow rate m in kg/h, the height b and horizontal distance l in meters, and (b) in U.S. customary units, for the power in hp, in terms of the material flow rate w in tons/h, and the height b and horizontal distance l in feet. SOLUTION (a) Material is lifted to a height b at a rate, 2( kg/h)( m/s ) [ (N/h)]m g mg= Thus, [ (N/h)][ ( )] N m/s (3600 s/h) 3600 U mg b m mgb t Δ  = = ⋅ Δ   1000 N m/s 1 kW⋅ = Thus, including motor efficiency, η (N m/s) (kw) 1000 N m/s (3600) ( ) kW mgb P η ⋅= ⋅      6(kW) 0.278 10 mgb P η −= ×  (b) [ (tons/h)(2000 lb/ton)][ (ft)] 3600 s/h U W b t Δ = Δ ft lb/s; 1hp 550 ft lb/s 1.8 Wb= ⋅ = ⋅ With ,η 1 hp 1(ft lb/s) 1.8 550 ft lb/s Wb hp η     = ⋅      ⋅     31.010 10 Wb hp η −×=  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 574 PROBLEM 13.52 The frictional resistance of a ship is known to vary directly as the 1.75 power of the speed v of the ship. A single tugboat at full power can tow the ship at a constant speed of 4.5 km/h by exerting a constant force of 300 kN. Determine (a) the power developed by the tugboat, (b) the maximum speed at which two tugboats, capable of delivering the same power, can tow the ship. SOLUTION (a) Power developed by tugboat at 4.5 km/h. 0 0 4.5 km/h 1.25 m/s 300 kN v F = = = 0 0 0 (300 kN)(1.25 m/s)P F v= = 0 375 kWP =  (b) Maximum speed. Power required to tow ship at speed v: 1.75 1.75 2.75 0 0 0 0 0 0 0 v v v F F P Fv F v F v v v v       = = = =            (1) Since we have two tugboats, the available power is twice maximum power 0 0F v developed by one tugboat. 2.75 0 0 0 0 0 2.75 1/2.75 0 0 0 2 2 (2) (1.2867) v F v F v v v v v v v   =       = = =    Recalling that 0 4.5 km/hv = (4.5 km/h)(1.2867) 5.7902 km/hv = = 5.79 km/hv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 575 PROBLEM 13.53 A train of total mass equal to 500 Mg starts from rest and accelerates uniformly to a speed of 90 km/h in 50 s. After reaching this speed, the train travels with a constant velocity. The track is horizontal and axle friction and rolling resistance result in a total force of 15 kN in a direction opposite to the direction of motion. Determine the power required as a function of time. SOLUTION Let FP be the driving force and FR be the resisting force due to axle friction and rolling resistance. Uniformly accelerated motion. ( 50 s):t < 0v v at= + 0 0v = At 50 ,t s= 90 km/h 25 m/sv = = 2 2 25 m/s 0 (50) 0.5 m/s (0.5 m/s ) a a v t = + = = Newton’s second law: P RF F ma− = where 3 3 2 15 kN 15 10 N 500 Mg 500 10 kg 0.5 m/s RF m a = = × = = × = 3 3 3 15 10 (500 10 )(0.5) 265 10 N 265 kN P RF F ma= + = × + × = × = Power: 3(265 10 )(0.5 )PF v t= × (0 50s)< Power (132.5 kW/s)t=  Uniform motion. ( 50 s):t > 0a = 315 10 N; 25 m/sP RF F v= = × = Power: 3 3(15 10 )(25 m/s) 375 10PF v W= × = × ( 50 s)t > Power 375 kW=  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 576 PROBLEM 13.54 The elevator E has a weight of 6600 lbs when fully loaded and is connected as shown to a counterweight W of weight of 2200 lb. Determine the power in hp delivered by the motor (a) when the elevator is moving down at a constant speed of 1 ft/s, (b) when it has an upward velocity of 1 ft/s and a deceleration of 20.18 ft/s . SOLUTION (a) Acceleration 0= Counterweight Elevator Motor 0: 0y W WF T WΣ = − = 0: 2 6600 0C WF T TΣ = + − = 2200 lbWT = 2200 lbCT = Kinematics: 2 , 2 , 2 2 ft/sE C E C C Ex x x x v v= = = =  (2200 lb)(2 ft/s) 4400 lb ft/s 8.00 hpC CP T v= ⋅ = = ⋅ = 8.00 hpP =  (b) 20.18 ft/sEa = , 1 ft/sEv = Counterweight Elevator Counterweight: : ( )W W W F Ma T W a g Σ = − = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 579 PROBLEM 13.CQ3 Block A is released from rest and slides down the frictionless ramp to the loop. The maximum height h of the loop is