SOLUTION MANUAL CHAPTER 11 PROBLEM 11.CQ1 A bus travels the 100 miles between A and B at 50 mi/h and then another 100 miles between B and C at 70 mi/h. The average speed of the bus for the entire 200-mile trip is: (a) more than 60 mi/h (b) equal to 60 mi/h (c) less than 60 mi/h SOLUTION The time required for the bus to travel from A to B is 2 h and from B to C is 100/70 = 1.43 h, so the total time is 3.43 h and the average speed is 200/3.43 = 58 mph. Answer: (c) 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. 3 PROBLEM 11CQ2 Two cars A and B race each other down a straight road. The position of each car as a function of time is shown. Which of the following statements are true (more than one answer can be correct)? (a) At time t2 both cars have traveled the same distance (b) At time t1 both cars have the same speed (c) Both cars have the same speed at some time t < t1 (d) Both cars have the same acceleration at some time t < t1 (e) Both cars have the same acceleration at some time t1 < t < t2 SOLUTION The speed is the slope of the curve, so answer c) is true. The acceleration is the second derivative of the position. Since A’s position increases linearly the second derivative will always be zero. The second derivative of curve B is zero at the pont of inflection which occurs between t1 and t2. Answers: (c) and (e) 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. 4 PROBLEM 11.1 The motion of a particle is defined by the relation x = t 4 − 10t 2 + 8t + 12 , where x and t are expressed in inches and seconds, respectively. Determine the position, the velocity, and the acceleration of the particle when t = 1 s. SOLUTION x = t 4 − 10t 2 + 8t + 12 At t = 1 s, v= dx = 4t 3 − 20t + 8 dt a= dv = 12t 2 − 20 dt x = 1 − 10 + 8 + 12 = 11 x = 11.00 in. v = 4 − 20 + 8 = −8 v = −8.00 in./s a = 12 − 20 = −8 a = −8.00 in./s 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. 5 PROBLEM 11.2 The motion of a particle is defined by the relation x = 2t 3 − 9t 2 + 12t + 10, where x and t are expressed in feet and seconds, respectively. Determine the time, the position, and the acceleration of the particle when v = 0. SOLUTION x = 2t 3 − 9t 2 + 12t + 10 Differentiating, v= a= dx = 6t 2 − 18t + 12 = 6(t 2 − 3t + 2) dt = 6(t − 2)(t − 1) dv = 12t − 18 dt So v = 0 at t = 1 s and t = 2 s. At t = 1 s, x1 = 2 − 9 + 12 + 10 = 15 a1 = 12 − 18 = −6 t = 1.000 s x1 = 15.00 ft a1 = −6.00 ft/s 2 At t = 2 s, x2 = 2(2)3 − 9(2) 2 + 12(2) + 10 = 14 t = 2.00 s x2 = 14.00 ft a2 = (12)(2) − 18 = 6 a2 = 6.00 ft/s 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. 6 PROBLEM 11.3 The vertical motion of mass A is defined by the relation x = 10 sin 2t + 15cos 2t + 100, where x and t are expressed in mm and seconds, respectively. Determine (a) the position, velocity and acceleration of A when t = 1 s, (b) the maximum velocity and acceleration of A. SOLUTION x = 10sin 2t + 15cos 2t + 100 v= dx = 20 cos 2t − 30sin 2t dt a= dv = −40sin 2t − 60 cos 2t dt For trigonometric functions set calculator to radians: (a) At t = 1 s. x1 = 10sin 2 + 15cos 2 + 100 = 102.9 v1 = 20cos 2 − 30sin 2 = −35.6 a1 = −40sin 2 − 60 cos 2 = −11.40 x1 = 102.9 mm v1 = −35.6 mm/s a1 = −11.40 mm/s 2 (b) Maximum velocity occurs when a = 0. −40sin 2t − 60cos 2t = 0 tan 2t = − 60 = −1.5 40 2t = tan −1 (−1.5) = −0.9828 and −0.9828 + π Reject the negative value. 2t = 2.1588 t = 1.0794 s t = 1.0794 s for vmax so vmax = 20cos(2.1588) − 30sin(2.1588) vmax = −36.1 mm/s = −36.056 Note that we could have also used vmax = 202 + 302 = 36.056 by combining the sine and cosine terms. For amax we can take the derivative and set equal to zero or just combine the sine and cosine terms. amax = 402 + 602 = 72.1 mm/s 2 amax = 72.1 mm/s 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. 7 PROBLEM 11.4 A loaded railroad car is rolling at a constant velocity when it couples with a spring and dashpot bumper system. After the coupling, the motion of the car is defined by the relation x = 60e−4.8t sin16t where x and t are expressed in mm and seconds, respectively. Determine the position, the velocity and the acceleration of the railroad car when (a) t = 0, (b) t = 0.3 s. SOLUTION x = 60e−4.8t sin16t dx = 60(−4.8)e −4.8t sin16t + 60(16)e−4.8t cos16t dt v = −288e−4.8t sin16t + 960e −4.8t cos16t v= a= dv = 1382.4e−4.8t sin16t − 4608e−4.8t cos16t dt − 4608e−4.8t cos16t − 15360e−4.8t sin16t a = −13977.6e −4.8t sin16t − 9216e−4.8 cos16t (a) At t = 0, x0 = 0 x0 = 0 mm v0 = 960 mm/s v0 = 960 mm/s a0 = −9216 mm/s 2 (b) At t = 0.3 s, a0 = 9220 mm/s 2 e−4.8t = e −1.44 = 0.23692 sin16t = sin 4.8 = −0.99616 cos16t = cos 4.8 = 0.08750 x0.3 = (60)(0.23692)(−0.99616) = −14.16 x0.3 = 14.16 mm v0.3 = 87.9 mm/s v0.3 = −(288)(0.23692)(−0.99616) + (960)(0.23692)(0.08750) = 87.9 a0.3 = −(13977.6)(0.23692)( −0.99616) − (9216)(0.23692)(0.08750) = 3108 a0.3 = 3110 mm/s 2 or 3.11 m/s 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. 8 PROBLEM 11.5 The motion of a particle is defined by the relation x = 6t 4 − 2t 3 − 12t 2 + 3t + 3, where x and t are expressed in meters and seconds, respectively. Determine the time, the position, and the velocity when a = 0. SOLUTION We have x = 6t 4 − 2t 3 − 12t 2 + 3t + 3 Then v= dx = 24t 3 − 6t 2 − 24t + 3 dt and a= dv = 72t 2 − 12t − 24 dt When a = 0: 72t 2 − 12t − 24 = 12(6t 2 − t − 2) = 0 (3t − 2)(2t + 1) = 0 or t= or At t = 2 s: 3 2 1 s and t = − s (Reject) 3 2 4 3 2 2 2 2 2 x2/3 = 6 − 2 − 12 + 3 + 3 3 3 3 3 3 t = 0.667 s or x2/3 = 0.259 m 2 2 2 2 v2/3 = 24 − 6 − 24 + 3 3 3 3 or v2/3 = −8.56 m/s 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. 9 PROBLEM 11.6 The motion of a particle is defined by the relation x = t 3 − 9t 2 + 24t − 8, where x and t are expressed in inches and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero. SOLUTION We have x = t 3 − 9t 2 + 24t − 8 Then v= dx = 3t 2 − 18t + 24 dt and a= dv = 6 t − 18 dt (a) When v = 0: 3 t 2 − 18t + 24 = 3(t 2 − 6t + 8) = 0 (t − 2)(t − 4) = 0 t = 2.00 s and t = 4.00 s (b) When a = 0: 6t − 18 = 0 or t = 3 s x3 = (3)3 − 9(3)2 + 24(3) − 8 At t = 3 s: First observe that 0 ≤ t < 2 s: v>0 2 s < t ≤ 3 s: v<0 or x3 = 10.00 in. Now At t = 0: x0 = −8 in. At t = 2 s: x2 = (2)3 − 9(2) 2 + 24(2) − 8 = 12 in. Then x2 − x0 = 12 − (−8) = 20 in. | x3 − x2 | = |10 − 12| = 2 in. Total distance traveled = (20 + 2) in. Total distance = 22.0 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. 10 PROBLEM 11.7 The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x is expressed in meters and t in seconds. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero. SOLUTION x = 2t 3 − 15t 2 + 24t + 4 dx = 6t 2 − 30t + 24 v= dt dv a= = 12t − 30 dt (a) v = 0 when 6t 2 − 30t + 24 = 0 6(t − 1)(t − 4) = 0 (b) a = 0 when t = 1.000 s and t = 4.00 s 12t − 30 = 0 t = 2.5 s x2.5 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4 For t = 2.5 s: x2.5 = +1.500 m To find total distance traveled, we note that x1 = 2(1)3 − 15(1)2 + 24(1) + 4 v = 0 when t = 1 s: x1 = +15 m For t = 0, x0 = +4 m Distance traveled From t = 0 to t = 1 s: From t = 1 s to t = 2.5 s: x1 − x0 = 15 − 4 = 11 m x2.5 − x1 = 1.5 − 15 = 13.5 m Total distance traveled = 11 m + 13.5 m Total distance = 24.5 m 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. 11 PROBLEM 11.8 The motion of a particle is defined by the relation x = t 3 − 6t 2 − 36t − 40, where x and t are expressed in feet and seconds, respectively. Determine (a) when the velocity is zero, (b) the velocity, the acceleration, and the total distance traveled when x = 0. SOLUTION We have x = t 3 − 6t 2 − 36t − 40 Then v= dx = 3t 2 − 12t − 36 dt and a= dv = 6t − 12 dt (a) When v = 0: 3t 2 − 12t − 36 = 3(t 2 − 4t − 12) = 0 (t + 2)(t − 6) = 0 or t = −2 s (Reject) and t = 6 s or (b) When x = 0: t = 6.00 s t 3 − 6t 2 − 36t − 40 = 0 Factoring (t − 10)(t + 2)(t + 2) = 0 or t = 10 s Now observe that 0 ≤ t < 6 s: v<0 6 s < t ≤ 10 s: v>0 and at t = 0: t = 6 s: x0 = −40 ft x6 = (6)3 − 6(6)2 − 36(6) − 40 = −256 ft t = 10 s: Then v10 = 3(10) 2 − 12(10) − 36 or v10 = 144.0 ft/s a10 = 6(10) − 12 or a10 = 48.0 ft/s 2 | x6 − x0 | = | − 256 − (−40)| = 216 ft x10 − x6 = 0 − (−256) = 256 ft Total distance traveled = (216 + 256) ft Total distance = 472 ft 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. 12 PROBLEM 11.9 The brakes of a car are applied, causing it to slow down at a rate of 10 m/s2. Knowing that the car stops in 100 m, determine (a) how fast the car was traveling immediately before the brakes were applied, (b) the time required for the car to stop. SOLUTION a = −10 ft/s 2 (a) Velocity at x = 0. v 0 v0 0− (b) dv = a = −10 dx vdv = − xf 0 (−10) dx v02 = −10 x f = −(10)(300) 2 v02 = 6000 v0 = 77.5 ft/s 2 Time to stop. dv = a = −10 dx 0 v0 dv = − tf −10dt 0 0 − v0 = −10t f tf = v0 77.5 = 10 10 t f = 7.75 s 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. 13 PROBLEM 11.10 The acceleration of a particle is directly proportional to the time t. At t = 0, the velocity of the particle is v = 16 in./s. Knowing that v = 15 in./s and that x = 20 in. when t = 1 s, determine the velocity, the position, and the total distance traveled when t = 7 s. SOLUTION a = kt We have dv = a = kt dt Now v At t = 0, v = 16 in./s: or v − 16 = dv = 16 0 kt dt 1 2 kt 2 1 2 kt (in./s) 2 15 in./s = 16 in./s + k = −2 in./s3 or 1 k (1 s) 2 2 and v = 16 − t 2 dx = v = 16 − t 2 dt Also At t = 1 s, x = 20 in.: t v = 16 + or At t = 1 s, v = 15 in./s: k = constant x 20 dx = t 1 (16 − t 2 ) dt t or or 1 x − 20 = 16t − t 3 3 1 1 13 x = − t 3 + 16t + (in.) 3 3 Then At t = 7 s: When v = 0: v7 = 16 − (7) 2 or v7 = −33.0 in./s 1 13 x7 = − (7)3 + 16(7) + 3 3 or x7 = 2.00 in. 16 − t 2 = 0 or t = 4 s 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. 14 PROBLEM 11.10 (Continued) At t = 0: x0 = 13 3 1 13 x4 = − (4)3 + 16(4) + = 47 in. 3 3 t = 4 s: Now observe that Then 0 ≤ t < 4 s: v>0 4 s < t ≤ 7 s: v<0 13 = 42.67 in. 3 | x7 − x4 | = |2 − 47| = 45 in. x4 − x0 = 47 − Total distance traveled = (42.67 + 45) in. Total distance = 87.7 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. 15 PROBLEM 11.11 The acceleration of a particle is directly proportional to the square of the time t. When t = 0, the particle is at x = 24 m. Knowing that at t = 6 s, x = 96 m and v = 18 m/s, express x and v in terms of t. SOLUTION a = kt 2 We have k = constant dv = a = kt 2 dt Now v or 1 v − 18 = k (t 3 − 216) 3 18 dv = t At t = 6 s, v = 18 m/s: 6 kt 2 dt 1 v = 18 + k (t 3 − 216)(m/s) 3 or dx 1 = v = 18 + k (t 3 − 216) dt 3 Also x 1 18 + k (t 3 − 216) dt 3 or 1 1 x − 24 = 18t + k t 4 − 216t 3 4 24 dx = t At t = 0, x = 24 m: 0 Now At t = 6 s, x = 96 m: or Then or and or 1 1 96 − 24 = 18(6) + k (6) 4 − 216(6) 3 4 k= 1 m/s 4 9 1 1 1 x − 24 = 18t + t 4 − 216t 3 9 4 x(t ) = 1 4 t + 10t + 24 108 11 v = 18 + (t 3 − 216) 3 9 v(t ) = 1 3 t + 10 27 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. 16 PROBLEM 11.12 The acceleration of a particle is defined by the relation a = kt 2 . (a) Knowing that v = −8 m/s when t = 0 and that v = +8 m/s when t = 2 s, determine the constant k. (b) Write the equations of motion, knowing also that x = 0 when t = 2 s. SOLUTION a = kt 2 dv = a = kt 2 dt (1) t = 0, v = −8 m/s and t = 2 s, v = +8 ft/s (a) 8 −8 dv = 2 0 kt 2 dt 1 8 − ( −8) = k (2)3 3 (b) k = 6.00 m/s 4 Substituting k = 6 m/s 4 into (1) dv = a = 6t 2 dt t = 0, v = −8 m/s: v −8 dv = t 0 a = 6t 2 6t 2 dt 1 v − (−8) = 6(t )3 3 v = 2t 3 − 8 dx = v = 2t 3 − 8 dt t = 2 s, x = 0: x 0 dx = t 2 (2t 3 − 8) dt ; x = 1 4 t − 8t 2 1 1 x = t 4 − 8t − (2) 4 − 8(2) 2 2 1 x = t 4 − 8t − 8 + 16 2 t 2 x= 1 4 t − 8t + 8 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. 17 PROBLEM 11.13 The acceleration of Point A is defined by the relation a = −1.8sin kt , where a and t are expressed in m/s 2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0 and v = 0.6 m/s when t = 0, determine the velocity and position of Point A when t = 0.5 s. SOLUTION Given: a = −1.8sin kt m/s 2 , v0 = 0.6 m/s, x0 = 0, t t v − v0 = 0 a dt = −1.8 0 sin kt dt = v − 0.6 = t 1.8 cos kt k 0 1.8 (cos kt − 1) = 0.6cos kt − 0.6 3 v = 0.6cos kt m/s Velocity: t t x − x0 = 0 v dt = 0.6 0 cos kt dt = x−0= 0.6 sin kt k t 0 0.6 (sin kt − 0) = 0.2sin kt 3 x = 0.2sin kt m Position: When t = 0.5 s, k = 3 rad/s kt = (3)(0.5) = 1.5 rad v = 0.6cos1.5 = 0.0424 m/s v = 42.4 mm/s x = 0.2sin1.5 = 0.1995 m x = 199.5 mm 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. 18 PROBLEM 11.14 The acceleration of Point A is defined by the relation a = −1.08sin kt − 1.44cos kt , where a and t are expressed in m/s2 and seconds, respectively, and k = 3 rad/s. Knowing that x = 0.16 m and v = 0.36 m/s when t = 0, determine the velocity and position of Point A when t = 0.5 s. SOLUTION Given: a = −1.08sin kt − 1.44 cos kt m/s 2 , x0 = 0.16 m, k = 3 rad/s v0 = 0.36 m/s t t t v − v0 = 0 a dt = −1.08 0 sin kt dt − 1.44 0 cos kt dt v − 0.36 = = 1.08 cos kt k t 0 − 1.44 sin kt k t 0 1.08 1.44 (cos kt − 1) − (sin kt − 0) 3 3 = 0.36 cos kt − 0.36 − 0.48sin kt Velocity: v = 0.36 cos kt − 0.48sin kt m/s t t t x − x0 = 0 v dt = 0.36 0 cos kt dt − 0.48 0 sin kt dt x − 0.16 = 0.36 sin kt k t 0 + 0.48 cos kt k t 0 0.36 0.48 (sin kt − 0) + (cos kt − 1) 3 3 = 0.12sin kt + 0.16cos kt − 0.16 = Position: When t = 0.5 s, x = 0.12sin kt + 0.16cos kt m kt = (3)(0.5) = 1.5 rad v = 0.36 cos1.5 − 0.48sin1.5 = −0.453 m/s v = −453 mm/s x = 0.12sin1.5 + 0.16 cos1.5 = 0.1310 m x = 131.0 mm 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. 19 PROBLEM 11.15 A piece of electronic equipment that is surrounded by packing material is dropped so that it hits the ground with a speed of 4 m/s. After contact the equipment experiences an acceleration of a = − kx, where k is a constant and x is the compression of the packing material. If the packing material experiences a maximum compression of 20 mm, determine the maximum acceleration of the equipment. SOLUTION a= vdv = −k x dx Separate and integrate. vf v0 vdv = − xf 0 k x dx 1 2 1 2 1 v f − v0 = − kx 2 2 2 2 xf 0 1 = − k x 2f 2 Use v0 = 4 m/s, x f = 0.02 m, and v f = 0. Solve for k. 1 1 0 − (4) 2 = − k (0.02) 2 2 2 k = 40,000 s −2 Maximum acceleration. amax = −kxmax : ( −40,000)(0.02) = −800 m/s 2 a = 800 m/s 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. 20 PROBLEM 11.16 A projectile enters a resisting medium at x = 0 with an initial velocity v0 = 900 ft/s and travels 4 in. before coming to rest. Assuming that the velocity of the projectile is defined by the relation v = v0 − kx, where v is expressed in ft/s and x is in feet, determine (a) the initial acceleration of the projectile, (b) the time required for the projectile to penetrate 3.9 in. into the resisting medium. SOLUTION First note When x = 4 0 = (900 ft/s) − k ft 12 4 ft, v = 0: 12 k = 2700 or (a) 1 s We have v = v0 − kx Then a= or a = −k (v0 − kx) At t = 0: 1 a = 2700 (900 ft/s − 0) s dv d = (v0 − kx) = −kv dt dt a0 = −2.43 × 106 ft/s 2 or (b) dx = v = v0 − kx dt We have At t = 0, x = 0: or x 0 dx = v0 − kx t dt 0 1 − [ln(v0 − kx)]0x = t k or t= 1 v0 1 1 ln = ln k v0 − kx k 1 − vk x 0 When x = 3.9 in.: t= 1 ln 1 2700 s 1 − 1 2700 1/s 900 ft/s ( 3.9 12 ft ) t = 1.366 × 10−3 s or 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. 21 PROBLEM 11.17 The acceleration of a particle is defined by the relation a = −k/x. It has been experimentally determined that v = 15 ft/s when x = 0.6 ft and that v = 9 ft/s when x = 1.2 ft. Determine (a) the velocity of the particle when x = 1.5 ft, (b) the position of the particle at which its velocity is zero. SOLUTION a= vdv − k = dx x Separate and integrate using x = 0.6 ft, v = 15 ft/s. v 15 vdv = − k x 0.6 v dx x 1 2 v = − k ln x 2 15 x 0.6 1 2 1 x v − (15) 2 = − k ln 2 2 0.6 (1) When v = 9 ft/s, x = 1.2 ft 1 2 1 1.2 (9) − (15) 2 = −k ln 2 2 0.6 Solve for k. k = 103.874 ft 2 /s 2 (a) Velocity when x = 65 ft. Substitute k = 103.874 ft 2 /s 2 and x = 1.5 ft into (1). 1 2 1 1.5 v − (15) 2 = −103.874 ln 2 2 0.6 v = 5.89 ft/s (b) Position when for v = 0, 1 x 0 − (15) 2 = −103.874 ln 2 0.6 x ln = 1.083 0.6 x = 1.772 ft 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. 22 PROBLEM 11.18 A brass (nonmagnetic) block A and a steel magnet B are in equilibrium in a brass tube under the magnetic repelling force of another steel magnet C located at a distance x = 0.004 m from B. The force is inversely proportional to the square of the distance between B and C. If block A is suddenly removed, the acceleration of block B is a = −9.81 + k /x 2 , where a and x are expressed in m/s2 and m, respectively, and k = 4 × 10−4 m3 /s 2 . Determine the maximum velocity and acceleration of B. SOLUTION The maximum veolocity occurs when a = 0. xm2 = 0 = −9.81 + k 4 × 10−4 = = 40.775 × 10−6 m 2 9.81 9.81 k xm2 xm = 0.0063855 m The acceleration is given as a function of x. v dv k = a = −9.81 + 2 dx x Separate variables and integrate: vdv = −9.81dx + v 0 vdv = −9.81 x x0 k dx x2 dx + k x x0 dx x2 1 1 1 2 v = −9.81( x − x0 ) − k − 2 x x0 1 1 2 1 vm = −9.81( xm − x0 ) − k − 2 xm x0 1 1 = −9.81(0.0063855 − 0.004) − (4 × 10−4 ) − 0.0063855 0.004 = −0.023402 + 0.037358 = 0.013956 m 2 /s2 Maximum velocity: vm = 0.1671 m/s vm = 167.1 mm/s The maximum acceleration occurs when x is smallest, that is, x = 0.004 m. am = −9.81 + 4 × 10−4 (0.004) 2 am = 15.19 m/s 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. 23 PROBLEM 11.19 Based on experimental observations, the acceleration of a particle is defined by the relation a = −(0.1 + sin x/b), where a and x are expressed in m/s2 and meters, respectively. Knowing that b = 0.8 m and that v = 1 m/s when x = 0, determine (a) the velocity of the particle when x = −1 m, (b) the position where the velocity is maximum, (c) the maximum velocity. SOLUTION We have When x = 0, v = 1 m/s: v v 1 dv x = a = − 0.1 + sin dx 0.8 vdv = x − 0.1 + sin dx 0 0.8 x x 1 2 x (v − 1) = − 0.1x − 0.8 cos 2 0.8 0 or x 1 2 v = −0.1x + 0.8 cos − 0.3 2 0.8 or (a) When x = −1 m: 1 2 −1 v = −0.1(−1) + 0.8 cos − 0.3 2 0.8 v = ± 0.323 m/s or (b) When v = vmax, a = 0: or (c) x − 0.1 + sin =0 0.8 x = −0.080 134 m x = −0.0801 m When x = −0.080 134 m: 1 2 −0.080 134 vmax = −0.1(−0.080 134) + 0.8 cos − 0.3 2 0.8 = 0.504 m 2 /s 2 vmax = 1.004 m/s or 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. 24 PROBLEM 11.20 A spring AB is attached to a support at A and to a collar. The unstretched length of the spring is l. Knowing that the collar is released from rest at x = x0 and has an acceleration defined by the relation a = −100( x − lx / l 2 + x 2 ) , determine the velocity of the collar as it passes through Point C. SOLUTION a=v Since a is function of x, dv lx = −100 x − dx l 2 + x2 Separate variables and integrate: vf v0 vdv = −100 lx x− 2 x0 l + x2 0 x2 1 2 1 2 − l l 2 + x2 v f − v0 = −100 2 2 2 dx 0 x0 x2 1 2 v f − 0 = −100 − 0 − l 2 + l l 2 + x02 2 2 1 2 100 2 (−l + x02 − l 2 − 2l l 2 + x02 ) vf = 2 2 100 ( l 2 + x02 − l )2 = 2 v f = 10( l 2 + x02 − l ) 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. 25 PROBLEM 11.21 The acceleration of a particle is defined by the relation a = −0.8v where a is expressed in m/s2 and v in m/s. Knowing that at t = 0 the velocity is 1 m/s, determine (a) the distance the particle will travel before coming to rest, (b) the time required for the particle’s velocity to be reduced by 50 percent of its initial value. SOLUTION (a) Determine relationship between x and v. a= vdv = −0.8v dx dv = −0.8dx Separate and integrate with v = 1 m/s when x = 0. v 1 dv = −0.8 x 0 dx v − 1 = −0.8 x Distance traveled. For v = 0, (b) x= −1 −0.8 x = 1.25 m Determine realtionship between v and t. a= v 1 dv = 0.8v dt dv =− v x 0 0.8dt v ln = −0.8t 1 1 t = 1.25ln v For v = 0.5(1 m/s) = 0.5 m/s, 1 t = 1.25ln 0.5 t = 0.866 s 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. 26 PROBLEM 11.22 Starting from x = 0 with no initial velocity, a particle is given an acceleration a = 0.1 v 2 + 16, where a and v are expressed in ft/s2 and ft/s, respectively. Determine (a) the position of the particle when v = 3ft/s, (b) the speed and acceleration of the particle when x = 4 ft. SOLUTION a= vdv = 0.1(v 2 + 16)1/ 2 dx (1) Separate and integrate. v 0 vdv v 2 + 16 = x 0 0.1 dx v (v 2 + 16)1/2 = 0.1 x 0 2 1/2 (v + 16) (a) (b) − 4 = 0.1 x x = 10[(v 2 + 16)1/2 − 4] (2) x = 10[(32 + 16)1/2 − 4] x = 10.00 ft v = 3 ft/s. x = 4 ft. From (2), (v 2 + 16)1/2 = 4 + 0.1x = 4 + (0.1)(4) = 4.4 v 2 + 16 = 19.36 v 2 = 3.36 ft 2 /s 2 From (1), a = 0.1(1.8332 + 16)1/2 v = 1.833 ft/s a = 0.440 ft/s 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. 27 PROBLEM 11.23 A ball is dropped from a boat so that it strikes the surface of a lake with a speed of 16.5 ft/s. While in the water the ball experiences an acceleration of a = 10 − 0.8v, where a and v are expressed in ft/s2 and ft/s, respectively. Knowing the ball takes 3 s to reach the bottom of the lake, determine (a) the depth of the lake, (b) the speed of the ball when it hits the bottom of the lake. SOLUTION a= dv = 10 − 0.8v dt Separate and integrate: dv = v0 10 − 0.8v v t 0 dt v − 1 ln(10 − 0.8v) = t 0.8 v0 10 − 0.8v ln = −0.8t 10 − 0.8v0 10 − 0.8v = (10 − 0.8v0 )e −0.8t or 0.8v = 10 − (10 − 0.8v0 )e−0.8t v = 12.5 − (12.5 − v0 )e −0.8t With v0 = 16.5ft/s v = 12.5 + 4e−0.8t 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. 28 PROBLEM 11.23 (Continued) Integrate to determine x as a function of t. dx = 12.5 + 4e−0.8t v= dt x 0 dx = t 0 (12.5 + 4e−0.8t )dt x = 12.5t − 5e −0.8t (a) t 0 = 12.5t − 5e−0.8t + 5 At t = 35 s, x = 12.5(3) − 5e −2.4 + 5 = 42.046 ft (b) x = 42.0 ft v = 12.5 + 4e −2.4 = 12.863 ft/s v = 12.86 ft/s 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. 29 PROBLEM 11.24 The acceleration of a particle is defined by the relation a = − k v , where k is a constant. Knowing that x = 0 and v = 81 m/s at t = 0 and that v = 36 m/s when x = 18 m, determine (a) the velocity of the particle when x = 20 m, (b) the time required for the particle to come to rest. SOLUTION (a) We have dv = a = −k v dx v v dv = − k dx so that v or 2 3/2 v [v ]81 = −kx 3 or 2 3/2 [v − 729] = −kx 3 When x = 18 m, v = 36 m/s: v dv = 81 x When x = 0, v = 81 m/s: 0 − k dx 2 (363/2 − 729) = − k (18) 3 k = 19 m/s 2 or Finally When x = 20 m: 2 3/2 (v − 729) = −19(20) 3 v3/2 = 159 or (b) We have dv = a = −19 v dt At t = 0, v = 81 m/s: v dv 81 v = t 0 −19dt or v 2[ v ]81 = −19t or 2( v − 9) = −19t When v = 0: v = 29.3 m/s 2(−9) = −19t t = 0.947 s or 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. 30 PROBLEM 11.25 A particle is projected to the right from the position x = 0 with an initial velocity of 9 m/s. If the acceleration of the particle is defined by the relation a = −0.6v3/ 2 , where a and v are expressed in m/s2 and m/s, respectively, determine (a) the distance the particle will have traveled when its velocity is 4 m/s, (b) the time when v = 1 m/s, (c) the time required for the particle to travel 6 m. SOLUTION (a) We have When x = 0, v = 9 m/s: v v 9 dv = a = −0.6v3/2 dx v − (1/2) dv = x 0 −0.6dx or 2[v1/2 ]9v = −0.6 x or x= 1 (3 − v1/ 2 ) 0.3 When v = 4 m/s: x= 1 (3 − 41/ 2 ) 0.3 x = 3.33 m or (b) dv = a = −0.6v3/2 dt We have v or −2[v − (1/2) ]9v = −0.6t When v = 1 m/s: 9 v − (3/2) dv = 1 v 1 1 t When t = 0, v = 9 m/s: or 0 −0.6dt − 1 = 0.3t 3 − 1 = 0.3t 3 t = 2.22 s or (c) We have (1) 1 v − 1 = 0.3t 3 2 or Now 9 3 v= = 1 0.9 t + (1 + 0.9t ) 2 dx 9 =v= dt (1 + 0.9t ) 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. 31 PROBLEM 11.25 (Continued) At t = 0, x = 0: x 0 dx = 9 dt 0 (1 + 0.9t ) 2 t t 1 1 x = 9 − 0.9 1 + 0.9t 0 or 1 = 10 1 − 1 + 0.9t 9t = 1 + 0.9t When x = 6 m: 6= 9t 1 + 0.9t t = 1.667 s or An alternative solution is to begin with Eq. (1). x= 1 (3 − v1/ 2 ) 0.3 dx = v = (3 − 0.3x) 2 dt Then Now At t = 0, x = 0: x 0 dx = (3 − 0.3x) 2 t 0 dt x or 1 1 x = t= 0.3 3 − 0.3x 0 9 − 0.9 x which leads to the same equation as above. 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. 32 PROBLEM 11.26 The acceleration of a particle is defined by the relation a = 0.4(1 − kv), where k is a constant. Knowing that at t = 0 the particle starts from rest at x = 4 m and that when t = 15 s, v = 4 m/s, determine (a) the constant k, (b) the position of the particle when v = 6 m/s, (c) the maximum velocity of the particle. SOLUTION (a) dv = a = 0.4(1 − kv) dt We have At t = 0, v = 0: dv = 0 1− kv v t 0 0.4dt 1 − [ln(1 − k v)]v0 = 0.4t k or or ln(1 − k v) = −0.4 kt At t = 15 s, v = 4 m/s: ln(1 − 4k ) = −0.4k (15) (1) = −6k k = 0.145703 s/m Solving yields k = 0.1457 s/m or (b) We have v When x = 4 m, v = 0: vdv = 0 1− kv v x 4 0.4 dx v 1 1/k =− + 1− kv k 1− kv Now Then dv = a = 0.4(1 − kv) dx v 1 1 − + dv = 0 k k (1 − k v) x 4 0.4 dx v or v 1 x − k − k 2 ln(1 − k v) = 0.4[ x]4 0 or v 1 − + 2 ln(1 − kv) = 0.4( x − 4) k k When v = 6 m/s: 6 1 ln(1 − 0.145 703 × 6) = 0.4( x − 4) − + 2 0.145 703 (0.145 703) or 0.4( x − 4) = 56.4778 x = 145.2 m or 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. 33 PROBLEM 11.26 (Continued) (c) The maximum velocity occurs when a = 0. a = 0: 0.4(1 − k vmax ) = 0 vmax = or 1 0.145 703 vmax = 6.86 m/s or An alternative solution is to begin with Eq. (1). ln(1 − k v) = −0.4 kt v= Then Thus, vmax is attained as t 1 (1 − k −0.4 kt ) k ∞ vmax = 1 k as above. 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. 34 PROBLEM 11.27 Experimental data indicate that in a region downstream of a given louvered supply vent the velocity of the emitted air is defined by v = 0.18v0 /x, where v and x are expressed in m/s and meters, respectively, and v0 is the initial discharge velocity of the air. For v0 = 3.6 m/s, determine (a) the acceleration of the air at x = 2 m, (b) the time required for the air to flow from x = 1 to x = 3 m. SOLUTION (a) dv dx 0.18v0 d 0.18v0 = x dx x a=v We have =− a=− When x = 2 m: 0.0324v02 x3 0.0324(3.6) 2 (2)3 a = −0.0525 m/s 2 or (b) 0.18v0 dx =v= dt x We have From x = 1 m to x = 3 m: 3 1 xdx = t3 t1 0.18v0 dt 3 or 1 2 2 x = 0.18v0 (t3 − t1 ) 1 or (t3 − t1 ) = 1 2 (9 − 1) 0.18(3.6) t3 − t1 = 6.17 s or 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. 35 PROBLEM 11.28 Based on observations, the speed of a jogger can be approximated by the relation v = 7.5(1 − 0.04 x)0.3 , where v and x are expressed in mi/h and miles, respectively. Knowing that x = 0 at t = 0, determine (a) the distance the jogger has run when t = 1 h, (b) the jogger’s acceleration in ft/s2 at t = 0, (c) the time required for the jogger to run 6 mi. SOLUTION (a) dx = v = 7.5(1 − 0.04 x)0.3 dt We have At t = 0, x = 0: or x 0 dx = (1 − 0.04 x)0.3 t 7.5dt 0 1 1 [(1 − 0.04 x)0.7 ]0x = 7.5t − 0.7 0.04 1 − (1 − 0.04 x)0.7 = 0.21t or or x= 1 [1 − (1 − 0.21t )1/0.7 ] 0.04 At t = 1 h: x= 1 {1 − [1 − 0.21(1)]1/0.7 } 0.04 (1) x = 7.15 mi or (b) We have a=v dv dx d [7.5(1 − 0.04 x)0.3 ] dx = 7.52 (1 − 0.04 x)0.3 [(0.3)(−0.04)(1 − 0.04 x) −0.7 ] = 7.5(1 − 0.04 x)0.3 = −0.675(1 − 0.04 x) −0.4 At t = 0, x = 0: a0 = −0.675 mi/h 2 × 5280 ft 1 h × 1 mi 3600 s a0 = −275 × 10−6 ft/s 2 or (c) 2 From Eq. (1) t= When x = 6 mi: t= 1 [1 − (1 − 0.04 x)0.7 ] 0.21 1 {1 − [1 − 0.04(6)]0.7 } 0.21 = 0.83229 h t = 49.9 min or 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. 36 PROBLEM 11.29 The acceleration due to gravity at an altitude y above the surface of the earth can be expressed as a= −32.2 [1 + ( y/20.9 × 106 )]2 where a and y are expressed in ft/s2 and feet, respectively. Using this expression, compute the height reached by a projectile fired vertically upward from the surface of the earth if its initial velocity is (a) 1800 ft/s, (b) 3000 ft/s, (c) 36,700 ft/s. SOLUTION We have v dv =a=− dy 32.2 (1 + y 20.9 × 106 When y = 0, provided that v does reduce to zero, y = ymax , v = 0 Then v0 v dv = v = v0 ymax 0 (1 + −32.2 y 20.9 × 106 ) 2 dy 1 v02 = 1345.96 × 106 1 − 1 + ymax 20.9 × 106 or ymax = or v0 = 1800 ft/s: ymax = v0 = 3000 ft/s: y max 0 v02 64.4 − v02 20.9 × 106 (1800)2 64.4 − (1800)2 20.9 × 106 ymax = 50.4 × 103 ft or (b) 2 1 1 − v02 = −32.2 −20.9 × 106 y 2 1+ × 106 20.9 or (a) 0 ) ymax = (3000) 2 64.4 − (3000)2 20.9 × 106 ymax = 140.7 × 103 ft or 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. 37 PROBLEM 11.29 (Continued) (c) v0 = 36,700 ft/s: ymax = (36, 700) 2 64.4 − (36,700)2 20.9 × 106 = −3.03 × 1010 ft This solution is invalid since the velocity does not reduce to zero. The velocity 36,700 ft/s is above the escape velocity vR from the earth. For vR and above. ymax ∞ 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. 38 PROBLEM 11.30 The acceleration due to gravity of a particle falling toward the earth is a = − gR 2 /r 2, where r is the distance from the center of the earth to the particle, R is the radius of the earth, and g is the acceleration due to gravity at the surface of the earth. If R = 3960 mi, calculate the escape velocity, that is, the minimum velocity with which a particle must be projected vertically upward from the surface of the earth if it is not to return to the earth. (Hint: v = 0 for r = ∞.) SOLUTION We have v dv gR 2 =a=− 2 dr r r = R, v = ve When r = ∞, v = 0 0 gR 2 dr r2 or 1 1 − ve2 = gR 2 2 r R or ve = 2 gR ve vdv = ∞ then R − ∞ 1/2 5280 ft = 2 × 32.2 ft/s 2 × 3960 mi × 1 mi ve = 36.7 × 103 ft/s or 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. 39 PROBLEM 11.31 The velocity of a particle is v = v0 [1 − sin (π t/T )]. Knowing that the particle starts from the origin with an initial velocity v0 , determine (a) its position and its acceleration at t = 3T , (b) its average velocity during the interval t = 0 to t = T . SOLUTION (a) dx π t = v = v0 1 − sin dt T We have At t = 0, x = 0: x 0 dx = π t v0 1 − sin dt 0 T t t T T π t πt T x = v0 t + cos = v0 t + cos − T 0 T π π π At t = 3T : T 2T π × 3T T − = v0 3T − x3T = v0 3T + cos π π T π a= At t = 3T : (b) (1) x3T = 2.36 v0T π πt dv d π t = v0 1 − sin = −v0 cos dt dt T T T a3T = −v0 π T cos π × 3T a3T = T π v0 T Using Eq. (1) At t = 0: At t = T : Now T T x0 = v0 0 + cos(0) − = 0 π π T πT xT = v0 T + cos π T vave = 2T T − π = v0 T − π xT − x0 0.363v0T − 0 = Δt T −0 = 0.363v0T vave = 0.363v0 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. 40 PROBLEM 11.32 The velocity of a slider is defined by the relation v = v′ sin (ωn t + φ ). Denoting the velocity and the position of the slider at t = 0 by v0 and x0 , respectively, and knowing that the maximum displacement of the slider is 2 x0 , show that (a) v′ = (v02 + x02ωn2 )/2 x0ωn , (b) the maximum value of the velocity occurs when x = x0 [3 − (v0 /x0ωn ) 2 ]/2. SOLUTION (a) v0 = v′ sin (0 + φ ) = v′ sin φ At t = 0, v = v0 : Then 2 cos φ = v′ − v02 v′ dx = v = v′ sin (ωn t + φ ) dt Now x or 1 x − x0 = v′ − cos (ωn t + φ ) ωn 0 x0 dx = t At t = 0, x = x0 : 0 v′ sin (ωn t + φ )dt t or x = x0 + v′ ωn [cos φ − cos (ωn t + φ )] Now observe that xmax occurs when cos (ωn t + φ ) = −1. Then xmax = 2 x0 = x0 + ωn [cos φ − ( −1)] 2 2 v′ v′ − v0 x0 = + 1 ωn v1 Substituting for cos φ or v′ x0ωn − v′ = v′2 − v02 Squaring both sides of this equation x02ωn2 − 2 x0ωn + v′2 = v′2 − v02 or v′ = v02 + x02ωn2 2 x0ωn Q. E. 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. 41 PROBLEM 11.32 (Continued) (b) First observe that vmax occurs when ωn t + φ = π2 . The corresponding value of x is v′ π cos φ − cos ωn 2 v′ = x0 + cos φ xvmax = x0 + ωn Substituting first for cos φ and then for v′ xvmax = x0 + 2 v′ − v02 v′ v′ ωn 1/2 2 1 v02 + x02ωn2 2 = x0 + − v0 ωn 2 x0ωn 1 v04 + 2v02 x02ωn2 + x04ωn4 − 4 x02ωn2 v02 = x0 + 2 2 x0ωn ( = x0 + = x0 + x = 0 2 1 ( x 2ω 2 − v 2 0 n 0 2 x0ωn2 ) 1/2 1/ 2 2 ) x02ωn2 − v02 2 x0ωn2 v 2 3 − 0 x0ωn Q. E. 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. 42 PROBLEM 11.33 A stone is thrown vertically upward from a point on a bridge located 40 m above the water. Knowing that it strikes the water 4 s after release, determine (a) the speed with which the stone was thrown upward, (b) the speed with which the stone strikes the water. SOLUTION Uniformly accelerated motion. Origin at water. 1 y = y0 + v0t + a t 2 2 v = v0 + a t where y0 = 40 m and a = −9.81 m/s2 . (a) Initial speed. y = 0 when t = 4 s. 1 0 = 40 + v0 (4) − (9.81)(4)2 2 v0 = 9.62 m/s (b) v0 = 9.62 m/s Speed when striking the water. (v at t = 4 s) v = 9.62 − (9.81)(4) = −29.62 m/s v = 29.6 m/s 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. 43 PROBLEM 11.34 A motorist is traveling at 54 km/h when she observes that a traffic light 240 m ahead of her turns red. The traffic light is timed to stay red for 24 s. If the motorist wishes to pass the light without stopping just as it turns green again, determine (a) the required uniform deceleration of the car, (b) the speed of the car as it passes the light. SOLUTION Uniformly accelerated motion: x0 = 0 v0 = 54 km/h = 15 m/s (a) x = x0 + v0t + 1 2 at 2 when t = 24s, x = 240 m: 240 m = 0 + (15 m/s)(24 s) + 1 a (24 s)2 2 a = −0.4167 m/s 2 (b) a = −0.417 m/s 2 v = v0 + a t when t = 24s: v = (15 m/s) + (−0.4167 m/s)(24 s) v = 5.00 m/s v = 18.00 km/h v = 18.00 km/h 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. 44 PROBLEM 11.35 A motorist enters a freeway at 30 mi/h and accelerates uniformly to 60 mi/h. From the odometer in the car, the motorist knows that she traveled 550 ft while accelerating. Determine (a) the acceleration of the car, (b) the time required to reach 60 mi/h. SOLUTION (a) Acceleration of the car. v12 = v02 + 2a( x1 − x0 ) a= Data: v12 − v02 2( x1 − x0 ) v0 = 30 mi/h = 44 ft/s v1 = 60 mi/h = 88 ft/s x0 = 0 x1 = 550 ft a= (b) (88) 2 − (44) 2 (2)(55 − 0) a = 5.28 ft/s 2 Time to reach 60 mi/h. v1 = v0 + a (t1 − t0 ) v1 − v0 a 88 − 44 = 5.28 = 8.333 s t1 − t0 = t1 − t0 = 8.33 s 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. 45 PROBLEM 11.36 A group of students launches a model rocket in the vertical direction. Based on tracking data, they determine that the altitude of the rocket was 89.6 ft at the end of the powered portion of the flight and that the rocket landed 16 s later. Knowing that the descent parachute failed to deploy so that the rocket fell freely to the ground after reaching its maximum altitude and assuming that g = 32.2 ft/s 2 , determine (a) the speed v1 of the rocket at the end of powered flight, (b) the maximum altitude reached by the rocket. SOLUTION (a) 1 2 at 2 We have y = y1 + v1t + At tland , y=0 Then 0 = 89.6 ft + v1 (16 s) + 1 (−32.2 ft/s 2 )(16 s) 2 2 v1 = 252 ft/s or (b) v 2 = v12 + 2a( y − y1 ) We have At y = ymax , v = 0 Then 0 = (252 ft/s) 2 + 2( −32.2 ft/s 2 )( ymax − 89.6) ft ymax = 1076 ft or 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. 46 PROBLEM 11.37 A small package is released from rest at A and moves along the skate wheel conveyor ABCD. The package has a uniform acceleration of 4.8 m/s 2 as it moves down sections AB and CD, and its velocity is constant between B and C. If the velocity of the package at D is 7.2 m/s, determine (a) the distance d between C and D, (b) the time required for the package to reach D. SOLUTION (a) For A B and C D we have v 2 = v02 + 2a( x − x0 ) 2 vBC = 0 + 2(4.8 m/s 2 )(3 − 0) m Then, at B = 28.8 m 2 /s 2 (vBC = 5.3666 m/s) 2 vD2 = vBC + 2aCD ( xD − xC ) and at D d = xD − xC (7.2 m/s)2 = (28.8 m 2 /s 2 ) + 2(4.8 m/s 2 )d or d = 2.40 m or (b) For A B and C D we have v = v0 + at Then A B 5.3666 m/s = 0 + (4.8 m/s 2 )t AB t AB = 1.11804 s or and C 7.2 m/s = 5.3666 m/s + (4.8 m/s 2 )tCD D tCD = 0.38196 s or Now, for B C, we have xC = xB + vBC t BC or 3 m = (5.3666 m/s)t BC or t BC = 0.55901 s Finally, t D = t AB + t BC + tCD = (1.11804 + 0.55901 + 0.38196) s t D = 2.06 s or 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. 47 PROBLEM 11.38 A sprinter in a 100-m race accelerates uniformly for the first 35 m and then runs with constant velocity. If the sprinter’s time for the first 35 m is 5.4 s, determine (a) his acceleration, (b) his final velocity, (c) his time for the race. SOLUTION Given: 0 ≤ x ≤ 35 m, a = constant 35 m < x ≤ 100 m, v = constant At t = 0, v = 0 when x = 35 m, t = 5.4 s Find: (a) a (b) v when x = 100 m (c) t when x = 100 m (a) We have At t = 5.4 s: or x = 0 + 0t + 35 m = 1 2 at 2 for 0 ≤ x ≤ 35 m 1 a(5.4 s) 2 2 a = 2.4005 m/s 2 a = 2.40 m/s 2 (b) First note that v = vmax for 35 m ≤ x ≤ 100 m. Now (c) v 2 = 0 + 2a( x − 0) for 0 ≤ x ≤ 35 m When x = 35 m: 2 vmax = 2(2.4005 m/s 2 )(35 m) or vmax = 12.9628 m/s We have When x = 100 m: vmax = 12.96 m/s x = x1 + v0 (t − t1 ) for 35 m < x ≤ 100 m 100 m = 35 m + (12.9628 m/s)(t2 − 5.4) s t2 = 10.41 s or 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. 48 PROBLEM 11.39 As relay runner A enters the 20-m-long exchange zone with a speed of 12.9 m/s, he begins to slow down. He hands the baton to runner B 1.82 s later as they leave the exchange zone with the same velocity. Determine (a) the uniform acceleration of each of the runners, (b) when runner B should begin to run. SOLUTION (a) For runner A: At t = 1.82 s: x A = 0 + (v A )0 t + 1 a At 2 2 20 m = (12.9 m/s)(1.82 s) + 1 a A (1.82 s) 2 2 a A = −2.10 m/s 2 or Also At t = 1.82 s: v A = (v A ) 0 + a A t (v A )1.82 = (12.9 m/s) + ( −2.10 m/s 2 )(1.82 s) = 9.078 m/s For runner B: vB2 = 0 + 2aB [ xB − 0] When xB = 20 m, vB = v A : (9.078 m/s) 2 = 2aB (20 m) or aB = 2.0603 m/s 2 aB = 2.06 m/s 2 (b) For runner B: vB = 0 + aB (t − t B ) where t B is the time at which he begins to run. At t = 1.82 s: or 9.078 m/s = (2.0603 m/s 2 )(1.82 − t B ) s t B = −2.59 s Runner B should start to run 2.59 s before A reaches the exchange zone. 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. 49 PROBLEM 11.40 In a boat race, boat A is leading boat B by 50 m and both boats are traveling at a constant speed of 180 km/h. At t = 0, the boats accelerate at constant rates. Knowing that when B passes A, t = 8 s and v A = 225 km/h, determine (a) the acceleration of A, (b) the acceleration of B. SOLUTION (a) We have v A = (v A ) 0 + a A t (v A )0 = 180 km/h = 50 m/s At t = 8 s: Then v A = 225 km/h = 62.5 m/s 62.5 m/s = 50 m/s + a A (8 s) a A = 1.563 m/s 2 or (b) We have x A = ( x A ) 0 + (v A )0 t + 1 1 a At 2 = 50 m + (50 m/s)(8 s) + (1.5625 m/s 2 )(8 s) 2 = 500 m 2 2 and xB = 0 + (vB )0 t + At t = 8 s: x A = xB 1 aB t 2 2 500 m = (50 m/s)(8 s) + (vB )0 = 50 m/s 1 aB (8 s) 2 2 aB = 3.13 m/s 2 or 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. 50 PROBLEM 11.41 A police officer in a patrol car parked in a 45 mi/h speed zone observes a passing automobile traveling at a slow, constant speed. Believing that the driver of the automobile might be intoxicated, the officer starts his car, accelerates uniformly to 60 mi/h in 8 s, and, maintaining a constant velocity of 60 mi/h, overtakes the motorist 42 s after the automobile passed him. Knowing that 18 s elapsed before the officer began pursuing the motorist, determine (a) the distance the officer traveled before overtaking the motorist, (b) the motorist’s speed. SOLUTION (vP )18 = 0 (a) (vP )26 = 60 mi/h = 88 ft/s (vP ) 42 = 90 mi/h = 88 ft/s Patrol car: For 18 s < t ≤ 26 s: At t = 26 s: vP = 0 + aP (t − 18) 88 ft/s = aP (26 − 18) s or aP = 11 ft/s 2 Also, xP = 0 + 0(t − 18) − At t = 26 s: For 26 s < t ≤ 42 s: At t = 42 s: (xP ) 26 = 1 aP (t − 18) 2 2 1 (11 ft/s 2 )(26 − 18) 2 = 352 ft 2 xP = ( xP )26 + (vP ) 26 (t − 26) (xP ) 42 = 352 m + (88 ft/s)(42 − 26) s = 1760 ft ( xP ) 42 = 1760 ft (b) For the motorist’s car: At t = 42 s, xM = xP : or xM = 0 + vM t 1760 ft = vM (42 s) vM = 41.9048 ft/s vM = 28.6 mi/h or 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. 51 PROBLEM 11.42 Automobiles A and B are traveling in adjacent highway lanes and at t = 0 have the positions and speeds shown. Knowing that automobile A has a constant acceleration of 1.8 ft/s 2 and that B has a constant deceleration of 1.2 ft/s 2 , determine (a) when and where A will overtake B, (b) the speed of each automobile at that time. SOLUTION a A = +1.8 ft/s 2 aB = −1.2 ft/s 2 (v A )0 = 24 mi/h = 35.2 ft/s A overtakes B (vB )0 = 36 mi/h = 52.8 ft/s Motion of auto A: v A = (v A )0 + a At = 35.2 + 1.8t x A = ( x A ) 0 + (v A ) 0 t + 1 1 a At 2 = 0 + 35.2t + (1.8)t 2 2 2 (1) (2) Motion of auto B: vB = (vB )0 + aB t = 52.8 − 1.2t x B = ( xB ) 0 + ( vB ) 0 t + (a) 1 1 aB t 2 = 75 + 52.8t + (−1.2)t 2 2 2 (3) (4) A overtakes B at t = t1 . x A = xB : 35.2t + 0.9t12 = 75 + 52.8t1 − 0.6t12 1.5t12 − 17.6t1 − 75 = 0 t1 = −3.22 s and t1 = 15.0546 Eq. (2): x A = 35.2(15.05) + 0.9(15.05) 2 t1 = 15.05 s x A = 734 ft 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. 52 PROBLEM 11.42 (Continued) (b) Velocities when t1 = 15.05 s Eq. (1): v A = 35.2 + 1.8(15.05) v A = 62.29 ft/s Eq. (3): v A = 42.5 mi/h vB = 23.7 mi/h vB = 52.8 − 1.2(15.05) vB = 34.74 ft/s 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. 53 PROBLEM 11.43 Two automobiles A and B are approaching each other in adjacent highway lanes. At t = 0, A and B are 3200 ft apart, their speeds are v A = 65 mi/h and vB = 40 mi/h, and they are at Points P and Q, respectively. Knowing that A passes Point Q 40 s after B was there and that B passes Point P 42 s after A was there, determine (a) the uniform accelerations of A and B, (b) when the vehicles pass each other, (c) the speed of B at that time. SOLUTION (a) x A = 0 + (v A ) 0 t + We have (x is positive 3200 m = (95.333 m/s)(40 s) + Also, xB = 0 + (vB )0 t + 1 aB t 2 2 Then (95.333 ft/s)t AB + (c) 1 aB (42 s) 2 2 aB = 0.83447 ft/s 2 When the cars pass each other Solving a A = −0.767 ft/s 2 (vB )0 = 40 mi/h = 58.667 ft/s 3200 ft = (58.667 ft/s)(42 s) + or or 1 a A (40 s) 2 2 ; origin at Q.) At t = 42 s: (b) (v A )0 = 65 mi/h = 95.33 ft/s ; origin at P.) At t = 40 s: (xB is positive 1 a At 2 2 aB = 0.834 ft/s 2 x A + xB = 3200 ft 1 1 2 2 + (58.667 ft/s)t AB + (0.83447 ft/s 2 )t AB = 3200 ft (−0.76667 ft/s)t AB 2 2 2 0.03390t AB + 154t AB − 3200 = 0 t = 20.685 s and t = −4563 s We have vB = ( vB ) 0 + a B t At t = t AB : vB = 58.667 ft/s + (0.83447 ft/s 2 )(20.685 s) = 75.927 ft/s t >0 t AB = 20.7 s vB = 51.8 mi/h 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. 