the same as the initial height of the block. Will A make it completely around the loop without losing contact with the track? (a) Yes (b) No (c) need more information SOLUTION Answer: (b) In order for A to not maintain contact with the track, the normal force must remain greater than zero, which requires a non-zero speed at the top of the loop. PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 580 PROBLEM 13.55 A force P is slowly applied to a plate that is attached to two springs and causes a deflection 0.x In each of the two cases shown, derive an expression for the constant ,ek in terms of 1k and 2 ,k of the single spring equivalent to the given system, that is, of the single spring which will undergo the same deflection 0x when subjected to the same force P. SOLUTION System is in equilibrium in deflected 0x position. Case (a) Force in both springs is the same P= 0 1 2 0 e x x x P x k = + = 1 1 P x k = 2 2 = P x k Thus, 1 2e P P P k k k = + 1 2 1 1 1 ek k k = + 1 2 1 2 e k k k k k = +  Case (b) Deflection in both springs is the same 0x= 1 0 2 0 1 2 0 0 ( ) e P k x k x P k k x P k x = + = + = Equating the two expressions for 1 2 0 0( ) eP k k x k x= + = 1 2ek k k= +  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 581 PROBLEM 13.56 A loaded railroad car of mass m is rolling at a constant velocity v0 when it couples with a massless bumper system. Determine the maximum deflection of the bumper assuming the two springs are (a) in series (as shown), (b) in parallel. SOLUTION Let position A be at the beginning of contact and position B be at maximum deflection. 2 0 2 2 1 1 2 2 1 2 0 (zero force in springs) 0 ( 0 at maximum deflection) 1 1 2 2 A A B B T mv V T v V k x k x = = = = = + where x1 is deflection of spring k1 and x2 is that of spring k2. Conservation of energy: A A B BT V T V+ = + 2 2 20 1 1 2 2 1 1 1 0 0 2 2 2 mv k x k x+ = + + 2 2 21 1 2 2 0k x k x mv+ = (1) (a) Springs are in series. Let F be the force carried by the two springs. Then, 1 2 1 2 and F F x x k k = = Eq. (1) becomes 2 20 1 2 1 1 F mv k k   + =    so that 0 1 2 1 1 /F v m k k   = +    The maximum deflection is 1 2 1 2 0 1 2 1 2 1 1 1 1 1 1 / x x F k k v m k k k k δ   = + = +        = + +        0 1 2 1 1 v m k k   = +    0 1 2 1 2( )/v m k k k kδ = +  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 584 PROBLEM 13.57 (Continued) (a) Speed of collar at B. Conservation of energy: A A B BT V T V+ = + 20 6.5767 0.300 1.2249Bv+ = + 2 2 217.839 m /sBv = 4.22 m/sBv =  (b) Speed of collar at D. Conservation of energy: A A D DT V T V+ = + 200 6.5767 0.300 0.7042v+ = + 2 2 219.575 m /sDv = 4.42 m/sDv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 585 PROBLEM 13.58 A 3-lb collar is attached to a spring and slides without friction along a circular rod in a horizontal plane. The spring has an undeformed length of 7 in. and a constant 1.5 lb/in.k = Knowing that the collar is in equilibrium at A and is given a slight push to get it moving, determine the velocity of the collar (a) as it passes through B, (b) as it passes through C. SOLUTION 0 7 in., 20 in.DAL L= = 2 2(8 6) 6 15.23 in.DBL = + + = 8 in.DCL = 20 7 13 in.DALΔ = − = 15.23 7 8.23 in.DBLΔ = − =  8 7 1 in.DCLΔ = − =  (a) 2 2 1 1 0, ( ) (1.5)(13) 126.75 lb in. 2 2A A DA T V k L= = Δ = = ⋅ 10.5625 lb ft= ⋅ 2 21 1.5 2B B B T mv v g = = 21 (1.5)(8.23) 50.8 lb in. 4.233 lb ft 2B V = = ⋅ = ⋅ 21.5 : 0 10.5625 4.233 32.2 B A A B B v T V T V+ = + + = + 11.66 ft/sBv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 586 PROBLEM 13.58 (Continued) (b) 2 1.5 0, 10.5625 lb ft, 32.2A A C C T V T v= = ⋅ = 2 1 (1.5)(1) 0.75 lb in. 0.0625 lb ft 2C V = = ⋅ = ⋅ 2 1.5 : 0 10.5625 0.0625 32.2A A C C c T V T V v+ = + + = +  15.01 ft/sCv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 