54 PROBLEM 11.44 An elevator is moving upward at a constant speed of 4 m/s. A man standing 10 m above the top of the elevator throws a ball upward with a speed of 3 m/s. Determine (a) when the ball will hit the elevator, (b) where the ball will hit the elevator with respect to the location of the man. SOLUTION Place the origin of the position coordinate at the level of the standing man, the positive direction being up. The ball undergoes uniformly accelerated motion. yB = ( yB )0 + (vB )0 t − 1 2 gt 2 with ( yB )0 = 0, (vB )0 = 3 m/s, and g = 9.81 m/s 2 . yB = 3t − 4.905t 2 The elevator undergoes uniform motion. y E = ( y E ) 0 + vE t with ( yE )0 = −10 m and vE = 4 m/s. (a) Set yB = yE Time of impact. 3t − 4.905t 2 = −10 + 4t 4.905t 2 + t − 10 = 0 t = 1.3295 and −1.5334 (b) t = 1.330 s Location of impact. yB = (3)(1.3295) − (4.905)(1.3295)2 = −4.68 m yE = −10 + (4)(1.3295) = −4.68 m (checks) 4.68 m below the man 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. 55 PROBLEM 11.45 Two rockets are launched at a fireworks display. Rocket A is launched with an initial velocity v0 = 100 m/s and rocket B is launched t1 seconds later with the same initial velocity. The two rockets are timed to explode simultaneously at a height of 300 m as A is falling and B is rising. Assuming a constant acceleration g = 9.81 m/s2 , determine (a) the time t1, (b) the velocity of B relative to A at the time of the explosion. SOLUTION Place origin at ground level. The motion of rockets A and B is Rocket A: v A = (v A )0 − gt = 100 − 9.81t y A = ( y A ) 0 + (v A ) 0 t − Rocket B: 1 2 gt = 100t − 4.905t 2 2 (1) (2) vB = (vB )0 − g (t − t1 ) = 100 − 9.81(t − t1 ) (3) 1 g (t − t1 ) 2 2 = 100(t − t1 ) − 4.905(t − t1 )2 (4) yB = ( yB )0 + (vB )0 (t − t1 ) − Time of explosion of rockets A and B. y A = yB = 300 ft From (2), 300 = 100t − 4.905t 2 4.905t 2 − 100t + 300 = 0 t = 16.732 s and 3.655 s From (4), 300 = 100(t − t1 ) − 4.905(t − t12 ) t − t1 = 16.732 s and 3.655 s Since rocket A is falling, Since rocket B is rising, (a) Time t1: (b) Relative velocity at explosion. t = 16.732 s t − t1 = 3.655 s t1 = t − (t − t1 ) From (1), v A = 100 − (9.81)(16.732) = −64.15 m/s From (3), vB = 100 − (9.81)(16.732 − 13.08) = 64.15 m/s Relative velocity: vB/A = vB − v A t1 = 13.08 s vB/A = 128.3 m/s 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. 56 PROBLEM 11.46 Car A is parked along the northbound lane of a highway, and car B is traveling in the southbound lane at a constant speed of 60 mi/h. At t = 0, A starts and accelerates at a constant rate a A , while at t = 5 s, B begins to slow down with a constant deceleration of magnitude a A /6. Knowing that when the cars pass each other x = 294 ft and v A = vB , determine (a) the acceleration a A , (b) when the vehicles pass each other, (c) the distance d between the vehicles at t = 0. SOLUTION For t ≥ 0: v A = 0 + a At xA = 0 + 0 + 1 a At 2 2 0 ≤ t < 5 s: xB = 0 + (vB )0 t (vB )0 = 60 mi/h = 88 ft/s At t = 5 s: xB = (88 ft/s)(5 s) = 440 ft For t ≥ 5 s: vB = (vB )0 + aB (t − 5) 1 aB = − a A 6 xB = ( xB ) S + (vB )0 (t − 5) + 1 aB (t − 5) 2 2 Assume t > 5 s when the cars pass each other. At that time (t AB ), v A = vB : a At AB = (88 ft/s) − x A = 294 ft: 294 ft = Then or a A( 76 t AB − 65 ) 1 2 2 a At AB = aA (t AB − 5) 6 1 2 a At AB 2 88 294 2 44t AB − 343t AB + 245 = 0 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. 57 PROBLEM 11.46 (Continued) Solving (a) With t AB > 5 s, t AB = 0.795 s and t AB = 7.00 s 294 ft = 1 a A (7.00 s) 2 2 a A = 12.00 ft/s 2 or (b) t AB = 7.00 s From above Note: An acceptable solution cannot be found if it is assumed that t AB ≤ 5 s. (c) We have d = x + ( xB )t AB = 294 ft + 440 ft + (88 ft/s)(2.00 s) 1 1 + − × 12.00 ft/s 2 (2.00 s)2 2 6 d = 906 ft or 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. 58 PROBLEM 11.47 The elevator shown in the figure moves downward with a constant velocity of 4 m/s. Determine (a) the velocity of the cable C, (b) the velocity of the counterweight W, (c) the relative velocity of the cable C with respect to the elevator, (d ) the relative velocity of the counterweight W with respect to the elevator. SOLUTION Choose the positive direction downward. (a) Velocity of cable C. yC + 2 yE = constant vC + 2vE = 0 vE = 4 m/s But, or (b) vC = −2vE = −8 m/s vC = 8.00 m/s Velocity of counterweight W. yW + yE = constant vW + vE = 0 vW = −vE = −4 m/s (c) vW = 4.00 m/s Relative velocity of C with respect to E. vC/E = vC − vE = (−8 m/s) − ( +4 m/s) = −12 m/s vC/E = 12.00 m/s (d ) Relative velocity of W with respect to E. vW /E = vW − vE = (−4 m/s) − (4 m/s) = −8 m/s vW /E = 8.00 m/s 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. 59 PROBLEM 11.48 The elevator shown starts from rest and moves upward with a constant acceleration. If the counterweight W moves through 30 ft in 5 s, determine (a) the acceleration of the elevator and the cable C, (b) the velocity of the elevator after 5 s. SOLUTION We choose positive direction downward for motion of counterweight. yW = At t = 5 s, 1 aW t 2 2 yW = 30 ft 30 ft = 1 aW (5 s) 2 2 aW = 2.4 ft/s 2 (a) Accelerations of E and C. Since Thus: Also, Thus: (b) aW = 2.4 ft/s 2 yW + yE = constant vW + vE = 0, and aW + aE = 0 aE = − aW = −(2.4 ft/s 2 ), a E = 2.40 ft/s 2 yC + 2 yE = constant, vC + 2vE = 0, and aC + 2aE = 0 aC = −2aE = −2(−2.4 ft/s 2 ) = +4.8 ft/s 2 , aC = 4.80 ft/s 2 Velocity of elevator after 5 s. vE = (vE )0 + aE t = 0 + (−2.4 ft/s 2 )(5 s) = −12 ft/s ( v E )5 = 12.00 ft/s 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. 60 PROBLEM 11.49 Slider block A moves to the left with a constant velocity of 6 m/s. Determine (a) the velocity of block B, (b) the velocity of portion D of the cable, (c) the relative velocity of portion C of the cable with respect to portion D. SOLUTION From the diagram, we have x A + 3 yB = constant Then v A + 3vB = 0 (1) and a A + 3aB = 0 (2) (a) Substituting into Eq. (1) 6 m/s + 3vB = 0 v B = 2.00 m/s or (b) From the diagram yB + yD = constant Then vB + vD = 0 v D = 2.00 m/s (c) From the diagram x A + yC = constant Then v A + vC = 0 Now vC = −6 m/s vC/D = vC − vD = (−6 m/s) − (2 m/s) = −8 m/s vC/D = 8.00 m/s 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. 61 PROBLEM 11.50 Block B starts from rest and moves downward with a constant acceleration. Knowing that after slider block A has moved 9 in. its velocity is 6 ft/s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 2 s. SOLUTION From the diagram, we have x A + 3 yB = constant Then v A + 3vB = 0 (1) and a A + 3aB = 0 (2) (a) Eq. (2): a A + 3aB = 0 and a B is constant and positive a A is constant and negative Also, Eq. (1) and (vB )0 = 0 (v A )0 = 0 v A2 = 0 + 2a A [ x A − ( x A )0 ] Then When |Δ x A | = 0.4 m: (6 ft/s) 2 = 2a A (9/12 ft) a A = 24.0 ft/s 2 or Then, substituting into Eq. (2): −24 ft/s 2 + 3aB = 0 aB = or 24 ft/s 2 3 a B = 8.00 ft/s 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. 62 PROBLEM 11.50 (Continued) (b) We have vB = 0 + a B t At t = 2 s: 24 vB = ft/s 2 (2 s) 3 v B = 16.00 ft/s or yB = ( yB )0 + 0 + Also At t = 2 s: or y B − ( y B )0 = 1 aB t 2 2 1 24 ft/s 2 (2 s) 2 2 3 y B − (y B )0 = 16.00 ft 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. 63 PROBLEM 11.51 Slider block B moves to the right with a constant velocity of 300 mm/s. Determine (a) the velocity of slider block A, (b) the velocity of portion C of the cable, (c) the velocity of portion D of the cable, (d ) the relative velocity of portion C of the cable with respect to slider block A. SOLUTION From the diagram xB + ( xB − x A ) − 2 x A = constant Then 2vB − 3v A = 0 (1) and 2aB − 3a A = 0 (2) Also, we have vD + v A = 0 Then (a) − xD − x A = constant Substituting into Eq. (1) 2(300 mm/s) − 3v A = 0 or (b) From the diagram Then Substituting (3) v A = 200 mm/s vC = 600 mm/s xB + ( xB − xC ) = constant 2vB − vC = 0 2(300 mm/s) − vC = 0 or 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. 64 PROBLEM 11.51 (Continued) (c) From the diagram Then Substituting ( xC − x A ) + ( xD − x A ) = constant vC − 2v A + vD = 0 600 mm/s − 2(200 mm/s) + vD = 0 v D = 200 mm/s or (d) We have vC/A = vC − v A = 600 mm/s − 200 mm/s vC/A = 400 mm/s or 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. 65 PROBLEM 11.52 At the instant shown, slider block B is moving with a constant acceleration, and its speed is 150 mm/s. Knowing that after slider block A has moved 240 mm to the right its velocity is 60 mm/s, determine (a) the accelerations of A and B, (b) the acceleration of portion D of the cable, (c) the velocity and change in position of slider block B after 4 s. SOLUTION xB + ( xB − x A ) − 2 x A = constant From the diagram Then 2vB − 3v A = 0 (1) and 2aB − 3a A = 0 (2) (a) First observe that if block A moves to the right, v A → and Eq. (1) v B → . Then, using Eq. (1) at t = 0 2(150 mm/s) − 3(v A )0 = 0 (v A )0 = 100 mm/s or Also, Eq. (2) and aB = constant a A = constant v A2 = (v A )02 + 2a A [ x A − ( x A )0 ] Then When x A − ( x A )0 = 240 mm: (60 mm/s) 2 = (100 mm/s) 2 + 2a A (240 mm) or aA = − 40 mm/s 2 3 a A = 13.33 mm/s 2 or 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. 66 PROBLEM 11.52 (Continued) Then, substituting into Eq. (2) 40 2aB − 3 − mm/s 2 = 0 3 aB = −20 mm/s 2 or (b) a B = 20.0 mm/s 2 From the diagram, − xD − x A = constant vD + v A = 0 Then Substituting aD + a A = 0 40 aD + − mm/s 2 = 0 3 or (c) We have vB = ( vB ) 0 + a B t At t = 4 s: vB = 150 mm/s + ( −20.0 mm/s 2 )(4 s) or Also At t = 4 s: x B = ( xB ) 0 + ( vB ) 0 t + a D = 13.33 mm/s 2 v B = 70.0 mm/s 1 aB t 2 2 xB − ( xB )0 = (150 mm/s)(4 s) + 1 (−20.0 mm/s 2 )(4 s) 2 2 x B − (x B )0 = 440 mm or 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. 67 PROBLEM 11.53 Collar A starts from rest and moves upward with a constant acceleration. Knowing that after 8 s the relative velocity of collar B with respect to collar A is 24 in./s, determine (a) the accelerations of A and B, (b) the velocity and the change in position of B after 6 s. SOLUTION From the diagram 2 y A + yB + ( yB − y A ) = constant Then v A + 2 vB = 0 (1) and a A + 2aB = 0 (2) (a) Eq. (1) and (v A )0 = 0 (vB )0 = 0 Also, Eq. (2) and a A is constant and negative a B is constant and positive. Then Now From Eq. (2) So that v A = 0 + a At vB = 0 + a B t vB/A = vB − v A = (aB − a A )t 1 aB = − a A 2 3 vB/A = − a At 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. 68 PROBLEM 11.53 (Continued) At t = 8 s: 3 24 in./s = − a A (8 s) 2 a A = 2.00 in./s 2 or and then 1 aB = − (−2 in./s 2 ) 2 a B = 1.000 in./s 2 or (b) At t = 6 s: vB = (1 in./s 2 )(6 s) v B = 6.00 in./s or Now At t = 6 s: yB = ( yB )0 + 0 + y B − ( y B )0 = 1 aB t 2 2 1 (1 in./s 2 )(6 s) 2 2 y B − (y B )0 = 18.00 in. or 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. 69 PROBLEM 11.54 The motor M reels in the cable at a constant rate of 100 mm/s. Determine (a) the velocity of load L, (b) the velocity of pulley B with respect to load L. SOLUTION Let xB and xL be the positions, respectively, of pulley B and load L measured downward from a fixed elevation above both. Let xM be the position of a point on the cable about to enter the reel driven by the motor. Then, considering the lengths of the two cables, xM + 3xB = constant vM + 3vB = 0 xL + ( xL − xB ) = constant 2v L + v B = 0 vM = 100 mm/s with vB = − vL = vM = −33.333 m/s 3 vB = −16.667 mm/s 2 v L = 16.67 mm/s (a) Velocity of load L. (b) Velocity of pulley B with respect to load L. vB/L = vB − vL = −33.333 − (−16.667) = −16.667 v B/L = 16.67 mm/s 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. 70 PROBLEM 11.55 Block C starts from rest at t = 0 and moves downward with a constant acceleration of 4 in./s2. Knowing that block B has a constant velocity of 3 in./s upward, determine (a) the time when the velocity of block A is zero, (b) the time when the velocity of block A is equal to the velocity of block D, (c) the change in position of block A after 5 s. SOLUTION From the diagram: Cord 1: 2 y A + 2 y B + yC = constant Then 2v A + 2vB + vC = 0 and 2a A + 2aB + aC = 0 Cord 2: (1) ( y D − y A ) + ( y D − y B ) = constant Then 2vD − v A − v B = 0 and 2aD − a A − aB = 0 (2) Use units of inches and seconds. Motion of block C: vC = vC 0 + aC t where aC = −4 in./s 2 = 0 + 4t Motion of block B: vB = −3 in./s; Motion of block A: From (1) and (2), aB = 0 1 1 v A = −vB − vC = 3 − (4t ) = 3 − 2t in./s 2 2 1 1 a A = −aB − aC = 0 − (4) = −2 in./s 2 2 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. 71 PROBLEM 11.55 (Continued) (a) Time when vB is zero. 3 − 2t = 0 Motion of block D: From (3), vD = (b) t = 1.500 s 1 1 1 1 v A + vB = (3 − 2t ) − (3) = −1t 2 2 2 2 Time when vA is equal to v0. 3 − 2t = −t (c) t = 3.00 s Change in position of block A (t = 5 s). 1 a At 2 2 1 = (3)(5) + (−2)(5)2 = −10 in. 2 Δy A = ( v A ) 0 t + Change in position = 10.00 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. 72 PROBLEM 11.56 Block A starts from rest at t = 0 and moves downward with a constant acceleration of 6 in./s2. Knowing that block B moves up with a constant velocity of 3 in./s, determine (a) the time when the velocity of block C is zero, (b) the corresponding position of block C. SOLUTION The cable lengths are constant. L1 = 2 yC + 2 yD + constant L 2 = y A + yB + ( yB − yD ) + constant Eliminate yD. L1 + 2 L 2 = 2 yC + 2 yD + 2 y A + 2 yB + 2( yB − yD ) + constant 2( yC + y A + 2 yB ) = constant Differentiate to obtain relationships for velocities and accelerations, positive downward. vC + v A + 2vB = 0 (1) aC + a A + 2aB = 0 (2) Use units of inches and seconds. Motion of block A: v A = a At + 6t Δy A = Motion of block B: 1 1 a At 2 = (6)t 2 = 3t 2 2 2 v B = 3 in./s vB = −3 in./s ΔyB = vB t = −3t Motion of block C: From (1), vC = −v A − 2vB = −6t − 2(−3) = 6 − 6t ΔyC = (a) Time when vC is zero. (b) Corresponding position. t 0 vC dt = 6t − 3t 2 6 − 6t = 0 t = 1.000 s ΔyC = (6)(1) − (3)(1)2 = 3 in. ΔyC = 3.00 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. 73 PROBLEM 11.57 Block B starts from rest, block A moves with a constant acceleration, and slider block C moves to the right with a constant acceleration of 75 mm/s 2 . Knowing that at t = 2 s the velocities of B and C are 480 mm/s downward and 280 mm/s to the right, respectively, determine (a) the accelerations of A and B, (b) the initial velocities of A and C, (c) the change in position of slider block C after 3 s. SOLUTION From the diagram 3 y A + 4 yB + xC = constant Then 3v A + 4vB + vC = 0 (1) and 3a A + 4aB + aC = 0 (2) (vB ) = 0, Given: a A = constent (aC ) = 75 mm/s 2 At t = 2 s, v B = 480 mm/s vC = 280 mm/s (a) Eq. (2) and a A = constant and aC = constant aB = constant vB = 0 + a B t Then At t = 2 s: 480 mm/s = aB (2 s) aB = 240 mm/s 2 or a B = 240 mm/s 2 or a A = 345 mm/s 2 Substituting into Eq. (2) 3a A + 4(240 mm/s 2 ) + (75 mm/s 2 ) = 0 a A = −345 mm/s 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. 74 PROBLEM 11.57 (Continued) (b) vC = (vC )0 + aC t We have At t = 2 s: 280 mm/s = (vC )0 + (75 mm/s)(2 s) vC = 130 mm/s or ( vC )0 = 130.0 mm/s Then, substituting into Eq. (1) at t = 0 3(v A )0 + 4(0) + (130 mm/s) = 0 v A = −43.3 mm/s (c) We have At t = 3 s: xC = ( xC )0 + (vC )0 t + or 1 aC t 2 2 xC − ( xC )0 = (130 mm/s)(3 s) + = 728 mm (v A )0 = 43.3 mm/s 1 (75 mm/s 2 )(3 s) 2 2 or xC − (xC )0 = 728 mm 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. 75 PROBLEM 11.58 Block B moves downward with a constant velocity of 20 mm/s. At t = 0, block A is moving upward with a constant acceleration, and its velocity is 30 mm/s. Knowing that at t = 3 s slider block C has moved 57 mm to the right, determine (a) the velocity of slider block C at t = 0, (b) the accelerations of A and C, (c) the change in position of block A after 5 s. SOLUTION From the diagram 3 y A + 4 yB + xC = constant Then 3v A + 4vB + vC = 0 (1) and 3a A + 4aB + aC = 0 (2) v B = 20 mm/s ; Given: ( v A )0 = 30 mm/s (a) Substituting into Eq. (1) at t = 0 3(−30 mm/s) + 4(20 mm/s) + (vC )0 = 0 (vC )0 = 10 mm/s (b) We have At t = 3 s: ( vC )0 = 10.00 mm/s or xC = ( xC )0 + (vC )0 t + 1 aC t 2 2 57 mm = (10 mm/s)(3 s) + 1 aC (3 s)2 2 aC = 6 mm/s 2 Now or aC = 6.00 mm/s 2 v B = constant → aB = 0 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. 76 PROBLEM 11.58 (Continued) Then, substituting into Eq. (2) 3a A + 4(0) + (6 mm/s 2 ) = 0 a A = −2 mm/s 2 (c) We have At t = 5 s: y A = ( y A )0 + (v A )0 t + or a A = 2.00 mm/s 2 1 a At 2 2 y A − ( y A )0 = (−30 mm/s)(5 s) + 1 (−2 mm/s 2 )(5 s)2 2 = −175 mm y A − (y A )0 = 175.0 mm or 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. 77 PROBLEM 11.59 The system shown starts from rest, and each component moves with a constant acceleration. If the relative acceleration of block C with respect to collar B is 60 mm/s 2 upward and the relative acceleration of block D with respect to block A is 110 mm/s 2 downward, determine (a) the velocity of block C after 3 s, (b) the change in position of block D after 5 s. SOLUTION From the diagram 2 y A + 2 yB + yC = constant Cable 1: Then 2v A + 2vB + vC = 0 (1) and 2a A + 2aB + aC = 0 (2) Cable 2: ( yD − y A ) + ( yD − yB ) = constant Then and − v A − v B + 2 vD = 0 (3) − a A − aB + 2aD = 0 (4) Given: At t = 0, v = 0; all accelerations constant; aC/B = 60 mm/s 2 , aD /A = 110 mm/s 2 (a) We have aC/B = aC − aB = −60 or aB = aC + 60 and aD/A = aD − a A = 110 or a A = aD − 110 Substituting into Eqs. (2) and (4) Eq. (2): 2(aD − 110) + 2( aC + 60) + aC = 0 3aC + 2aD = 100 or Eq. (4): −(aD − 110) − ( aC + 60) + 2aD = 0 − aC + aD = −50 or (5) (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. 78 PROBLEM 11.59 (Continued) Solving Eqs. (5) and (6) for aC and aD aC = 40 mm/s 2 aD = −10 mm/s 2 Now vC = 0 + aC t At t = 3 s: vC = (40 mm/s 2 )(3 s) vC = 120.0 mm/s or (b) We have At t = 5 s: yD = ( yD )0 + (0)t + yD − ( yD )0 = 1 aD t 2 2 1 ( −10 mm/s 2 )(5 s) 2 2 y D − (y D )0 = 125.0 mm or 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. 79 PROBLEM 11.60* The system shown starts from rest, and the length of the upper cord is adjusted so that A, B, and C are initially at the same level. Each component moves with a constant acceleration, and after 2 s the relative change in position of block C with respect to block A is 280 mm upward. Knowing that when the relative velocity of collar B with respect to block A is 80 mm/s downward, the displacements of A and B are 160 mm downward and 320 mm downward, respectively, determine (a) the accelerations of A and B if aB > 10 mm/s 2 , (b) the change in position of block D when the velocity of block C is 600 mm/s upward. SOLUTION From the diagram 2 y A + 2 yB + yC = constant Cable 1: Then 2v A + 2vB + vC = 0 (1) and 2a A + 2aB + aC = 0 (2) Cable 2: ( yD − y A ) + ( yD − yB ) = constant Then − v A − vB − 2 vD = 0 (3) and − a A − aB + 2aD = 0 (4) t =0 Given: At v=0 ( y A )0 = ( yB )0 = ( yC )0 All accelerations constant. At t = 2 s yC /A = 280 mm vB /A = 80 mm/s When y A − ( y A )0 = 160 mm yB − ( yB )0 = 320 mm aB > 10 mm/s 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. 80 PROBLEM 11.60* (Continued) (a) We have y A = ( y A )0 + (0)t + 1 a At 2 2 and yC = ( yC )0 + (0)t + 1 aC t 2 2 Then yC/A = yC − y A = 1 (aC − a A )t 2 2 At t = 2 s, yC/A = −280 mm: −280 mm = or 1 (aC − a A )(2 s) 2 2 aC = a A − 140 (5) Substituting into Eq. (2) 2a A + 2aB + (a A − 140) = 0 or 1 a A = (140 − 2aB ) 3 Now vB = 0 + a B t (6) v A = 0 + a At vB/A = vB − v A = (aB − a A )t Also When yB = ( yB )0 + (0)t + v B /A = 80 mm/s : 1 aB t 2 2 80 = (aB − a A )t Δy A = 160 mm : 160 = 1 a At 2 2 ΔyB = 320 mm : 320 = 1 aB t 2 2 (7) 1 (aB − a A )t 2 2 Then 160 = Using Eq. (7) 320 = (80)t or t = 4 s Then 160 = 1 a A (4)2 2 or a A = 20.0 mm/s 2 and 320 = 1 aB (4) 2 2 or a B = 40.0 mm/s 2 Note that Eq. (6) is not used; thus, the problem is over-determined. 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. 81 PROBLEM 11.60* (Continued) (b) Substituting into Eq. (5) aC = 20 − 140 = −120 mm/s 2 and into Eq. (4) −(20 mm/s 2 ) − (40 mm/s 2 ) + 2aD = 0 or aD = 30 mm/s 2 Now vC = 0 + aC t When vC = −600 mm/s: or Also At t = 5 s: −600 mm/s = ( −120 mm/s 2 )t t =5s yD = ( yD )0 + (0)t + yD − ( yD )0 = 1 aD t 2 2 1 (30 mm/s 2 )(5 s)2 2 y D − (y D )0 = 375 mm or 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. 82 PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate. SOLUTION v0 = −14 ft/s Change in v = area under a−t curve. t = 0 to t = 2 s: v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s v2 = −8 ft/s t = 2 s to t = 5 s: v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s v5 = +16 ft/s t = 5 s to t = 8 s: v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s v8 = +25 ft/s t = 8 s to t = 10 s: v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s v10 = +15 ft/s t = 10 s to t = 15 s: v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s v15 = −10 ft/s 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. 83 PROBLEM 11.61 (Continued) Plot v−t curve. Then by similar triangles Δ’s find t for v = 0. x0 = 0 Change in x = area under v−t curve t = 0 to t = 2 s: 1 x2 − x0 = (−14 − 8)(2) = −22 ft 2 x2 = −22 ft t = 2 s to t = 3 s: 1 x3 − x2 = (−8)(1) = −4 ft 2 x8 = −26 ft t = 3 s to t = 5 s: 1 x5 − x3 = (+16)(2) = +16 ft 2 x5 = −10 ft t = 5 s to t = 8 s: 1 x8 − x5 = (+16 + 25)(3) = +61.5 ft 2 x8 = +51.6 ft t = 8 s to t = 10 s: 1 x10 − x8 = (+25 + 15)(2) = + 40 ft 2 x10 = +91.6 ft t = 10 s to t = 13 s: 1 x13 − x10 = (+15)(3) = +22.5 ft 2 x13 = +114 ft t = 13 s to t = 15 s: 1 x15 − x13 = (−10)(2) = −10 ft 2 x15 = +94 ft (a) Maximum velocity: When t = 8 s, (b) Maximum x: When t = 13 s, vm = 25.0 ft/s xm = 114.0 ft 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. 84 PROBLEM 11.62 For the particle and motion of Problem 11.61, plot the v−t and x−t curves for 0 < t < 15 s and determine the velocity of the particle, its position, and the total distance traveled after 10 s. PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate. SOLUTION v0 = −14 ft/s Change in v = area under a−t curve. t = 0 to t = 2 s: v2 − v0 = (3 ft/s 2 )(2 s) = +6 ft/s v2 = −8 ft/s t = 2 s to t = 5 s: v5 − v2 = (8 ft/s2 )(3 s) = +24 ft/s v5 = +16 ft/s t = 5 s to t = 8 s: v8 − v5 = (3 ft/s 2 )(3 s) = +9 ft/s v8 = +25 ft/s t = 8 s to t = 10 s: v10 − v8 = (−5 ft/s2 )(2 s) = −10 ft/s v10 = +15 ft/s t = 10 s to t = 15 s: v15 − v10 = (−5 ft/s2 )(5 s) = −25 ft/s v15 = −10 ft/s 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. 85 PROBLEM 11.62 (Continued) Plot v−t curve. Then by similar triangles Δ’s find t for v = 0. x0 = 0 Change in x = area under v−t curve t = 0 to t = 2 s: 1 x2 − x0 = (−14 − 8)(2) = −22 ft 2 x2 = −22 ft t = 2 s to t = 3 s: 1 x3 − x2 = (−8)(1) = −4 ft 2 x8 = −26 ft t = 3 s to t = 5 s: 1 x5 − x3 = (+16)(2) = +16 ft 2 x5 = −10 ft t = 5 s to t = 8 s: 1 x8 − x5 = (+16 + 25)(3) = +61.5 ft 2 x8 = +51.6 ft t = 8 s to t = 10 s: 1 x10 − x8 = (+25 + 15)(2) = + 40 ft 2 x10 = +91.6 ft t = 10 s to t = 13 s: 1 x13 − x10 = (+15)(3) = +22.5 ft 2 x13 = +114 ft t = 13 s to t = 15 s: 1 x15 − x13 = (−10)(2) = −10 ft 2 x15 = +94 ft when t = 10 s: v10 = +15 ft/s x10 = +91.5 ft/s Distance traveled: t = 0 to t = 105 t = 0 to t = 3 s: Distance traveled = 26 ft t = 3 s to t = 10 s Distance traveled = 26 ft + 91.5 ft = 117.5 ft Total distance traveled = 26 + 117.5 = 143.5 ft 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. 86 PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a −t and x−t curves for 0 < t < 50 s, and determine (b) the total distance traveled by the particle when t = 50 s, (c) the two times at which x = 0. SOLUTION (a) at = slope of v −t curve at time t From t = 0 to t = 10 s: v = constant a = 0 −20 − 60 = −5 m/s 2 26 − 10 t = 10 s to t = 26 s: a= t = 26 s to t = 41 s: v = constant a = 0 t = 41 s to t = 46 s: a= t = 46 s: −5 − (−20) = 3 m/s 2 46 − 41 v = constant a = 0 x2 = x1 + (area under v −t curve from t1 to t2 ) At t = 10 s: x10 = −540 + 10(60) + 60 m Next, find time at which v = 0. Using similar triangles tv = 0 − 10 60 At t = 22 s: t = 26 s: t = 41 s: t = 46 s: t = 50 s: = 26 − 10 80 or tv = 0 = 22 s 1 x22 = 60 + (12)(60) = 420 m 2 1 x26 = 420 − (4)(20) = 380 m 2 x41 = 380 − 15(20) = 80 m 20 + 5 x46 = 80 − 5 = 17.5 m 2 x50 = 17.5 − 4(5) = −2.5 m 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. 87 PROBLEM 11.63 (Continued) (b) From t = 0 to t = 22 s: Distance traveled = 420 − (−540) = 960 m t = 22 s to t = 50 s: Distance traveled = |− 2.5 − 420| = 422.5 m Total distance traveled = (960 + 422.5) ft = 1382.5 m Total distance traveled = 1383 m (c) Using similar triangles Between 0 and 10 s: (t x = 0 )1 − 0 10 = 540 600 (t x = 0 )1 = 9.00 s Between 46 s and 50 s: (t x = 0 )2 − 46 4 = 17.5 20 (t x =0 )2 = 49.5 s 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. 88 PROBLEM 11.64 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a −t and x −t curves for 0 < t < 50 s, and determine (b) the maximum value of the position coordinate of the particle, (c) the values of t for which the particle is at x = 100 m. SOLUTION (a) at = slope of v −t curve at time t v = constant a = 0 From t = 0 to t = 10 s: −20 − 60 = −5 m/s 2 26 − 10 t = 10 s to t = 26 s: a= t = 26 s to t = 41 s: v = constant a = 0 t = 41 s to t = 46 s: a= −5 − (−20) = 3 m/s 2 46 − 41 v = constant a = 0 t = 46 s: x2 = x1 + (area under v −t curve from t1 to t2 ) At t = 10 s: x10 = −540 + 10(60) = 60 m Next, find time at which v = 0. Using similar triangles tv = 0 − 10 60 At t = 22 s: t = 26 s: t = 41 s: t = 46 s: t = 50 s: = 26 − 10 80 or tv = 0 = 22 s 1 x22 = 60 + (12)(60) = 420 m 2 1 x26 = 420 − (4)(20) = 380 m 2 x41 = 380 − 15(20) = 80 m 20 + 5 x46 = 80 − 5 = 17.5 m 2 x50 = 17.5 − 4(5) = −2.5 m 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. 89 PROBLEM 11.64 (Continued) (b) Reading from the x −t curve (c) Between 10 s and 22 s xmax = 420 m 100 m = 420 m − (area under v −t curve from t , to 22 s) m 100 = 420 − 1 (22 − t1 )(v1 ) 2 (22 − t1 )(v1 ) = 640 Using similar triangles v1 60 = 22 − t1 12 Then v1 = 5(22 − t1 ) or (22 − t1 )[5(22 − t1 )] = 640 t1 = 10.69 s and t1 = 33.3 s 10 s < t1 < 22 s We have t1 = 10.69 s Between 26 s and 41 s: Using similar triangles 41 − t2 15 = 20 300 t2 = 40.0 s 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. 90 PROBLEM 11.65 During a finishing operation the bed of an industrial planer moves alternately 30 in. to the right and 30 in. to the left. The velocity of the bed is limited to a maximum value of 6 in./s to the right and 12 in./s to the left; the acceleration is successively equal to 6 in./s2 to the right, zero 6 in./s2 to the left, zero, etc. Determine the time required for the bed to complete a full cycle, and draw the v−t and x−t curves. SOLUTION We choose positive to the right, thus the range of permissible velocities is −12 in./s < v < 6 in./s since acceleration is −6 in./s 2 , 0, or + 6 in./s 2 . The slope the v−t curve must also be −6 in./s2, 0, or +6 in./s2. Planer moves = 30 in. to right: +30 in. = 3 + 6t1 + 3 t1 = 4.00 s Planer moves = 30 in. to left: −30 in. = −12 − 12t2 + 12 t2 = 0.50 s Total time = 1 s + 4 s + 1 s + 2 s + 0.5 s + 2 s = 10.5 s ttotal = 10.50 s 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. 91 PROBLEM 11.66 A parachutist is in free fall at a rate of 200 km/h when he opens his parachute at an altitude of 600 m. Following a rapid and constant deceleration, he then descends at a constant rate of 50 km/h from 586 m to 30 m, where he maneuvers the parachute into the wind to further slow his descent. Knowing that the parachutist lands with a negligible downward velocity, determine (a) the time required for the parachutist to land after opening his parachute, (b) the initial deceleration. SOLUTION Assume second deceleration is constant. Also, note that 200 km/h = 55.555 m/s, 50 km/h = 13.888 m/s 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. 92 PROBLEM 11.66 (Continued) (a) Now Δ x = area under v −t curve for given time interval Then 55.555 + 13.888 (586 − 600) m = −t1 m/s 2 t1 = 0.4032 s (30 − 586) m = −t2 (13.888 m/s) t2 = 40.0346 s 1 (0 − 30) m = − (t3 )(13.888 m/s) 2 t3 = 4.3203 s ttotal = (0.4032 + 40.0346 + 4.3203) s (b) We have ainitial = = ttotal = 44.8 s Δ vinitial t1 [−13.888 − (−55.555)] m/s 0.4032 s = 103.3 m/s 2 ainitial = 103.3 m/s 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. 93 PROBLEM 11.67 A commuter train traveling at 40 mi/h is 3 mi from a station. The train then decelerates so that its speed is 20 mi/h when it is 0.5 mi from the station. Knowing that the train arrives at the station 7.5 min after beginning to decelerate and assuming constant decelerations, determine (a) the time required for the train to travel the first 2.5 mi, (b) the speed of the train as it arrives at the station, (c) the final constant deceleration of the train. SOLUTION Given: At t = 0, v = 40 mi/h, x = 0; when x = 2.5 mi, v = 20 mi/h; at t = 7.5 min, x = 3 mi; constant decelerations. The v −t curve is first drawn as shown. (a) A1 = 2.5 mi We have 1h 40 + 20 (t1 min) mi/h × = 2.5 mi 60 min 2 t1 = 5.00 min (b) A2 = 0.5 mi We have 20 + v2 (7.5 − 5) min × 2 1h mi/h × 60 min = 0.5 mi v2 = 4.00 mi/h (c) We have afinal = a12 = (4 − 20) mi/h 5280 ft 1 min 1h × × × (7.5 − 5) min mi 60 s 3600 s afinal = −0.1564 ft/s 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. 94 PROBLEM 11.68 A temperature sensor is attached to slider AB which moves back and forth through 60 in. The maximum velocities of the slider are 12 in./s to the right and 30 in./s to the left. When the slider is moving to the right, it accelerates and decelerates at a constant rate of 6 in./s2; when moving to the left, the slider accelerates and decelerates at a constant rate of 20 in./s2. Determine the time required for the slider to complete a full cycle, and construct the v – t and x −t curves of its motion. SOLUTION The v −t curve is first drawn as shown. Then ta = vright aright = 12 in./s =2s 6 in./s 2 vleft 30 in./s = aleft 20 in./s = 1.5 s td = A1 = 60 in. Now [(t1 − 2) s](12 in./s) = 60 in. or t1 = 7 s or A2 = 60 in. and or {[(t2 − 7) − 1.5] s}(30 in./s) = 60 in. or t2 = 10.5 s tcycle = t2 Now tcycle = 10.5 s We have xii = xi + (area under v −t curve from ti to tii ) At t = 5 s: 1 (2) (12) = 12 in. 2 x5 = 12 + (5 − 2)(12) t = 7 s: = 48 in. x7 = 60 in. t = 2 s: t = 8.5 s: t = 9 s: t = 10.5 s: x2 = 1 x8.5 = 60 − (1.5)(30) 2 = 37.5 in. x9 = 37.5 − (0.5)(30) = 22.5 in. x10.5 = 0 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. 95 PROBLEM 11.69 In a water-tank test involving the launching of a small model boat, the model’s initial horizontal velocity is 6 m/s, and its horizontal acceleration varies linearly from −12 m/s 2 at t = 0 to −2 m/s 2 at t = t1 and then remains equal to −2 m/s 2 until t = 1.4 s. Knowing that v = 1.8 m/s when t = t1 , determine (a) the value of t1 , (b) the velocity and the position of the model at t = 1.4 s. SOLUTION Given: v0 = 6 m/s; for 0 < t < t1 , for t1 < t < 1.4 s a = −2 m/s 2 ; at t = 0 a = −12 m/s 2 ; at t = t1 a = −2 m/s 2 , v = 1.8 m/s 2 The a −t and v −t curves are first drawn as shown. The time axis is not drawn to scale. (a) vt1 = v0 + A1 We have 12 + 2 1.8 m/s = 6 m/s − (t1 s) m/s2 2 t1 = 0.6 s (b) v1.4 = vt1 + A2 We have v1.4 = 1.8 m/s − (1.4 − 0.6) s × 2 m/s 2 v1.4 = 0.20 m/s Now x1.4 = A3 + A4 , where A3 is most easily determined using integration. Thus, a= for 0 < t < t1: −2 − (−12) 50 t − 12 = t − 12 0.6 3 dv 50 = a = t − 12 dt 3 Now 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. 96 PROBLEM 11.69 (Continued) At t = 0, v = 6 m/s: or v 6 dv = t 50 t − 12 dt 0 3 v=6+ 25 2 t − 12t 3 We have dx 25 = v = 6 − 12t + t 2 dt 3 Then A3 = xt1 0 dx = 0.6 0 (6 − 12t + 25 2 t )dt 3 0.6 25 = 6t − 6t 2 + t 3 = 2.04 m 9 0 Also 1.8 + 0.2 A4 = (1.4 − 0.6) = 0.8 m 2 Then x1.4 = (2.04 + 0.8) m x1.4 = 2.84 m or 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. 97 PROBLEM 11.70 The acceleration record shown was obtained for a small airplane traveling along a straight course. Knowing that x = 0 and v = 60 m/s when t = 0, determine (a) the velocity and position of the plane at t = 20 s, (b) its average velocity during the interval 6 s < t < 14 s. SOLUTION Geometry of “bell-shaped” portion of v −t curve The parabolic spandrels marked by * are of equal area. Thus, total area of shaded portion of v −t diagram is: = Δx = 6 m (a) When t = 20 s: v20 = 60 m/s x20 = (60 m/s) (20 s) − (shaded area) = 1200 m − 6 m (b) From t = 6 s to t = 14 s: x20 = 1194 m Δt = 8 s Δ x = (60 m/s)(14 s − 6 s) − (shaded area) = (60 m/s)(8 s) − 6 m = 480 m − 6 m = 474 m vaverage = Δ x 474 m = Δt 8s vaverage = 59.25 m/s 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. 98 PROBLEM 11.71 In a 400-m race, runner A reaches her maximum velocity v A in 4 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25 s. Runner B reaches her maximum velocity vB in 5 s with constant acceleration and maintains that velocity until she reaches the half-way point with a split time of 25.2 s. Both runners then run the second half of the race with the same constant deceleration of 0.1 m/s 2. Determine (a) the race times for both runners, (b) the position of the winner relative to the loser when the winner reaches the finish line. SOLUTION Sketch v −t curves for first 200 m. Runner A: t1 = 4 s, t2 = 25 − 4 = 21 s A1 = 1 (4)(v A )max = 2(vA ) max 2 A2 = 21(v A )max A1 + A2 = Δ x = 200 m 23(v A ) max = 200 Runner B: or t1 = 5 s, A1 = (v A ) max = 8.6957 m/s t2 = 25.2 − 5 = 20.2 s 1 (5)(vB ) max = 2.5(vB ) max 2 A2 = 20.2(vB ) max A1 + A2 = Δ x = 200 m 22.7(vB ) max = 200 Sketch v −t curve for second 200 m. A3 = vmaxt3 − t3 = Runner A: or (vB ) max = 8.8106 m/s Δv = | a |t3 = 0.1t3 1 Δvt3 = 200 2 or 0.05t32 − vmaxt3 + 200 = 0 ( vmax ± (vmax ) 2 − (4)(0.05)(200) = 10 vmax ± (vmax )2 − 40 (2)(0.05) (vmax ) A = 8.6957, Reject the larger root. Then total time (t3 ) A = 146.64 s and ) 27.279 s t A = 25 + 27.279 = 52.279 s t A = 52.2 s 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. 99 PROBLEM 11.71 (Continued) Runner B: (vmax ) B = 8.8106, (t3 ) B = 149.45 s and 26.765 s t B = 25.2 + 26.765 = 51.965 s Reject the larger root. Then total time t B = 52.0 s Velocity of A at t = 51.965 s: v1 = 8.6957 − (0.1)(51.965 − 25) = 5.999 m/s Velocity of A at t = 51.279 s: v2 = 8.6957 − (0.1)(52.279 − 25) = 5.968 m/s Over 51.965 s ≤ t ≤ 52.965 s, runner A covers a distance Δ x Δ x = vave (Δt ) = 1 (5.999 + 5.968)(52.279 − 51.965) 2 Δ x = 1.879 m 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. 100 PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h? Draw the v −t curve. SOLUTION Relative to truck, car must move a distance: Allowable increase in speed: Δ x = 16 + 40 + 50 + 40 = 146 ft Δvm = 50 − 35 = 15 mi/h = 22 ft/s Acceleration Phase: t1 = 22/5 = 4.4 s A1 = 1 (22)(4.4) = 48.4 ft 2 Deceleration Phase: t3 = 22/20 = 1.1 s A3 = 1 (22)(1.1) = 12.1 ft 2 But: Δ x = A1 + A2 + A3 : 146 ft = 48.4 + (22)t2 + 12.1 ttotal = t1 + t2 + t3 = 4.4 s + 3.89 s + 1.1 s = 9.39 s t2 = 3.89 s t B = 9.39 s 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. 101 PROBLEM 11.73 Solve Problem 11.72, assuming that the driver of the car does not pay any attention to the speed limit while passing and concentrates on reaching position B and resuming a speed of 35 mi/h in the shortest possible time. What is the maximum speed reached? Draw the v −t curve. PROBLEM 11.72 A car and a truck are both traveling at the constant speed of 35 mi/h; the car is 40 ft behind the truck. The driver of the car wants to pass the truck, i.e., he wishes to place his car at B, 40 ft in front of the truck, and then resume the speed of 35 mi/h. The maximum acceleration of the car is 5 ft/s2 and the maximum deceleration obtained by applying the brakes is 20 ft/s 2 . What is the shortest time in which the driver of the car can complete the passing operation if he does not at any time exceed a speed of 50 mi/h? Draw the v −t curve. SOLUTION Relative to truck, car must move a distance: Δ x = 16 + 40 + 50 + 40 = 146 ft Δvm = 5t1 = 20t2 ; Δ x = A1 + A2 : t2 = 146 ft = 1 (Δvm )(t1 + t2 ) 2 146 ft = 1 1 (5t1 ) t1 + t1 2 4 t12 = 46.72 1 t1 4 t1 = 6.835 s t2 = 1 t1 = 1.709 4 ttotal = t1 + t2 = 6.835 + 1.709 t B = 8.54 s Δvm = 5t1 = 5(6.835) = 34.18 ft/s = 23.3 mi/h Speed vtotal = 35 mi/h, vm = 35 mi/h + 23.3 mi/h vm = 58.3 mi/h 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. 102 PROBLEM 11.74 Car A is traveling on a highway at a constant speed (v A )0 = 60 mi/h, and is 380 ft from the entrance of an access ramp when car B enters the acceleration lane at that point at a speed (vB )0 = 15 mi/h. Car B accelerates uniformly and enters the main traffic lane after traveling 200 ft in 5 s. It then continues to accelerate at the same rate until it reaches a speed of 60 mi/h, which it then maintains. Determine the final distance between the two cars. SOLUTION Given: (v A )0 = 60 mi/h, (vB )0 = 1.5 mi/h; at t = 0, ( x A )0 = −380 ft, ( xB ) 0 = 0; at t = 5 s, xB = 200 ft; for 15 mi/h < vB ≤ 60 mi/h, aB = constant; for vB = 60 mi/h, aB = 0 First note 60 mi/h = 88 ft/s 15 mi/h = 22 ft/s The v −t curves of the two cars are then drawn as shown. Using the coordinate system shown, we have at t = 5 s, xB = 200 ft: 22 +(vB )5 (5 s) ft/s = 200 ft 2 (vB )5 = 58 ft/s or Then, using similar triangles, we have (88 − 22) ft/s (58 − 22) ft/s = ( = aB ) t1 5s or t1 = 9.16 67 s Finally, at t = t1 22 + 88 ft/s xB/A = xB − x A = (9.1667 s) 2 − [−380 ft + (9.1667 s) (88 ft/s)] xB/A = 77.5 ft or 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. 103 PROBLEM 11.75 An elevator starts from rest and moves upward, accelerating at a rate of 1.2 m/s2 until it reaches a speed of 7.8 m/s, which it then maintains. Two seconds after the elevator begins to move, a man standing 12 m above the initial position of the top of the elevator throws a ball upward with an initial velocity of 20 m/s. Determine when the ball will hit the elevator. SOLUTION Given: At t = 0 vE = 0; For 0 < vE ≤ 7.8 m/s, aE = 1.2 m/s 2 ↑; For vE = 7.8 m/s, aE = 0; At t = 2 s, vB = 20m/s ↑ The v −t curves of the ball and the elevator are first drawn as shown. Note that the initial slope of the curve for the elevator is 1.2 m/s2 , while the slope of the curve for the ball is − g (−9.81 m/s 2 ). The time t1 is the time when vE reaches 7.8 m/s. Thus, vE = (0) + aE t or 7.8 m/s = (1.2 m/s 2 )t1 or t1 = 6.5 s The time ttop is the time at which the ball reaches the top of its trajectory. Thus, or or vB = (vB )0 − g (t − 2) 0 = 20 m/s − (9.81 m/s 2 ) (t top − 2) s ttop = 4.0387 s 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. 104 PROBLEM 11.75 (Continued) Using the coordinate system shown, we have 0 < t < t1 : 1 yE = −12 m + aE t 2 m 2 At t = ttop : yB = and y E = −12 m + At and at t = t1 , 1 (4.0387 − 2) s × (20 m/s) 2 = 20.387 m 1 (1.2 m/s 2 )(4.0387 s) 2 2 = −2.213 m t = [2 + 2(4.0387 − 2)] s = 6.0774 s, yB = 0 yE = −12 m + 1 (6.5 s) (7.8 m/s) = 13.35 m 2 The ball hits the elevator ( yB = yE ) when ttop ≤ t ≤ t1. For t ≥ ttop : 1 yB = 20.387 m − g (t − ttop )2 m 2 Then, when yB = yE 1 20.387 m − (9.81 m/s 2 ) (t − 4.0387)2 2 1 = −12 m + (1.2 m/s 2 ) (t s)2 2 or Solving 5.505t 2 − 39.6196t + 47.619 = 0 t = 1.525 s and t = 5.67 s t = 5.67 s Since 1.525 s is less than 2 s, 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. 105 PROBLEM 11.76 Car A is traveling at 40 mi/h when it enters a 30 mi/h speed zone. The driver of car A decelerates at a rate of 16 ft/s2 until reaching a speed of 30 mi/h, which she then maintains. When car B, which was initially 60 ft behind car A and traveling at a constant speed of 45 mi/h, enters the speed zone, its driver decelerates at a rate of 20 ft/s2 until reaching a speed of 28 mi/h. Knowing that the driver of car B maintains a speed of 28 mi/h, determine (a) the closest that car B comes to car A, (b) the time at which car A is 70 ft in front of car B. SOLUTION (v A )0 = 40 mi/h; For 30 mi/h < v A ≤ 40 mi/h, a A = −16 ft/s 2 ; For v A = 30 mi/h, a A = 0; Given: ( x A /B )0 = 60 ft; (vB )0 = 45 mi/h; When xB = 0, aB = −20 ft/s 2 ; For vB = 28 mi/h, aB = 0 First note 40 mi/h = 58.667 ft/s 30 mi/h = 44 ft/s 45 mi/h = 66 ft/s 28 mi/h = 41.067 ft/s At t = 0 The v −t curves of the two cars are as shown. At t = 0: Car A enters the speed zone. t = (tB )1: Car B enters the speed zone. t = tA: Car A reaches its final speed. t = tmin : v A = vB t = (t B )2 : Car B reaches its final speed. 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. 