589 PROBLEM 13.60 A 500-g collar can slide without friction on the curved rod BC in a horizontal plane. Knowing that the undeformed length of the spring is 80 mm and that 400k = kN/m, determine (a) the velocity that the collar should be given at A to reach B with zero velocity, (b) the velocity of the collar when it eventually reaches C. SOLUTION (a) Velocity at A: 2 2 2 2 3 2 2 3 2 1 0.5 kg 2 2 (0.25) 0.150 m 0.080 m 0.070 m 1 ( ) 2 1 (400 10 N/m)(0.070 m) 2 980 J 0 0 0.200 m 0.080 m 0.120 m 1 1 ( ) (400 10 N/m)(0.120 m) 2 2 28 A A A A A A A A A A A B B B B B B T mv v T v L L V k L V V v T L V k L V  = =     = Δ = − Δ = = Δ = × = = = Δ = − = = Δ = × = 80 J Substitute into conservation of energy. 2 2 2 2 2 0.25 980 0 2880 (2880 980) (0.25) 7600 m /s A A B B A A A T V T V v v v + = + + = + −= = 87.2 m/sAv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 590 PROBLEM 13.60 (Continued) (b) Velocity at C: Since slope at B is positive, the component of the spring force ,PF parallel to the rod, causes the block to move back toward A. 2 2 2 2 3 2 0, 2880 J [from part ( )] 1 (0.5 kg) 0.25 2 2 0.100 m 0.080 m 0.020 m 1 1 ( ) (400 10 N/m)(0.020 m) 80.0 J 2 2 B B C C C C C C C T V a T mv v v L V k L = = = = = Δ = − = = Δ = × = Substitute into conservation of energy. 2 2 2 2 0 2880 0.25 80.0 11,200 m /s B B C C C C T V T V v v + = + + = + = 105.8 m/sCv =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 591 PROBLEM 13.61 An elastic cord is stretched between two Points A and B, located 800 mm apart in the same horizontal plane. When stretched directly between A and B, the tension is 40 N. The cord is then stretched as shown until its midpoint C has moved through 300 mm to ;C′ a force of 240 N is required to hold the cord at .C′ A 0.1 kg pellet is placed at ,C′ and the cord is released. Determine the speed of the pellet as it passes through C. SOLUTION Let  = undeformed length of cord. Position 1. 1Length 1.0 m; Elongation 1.0AC B x′ = = = −  1 1 3 0: 2 240 N 0 200 N 5x F F F  Σ = − = =    Position 2. 2 2 Length 0.8 m; Elongation 0.8 Given 40 N ACB x F = = = − =  1 2 2,F kx F k x= = 1 2 1 2 ( ) 200 40 [(1.0 ) (0.8 )] 0.2 F F k x x k k − = − − = − − − =  160 800 N/m 0.2 k = = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 594 PROBLEM 13.62 (Continued) d = distance from the ground 1 1 2 2T V T V+ = + 2 2 0 (186)(130 ) 0 (266.67)(45 ) 266.7 23815 515827 0 d d d d + − = + − − + = 2 36.99 ft23815 (23815) 4(266.7)(515827) 52.3 ft(2)(266.7) d − = =  Discard 52.3 ft (since the cord acts in compression when rebound occurs). 37.0 ftd =  PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 595 PROBLEM 13.63 It is shown in mechanics of materials that the stiffness of an elastic cable is k = AE/L where A is the cross sectional area of the cable, E is the modulus of elasticity and L is the length of the cable. A winch is lowering a 4000-lb piece of machinery using at a constant speed of 3ft/s when the winch suddenly stops. Knowing that the steel cable has a diameter of 0.4 in., E = 29 × 106 lb/in2, and when the winch stops L = 30 ft, determine the maximum downward displacement of the piece of machinery from the point it was when the winch stopped. SOLUTION Mass of machinery: 2 4000 124.22 lb s /ft 32.2 W m g = = = ⋅ Let position 1 be the state just before the winch stops and the gravitational potential Vg be equal to zero at this state. For the cable, 2 2 2(diameter) (0.4 in.) 0.12566 in 4 4 A π π= = = 2 6 2 6(0.12556 in. )(29 10 lb/in ) 3.6442 10 lbAE = × = × For 30 ft,L = 6 33.6442 10 lb 121.47 10 lb/ft 30 ft AE k L ×= = = × Initial force in cable (equilibrium): 1 4000 lb.F W= = Elongation in position 1: 11 3 4000 0.03293 ft 121.47 10 F x k = = = × Potential energy: 2 2 1 1 1 1 2 2 