106 PROBLEM 11.76 (Continued) (a) (v A )final − (v A )0 tA aA = We have (44 − 58.667) ft/s tA −16 ft/s 2 = or t A = 0.91669 s or Also 60 ft = (tB )1 (vB )0 or 60 ft = (tB )1 (66 ft/s) aB = and −20 ft/s 2 = or or (t B )1 = 0.90909 s (vB )final − (vB )0 (t B ) 2 − (t B )1 (41.067 − 66) ft/s [(t B )2 − 0.90909] s Car B will continue to overtake car A while vB > v A . Therefore, ( x A/B ) min will occur when v A = vB , which occurs for (tB )1 < tmin < (tB ) 2 For this time interval v A = 44 ft/s vB = (vB )0 + aB [t − (tB )1 ] Then at t = tmin : 44 ft/s = 66 ft/s + (−20 ft/s2 )(tmin − 0.90909) s tmin = 2.00909 s or Finally ( x A/B )min = ( x A )tmin − ( xB )tmin (v ) + (v A )final = t A A 0 + (tmin − t A )(v A )final 2 (v ) + (v A )final − ( xB )0 + (t B )1 (vB )0 + [tmin − (t B )1 ] B 0 2 58.667 + 44 = (0.91669 s) ft/s + (2.00909 − 0.91669) s × (44 ft/s) 2 66 + 44 − −60 ft + (0.90909 s)(66 ft/s) + (2.00909 − 0.90909) s × ft/s 2 = (47.057 + 48.066) ft − (−60 + 60.000 + 60.500) ft = 34.623 ft or ( x A/B ) min = 34.6 ft 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. 107 PROBLEM 11.76 (Continued) (b) Since ( x A/B ) ≤ 60 ft for t ≤ tmin , it follows that x A/B = 70 ft for t > (tB ) 2 [Note (tB ) 2 tmin ]. Then, for t > (tB ) 2 x A/B = ( x A/B )min + [(t − tmin )(vA )final ] + (vB )final (v ) − [(t B )2 − (tmin )] A final + [t − (t B )2 ](vB )final 2 or 70 ft = 34.623 ft + [(t − 2.00909) s × (44 ft/s)] 44 + 41.06 − (2.15574 − 2.00909) s × ft/s + (t − 2.15574) s × (41.067) ft/s 2 t = 14.14 s or 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. 108 PROBLEM 11.77 An accelerometer record for the motion of a given part of a mechanism is approximated by an arc of a parabola for 0.2 s and a straight line for the next 0.2 s as shown in the figure. Knowing that v = 0 when t = 0 and x = 0.8 ft when t = 0.4 s, (a) construct the v −t curve for 0 ≤ t ≤ 0.4 s, (b) determine the position of the part at t = 0.3 s and t = 0.2 s. SOLUTION Divide the area of the a −t curve into the four areas A1, A2 , A3 and A4. 2 (8)(0.2) = 1.0667 ft/s 3 A2 = (16)(0.2) = 3.2 ft/s A1 = 1 (16 + 8)(0.1) = 1.2 ft/s 2 1 A4 = (8)(0.1) = 0.4 ft/s 2 A3 = Velocities: v0 = 0 v0.2 = v0 + A1 + A2 v0.2 = 4.27 ft/s v0.3 = v0.2 + A3 v0.3 = 5.47 ft/s v0.4 = v0.3 + A4 v0.4 = 5.87 ft/s Sketch the v −t curve and divide its area into A5 , A6 , and A7 as shown. 0.8 0.4 x dx = 0.8 − x = t vdt 2 (0.4)(0.1) = 0.0267 ft, 3 x0.3 = 0.227 ft x0.2 = 0.8 − ( A5 + A6 ) − A7 At t = 0.2 s, With A5 + A6 = and 0.4 x = 0.8 − t vdt x0.3 = 0.8 − A5 − (5.47)(0.1) At t = 0.3 s, With A5 = or 2 (1.6)(0.2) = 0.2133 ft, 3 A7 = (4.27)(0.2) = 0.8533 ft x0.2 = 0.8 − 0.2133 − 0.8533 x0.2 = −0.267 ft 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. 109 PROBLEM 11.78 A car is traveling at a constant speed of 54 km/h when its driver sees a child run into the road. The driver applies her brakes until the child returns to the sidewalk and then accelerates to resume her original speed of 54 km/h; the acceleration record of the car is shown in the figure. Assuming x = 0 when t = 0, determine (a) the time t1 at which the velocity is again 54 km/h, (b) the position of the car at that time, (c) the average velocity of the car during the interval 1 s ≤ t ≤ t1. SOLUTION Given: At t = 0, x = 0, v = 54 km/h; For t = t1 , v = 54 km/h First note (a) 54 km/h = 15 m/s vb = va + (area under a −t curve from ta to tb ) We have Then at t = 2 s: v = 15 − (1)(6) = 9 m/s t = 4.5 s: 1 v = 9 − (2.5)(6) = 1.5 m/s 2 t = t1 : 15 = 1.5 + 1 (t1 − 4.5)(2) 2 t1 = 18.00 s or (b) Using the above values of the velocities, the v −t curve is drawn as shown. 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. 110 PROBLEM 11.78 (Continued) Now x at t = 18 s x18 = 0 + Σ (area under the v −t curve from t = 0 to t = 18 s) 15 + 9 = (1 s)(15 m/s) + (1 s) m/s 2 1 + (2.5 s)(1.5 m/s)+ (2.5 s)(7.5 m/s) 3 2 + (13.5 s)(1.5 m/s) + (13.5 s)(13.5 m/s) 3 = [15 + 12 + (3.75 + 6.25) + (20.25 + 121.50)] m = 178.75 m (c) First note or x18 = 178.8 m x1 = 15 m x18 = 178.75 m Now vave = Δ x (178.75 − 15) m = = 9.6324 m/s Δt (18 − 1) s vave = 34.7 km/h or 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. 111 PROBLEM 11.79 An airport shuttle train travels between two terminals that are 1.6 mi apart. To maintain passenger comfort, the acceleration of the train is limited to ±4 ft/s2, and the jerk, or rate of change of acceleration, is limited to ±0.8 ft/s2 per second. If the shuttle has a maximum speed of 20 mi/h, determine (a) the shortest time for the shuttle to travel between the two terminals, (b) the corresponding average velocity of the shuttle. SOLUTION xmax = 1.6 mi; | amax | = 4 ft/s 2 Given: da = 0.8 ft/s2 /s; vmax = 20 mi/h dt max First note (a) 20 mi/h = 29.333 ft/s 1.6 mi = 8448 ft To obtain tmin , the train must accelerate and decelerate at the maximum rate to maximize the time for which v = vmax . The time Δ t required for the train to have an acceleration of 4 ft/s2 is found from a da = max dt Δt max or or 4 ft/s2 0.8 ft/s2 /s Δt = 5 s Δt = Now, da since dt = constant after 5 s, the speed of the train is v5 = 1 (Δt )(amax ) 2 or v5 = 1 (5 s)(4 ft/s 2 ) = 10 ft/s 2 Then, since v5 < vmax , the train will continue to accelerate at 4 ft/s2 until v = vmax . The a −t curve must then have the shape shown. Note that the magnitude of the slope of each inclined portion of the curve is 0.8 ft/s2/s. 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. 112 PROBLEM 11.79 (Continued) Now at t = (10 + Δt1 ) s, v = vmax : 1 2 (5 s)(4 ft/s2 ) + (Δt1 )(4 ft/s2 ) = 29.333 ft/s 2 or Δt1 = 2.3333 s Then at t = 5 s: t = 7.3333 s: 1 (5)(4) = 10 ft/s 2 v = 10 + (2.3333)(4) = 19.3332 ft/s v =0+ 1 t = 12.3333 s: v = 19.3332 + (5)(4) = 29.3332 ft/s 2 Using symmetry, the v −t curve is then drawn as shown. Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have 10 + 19.3332 2 (2.3333 s) ft/s 2 + (10 + Δt2 ) s × (29.3332 ft/s) = 8448 ft or Δt2 = 275.67 s Then tmin = 4(5 s) + 2(2.3333 s) + 275.67 s = 300.34 s tmin = 5.01 min or (b) We have vave = Δx 1.6 mi 3600 s = × Δt 300.34 s 1h vave = 19.18 mi/h or 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. 113 PROBLEM 11.80 During a manufacturing process, a conveyor belt starts from rest and travels a total of 1.2 ft before temporarily coming to rest. Knowing that the jerk, or rate of change of acceleration, is limited to ±4.8 ft/s2 per second, determine (a) the shortest time required for the belt to move 1.2 ft, (b) the maximum and average values of the velocity of the belt during that time. SOLUTION Given: (a) At t = 0, x = 0, v = 0; xmax = 1.2 ft; when da x = xmax , v = 0; = 4.8 ft/s 2 dt max Observing that vmax must occur at t = 12 tmin , the a −t curve must have the shape shown. Note that the magnitude of the slope of each portion of the curve is 4.8 ft/s2/s. We have at t = Δt : t = 2Δt : 1 1 v = 0 + (Δt )(amax ) = amax Δt 2 2 vmax = 1 1 amax Δt + ( Δt )( amax ) = amax Δt 2 2 Using symmetry, the v −t is then drawn as shown. Noting that A1 = A2 = A3 = A4 and that the area under the v −t curve is equal to xmax , we have (2Δt )(vmax ) = xmax vmax = amax Δt 2amax Δt 2 = xmax 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. 114 PROBLEM 11.80 (Continued) Now amax = 4.8 ft/s2 /s so that Δt 2(4.8Δt ft/s3 )Δt 2 = 1.2 ft or Then Δt = 0.5 s tmin = 4Δt tmin = 2.00 s or (b) We have vmax = amax Δt = (4.8 ft/s 2 /s × Δt)Δt = 4.8 ft/s 2 /s × (0.5 s) 2 vmax = 1.2 ft/s or Also vave = Δx 1.2 ft = Δttotal 2.00 s vave = 0.6 ft/s or 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. 115 PROBLEM 11.81 Two seconds are required to bring the piston rod of an air cylinder to rest; the acceleration record of the piston rod during the 2 s is as shown. Determine by approximate means (a) the initial velocity of the piston rod, (b) the distance traveled by the piston rod as it is brought to rest. SOLUTION a −t curve; at Given: 1. t = 2 s, v = 0 The a −t curve is first approximated with a series of rectangles, each of width Δt = 0.25 s. The area (Δt)(aave) of each rectangle is approximately equal to the change in velocity Δv for the specified interval of time. Thus, Δv ≅ aave Δt where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table. 2. v(2) = v0 + Now and approximating the area 2 0 2 0 a dt = 0 a dt under the a −t curve by Σaave Δt ≈ ΣΔv, the initial velocity is then equal to v0 = −ΣΔv Finally, using v2 = v1 + Δv12 where Δv12 is the change in velocity between times t1 and t2, the velocity at the end of each 0.25 interval can be computed; see column 3 of the table and the v −t curve. 3. The v −t curve is then approximated with a series of rectangles, each of width 0.25 s. The area (Δt )(vave ) of each rectangle is approximately equal to the change in position Δ x for the specified interval of time. Thus Δ x ≈ vave Δt where vave and Δ x are given in columns 4 and 5, respectively, of the table. 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. 116 PROBLEM 11.81 (Continued) 4. With x0 = 0 and noting that x2 = x1 + Δ x12 where Δ x12 is the change in position between times t1 and t2, the position at the end of each 0.25 s interval can be computed; see column 6 of the table and the x−t curve. (a) We had found (b) At t = 2 s v0 = 1.914 m/s x = 0.840 m 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. 117 PROBLEM 11.82 The acceleration record shown was obtained during the speed trials of a sports car. Knowing that the car starts from rest, determine by approximate means (a) the velocity of the car at t = 8 s, (b) the distance the car has traveled at t = 20 s. SOLUTION Given: a −t curve; at 1. t = 0, x = 0, v = 0 The a −t curve is first approximated with a series of rectangles, each of width Δt = 2 s. The area (Δt )(aave ) of each rectangle is approximately equal to the change in velocity Δv for the specified interval of time. Thus, Δv ≅ aave Δt where the values of aave and Δv are given in columns 1 and 2, respectively, of the following table. 2. Noting that v0 = 0 and that v2 = v1 + Δv12 where Δv12 is the change in velocity between times t1 and t2, the velocity at the end of each 2 s interval can be computed; see column 3 of the table and the v −t curve. 3. The v −t curve is next approximated with a series of rectangles, each of width Δt = 2 s. The area (Δt )(vave ) of each rectangle is approximately equal to the change in position Δx for the specified interval of time. Thus, Δx ≅ vave Δt where vave and Δx are given in columns 4 and 5, respectively, of the table. 4. With x0 = 0 and noting that x2 = x1 + Δ x12 where Δ x12 is the change in position between times t1 and t2, the position at the end of each 2 s interval can be computed; see column 6 of the table and the x −t curve. 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. 118 PROBLEM 11.82 (Continued) (a) At t = 8 s, v = 32.58 m/s (b) At t = 20 s v = 117.3 km/h or x = 660 m 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. 119 PROBLEM 11.83 A training airplane has a velocity of 126 ft/s when it lands on an aircraft carrier. As the arresting gear of the carrier brings the airplane to rest, the velocity and the acceleration of the airplane are recorded; the results are shown (solid curve) in the figure. Determine by approximate means (a) the time required for the airplane to come to rest, (b) the distance traveled in that time. SOLUTION Given: a −v curve: v0 = 126 ft/s The given curve is approximated by a series of uniformly accelerated motions (the horizontal dashed lines on the figure). For uniformly accelerated motion v22 = v12 + 2a( x2 − x1 ) v2 = v1 + a(t2 − t1 ) v22 − v12 2a v −v Δt = 2 1 a Δx = or For the five regions shown above, we have Region v1 , ft/s v2 , ft/s a, ft/s 2 Δx, ft Δt , s 1 126 120 −12.5 59.0 0.480 2 120 100 −33 66.7 0.606 3 100 80 −45.5 39.6 0.440 4 80 40 −54 44.4 0.741 5 40 0 −58 13.8 0.690 223.5 2.957 Σ (a) From the table, when v = 0 t = 2.96 s (b) From the table and assuming x0 = 0, when v = 0 x = 224 ft 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. 120 PROBLEM 11.84 Shown in the figure is a portion of the experimentally determined v − x curve for a shuttle cart. Determine by approximate means the acceleration of the cart (a) when x = 10 in., (b) when v = 80 in./s. SOLUTION Given: v − x curve First note that the slope of the above curve is dv . dx a=v (a) Now dv dx When x = 10 in., v = 55 in./s Then 40 in./s a = 55 in./s 13.5 in. a = 163.0 in./s 2 or (b) When v = 80 in./s, we have 40 in./s a = 80 in./s 28 in. a = 114.3 in./s 2 or Note: To use the method of measuring the subnormal outlined at the end of Section 11.8, it is necessary that the same scale be used for the x and v axes (e.g., 1 in. = 50 in., 1 in. = 50 in./s). In the above solution, Δv and Δ x were measured directly, so different scales could be used. 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. 121 PROBLEM 11.85 Using the method of Section 11.8, derive the formula x = x0 + v0t + 12 at 2 for the position coordinate of a particle in uniformly accelerated rectilinear motion. SOLUTION The a −t curve for uniformly accelerated motion is as shown. Using Eq. (11.13), we have x = x0 + v0t + (area under a −t curve) (t − t ) 1 = x0 + v0t + (t × a) t − t 2 1 = x0 + v0 t + at 2 Q.E.D. 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. 122 PROBLEM 11.86 Using the method of Section 11.8 determine the position of the particle of Problem 11.61 when t = 8 s. PROBLEM 11.61 A particle moves in a straight line with the acceleration shown in the figure. Knowing that it starts from the origin with v0 = −14 ft/s, plot the v−t and x−t curves for 0 < t < 15 s and determine (a) the maximum value of the velocity of the particle, (b) the maximum value of its position coordinate. SOLUTION x0 = 0 v0 = −14 ft/s when t = 8s: x = x0 + v0t + Σ A(t1 − t ) = 0 − (14 ft/s)(8 s) + [(3 ft/s2 )(2 s)](7 s) + [(8 ft/s2 )(3 s)](4.5 s) + [(3 ft/s)(3 s)](1.5 s) x8 = −112 ft + 42 ft + 108 ft + 13.5 ft x8 = 51.5 ft 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. 123 PROBLEM 11.87 The acceleration of an object subjected to the pressure wave of a large explosion is defined approximately by the curve shown. The object is initially at rest and is again at rest at time t1. Using the method of section 11.8, determine (a) the time t1, (b) the distance through which the object is moved by the pressure wave. SOLUTION (a) Since v = 0 when t = 0 and when t = t1 the change in v between t = 0 and t = t1 is zero. Thus, area under a−t curve is zero A1 + A2 + A3 = 0 1 1 1 (30)(0.6) + ( −10)(0.2) + ( −10)(t1 − 0.8) = 0 2 2 2 9 − 1 − 5t1 + 4 = 0 (b) t1 = 2.40 s Position when t = t1 = 2.4 s 2 x = x0 + v0t1 + A1 (t1 − 0.2) + A2 (t1 − 0.733) + A3 (t1 − 0.8) 3 1 2 = 0 + 0 + (9)(2.4 − 0.2) + ( −1)(2.4 − 0.733) + ( −10)(2.4 − 0.8) (2.4 − 0.8) 2 3 = 19.8 m − 1.667 m − 8.533 m x = 9.60 m 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. 124 PROBLEM 11.88 For the particle of Problem 11.63, draw the a −t curve and determine, using the method of Section 11.8, (a) the position of the particle when t = 52 s, (b) the maximum value of its position coordinate. PROBLEM 11.63 A particle moves in a straight line with the velocity shown in the figure. Knowing that x = −540 m at t = 0, (a) construct the a −t and x −t curves for 0 < t < 50 s, and determine (b) the total distance traveled by the particle when t = 50 s, (c) the two times at which x = 0. SOLUTION a= We have where dv dt dv dt is the slope of the v −t curve. Then t=0 from to t = 10 s: v = constant a = 0 t = 10 s to t = 26 s: a = t = 26 s to t = 41 s: v = constant a = 0 t = 41 s to t = 46 s: a = t > 46 s: −20 − 60 = −5 m/s 2 26 − 10 −5 − (−20) = 3 m/s 2 46 − 41 v = constant a = 0 The a −t curve is then drawn as shown. (a) From the discussion following Eq. (11.13), we have x = x0 + v0t1 + ΣA(t1 − t ) where A is the area of a region and t is the distance to its centroid. Then, for t1 = 52 s x = −540 m + (60 m/s)(52 s) + {−[(16 s)(5 m/s 2 )](52 − 18) s + [(5 s)(3 m/s 2 )](52 − 43.5)s} = [−540 + (3120) + (−2720 + 127.5)] m x = −12.50 m or 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. 125 PROBLEM 11.88 (Continued) (b) Noting that xmax occurs when v = 0 ( dx = 0 ), it is seen from the v–t curve that xmax occurs for dt 10 s < t < 26 s. Although similar triangles could be used to determine the time at which x = xmax (see the solution to Problem 11.63), the following method will be used. For 10 s < t1 < 26 s, we have x = −540 + 60t1 1 − [(t1 − 10)(5)] (t1 − 10) m 2 5 = −540 + 60t1 − (t1 − 10) 2 2 When x = xmax : or Then dx = 60 − 5(t1 − 10) = 0 dt (t1 ) xmax = 22 s 5 xmax = −540 + 60(22) − (22 − 10)2 2 xmax = 420 m or 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. 126 PROBLEM 11.CQ3 Two model rockets are fired simultaneously from a ledge and follow the trajectories shown. Neglecting air resistance, which of the rockets will hit the ground first? (a) A (b) B (c) They hit at the same time. (d) The answer depends on h. SOLUTION The motion in the vertical direction depends on the initial velocity in the y-direction. Since A has a larger initial velocity in this direction it will take longer to hit the ground. Answer: (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. 127 PROBLEM 11.CQ4 Ball A is thrown straight up. Which of the following statements about the ball are true at the highest point in its path? (a) The velocity and acceleration are both zero. (b) The velocity is zero, but the acceleration is not zero. (c) The velocity is not zero, but the acceleration is zero. (d) Neither the velocity nor the acceleration are zero. SOLUTION At the highest point the velocity is zero. The acceleration is never zero. Answer: (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. 128 PROBLEM 11.CQ5 Ball A is thrown straight up with an initial speed v0 and reaches a maximum elevation h before falling back down. When A reaches its maximum elevation, a second ball is thrown straight upward with the same initial speed v0. At what height, y, will the balls cross paths? (a) y=h (b) y > h/2 (c) y = h/2 (d) y < h/2 (e) y=0 SOLUTION When the ball is thrown up in the air it will be constantly slowing down until it reaches its apex, at which point it will have a speed of zero. So, the time it will take to travel the last half of the distance to the apex will be longer than the time it takes for the first half. This same argument can be made for the ball falling from the maximum elevation. It will be speeding up, so the first half of the distance will take longer than the second half. Therefore, the balls should cross above the half-way point. Answer: (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. 129 PROBLEM 11.CQ6 Two cars are approaching an intersection at constant speeds as shown. What velocity will car B appear to have to an observer in car A? (a) (b) (c) (d ) (e) SOLUTION Since vB = vA+ vB/A we can draw the vector triangle and see v B = v A + v B/A Answer: (e) 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. 130 PROBLEM 11.CQ7 Blocks A and B are released from rest in the positions shown. Neglecting friction between all surfaces, which figure below best indicates the direction α of the acceleration of block B? (a) (b) (c) (d) (e) SOLUTION Since aB = a A + aB/A we get Answer: (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. 131 PROBLEM 11.89 A ball is thrown so that the motion is defined by the equations x = 5t and y = 2 + 6t − 4.9t 2 , where x and y are expressed in meters and t is expressed in seconds. Determine (a) the velocity at t = 1 s, (b) the horizontal distance the ball travels before hitting the ground. SOLUTION Units are meters and seconds. Horizontal motion: vx = dx =5 dt Vertical motion: vy = dy = 6 − 9.8t dt (a) Velocity at t = 1 s. vx = 5 v y = 6 − 9.8 = −3.8 v = vx2 + v y2 = 52 + 3.82 = 6.28 m/s tan θ = (b) vy vx = −3.8 5 Horizontal distance: θ = −37.2° v = 6.28 m/s 37.2° ( y = 0) y = 2 + 6t − 4.9t 2 t = 1.4971 s x = (5)(1.4971) = 7.4856 m x = 7.49 m 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. 132 PROBLEM 11.90 The motion of a vibrating particle is defined by the position vector r = 10(1 − e−3t )i + (4e −2t sin15t ) j, where r and t are expressed in millimeters and seconds, respectively. Determine the velocity and acceleration when (a) t = 0, (b) t = 0.5 s. SOLUTION r = 10(1 − e−3t )i + (4e−2t sin15t ) j Then v= and a= (a) dr = 30e−3t i + [60e−2t cos15t − 8e−2t sin15t ]j dt dv = −90e −3t i + [−120e −2t cos15t − 900e−2t sin15t − 120e−2t cos15t + 16e−2t sin15t ]j dt = −90e −3t i + [−240e −2t cos15t − 884e −2t sin15t ] j When t = 0: v = 30i + 60 j mm/s v = 67.1 mm/s 63.4° a = −90i − 240 j mm/s 2 a = 256 mm/s 2 69.4° v = 8.29 mm/s 36.2° a = 336 mm/s 2 86.6° When t = 0.5 s: v = 30e −1.5 i + [60e−1 cos 7.5 − 8e−1 sin 7.5] = 6.694i + 4.8906 j mm/s a = 90e−1.5 i + [ −240e −1 cos 7.5 − 884e−1 sin 7.5 j] = −20.08i − 335.65 j mm/s 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. 133 PROBLEM 11.91 The motion of a vibrating particle is defined by the position vector r = (4sin π t )i − (cos 2π t ) j, where r is expressed in inches and t in seconds. (a) Determine the velocity and acceleration when t = 1 s. (b) Show that the path of the particle is parabolic. SOLUTION r = (4sin π t )i − (cos 2π t ) j v = (4π cos π t )i + (2π sin 2π t ) j a = −(4π 2 sin π t )i + (4π 2 cos 2π t ) j (a) When t = 1 s: v = (4π cos π )i + (2π sin 2π ) j v = −(4π in/s)i a = −(4π 2 sin π )i − (4π 2 cos π ) j (b) a = −(4π 2 in/s 2 ) j Path of particle: Since r = xi + y j ; x = 4sin π t , y = − cos 2π t Recall that cos 2θ = 1 − 2sin 2 θ and write y = − cos 2π t = −(1 − 2sin 2 π t ) But since x = 4sin π t or sin π t = (1) 1 x, Eq.(1) yields 4 2 1 y = − 1 − 2 x 4 y= 1 2 x − 1 (Parabola) 8 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. 134 PROBLEM 11.92 The motion of a particle is defined by the equations x = 10t − 5sin t and y = 10 − 5cos t , where x and y are expresed in feet and t is expressed in seconds. Sketch the path of the particle for the time interval 0 ≤ t ≤ 2π , and determine (a) the magnitudes of the smallest and largest velocities reached by the particle, (b) the corresponding times, positions, and directions of the velocities. SOLUTION Sketch the path of the particle, i.e., plot of y versus x. Using x = 10t − 5sin t , and y = 10 − 5cos t obtain the values in the table below. Plot as shown. t(s) x(ft) y(ft) 0 0.00 5 10.71 10 31.41 15 52.12 10 62.83 5 π 2 π 3 π 2 2π 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. 135 PROBLEM 11.92 (Continued) (a) Differentiate with respect to t to obtain velocity components. vx = dx = 10 − 5cos t and v y = 5sin t dt v 2 = vx2 + v y2 = (10 − 5cos t )2 + 25sin 2 t = 125 − 100cos t d (v ) 2 = 100sin t = 0 t = 0. ± π . ± 2π ± N π dt (b) When t = 2 N π . cos t = 1. and v2 is minimum. When t = (2 N + 1)π . cos t = −1. and v2 is maximum. (v 2 )min = 125 − 100 = 25(ft/s) 2 vmin = 5 ft/s (v 2 )max = 125 + 100 = 225(ft/s) 2 vmax = 15 ft/s When v = vmin . When N = 0,1, 2, x = 10(2π N ) − 5sin(2π N ) x = 20π N ft y = 10 − 5cos(2π N ) y = 5 ft vx = 10 − 5cos(2π N ) vx = 5 ft/s v y = 5sin(2π N ) tan θ = vy vx vy = 0 θ =0 = 0, 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. 136 PROBLEM 11.92 (Continued) t = (2 N + 1)π s When v = vmax . x = 10[2π ( N − 1)] − 5sin[2π ( N + 1)] x = 20π ( N + 1) ft y = 10 − 5cos[2π ( N + 1)] y = 15 ft vx = 10 − 5cos[2π ( N + 1)] vx = 15 ft/s v y = 5sin[2π ( N + 1)] tan θ = vy vx vy = 0 θ =0 = 0, 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. 137 PROBLEM 11.93 The damped motion of a vibrating particle is defined by the position vector r = x1[1 − 1/(t + 1)]i + ( y1e−π t/2 cos 2π t ) j, where t is expressed in seconds. For x1 = 30 mm and y1 = 20 mm, determine the position, the velocity, and the acceleration of the particle when (a) t = 0, (b) t = 1.5 s. SOLUTION We have 1 −π t/2 r = 30 1 − cos 2π t ) j i + 20(e t +1 Then v= = 30 1 π i + 20 − e −π t/2 cos 2π t − 2π e−π t/2 sin 2π t j 2 (t + 1) 2 = 30 1 1 i − 20π e −π t/ 2 cos 2π t + 2 sin 2π t j 2 2 (t + 1) a= and dr dt dv dt π 2 1 i − 20π − e−π t/2 cos 2π t + 2 sin 2π t + e−π t/2 (−π sin 2π t + 4 cos 2π t ) j 3 (t + 1) 2 2 60 =− i + 10π 2 e−π t/2 (4 sin 2π t − 7.5 cos 2π t ) j (t + 1)3 = −30 (a) At t = 0: 1 r = 30 1 − i + 20(1) j 1 r = 20 mm or 1 1 v = 30 i − 20π (1) + 0 j 1 2 v = 43.4 mm/s or a=− 46.3° 60 i + 10π 2 (1)(0 − 7.5) j (1) a = 743 mm/s 2 or 85.4° 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. 138 PROBLEM 11.93 (Continued) (b) At t = 1.5 s: 1 −0.75π r = 30 1 − (cos 3π ) j i + 20e 2.5 = (18 mm)i + ( −1.8956 mm) j or r = 18.10 mm 6.01° v = 5.65 mm/s 31.8° a = 70.3 mm/s 2 86.9° 30 1 i − 20π e−0.75π cos 3π + 0 j 2 (2.5) 2 = (4.80 mm/s)i + (2.9778 mm/s) j v= or a=− 60 i + 10π 2 e−0.75π (0 − 7.5 cos 3π ) j (2.5)3 = (−3.84 mm/s 2 )i + (70.1582 mm/s 2 ) j or 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. 139 PROBLEM 11.94 The motion of a particle is defined by the position vector r = A(cos t + t sin t )i + A(sin t − t cos t ) j, where t is expressed in seconds. Determine the values of t for which the position vector and the acceleration are (a) perpendicular, (b) parallel. SOLUTION We have r = A(cos t + t sin t )i + A(sin t − t cos t ) j Then v= and a= (a) dr = A( − sin t + sin t + t cos t )i dt + A(cos t − cos t + t sin t ) j = A(t cos t )i + A(t sin t ) j dv = A(cos t − t sin t )i + A(sin t + t cos t ) j dt When r and a are perpendicular, r ⋅ a = 0 A[(cos t + t sin t )i + (sin t − t cos t ) j] ⋅ A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0 (b) or (cos t + t sin t )(cos t − t sin t ) + (sin t − t cos t )(sin t + t cos t ) = 0 or (cos 2 t − t 2 sin 2 t ) + (sin 2 t − t 2 cos 2 t ) = 0 or 1− t2 = 0 or t =1 s or t=0 When r and a are parallel, r × a = 0 A[(cos t + t sin t )i + (sin t − t cos t ) j] × A[(cos t − t sin t )i + (sin t + t cos t ) j] = 0 or Expanding [(cos t + t sin t )(sin t + t cos t ) − (sin t − t cos t )(cos t − t sin t )]k = 0 (sin t cos t + t + t 2 sin t cos t ) − (sin t cos t − t + t 2 sin t cos t ) = 0 2t = 0 or 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. 140 PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.) SOLUTION We have r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k Then v= and a= dr = R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k dt dv dt = R(−ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t )i + R(ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t )k = R(−2ωn sin ωn t − ωn2 t cos ωn t )i + R(2ωn cos ωn t − ωn2 t sin ωn t )k Now v 2 = vx2 + v 2y + vz2 = [ R(cos ωn t − ωn t sin ωn t )]2 + (c)2 + [ R(sin ωn t + ωn t cos ωn t )]2 ( ) = R 2 cos 2 ωn t − 2ωn t sin ωn t cos ωn t + ωn2 t 2 sin 2 ωn t + sin 2 ωn t + 2ωn t sin ωn t cos ωn t + ωn2 t 2 cos 2 ωn t + c 2 ( ( ) ) = R 2 1 + ωn2 t 2 + c 2 ( Also, ) v = R 2 1 + ωn2 t 2 + c 2 or a 2 = ax2 + a 2y + az2 ( ) cos ω t − ω t sin ω t ) 2 = R −2ωn sin ωn t − ωn2 t cos ωn t + (0) 2 ( ( 2 2 + R 2ωn n n n = R 2 4ωn2 sin 2 ωn t + 4ωn3t sin ωn t cos ωn t + ωn4t 2 cos 2 ωn t + 4ωn2 cos 2 ωn t − 4ωn3t sin ωn t cos ωn t + ωn4 t 2 sin 2 ωn t ( ( = R 2 4ωn2 + ωn4 t 2 ) ) ) a = Rωn 4 + ωn2 t 2 or 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. 141 PROBLEM 11.96 The three-dimensional motion of a particle is defined by the position vector r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k , where r and t are expressed in feet and seconds, respectively. Show that the curve described by the particle lies on the hyperboloid (y/A)2 − (x/A)2 − (z/B)2 = 1. For A = 3 and B = 1, determine (a) the magnitudes of the velocity and acceleration when t = 0, (b) the smallest nonzero value of t for which the position vector and the velocity are perpendicular to each other. SOLUTION We have r = ( At cos t )i + ( A t 2 + 1) j + ( Bt sin t )k or x = At cos t cos t = Then x At y = A t 2 + 1 z = Bt sin t sin t = 2 y t2 = −1 A z Bt 2 2 2 2 2 x z t2 = + A B or 2 Then y x z A −1 = A + B or y x z A − A − B =1 2 (a) 2 x z cos 2 t + sin 2 t = 1 + = 1 At Bt Now 2 2 Q.E.D. With A = 3 and B = 1, we have v= dr t j + (sin t + t cos t )k = 3(cos t − t sin t )i + 3 2 dt t +1 ( ) t and t 2 + 1 − t t 2 +1 dv a= j = 3( − sin t − sin t − t cos t )i + 3 dt (t 2 + 1) + (cos t + cos t − t sin t )k 1 j + (2 cos t − t sin t )k = −3(2 sin t + t cos t )i + 3 2 (t + 1)3/2 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. 142 PROBLEM 11.96 (Continued) At t = 0: v = 3(1 − 0)i + (0) j + (0)k v = vx2 + v y2 + vz2 v = 3 ft/s or and Then a = −3(0)i + 3(1) j + (2 − 0)k a 2 = (0) 2 + (3) 2 + (2) 2 = 13 a = 3.61 ft/s 2 or (b) If r and v are perpendicular, r ⋅ v = 0 t [(3t cos t )i + (3 t 2 + 1) j + (t sin t )k ] ⋅ [3(cos t − t sin t )i + 3 j + (sin t + t cos t )k ] = 0 2 t +1 or Expanding t (3t cos t )[3(cos t − t sin t )] + (3 t 2 + 1) 3 + (t sin t )(sin t + t cos t ) = 0 2 t 1 + (9t cos 2 t − 9t 2 sin t cos t ) + (9t ) + (t sin 2 t + t 2 sin t cos t ) = 0 10 + 8 cos 2 t − 8t sin t cos t = 0 or (with t ≠ 0) 7 + 2 cos 2t − 2t sin 2t = 0 or Using “trial and error” or numerical methods, the smallest root is t = 3.82 s Note: The next root is t = 4.38 s. 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. 143 PROBLEM 11.97 An airplane used to drop water on brushfires is flying horizontally in a straight line at 180 mi/h at an altitude of 300 ft. Determine the distance d at which the pilot should release the water so that it will hit the fire at B. SOLUTION First note v0 = 180 km/h = 264 ft/s Place origin of coordinates at Point A. Vertical motion. (Uniformly accelerated motion) y = 0 + (0)t − At B: or 1 2 gt 2 1 −300 ft = − (32.2 ft/s 2 )t 2 2 t B = 4.31666 s Horizontal motion. (Uniform) x = 0 + (v x ) 0 t At B: d = (264 ft/s)(4.31666 s) d = 1140 ft or 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. 144 PROBLEM 11.98 A helicopter is flying with a constant horizontal velocity of 180 km/h and is directly above Point A when a loose part begins to fall. The part lands 6.5 s later at Point B on an inclined surface. Determine (a) the distance d between Points A and B, (b) the initial height h. SOLUTION Place origin of coordinates at Point A. Horizontal motion: (vx )0 = 180 km/h = 50 m/s x = x0 + (vx )0 t = 0 + 50t m At Point B where t B = 6.5 s, (a) Distance AB. 325 cos10° From geometry d= Vertical motion: y = y0 + ( v y ) 0 t − At Point B (b) xB = (50)(6.5) = 325 m d = 330 m 1 2 gt 2 1 − xB tan10° = h + 0 − (9.81)(6.5) 2 2 h = 149.9 m Initial height. 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. 145 PROBLEM 11.99 A baseball pitching machine “throws” baseballs with a horizontal velocity v0. Knowing that height h varies between 788 mm and 1068 mm, determine (a) the range of values of v0, (b) the values of α corresponding to h = 788 mm and h = 1068 mm. SOLUTION (a) y0 = 1.5 m, (v y )0 = 0 Vertical motion: t = 2( y0 − y ) g or tB = 2( y0 − h) g When h = 788 mm = 0.788 m, tB = (2)(1.5 − 0.788) = 0.3810 s 9.81 When h = 1068 mm = 1.068 m, tB = (2)(1.5 − 1.068) = 0.2968 s 9.81 y = y0 + (v y )0 t − At Point B, Horizontal motion: 1 2 gt 2 or y =h x0 = 0, (vx )0 = v0 , x = v0t or v0 = x x = B t tB 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. 146 PROBLEM 11.99 (Continued) v0 = 12.2 = 32.02 m/s 0.3810 and v0 = 12.2 = 41.11 m/s 0.2968 32.02 m/s ≤ v0 ≤ 41.11 m/s or Vertical motion: v y = (v y )0 − gt = − gt Horizontal motion: vx = v0 With xB = 12.2 m, (b) we get tan α = − 115.3 km/h ≤ v0 ≤ 148.0 km/h (v y ) B dy gt =− = B dx (vx ) B v0 For h = 0.788 m, tan α = (9.81)(0.3810) = 0.11673, 32.02 α = 6.66° For h = 1.068 m, tan α = (9.81)(0.2968) = 0.07082, 41.11 α = 4.05° 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. 147 PROBLEM 11.100 While delivering newspapers, a girl throws a newspaper with a horizontal velocity v0. Determine the range of values of v0 if the newspaper is to land between Points B and C. SOLUTION Vertical motion. (Uniformly accelerated motion) y = 0 + (0)t − 1 2 gt 2 Horizontal motion. (Uniform) x = 0 + (vx )0 t = v0 t At B: y: t B = 0.455016 s or Then x: y: or 1 −2 ft = − (32.2 ft/s 2 )t 2 2 tC = 0.352454 s or Then 7 ft = (v0 ) B (0.455016 s) (v0 ) B = 15.38 ft/s or At C: 1 1 −3 ft = − (32.2 ft/s 2 )t 2 3 2 x: 1 12 ft = (v0 )C (0.352454 s) 3 (v0 )C = 35.0 ft/s 15.38 ft/s < v0 < 35.0 ft/s 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. 148 PROBLEM 11.101 Water flows from a drain spout with an initial velocity of 2.5 ft/s at an angle of 15° with the horizontal. Determine the range of values of the distance d for which the water will enter the trough BC. SOLUTION First note (vx )0 = (2.5 ft/s) cos 15° = 2.4148 ft/s (v y )0 = −(2.5 ft/s) sin 15° = −0.64705 ft/s Vertical motion. (Uniformly accelerated motion) y = 0 + (v y ) 0 t − 1 2 gt 2 At the top of the trough 1 −8.8 ft = (−0.64705 ft/s) t − (32.2 ft/s 2 ) t 2 2 or t BC = 0.719491 s (the other root is negative) Horizontal motion. (Uniform) x = 0 + (v x ) 0 t In time t BC xBC = (2.4148 ft/s)(0.719491 s) = 1.737 ft Thus, the trough must be placed so that xB < 1.737 ft or xC ≥ 1.737 ft 0 < d < 1.737 ft Since the trough is 2 ft wide, it then follows that 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. 149 PROBLEM 11.102 Milk is poured into a glass of height 140 mm and inside diameter 66 mm. If the initial velocity of the milk is 1.2 m/s at an angle of 40° with the horizontal, determine the range of values of the height h for which the milk will enter the glass. SOLUTION First note (vx )0 = (1.2 m/s) cos 40° = 0.91925 m/s (v y )0 = −(1.2 m/s) sin 40° = −0.77135 m/s Horizontal motion. (Uniform) x = 0 + (v x ) 0 t Vertical motion. (Uniformly accelerated motion) y = y0 + ( v y ) 0 t − 1 2 gt 2 Milk enters glass at B. x: 0.08 m = (0.91925 m/s) t or tB = 0.087028 s y : 0.140 m = hB + (−0.77135 m/s)(0.087028 s) 1 − (9.81 m/s 2 )(0.087028 s) 2 2 or hB = 0.244 m Milk enters glass at C. x : 0.146 m = (0.91925 m/s) t or tC = 0.158825 s y : 0.140 m = hC + (−0.77135 m/s)(0.158825 s) 1 − (9.81 m/s 2 )(0.158825 s) 2 2 or hC = 0.386 m 0.244 m < h < 0.386 m 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. 150 PROBLEM 11.103 A volleyball player serves the ball with an initial velocity v0 of magnitude 13.40 m/s at an angle of 20° with the horizontal. Determine (a) if the ball will clear the top of the net, (b) how far from the net the ball will land. SOLUTION First note (vx )0 = (13.40 m/s) cos 20° = 12.5919 m/s (v y )0 = (13.40 m/s) sin 20° = 4.5831 m/s (a) Horizontal motion. (Uniform) x = 0 + (v x ) 0 t At C 9 m = (12.5919 m/s) t or tC = 0.71475 s Vertical motion. (Uniformly accelerated motion) y = y0 + ( v y ) 0 t − At C: 1 2 gt 2 yC = 2.1 m + (4.5831 m/s)(0.71475 s) 1 − (9.81 m/s 2 )(0.71475 s)2 2 = 2.87 m yC > 2.43 m (height of net) ball clears net (b) At B, y = 0: 1 0 = 2.1 m + (4.5831 m/s)t − (9.81 m/s 2 )t 2 2 Solving t B = 1.271175 s (the other root is negative) Then d = (vx )0 t B = (12.5919 m/s)(1.271175 s) = 16.01 m b = (16.01 − 9.00) m = 7.01 m from the net The ball lands 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. 151 PROBLEM 11.104 A golfer hits a golf ball with an initial velocity of 160 ft/s at an angle of 25° with the horizontal. Knowing that the fairway slopes downward at an average angle of 5°, determine the distance d between the golfer and Point B where the ball first lands. SOLUTION (vx )0 = (160 ft/s) cos 25° First note (v y )0 = (160 ft/s) sin 25° xB = d cos 5° and at B Now yB = − d sin 5° Horizontal motion. (Uniform) x = 0 + (v x ) 0 t d cos 5° = (160 cos 25°)t or t B = At B cos 5° d 160 cos 25° Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t − 1 2 gt 2 ( g = 32.2 ft/s 2 ) 1 2 gt B 2 At B: − d sin 5° = (160 sin 25°)t B − Substituting for t B cos 5° 1 cos 5° 2 − d sin 5° = (160 sin 25°) d − g d 2 160 cos 25° 160 cos 25° 2 or 2 (160 cos 25°) 2 (tan 5° + tan 25°) 32.2 cos 5° = 726.06 ft d= d = 242 yd or 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. 152 PROBLEM 11.105 A homeowner uses a snowblower to clear his driveway. Knowing that the snow is discharged at an average angle of 40° with the horizontal, determine the initial velocity v0 of the snow. SOLUTION First note (vx )0 = v0 cos 40° (v y )0 = v0 sin 40° Horizontal motion. (Uniform) x = 0 + (v x ) 0 t At B: 14 = (v0 cos 40°) t or t B = 14 v0 cos 40° Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t − At B: 1 2 gt 2 1.5 = (v0 sin 40°) t B − ( g = 32.2 ft/s 2 ) 1 2 gt B 2 Substituting for t B 1 14 14 1.5 = (v0 sin 40°) − g v0 cos 40° 2 v0 cos 40° or v02 = 1 2 2 (32.2)(196)/ cos 2 40° −1.5 + 14 tan 40° v0 = 22.9 ft/s or 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. 153 PROBLEM 11.106 At halftime of a football game souvenir balls are thrown to the spectators with a velocity v0. Determine the range of values of v0 if the balls are to land between Points B and C. SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below Point A. The coordinates of Point A are x0 = 0, y0 = 2m. The components of initial velocity are (vx )0 = v0 cos 40° m/s and (v y )0 = v0 sin 40°. Horizontal motion: x = x0 + (vx )0 t = (v0 cos 40°)t Vertical motion: y = y0 + ( v y ) 0 t = From (1), Then 1 2 gt 2 1 = 2 + (v0 sin 40°) = − (9.81)t 2 2 v0 t = (2) x cos 40° (3) y = 2 + x tan 40° − 4.905t 2 t2 = Point B: (1) 2 + x tan 40° − y 4.905 (4) x = 8 + 10 cos 35° = 16.1915 m y = 1.5 + 10sin 35° = 7.2358 m 16.1915 v0 t = = 21.1365 m cos 40° 2 + 16.1915 tan 40° − 7.2358 t2 = 4.905 21.1365 v0 = 1.3048 t = 1.3048 s v0 = 16.199 m/s 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. 154 PROBLEM 11.106 (Continued) Point C: x = 8 + (10 + 7) cos 35° = 21.9256 m y = 1.5 + (10 + 7)sin 35° = 11.2508 m v0 t = 21.9256 = 28.622 m cos 40° t2 = 2 + 21.9256 tan 40° − 11.2508 4.905 v0 = 28.622 1.3656 t = 1.3656 s v0 = 20.96 m/s 16.20 m/s < v0 < 21.0 m/s Range of values of v0. 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. 155 PROBLEM 11.107 A basketball player shoots when she is 16 ft from the backboard. Knowing that the ball has an initial velocity v0 at an angle of 30° with the horizontal, determine the value of v0 when d is equal to (a) 9 in., (b) 17 in. SOLUTION First note (vx )0 = v0 cos 30° Horizontal motion. (Uniform) (v y )0 = v0 sin 30° x = 0 + (v x ) 0 t (16 − d ) = (v0 cos 30°) t or t B = At B: 16 − d v0 cos 30° y = 0 + (v y )0 t − Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 2 gt B 2 At B: 3.2 = (v0 sin 30°) t B − Substituting for tB 16 − d 1 16 − d 3.2 = (v0 sin 30°) − g v0 cos 30° 2 v0 cos 30° or v02 = (a) d = 9 in.: v02 d = 17 in.: v02 (b) = = ( g = 32.2 ft/s 2 ) 2 2 g (16 − d ) 2 3 1 3 (16 − d ) − 3.2 2(32.2) (16 − 129 ) 3 1 3 (16 − 129 ) − 3.2 2(32.2) (16 − 17 12 ) 3 1 3 2 v0 = 29.8 ft/s 2 (16 − 1217 ) − 3.2 v0 = 29.6 ft/s 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. 156 PROBLEM 11.108 A tennis player serves the ball at a height h = 2.5 m with an initial velocity of v0 at an angle of 5° with the horizontal. Determine the range for which of v0 for which the ball will land in the service area which extends to 6.4 m beyond the net. SOLUTION The motion is projectile motion. Place the origin of the xy-coordinate system at ground level just below the point where the racket impacts the ball. The coordinates of this impact point are x0 = 0, y0 = h = 2.5 m. The components of initial velocity are (vx )0 = v0 cos 5° and (v y )0 = v0 sin 5°. Horizontal motion: x = x0 + (vx )0 t = (v0 cos 5°)t Vertical motion: y = y0 + ( v y ) 0 t = (1) 1 2 gt 2 1 = 2.5 − (v0 sin 5°)t = − (9.81)t 2 2 From (1), Then v0 t = x cos 5° (2) (3) y = 2.5 − x tan 5° − 4.905t 2 t2 = 2.5 − x tan 5° − y 4.905 (4) At the minimum speed the ball just clears the net. x = 12.2 m, y = 0.914 m 12.2 = 12.2466 m cos 5° 2.5 − 12.2 tan 5° − 0.914 t2 = 4.905 12.2466 v0 = 0.32517 v0 t = t = 0.32517 s v0 = 37.66 m/s 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. 157 PROBLEM 11.108 (Continued) At the maximum speed the ball lands 6.4 m beyond the net. x = 12.2 + 6.4 = 18.6 m y=0 18.6 v0 t = = 18.6710 m cos 5° 2.5 − 18.6 tan 5° − 0 t2 = t = 0.42181 s 4.905 18.6710 v0 = v0 = 44.26 m/s 0.42181 37.7 m/s < v0 < 44.3 m/s Range for v0. 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. 158 PROBLEM 11.109 The nozzle at A discharges cooling water with an initial velocity v0 at an angle of 6° with the horizontal onto a grinding wheel 350 mm in diameter. Determine the range of values of the initial velocity for which the water will land on the grinding wheel between Points B and C. SOLUTION First note (vx )0 = v0 cos 6° (v y )0 = −v0 sin 6° Horizontal motion. (Uniform) x = x0 + (vx )0 t Vertical motion. (Uniformly accelerated motion) y = y0 + (v y )0 t − 1 2 gt 2 ( g = 9.81 m/s 2 ) x = (0.175 m) sin 10° y = (0.175 m) cos 10° At Point B: x : 0.175 sin 10° = −0.020 + (v0 cos 6°)t tB = or 0.050388 v0 cos 6° y : 0.175 cos 10° = 0.205 + (−v0 sin 6°)t B − 1 2 gtB 2 Substituting for t B 0.050388 1 0.050388 −0.032659 = (−v0 sin 6°) − (9.81) v0 cos 6° 2 v0 cos 6° 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. 159 PROBLEM 11.109 (Continued) v02 = or 1 2 (9.81)(0.050388)2 cos 2 6°(0.032659 − 0.050388 tan 6°) (v0 ) B = 0.678 m/s or x = (0.175 m) cos 30° y = (0.175 m) sin 30° At Point C: x : 0.175 cos 30° = −0.020 + (v0 cos 6°)t tC = or 0.171554 v0 cos 6° y : 0.175 sin 30° = 0.205 + (−v0 sin 6°)tC − 1 2 gtC 2 Substituting for tC 0.171554 1 0.171554 −0.117500 = (−v0 sin 6°) − (9.81) v0 cos 6° 2 v0 cos 6° or or v02 = 1 2 2 (9.81)(0.171554) 2 cos 2 6°(0.117500 − 0.171554 tan 6°) (v0 )C = 1.211 m/s 0.678 m/s < v0 < 1.211 m/s 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. 