F V kx k = = 2 1 3 (4000 lb) 65.860 ft lb (2)(121.47 10 lb/ft) V = = ⋅ × Kinetic energy: 21 1 1 2 T mv= 2 21 1 (124.22 lb s /ft)(3 ft/s ) 558.99 ft lb 2 T = ⋅ = ⋅ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 596 PROBLEM 13.63 (Continued) Let position 2 be the position of maximum downward displacement. Let x2 be the elongation in this position. Potential energy: 22 2 2 1 1 ( ) 2 V kx W x x= − − 3 2 2 2 2 3 2 2 1 (121.47 10 ) (4000)( 0.03293) 2 60.735 10 4000 131.72 V x x x x = × − − = × − + Kinetic energy: 2 20 (since 0)T v= = Principle of work and energy: 1 1 2 2T V T V+ = + 3 2 2 2 3 2 2 2 2 558.99 65.860 60.735 10 4000 131.72 60.735 10 4000 493.13 0 0.12887 ft x x x x x + = × − + × − − = = Maximum displacement: 2 1 0.09594 ftx xδ = − = 1.151 in.δ = PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 599 PROBLEM 13.65 A 1-kg collar can slide along the rod shown. It is attached to an elastic cord anchored at F, which has an undeformed length of 250 mm and a spring constant of 75 N/m. Knowing that the collar is released from rest at A and neglecting friction, determine the speed of the collar (a) at B, (b) at E. SOLUTION 2 2 2 2 2 2 2 (0.5) (0.4) (0.3) 0.70711 m (0.4) (0.3) 0.5 m (0.5) (0.3) 0.58309 m AF AF BF BF FE FE e g L L L L L L V V V = + + = = + = = + = = + (a) Speed at B: 0, 0A Av T= = Point A: 2 0 1 ( ) ( ) 0.70711 0.25 2A e AF AF AF V k L L L L= Δ Δ = − = − 0.45711 mAFLΔ = 2 2 1 ( ) (75 N/m)(0.45711 m) 2 ( ) 7.8355 N m ( ) ( )(0.4) (1.0 kg)(9.81 m/s )(0.4 m) 3.9240 N m ( ) ( ) 7.8355 3.9240 11.7595 N m A e A e A g A A e A g V V V mg V V V = = ⋅ = = = ⋅ = + = + = ⋅ PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 600 PROBLEM 13.65 (Continued) Point B: 2 2 2 1 1 (1.0 kg) 2 2 0.5 B B B B B T mv v T v = = = 2 0 1 ( ) ( ) 0.5 0.25 2B e BF BF BF V k L L L L= Δ Δ = − = − 0.25 mBFLΔ = 2 2 1 ( ) (75 N/m)(0.25 m) 2.3438 N m 2 ( ) ( )(0.4) (1.0 kg)(9.81 m/s )(0.4 m) 3.9240 N m ( ) ( ) 2.3438 3.9240 6.2678 N m B e B g B B e B g V V mg V v V = = ⋅ = = = ⋅ = + = + = ⋅ 2 2 2 2 2 0 11.7595 0.5 6.2678 (5.49169)/(0.5) 10.983 m /s A A B B B B B T V T V v v v + = + + = + = = 3.31 m/sBv =  (b) Speed at E: Point A: 0 11.7595 N mA AT V= = ⋅ (from part (a)) Point E: 2 2 2 2 0 1 1 (1.0 kg) 0.5 2 2 1 ( ) ( ) 0.5831 0.25 2 E E E E E e FE FE FE T mv v v V k L L L L = = = = Δ Δ = − = − 0.3331 mFELΔ = 2 2 2 2 2 2 1 ( ) (75 N/m)(0.3331 m) 4.1607 N m 2 ( ) 0 4.1607 N m 0 11.7595 0.5 4.1607 7.5988/0.5 15.1976 m /s E e E g E A A E E E E E V V V T V T V v v v = = ⋅ = = ⋅ + = + + = + = = 3.90 m/sEv =   PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. 601 PROBLEM 13.66 A thin circular rod is supported in a vertical plane by a bracket at A. Attached to the bracket and loosely wound around the rod is a spring of constant 3 lb/ftk = and undeformed length equal to the arc of circle AB. An 8-oz collar C, not attached to the spring, can slide without friction along the rod. Knowing that the collar is released from rest at an angle θ with the vertical, determine (a) the smallest value of θ for which the collar will pass through D and reach Point A, (b) the velocity of the collar as it reaches Point A. SOLUTION (a) Smallest angle θ occurs when the velocity at D is close to zero. 0 0 0 0 C D C D e g v v T T V V V = = = = = + Point C: 2 2 (1 ft)( ) ft 1 ( ) ( ) 2 3 ( ) 2 BC C e BC C e L V k L V θ θ θ Δ = = = Δ = 12 in. 1 ftR = = ( ) (1 cos ) 8 oz ( ) (1 ft)(1 cos ) 16 oz/lb C g C g V WR V θ θ = −  = −    2 1 ( ) (1 cos ) 2 3 1 ( ) ( ) (1 cos ) 2 2 C g C C e C g V V V V θ θ θ = − = + = + − Point D: 2 ( ) 0 (spring is unattached) ( ) (2 ) (2)(0.5 lb)(1 ft) 1 lb ft 3 1 0 (1 cos ) 1 2 2 D e D g C C D D V V W R T V T V θ θ = = = = ⋅ + = + + + − = 2(1.5) (0.5)cos 0.5θ θ− = By trial, 0.7592 radθ = 43.5θ = ° 