160 PROBLEM 11.110 While holding one of its ends, a worker lobs a coil of rope over the lowest limb of a tree. If he throws the rope with an initial velocity v0 at an angle of 65° with the horizontal, determine the range of values of v0 for which the rope will go over only the lowest limb. SOLUTION First note (vx )0 = v0 cos 65° (v y )0 = v0 sin 65° Horizontal motion. (Uniform) x = 0 + (v x ) 0 t At either B or C, x = 5 m s = (v0 cos 65°)t B,C or t B ,C = 5 (v0 cos 65°) Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t − 1 2 gt 2 ( g = 9.81 m/s 2 ) At the tree limbs, t = tB ,C 1 5 5 yB ,C = (v0 sin 65°) − g v0 cos 65° 2 v0 cos 65° 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. 161 PROBLEM 11.110 (Continued) or v02 = = 1 2 (9.81)(25) 2 cos 65°(5 tan 65° − yB , C ) 686.566 5 tan 65° − yB, C At Point B: v02 = 686.566 5 tan 65° − 5 or (v0 ) B = 10.95 m/s At Point C: v02 = 686.566 5 tan 65° − 5.9 or (v0 )C = 11.93 m/s 10.95 m/s < v0 < 11.93 m/s 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. 162 PROBLEM 11.111 The pitcher in a softball game throws a ball with an initial velocity v0 of 72 km/h at an angle α with the horizontal. If the height of the ball at Point B is 0.68 m, determine (a) the angle α, (b) the angle θ that the velocity of the ball at Point B forms with the horizontal. SOLUTION First note v0 = 72 km/h = 20 m/s (vx )0 = v0 cos α = (20 m/s) cos α and (v y )0 = v0 sin α = (20 m/s) sin α (a) Horizontal motion. (Uniform) x = 0 + (vx )0 t = (20 cos α ) t At Point B: 14 = (20 cos α )t or t B = 7 10 cos α Vertical motion. (Uniformly accelerated motion) y = 0 + (v y )0 t − At Point B: 1 2 1 gt = (20 sin α )t − gt 2 2 2 0.08 = (20 sin α )t B − ( g = 9.81 m/s 2 ) 1 2 gt B 2 Substituting for t B 1 7 7 0.08 = (20 sin α ) − g 10 cos α 2 10 cos α or Now 8 = 1400 tan α − 2 1 49 g 2 cos 2 α 1 = sec2 α = 1 + tan 2 α 2 cos α 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. 163 PROBLEM 11.111 (Continued) Then or Solving 8 = 1400 tan α − 24.5 g (1 + tan 2 α ) 240.345 tan 2 α − 1400 tan α + 248.345 = 0 α = 10.3786° and α = 79.949° Rejecting the second root because it is not physically reasonable, we have α = 10.38° (b) We have vx = (vx )0 = 20 cos α and v y = (v y )0 − gt = 20 sin α − gt At Point B: (v y ) B = 20 sin α − gt B = 20 sin α − 7g 10 cos α Noting that at Point B, v y < 0, we have tan θ = = = |(v y ) B | vx 7g 10 cos α − 20 sin α 20 cos α 7 9.81 200 cos 10.3786° − sin 10.3786° cos 10.3786° θ = 9.74° or 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. 164 PROBLEM 11.112 A model rocket is launched from Point A with an initial velocity v0 of 75 m/s. If the rocket’s descent parachute does not deploy and the rocket lands a distance d = 100 m from A, determine (a) the angle α that v0 forms with the vertical, (b) the maximum height above Point A reached by the rocket, and (c) the duration of the flight. SOLUTION Set the origin at Point A. x0 = 0, y0 = 0 Horizontal motion: x = v0 t sin α Vertical motion: y = v0t cos α − cos α = sin 2 α + cos 2 α = sin α = x v0t 1 2 gt 2 1 1 y + gt 2 v0 t 2 (2) 2 1 2 1 2 x y gt + + =1 2 (v0 t )2 x 2 + y 2 + gyt 2 + 1 24 g t = v02t 2 4 1 24 g t − v02 − gy t 2 + ( x 2 + y 2 ) = 0 4 ( At Point B, (1) ) (3) x 2 + y 2 = 100 m, x = 100 cos 30° m y = −100 sin 30° = −50 m 1 (9.81)2 t 4 − [752 − (9.81)(−50)] t 2 + 1002 = 0 4 24.0590 t 4 − 6115.5 t 2 + 10000 = 0 t 2 = 252.54 s 2 t = 15.8916 s and 1.6458 s 2 and 1.2829 s 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. 165 PROBLEM 11.112 (Continued) Restrictions on α : 0 < α < 120° tan α = x 2 y + gt 1 2 = 100 cos 30° = 0.0729 −50 + (4.905)(15.8916) 2 α = 4.1669° 100 cos 30° = −2.0655 −50 + (4.905)(1.2829)2 α = 115.8331° and Use α = 4.1669° corresponding to the steeper possible trajectory. (a) Angle α . (b) Maximum height. α = 4.17° v y = 0 at y = ymax v y = v0 cos α − gt = 0 t= v0 cos α g ymax = v0 t cos α − = (c) Duration of the flight. v 2 cos 2 α 1 gt = 0 2 2g (75) 2 cos 2 4.1669° (2)(9.81) ymax = 285 m (time to reach B) t = 15.89 s 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. 166 PROBLEM 11.113 The initial velocity v0 of a hockey puck is 105 mi/h. Determine (a) the largest value (less than 45°) of the angle α for which the puck will enter the net, (b) the corresponding time required for the puck to reach the net. SOLUTION First note v0 = 105 mi/h = 154 ft/s (vx )0 = v0 cos α = (154 ft/s) cos α and (v y )0 = v0 sin α = (154 ft/s) sin α (a) Horizontal motion. (Uniform) x = 0 + (vx )0 t = (154 cos α )t At the front of the net, Then or x = 16 ft 16 = (154 cos α )t tenter = 8 77 cos α Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (154 sin α ) t − gt 2 2 y = 0 + (v y )0 t − ( g = 32.2 ft/s 2 ) At the front of the net, yfront = (154 sin α ) tenter − 1 2 gtenter 2 1 8 8 = (154 sin α ) − g 77 cos 2 77 cos α α 32 g = 16 tan α − 5929 cos 2 α 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. 167 PROBLEM 11.113 (Continued) Now Then or 1 = sec2 α = 1 + tan 2 α cos 2 α yfront = 16 tan α − tan 2 α − 32 g (1 + tan 2 α ) 5929 5929 5929 tan α + 1 + yfront = 0 2g 32 g 5929 2g ( ± − 5929 2g ) 2 ( ) 1/2 − 4 1 + 5929 y 32 g front 2 Then tan α = or 5929 2 5929 5929 ± − tan α = yfront − 1 + 4 × 32.2 4 × 32.2 32 × 32.2 or tan α = 46.0326 ± [(46.0326) 2 − (1 + 5.7541 yfront )]1/2 1/2 Now 0 < yfront < 4 ft so that the positive root will yield values of α > 45° for all values of yfront. When the negative root is selected, α increases as yfront is increased. Therefore, for α max , set yfront = yC = 4 ft (b) Then tan α = 46.0326 − [(46.0326) 2 − (1 + 5.7541 + 4)]1/ 2 or α max = 14.6604° We had found α max = 14.66° 8 77 cos α 8 = 77 cos 14.6604° tenter = tenter = 0.1074 s or 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. 168 PROBLEM 11.114 A worker uses high-pressure water to clean the inside of a long drainpipe. If the water is discharged with an initial velocity v0 of 11.5 m/s, determine (a) the distance d to the farthest Point B on the top of the pipe that the worker can wash from his position at A, (b) the corresponding angle α. SOLUTION First note (vx )0 = v0 cos α = (11.5 m/s) cos α (v y )0 = v0 sin α = (11.5 m/s) sin α By observation, d max occurs when ymax = 1.1 m. Vertical motion. (Uniformly accelerated motion) v y = (v y )0 − gt = (11.5 sin α ) − gt y = ymax When Then at 1 2 gt 2 1 = (11.5 sin α ) t − gt 2 2 y = 0 + (v y ) 0 t − B , (v y ) B = 0 (v y ) B = 0 = (11.5 sin α ) − gt 11.5 sin α g ( g = 9.81 m/s 2 ) or tB = and yB = (11.5 sin α ) t B − 1 2 gt B 2 Substituting for t B and noting yB = 1.1 m 11.5 sin α 1.1 = (11.5 sin α ) g 1 = (11.5) 2 sin 2 α 2g or sin 2 α = 2.2 × 9.81 11.52 1 11.5 sin α − g g 2 2 α = 23.8265° 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. 169 PROBLEM 11.114 (Continued) (a) Horizontal motion. (Uniform) x = 0 + (vx )0 t = (11.5 cos α ) t At Point B: where Then x = d max tB = and t = t B 11.5 sin 23.8265° = 0.47356 s 9.81 d max = (11.5)(cos 23.8265°)(0.47356) d max = 4.98 m or (b) α = 23.8° From above 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. 170 PROBLEM 11.115 An oscillating garden sprinkler which discharges water with an initial velocity v0 of 8 m/s is used to water a vegetable garden. Determine the distance d to the farthest Point B that will be watered and the corresponding angle α when (a) the vegetables are just beginning to grow, (b) the height h of the corn is 1.8 m. SOLUTION First note (vx )0 = v0 cos α = (8 m/s) cos α (v y )0 = v0 sin α = (8 m/s) sin α Horizontal motion. (Uniform) x = 0 + (vx )0 t = (8 cos α ) t At Point B: or x = d : d = (8 cos α ) t d tB = 8 cos α Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (8 sin α ) t − gt 2 ( g = 9.81 m/s 2 ) 2 1 0 = (8 sin α ) t B − gt B2 2 y = 0 + (v y )0 t − At Point B: Simplifying and substituting for t B 0 = 8 sin α − d= or (a) 1 d g 2 8 cos α 64 sin 2α g (1) When h = 0, the water can follow any physically possible trajectory. It then follows from Eq. (1) that d is maximum when 2α = 90° α = 45° or Then d= 64 sin (2 × 45°) 9.81 d max = 6.52 m or 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. 171 PROBLEM 11.115 (Continued) (b) Based on Eq. (1) and the results of Part a, it can be concluded that d increases in value as α increases in value from 0 to 45° and then d decreases as α is further increased. Thus, d max occurs for the value of α closest to 45° and for which the water just passes over the first row of corn plants. At this row, xcom = 1.5 m tcorn = so that 1.5 8 cos α Also, with ycorn = h, we have h = (8 sin α ) tcorn − 1 2 gtcorn 2 Substituting for tcorn and noting h = 1.8 m, 1.5 1 1.5 1.8 = (8 sin α ) − g 8 cos α 2 8 cos α or Now Then or 1.8 = 1.5 tan α − 2 2.25 g 128 cos 2 α 1 = sec2 α = 1 + tan 2 α cos 2 α 1.8 = 1.5 tan α − 2.25(9.81) (1 + tan 2 α ) 128 0.172441 tan 2 α − 1.5 tan α + 1.972441 = 0 α = 58.229° and α = 81.965° Solving From the above discussion, it follows that d = d max when α = 58.2° Finally, using Eq. (1) d= 64 sin (2 × 58.229°) 9.81 d max = 5.84 m or 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. 172 PROBLEM 11.116* A mountain climber plans to jump from A to B over a crevasse. Determine the smallest value of the climber’s initial velocity v0 and the corresponding value of angle α so that he lands at B. SOLUTION First note (vx )0 = v0 cos α (v y )0 = v0 sin α Horizontal motion. (Uniform) x = 0 + (vx )0 t = (v0 cos α ) t At Point B: or 1.8 = (v0 cos α )t tB = 1.8 v0 cos α Vertical motion. (Uniformly accelerated motion) 1 2 gt 2 1 = (v0 sin α ) t − gt 2 2 y = 0 + (v y )0 t − At Point B: −1.4 = (v0 sin α ) t B − ( g = 9.81 m/s 2 ) 1 2 gt B 2 Substituting for t B 1.8 −1.4 = (v0 sin α ) v0 cos α or 1 1.8 − g 2 v0 cos α v02 = 1.62 g cos α (1.8 tan α + 1.4) = 1.62 g 0.9 sin 2α + 1.4 cos 2 α 2 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. 173 PROBLEM 11.116* (Continued) Now minimize v02 with respect to α. We have dv02 −(1.8 cos 2α − 2.8 cos α sin α ) = 1.62 g =0 dα (0.9 sin 2α + 1.4 cos 2 α ) 2 1.8 cos 2α − 1.4 sin 2α = 0 or tan 2α = or 18 14 α = 26.0625° and α = 206.06° or Rejecting the second value because it is not physically possible, we have α = 26.1° Finally, v02 = 1.62 × 9.81 cos 26.0625°(1.8 tan 26.0625° + 1.4) 2 (v0 )min = 2.94 m/s or 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. 174 PROBLEM 11.117 The velocities of skiers A and B are as shown. Determine the velocity of A with respect to B. SOLUTION We have vA = vB + vA/B The graphical representation of this equation is then as shown. Then v 2A/B = 302 + 452 − 2(30)(45) cos 15° or vA /B = 17.80450 ft/s and or 30 17.80450 = sin α sin 15° α = 25.8554° α + 25° = 50.8554° vA /B = 17.8 ft/s 50.9° Alternative solution. vA /B = v A − vB = 30 cos 10° i − 30 sin 10° j − (45 cos 25° i − 45° sin 25° j) = 11.2396i + 13.8084 j = 5.05 m/s = 17.8 ft/s 50.9° 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. 175 PROBLEM 11.118 The three blocks shown move with constant velocities. Find the velocity of each block, knowing that the relative velocity of A with respect to C is 300 mm/s upward and that the relative velocity of B with respect to A is 200 mm/s downward. SOLUTION From the diagram Cable 1: y A + yD = constant Then v A + vD = 0 Cable 2: (1) ( yB − yD ) + ( yC − yD ) = constant vB + vC − 2vD = 0 Then (2) Combining Eqs. (1) and (2) to eliminate vD , 2v A + vB + vC = 0 (3) Now v A/C = v A − vC = −300 mm/s (4) and vB/A = vB − v A = 200 mm/s (5) (3) + (4) − (5) Then (2v A + vB + vC ) + (v A − vC ) − (vB − v A ) = (−300) − (200) v A = 125 mm/s or vB − (−125) = 200 and using Eq. (5) v B = 75 mm/s or −125 − vC = −300 Eq. (4) vC = 175 mm/s or 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. 176 PROBLEM 11.119 Three seconds after automobile B passes through the intersection shown, automobile A passes through the same intersection. Knowing that the speed of each automobile is constant, determine (a) the relative velocity of B with respect to A, (b) the change in position of B with respect to A during a 4-s interval, (c) the distance between the two automobiles 2 s after A has passed through the intersection. SOLUTION v A = 45 mi/h = 66 ft/s vB = 30 mi/h = 44 ft/s Law of cosines vB2/A = 662 + 442 − 2(66)(44) cos110° vB/A = 90.99 ft/s vB = v A + vB/A Law of sines sin β sin110° = 66 90.99 β = 42.97° α = 90° − β = 90° − 42.97° = 47.03° v B/A = 91.0 ft/s (a) Relative velocity: (b) Change in position for Δt = 4 s. ΔrB/A = vB/A Δt = (91.0 ft/s)(4 s) (c) rB/A = 364 ft 47.0° 47.0° Distance between autos 2 seconds after auto A has passed intersection. Auto A travels for 2 s. v A = 66 ft/s 20° rA = v At = (66 ft/s)(2 s) = 132 ft rA = 132 ft 20° 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. 177 PROBLEM 11.119 (Continued) v B = 44 ft/s Auto B rB = v B t = (44 ft/s)(5 s) = 220 ft rB = rA + rB/A Law of cosines rB2/A = (132) 2 + (220) 2 − 2(132)(220) cos110° rB/A = 292.7 ft Distance between autos = 293 ft 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. 178 PROBLEM 11.120 Shore-based radar indicates that a ferry leaves its slip with a velocity v = 18 km/h 70°, while instruments aboard the ferry indicate a speed of 18.4 km/h and a heading of 30° west of south relative to the river. Determine the velocity of the river. SOLUTION We have v F = v R + v F/R or v F = v F/R + v R The graphical representation of the second equation is then as shown. We have vR2 = 182 + 18.42 − 2(18)(18.4) cos 10° or vR = 3.1974 km/h and or 18 3.1974 = sin α sin 10° α = 77.84° Noting that v R = 3.20 km/h 17.8° Alternatively one could use vector algebra. 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. 179 PROBLEM 11.121 Airplanes A and B are flying at the same altitude and are tracking the eye of hurricane C. The relative velocity of C with respect to A is vC/A = 350 km/h 75°, and the relative velocity of C with respect to B is vC/B = 400 km/h 40°. Determine (a) the relative velocity of B with respect to A, (b) the velocity of A if ground-based radar indicates that the hurricane is moving at a speed of 30 km/h due north, (c) the change in position of C with respect to B during a 15-min interval. SOLUTION (a) We have v C = v A + v C/ A and v C = v B + v C/ B Then v A + vC/A = v B + v C/B or v B − v A = v C/A − vC/B Now v B − v A = v B/A so that v B/A = vC/A − vC/B or vC/A = vC/B + v B/A The graphical representation of the last equation is then as shown. We have vB2/A = 3502 + 4002 − 2(350)(400) cos 65° or vB/A = 405.175 km/h and or 400 405.175 = sin α sin 65° α = 63.474° 75° − α = 11.526° v B/A = 405 km/h 11.53° 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. 180 PROBLEM 11.121 (Continued) (b) We have v C = v A + v C/ A or v A = (30 km/h) j − (350 km/h)(− cos 75°i − sin 75° j ) v A = (90.587 km/h)i + (368.07 km/h) j v A = 379 km/h or (c) 76.17° Noting that the velocities of B and C are constant, we have rB = (rB )0 + v B t Now rC = (rC )0 + v C t rC/B = rC − rB = [(rC )0 − (rB )0 ] + ( vC − v B )t = [(rC )0 − (rB )0 ] + v C/B t Then ΔrC/B = (rC/B )t2 − (rC/B )t1 = v C/B (t2 − t1 ) = vC/B Δt For Δt = 15 min: 1 ΔrC/B = (400 km/h) h = 100 km 4 ΔrC/B = 100 km 40° 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. 181 PROBLEM 11.122 Pin P moves at a constant speed of 150 mm/s in a counterclockwise sense along a circular slot which has been milled in the slider block A shown. Knowing that the block moves downward at a constant speed 100 mm/s determine the velocity of pin P when (a) θ = 30°, (b) θ = 120°. SOLUTION v P = v A + v P/A v P = 100 ms (− j) + 150(cos θ i + sin θ j)mm/s (a) For θ = 30° v P = −100 mm/s ( j) + 150( − cos(30°)i + sin(30°) j)mm/s v P = (−75i + 29.9038 j)mm/s v P = 80.7 mm/s (b) For θ = 120° 21.7° v P = −100 mm/s ( j) + 150(− cos(120°)i + sin(120°) j)mm/s v P = (−129.9038i + −175 j) mm/s v P = 218 mm/s 53.4° Alternative Solution (a) For θ = 30°, vP /A = 7.5 in./s 30° vP = v A + vP/A Law of cosines vP2 = (150) 2 + (100) 2 − 2(100)(150) cos 30° vP = 80.7418 mm/s Law of sines sin β sin 30° = 150 80.7418 β = 111.7° vP = 80.7 mm/s 21.7° 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. 182 PROBLEM 11.122 (Continued) (b) For θ = 120°, vP/A = 150 mm/s 30° Law of cosines vP2 = (150) 2 + (100) 2 − 2(100)(150) cos120° vP = 217.9449 mm/s Law of sines sin β sin120° = β = 36.6° 150 217.9449 α = 90 − β = 90° − 36.6 = 53.4° vP = 218 mm/s 53.4° 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. 183 PROBLEM 11.123 Knowing that at the instant shown assembly A has a velocity of 9 in./s and an acceleration of 15 in./s2 both directed downward, determine (a) the velocity of block B, (b) the acceleration of block B. SOLUTION Length of cable = constant L = x A + 2 xB/A = constant v A + 2vB/A = 0 (1) a A + 2aB/A = 0 (2) a A = 15 in./s 2 Data: v A = 9 in./s Eqs. (1) and (2) a A = −2aB/A v A = −2vB/A 15 = −2aB/A 9 = −2vB/A aB/A = −7.5 in./s 2 a B/A = 7.5 in./s 2 (a) vB/A = −4.5 in./s 40° v B/A = −4.5 in./s Velocity of B. v B = v A + v B/A Law of cosines: vB2 = (9) 2 + (4.5) 2 − 2(9)(4.5) cos 50° 40° vB = 7.013 in./s Law of sines: sin β sin 50° = 4.5 7.013 β = 29.44° α = 90° − β = 90° − 29.44° = 60.56° v B = 7.01 in./s 60.6° 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. 184 PROBLEM 11.123 (Continued) (b) Acceleration of B. a B may be found by using analysis similar to that used above for vB . An alternate method is a B = a A + a B/A a B = 15 in./s 2 ↓ +7.5 in./s 2 40° = −15 j − (7.5 cos 40°)i + (7.5 sin 40°) j = −15 j − 5.745i + 4.821j a B = −5.745i − 10.179 j a B = 11.69 in./s 2 60.6° 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. 185 PROBLEM 11.124 Knowing that at the instant shown block A has a velocity of 8 in./s and an acceleration of 6 in./s2 both directed down the incline, determine (a) the velocity of block B, (b) the acceleration of block B. SOLUTION From the diagram 2 x A + xB/A = constant Then 2v A + vB/A = 0 | vB/A | = 16 in./s or 2a A + aB/A = 0 and | aB/A | = 12 in./s 2 or Note that v B/A and a B/A must be parallel to the top surface of block A. (a) We have v B = v A + v B/A The graphical representation of this equation is then as shown. Note that because A is moving downward, B must be moving upward relative to A. We have vB2 = 82 + 162 − 2(8)(16) cos 15° or vB = 8.5278 in./s and or 8 8.5278 = sin α sin 15° α = 14.05° v B = 8.53 in./s (b) 54.1° The same technique that was used to determine v B can be used to determine a B . An alternative method is as follows. We have a B = a A + a B/A = (6i ) + 12(− cos 15°i + sin 15° j)* = −(5.5911 in./s 2 )i + (3.1058 in./s 2 ) j a B = 6.40 in./s 2 or 54.1° * Note the orientation of the coordinate axes on the sketch of the system. 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. 186 PROBLEM 11.125 A boat is moving to the right with a constant deceleration of 0.3 m/s2 when a boy standing on the deck D throws a ball with an initial velocity relative to the deck which is vertical. The ball rises to a maximum height of 8 m above the release point and the boy must step forward a distance d to catch it at the same height as the release point. Determine (a) the distance d, (b) the relative velocity of the ball with respect to the deck when the ball is caught. SOLUTION Horizontal motion of the ball: vx = (vx )0 , Vertical motion of the ball: v y = (v y )0 − gt yB = (v y )0 t − At maximum height, xball = (vx )0 t 1 2 gt , (v y ) 2 − (v y )02 = −2 gy 2 vy = 0 and y = ymax (v y )2 = 2 gymax = (2)(9.81)(8) = 156.96 m 2 /s 2 (v y )0 = 12.528 m/s At time of catch, tcatch = 2.554 s or Motion of the deck: 1 (9.81)t 2 2 and v y = 12.528 m/s y = 0 = 12.528 − vx = (vx )0 + aDt , xdeck = (vx )0 t + 1 a Dt 2 2 Motion of the ball relative to the deck: (vB/D ) x = (vx )0 − [(vx )0 + aDt ] = −aDt 1 1 xB/D = (vx )0 t − (vx )0 t + aDt 2 = − aDt 2 2 2 (vB/D ) y = (v y )0 − gt , yB/D = yB (a) (b) At time of catch, 1 d = xD/B = − (− 0.3)(2.554) 2 2 (vB/D ) x = −(− 0.3)(2.554) = + 0.766 m/s (vB/D ) y = 12.528 m/s d = 0.979 m or 0.766 m/s v B/D = 12.55 m/s 86.5° 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. 187 PROBLEM 11.126 The assembly of rod A and wedge B starts from rest and moves to the right with a constant acceleration of 2 mm/s2. Determine (a) the acceleration of wedge C, (b) the velocity of wedge C when t = 10 s. SOLUTION (a) We have a C = a B + a C/ B The graphical representation of this equation is then as shown. First note α = 180° − (20° + 105°) = 55° Then aC 2 = sin 20° sin 55° aC = 0.83506 mm/s 2 (b) aC = 0.835 mm/s 2 75° vC = 8.35 mm/s 75° For uniformly accelerated motion vC = 0 + aC t At t = 10 s: vC = (0.83506 mm/s 2 )(10 s) = 8.3506 mm/s or 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. 188 PROBLEM 11.127 Determine the required velocity of the belt B if the relative velocity with which the sand hits belt B is to be (a) vertical, (b) as small as possible. SOLUTION A grain of sand will undergo projectile motion. vsx = vsx = constant = −5 ft/s 0 vs y = 2 gh = (2)(32.2 ft/s 2 )(3 ft) = 13.90 ft/s ↓ y-direction. v S/B = v S − v B Relative velocity. (a) (1) If vS/B is vertical, −vS /B j = −5i − 13.9 j − ( −vB cos 15°i + vB sin 15° j) = −5i − 13.9 j + vB cos 15°i − vB sin 15° j Equate components. i : 0 = −5 + vB cos 15° vB = 5 = 5.176 ft/s cos 15° v B = 5.18 ft/s (b) 15° vS/C is as small as possible, so make vS/B ⊥ to vB into (1). −vS/B sin 15°i − vS/B cos 15° j = − 5i − 13.9 j + vB cos 15°i − vB sin 15° j Equate components and transpose terms. (sin 15°) vS/B + (cos 15°) vB = 5 (cos 15°) vS/B − (sin 15°) vB = 13.90 Solving, vS/B = 14.72 ft/s vB = 1.232 ft/s v B = 1.232 ft/s 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. 189 PROBLEM 11.128 Conveyor belt A, which forms a 20° angle with the horizontal, moves at a constant speed of 4 ft/s and is used to load an airplane. Knowing that a worker tosses duffel bag B with an initial velocity of 2.5 ft/s at an angle of 30° with the horizontal, determine the velocity of the bag relative to the belt as it lands on the belt. SOLUTION First determine the velocity of the bag as it lands on the belt. Now [(vB ) x ]0 = (vB )0 cos 30° = (2.5 ft/s) cos 30° [(vB ) y ]0 = (vB )0 sin 30° = (2.5 ft/s) sin 30° Horizontal motion. (Uniform) x = 0 + [(vB ) x ]0 t (vB ) x = [(vB ) x ]0 = (2.5 cos 30°) t = 2.5 cos 30° Vertical motion. (Uniformly accelerated motion) y = y0 + [(vB ) y ]0 t − 1 2 gt 2 = 1.5 + (2.5 sin 30°) t − (vB ) y = [(vB ) y ]0 − gt 1 2 gt 2 = 2.5 sin 30° − gt The equation of the line collinear with the top surface of the belt is y = x tan 20° Thus, when the bag reaches the belt 1.5 + (2.5 sin 30°) t − 1 2 gt = [(2.5 cos 30°) t ] tan 20° 2 or 1 (32.2) t 2 + 2.5(cos 30° tan 20° − sin 30°) t − 1.5 = 0 2 or 16.1t 2 − 0.46198t − 1.5 = 0 Solving t = 0.31992 s and t = −0.29122 s (Reject) 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. 190 PROBLEM 11.128 (Continued) The velocity v B of the bag as it lands on the belt is then v B = (2.5 cos 30°)i + [2.5 sin 30° − 32.2(0.319 92)] j = (2.1651 ft/s)i − (9.0514 ft/s) j Finally or v B = v A + v B/A v B/A = (2.1651i − 9.0514 j) − 4(cos 20°i + sin 20° j) = −(1.59367 ft/s)i − (10.4195 ft/s) j v B/A = 10.54 ft/s or 81.3° 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. 191 PROBLEM 11.129 During a rainstorm the paths of the raindrops appear to form an angle of 30° with the vertical and to be directed to the left when observed from a side window of a train moving at a speed of 15 km/h. A short time later, after the speed of the train has increased to 24 km/h, the angle between the vertical and the paths of the drops appears to be 45°. If the train were stopped, at what angle and with what velocity would the drops be observed to fall? SOLUTION vrain = vtrain + vrain/train Case : vT = 15 km/h ; vR / T 30° Case : vT = 24 km/h ; vR / T 45° Case : (vR ) y tan 30 = 15 − (vR ) x (1) Case : (vR ) y tan 45° = 24 − (vR ) x (2) Substract (1) from (2) (vR ) y (tan 45° − tan 30°) = 9 (vR ) y = 21.294 km/h Eq. (2): 21.294 tan 45° = 25 − (vR ) x (vR ) x = 2.706 km/h 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. 192 PROBLEM 11.129 (Continued) 3.706 21.294 β = 7.24° 21.294 = 21.47 km/h = 5.96 m/s vR = cos 7.24° tan β = vR = 5.96 m/s 82.8° Alternate solution Alternate, vector equation v R = vT + v R / T For first case, v R = 15i + vR / T −1 (− sin 30°i − cos 30° j) For second case, v R = 24i + vR /T − 2 (− sin 45°i − cos 45° j) Set equal 15i + vR /T −1 (− sin 30°i − cos 30° j) = 24i + vR /T − 2 (− sin 45°i − cos 45° j) Separate into components: i: 15 − vR /T −1 sin 30° = 24 − vR /T − 2 sin 45° −vR /T −1 sin 30° + vR /T − 2 sin 45° = 9 (3) −vR /T −1 cos 30° = −vR / T − 2 cos 45° j: vR / T −1 cos 30° + vR /T − 2 cos 45° = 0 (4) Solving Eqs. (3) and (4) simultaneously, vR /T −1 = 24.5885 km/h vR /T − 2 = 30.1146 km/h Substitute v R /T − 2 back into equation for v R . v R = 24i + 30.1146(− sin 45°i − cos 45° j) v R = 2.71i − 21.29 j −21.29 = −82.7585° 2.71 θ = tan −1 v R = 21.4654 km/hr = 5.96 m/s v R = 5.96 m/s 82.8° 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. 193 PROBLEM 11.130 As observed from a ship moving due east at 9 km/h, the wind appears to blow from the south. After the ship has changed course and speed, and as it is moving north at 6 km/h, the wind appears to blow from the southwest. Assuming that the wind velocity is constant during the period of observation, determine the magnitude and direction of the true wind velocity. SOLUTION v wind = v ship + v wind/ship v w = v s + v w/s Case v s = 9 km/h →; v w /s Case v s = 6 km/h ↑ ; v w /s 15 = 1.6667 9 α = 59.0° tan α = vw = 92 + 152 = 17.49 km/h v w = 17.49 km/h 59.0° 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. 194 PROBLEM 11.131 When a small boat travels north at 5 km/h, a flag mounted on its stern forms an angle θ = 50° with the centerline of the boat as shown. A short time later, when the boat travels east at 20 km/h, angle θ is again 50°. Determine the speed and the direction of the wind. SOLUTION We have vW = v B + vW/B Using this equation, the two cases are then graphically represented as shown. With vW now defined, the above diagram is redrawn for the two cases for clarity. Noting that θ = 180° − (50° + 90° + α ) = 40° − α We have vW 5 = sin 50° sin (40° − α ) φ = 180° − (50° + α ) = 130° − α vW 20 = sin 50° sin (130° − α ) 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. 195 PROBLEM 11.131 (Continued) Therefore or or 5 20 = sin (40° − α ) sin (130° − α ) sin 130° cos α − cos 130° sin α = 4(sin 40° cos α − cos 40° sin α ) tan α = sin 130° − 4 sin 40° cos 130° − 4 cos 40° or α = 25.964° Then vW = 5 sin 50° = 15.79 km/h sin (40° − 25.964°) vW = 15.79 km/h 26.0° 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. 196 PROBLEM 11.132 As part of a department store display, a model train D runs on a slight incline between the store’s up and down escalators. When the train and shoppers pass Point A, the train appears to a shopper on the up escalator B to move downward at an angle of 22° with the horizontal, and to a shopper on the down escalator C to move upward at an angle of 23° with the horizontal and to travel to the left. Knowing that the speed of the escalators is 3 ft/s, determine the speed and the direction of the train. SOLUTION v D = v B + v D/B We have v D = vC + v D/C The graphical representations of these equations are then as shown. Then vD 3 = sin 8° sin (22° + α ) Equating the expressions for vD 3 = sin 7° sin (23° − α ) vD 3 sin 8° sin 7° = sin (22° + α ) sin (23° − α ) or or or Then sin 8° (sin 23° cos α − cos 23° sin α ) = sin 7° (sin 22° cos α + cos 22° sin α ) tan α = sin 8° sin 23° − sin 7° sin 22° sin 8° cos 23° + sin 7° cos 22° α = 2.0728° vD = 3 sin 8° = 1.024 ft/s sin (22° + 2.0728°) v D = 1.024 ft/s 2.07° 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. 197 PROBLEM 11.132 (Continued) Alternate solution using components. v B = (3 ft/s) 30° = (2.5981 ft/s)i + (1.5 ft/s) j vC = (3 ft/s) 30° = (2.5981 ft/s)i − (1.5 ft/s) j v D/B = u1 22° = −(u1 cos 22°)i − (u1 sin 22°) j v D/C = u2 23° = −(u2 cos 23°)i + (u2 sin 23°) j v D = vD α = −(vD cos α )i + (vD sin α ) j v D = v B + v D/B = vC + v D/C 2.5981i + 1.5 j − (u1 cos 22°)i − (u1 sin 22°) j = 2.5981i − 1.5 j − (u2 cos 23°)i + (u2 sin 23°) j Separate into components, transpose, and change signs. u1 cos 22° − u2 cos 23° = 0 u1 sin 22° + u1 sin 23° = 3 Solving for u1 and u2 , u1 = 3.9054 ft/s u2 = 3.9337 ft/s v D = 2.5981i + 1.5 j − (3.9054 cos 22°)i − (3.9054 sin 22°) j = −(1.0229 ft/s)i + (0.0370 ft/s) j or v D = 2.5981i − 1.5 j − (3.9337 cos 23°)i + (3.9337 sin 23°) j = −(1.0229 ft/s)i + (0.0370 ft/s) j v D = 1.024 ft/s 2.07° 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. 198 PROBLEM 11.CQ8 The Ferris wheel is rotating with a constant angular velocity ω. What is the direction of the acceleration of Point A? (a) (b) (c) (d ) (e) The acceleration is zero. SOLUTION The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration pointed upwards. Answer: (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. 199 PROBLEM 11.CQ9 A racecar travels around the track shown at a constant speed. At which point will the racecar have the largest acceleration? (a) A (b) B (c) C (d ) The acceleration will be zero at all the points. SOLUTION The tangential acceleration is zero since the speed is constant, so there will only be normal acceleration. The normal acceleration will be maximum where the radius of curvature is a minimum, that is at Point A. Answer: (a) 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. 200 PROBLEM 11.CQ10 A child walks across merry-go-round A with a constant speed u relative to A. The merry-go-round undergoes fixed axis rotation about its center with a constant angular velocity ω counterclockwise.When the child is at the center of A, as shown, what is the direction of his acceleration when viewed from above. (a) (b) (c) (d ) (e) The acceleration is zero. SOLUTION Polar coordinates are most natural for this problem, that is, a = ( r − rθ 2 )er + (rθ + 2rθ)eθ (1) r = 0, θ = 0, r = 0, θ = ω, r = -u. When we substitute From the information given, we know these values into (1), we will only have a term in the −θ direction. Answer: (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. 201 PROBLEM 11.133 Determine the smallest radius that should be used for a highway if the normal component of the acceleration of a car traveling at 72 km/h is not to exceed 0.8 m/s 2 . SOLUTION an = v2 an = 0.8 m/s 2 ρ v = 72 km/h = 20 m/s 0.8 m/s 2 = (20 m/s) 2 ρ ρ = 500 m 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. 202 PROBLEM 11.134 Determine the maximum speed that the cars of the roller-coaster can reach along the circular portion AB of the track if ρ is 25 m and the normal component of their acceleration cannot exceed 3 g. SOLUTION We have an = v2 ρ Then (vmax ) 2AB = (3 × 9.81 m/s 2 )(25 m) or (vmax ) AB = 27.124 m/s (vmax ) AB = 97.6 km/h or 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. 203 PROBLEM 11.135 A bull-roarer is a piece of wood that produces a roaring sound when attached to the end of a string and whirled around in a circle. Determine the magnitude of the normal acceleration of a bull-roarer when it is spun in a circle of radius 0.9 m at a speed of 20 m/s. SOLUTION an = v2 ρ = (20 m/s) 2 = 444.4 m/s 2 0.9 m an = 444 m/s 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. 204 PROBLEM 11.136 To test its performance, an automobile is driven around a circular test track of diameter d. Determine (a) the value of d if when the speed of the automobile is 45 mi/h, the normal component of the acceleration is 11 ft/s 2 , (b) the speed of the automobile if d = 600 ft and the normal component of the acceleration is measured to be 0.6 g. SOLUTION (a) First note Now v = 45 mi/h = 66 ft/s an = ρ= v2 ρ (66 ft/s) 2 = 396 ft 11 ft/s 2 d = 2ρ (b) d = 792 ft v2 We have an = Then 1 v 2 = (0.6 × 32.2 ft/s 2 ) × 600 ft 2 ρ v = 76.131 ft/s v = 51.9 mi/h 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. 205 PROBLEM 11.137 An outdoor track is 420 ft in diameter. A runner increases her speed at a constant rate from 14 to 24 ft/s over a distance of 95 ft. Determine the magnitude of the total acceleration of the runner 2 s after she begins to increase her speed. SOLUTION We have uniformly accelerated motion v22 = v12 + 2at Δs12 Substituting or (24 ft/s)2 = (14 ft/s)2 + 2at (95 ft) at = 2 ft/s 2 Also v = v1 + at t At t = 2 s: v = 14 ft/s + (2 ft/s 2 )(2 s) = 18 ft/s v2 Now an = At t = 2 s: an = Finally a 2 = at2 + an2 At t = 2 s: a 2 = 22 + 1.542862 ρ (18 ft/s) 2 = 1.54286 ft/s 2 210 ft a = 2.53 ft/s 2 or 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. 206 PROBLEM 11.138 A robot arm moves so that P travels in a circle about Point B, which is not moving. Knowing that P starts from rest, and its speed increases at a constant rate of 10 mm/s2, determine (a) the magnitude of the acceleration when t = 4 s, (b) the time for the magnitude of the acceleration to be 80 mm/s2. SOLUTION Tangential acceleration: Speed: at = 10 mm/s 2 v = at t v2 an = where ρ = 0.8 m = 800 mm (a) When t = 4 s ρ v = (10)(4) = 40 mm/s an = Acceleration: ρ = at2 t 2 Normal acceleration: (40)2 = 2 mm/s 2 800 a = at2 + an2 = (10) 2 + (2) 2 a = 10.20 mm/s 2 (b) Time when a = 80 mm/s 2 a 2 = an2 + at2 2 (10) 2 t 2 2 (80) = + 10 800 2 t 4 = 403200 s 4 t = 25.2 s 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. 207 PROBLEM 11.139 A monorail train starts from rest on a curve of radius 400 m and accelerates at the constant rate at . If the maximum total acceleration of the train must not exceed 1.5 m/s 2 , determine (a) the shortest distance in which the train can reach a speed of 72 km/h, (b) the corresponding constant rate of acceleration at . SOLUTION When v = 72 km/h = 20 m/s and ρ = 400 m, an = v2 ρ = (20)2 = 1.000 m/s 2 400 a = an2 + at2 But at = a 2 − an2 = (1.5) 2 − (1.000) 2 = ± 1.11803 m/s 2 Since the train is accelerating, reject the negative value. (a) Distance to reach the speed. v0 = 0 Let x0 = 0 v12 = v02 + 2at ( x1 − x0 ) = 2at x1 x1 = (b) v12 (20) 2 = 2at (2)(1.11803) x1 = 178.9 m Corresponding tangential acceleration. at = 1.118 m/s 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. 208 PROBLEM 11.140 A motorist starts from rest at Point A on a circular entrance ramp when t = 0, increases the speed of her automobile at a constant rate and enters the highway at Point B. Knowing that her speed continues to increase at the same rate until it reaches 100 km/h at Point C, determine (a) the speed at Point B, (b) the magnitude of the total acceleration when t = 20 s. SOLUTION v0 = 0 Speeds: v1 = 100 km/h = 27.78 m/s s = Distance: π 2 (150) + 100 = 335.6 m v12 = v02 + 2at s Tangential component of acceleration: at = At Point B, v12 − v02 (27.78) 2 − 0 = = 1.1495 m/s 2 2s (2)(335.6) vB2 = v02 + 2at sB where sB = π 2 (150) = 235.6 m vB2 = 0 + (2)(1.1495)(235.6) = 541.69 m 2 /s 2 vB = 23.27 m/s (a) At t = 20 s, vB = 83.8 km/h v = v0 + at t = 0 + (1.1495)(20) = 22.99 m/s ρ = 150 m Since v < vB , the car is still on the curve. (b) Normal component of acceleration: an = Magnitude of total acceleration: |a| = v2 ρ = (22.99)2 = 3.524 m/s 2 150 at2 + an2 = (1.1495) 2 + (3.524) 2 | a | = 3.71 m/s 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. 209 PROBLEM 11.141 Racecar A is traveling on a straight portion of the track while racecar B is traveling on a circular portion of the track. At the instant shown, the speed of A is increasing at the rate of 10 m/s2, and the speed of B is decreasing at the rate of 6 m/s2. For the position shown, determine (a) the velocity of B relative to A, (b) the acceleration of B relative to A. SOLUTION v A = 240 km/h = 66.67 m/s Speeds: vB = 200 km/h = 55.56 m/s vA = 66.67 m/s Velocities: vB = 55.56 m/s (a) 50° vB/A = vB − vA Relative velocity: vB/A = (55.56 cos 50°) ← + 55.56sin 50° ↓ + 66.67 = 30.96 → + 42.56 = 52.63 m/s 53.96° vB /A = 189.5 km/h Tangential accelerations: (aA )t = 10 m/s 2 (aB )t = 6 m/s 2 an = Normal accelerations: ρ ( ρ = ∞) Car B: ( ρ = 300 m) Acceleration of B relative to A: 50° v2 Car A: (aB ) n = (b) 54.0° (aA ) n = 0 (55.56)2 = 10.288 300 (aB ) n = 10.288 m/s 2 40° aB/A = aB − aA a B/A = (a B )t + (a B ) n − (a A )t − (a A ) n =6 50° + 10.288 40° + 10 → + 0 = (6cos 50° + 10.288cos 40° + 10) + (6sin 50° − 10.288sin 40°) = 21.738 → + 2.017 a B/A = 21.8 m/s 2 5.3° 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. 210 PROBLEM 11.142 At a given instant in an airplane race, airplane A is flying horizontally in a straight line, and its speed is being increased at the rate of 8 m/s 2 . Airplane B is flying at the same altitude as airplane A and, as it rounds a pylon, is following a circular path of 300-m radius. Knowing that at the given instant the speed of B is being decreased at the rate of 3 m/s 2 , determine, for the positions shown, (a) the velocity of B relative to A, (b) the acceleration of B relative to A. SOLUTION First note v A = 450 km/h vB = 540 km/h = 150 m/s (a) v B = v A + v B/A We have The graphical representation of this equation is then as shown. We have vB2 /A = 4502 + 5402 − 2(450)(540) cos 60° vB/A = 501.10 km/h and 540 501.10 = sin α sin 60° α = 68.9° v B/A = 501 km/h (b) First note Now a A = 8 m/s 2 ( aB ) n = v 2B ρB = Then 60° (150 m/s) 2 300 m (a B ) n = 75 m/s 2 (a B )t = 3 m/s 2 68.9° 30° a B = (a B )t + (a B )n = 3(− cos 60° i + sin 60° j) + 75(−cos 30° i − sin 30° j) = −(66.452 m/s 2 )i − (34.902 m/s 2 ) j Finally a B = a A + a B/A a B/A = ( −66.452i − 34.902 j) − (8i ) = −(74.452 m/s 2 )i − (34.902 m/s 2 ) j a B/A = 82.2 m/s 2 25.1° 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. 211 PROBLEM 11.143 From a photograph of a homeowner using a snowblower, it is determined that the radius of curvature of the trajectory of the snow was 30 ft as the snow left the discharge chute at A. Determine (a) the discharge velocity v A of the snow, (b) the radius of curvature of the trajectory at its maximum height. SOLUTION (a) The acceleration vector is 32.2 ft/s . At Point A, tangential and normal components of a are as shown in the sketch. an = a cos 40° = 32.2 cos 40° = 24.67 ft/s 2 v A2 = ρ A (a A ) n = (30)(24.67) = 740.0 ft 2 /s 2 v A = 27.2 ft/s 40° vx = 27.20 cos 40° = 20.84 ft/s (b) At maximum height, v = vx = 20.84 ft/s an = g = 32.2 ft/s 2 , ρ = v2 (20.84) 2 = 32.2 an ρ = 13.48 ft 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. 212 PROBLEM 11.144 A basketball is bounced on the ground at Point A and rebounds with a velocity v A of magnitude 2.5 m/s as shown. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory. SOLUTION (a) (a A ) n = We have ρA = or v 2A ρA (2.5 m/s) 2 (9.81 m/s 2 ) sin 15° ρ A = 2.46 m or (b) ( aB ) n = We have vB2 ρB where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A sin 15° ρB = Then [(2.5 m/s) sin 15°]2 = 0.0427 m 9.81 m/s 2 ρ B = 42.7 mm or 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. 213 PROBLEM 11.145 A golfer hits a golf ball from Point A with an initial velocity of 50 m/s at an angle of 25° with the horizontal. Determine the radius of curvature of the trajectory described by the ball (a) at Point A, (b) at the highest point of the trajectory. SOLUTION (a) We have or (a A ) n = ρA = v 2A ρA (50 m/s)2 (9.81 m/s 2 ) cos 25° ρ A = 281 m or (b) We have ( aB ) n = vB2 ρB where Point B is the highest point of the trajectory, so that vB = (v A ) x = v A cos 25° Then ρB = [(50 m/s) cos 25°]2 9.81 m/s 2 ρ B = 209 m or 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. 214 PROBLEM 11.146 Three children are throwing snowballs at each other. Child A throws a snowball with a horizontal velocity v0. If the snowball just passes over the head of child B and hits child C, determine the radius of curvature of the trajectory described by the snowball (a) at Point B, (b) at Point C. SOLUTION The motion is projectile motion. Place the origin at Point A. Horizontal motion: v x = v0 x = v0 t Vertical motion: y0 = 0, (v y ) = 0 v y = − gt 1 y = − gt 2 2 2h , g t= where h is the vertical distance fallen. | v y| = 2 gh Speed: v 2 = vx2 + v 2y = v02 + 2 gh Direction of velocity. cos θ = v0 v Direction of normal acceleration. an = g cos θ = Radius of curvature: At Point B, ρ= gv0 v 2 = v ρ v3 gv0 hB = 1 m; xB = 7 m tB = (2)(1 m) = 0.45152 s 9.81 m/s 2 xB = v0t B v0 = xB 7m = = 15.504 m/s t B 0.45152 s vB2 = (15.504) 2 + (2)(9.81)(1) = 259.97 m 2 /s 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. 215 PROBLEM 11.146 (Continued) (a) Radius of curvature at Point B. ρB = At Point C (259.97 m 2 /s 2 )3/ 2 (9.81 m/s 2 )(15.504 m/s) ρ B = 27.6 m hC = 1 m + 2 m = 3 m vC2 = (15.504) 2 + (2)(9.81)(3) = 299.23 m 2 /s 2 (b) Radius of curvature at Point C. ρC = (299.23 m 2 /s 2 )3/2 (9.81 m/s 2 )(15.504 m/s) ρC = 34.0 m 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. 216 PROBLEM 11.147 Coal is discharged from the tailgate A of a dump truck with an initial velocity v A = 2 m/s 50° . Determine the radius of curvature of the trajectory described by the coal (a) at Point A, (b) at the point of the trajectory 1 m below Point A. SOLUTION a A = g = 9.81 m/s 2 (a) At Point A. Sketch tangential and normal components of acceleration at A. (a A ) n = g cos 50° ρA = vA2 (2) 2 = (a A ) n 9.81cos 50° ρ A = 0.634 m (b) At Point B, 1 meter below Point A. Horizontal motion: (vB ) x = (v A ) x = 2 cos 50° = 1.286 m/s Vertical motion: (vB ) 2y = (v A ) 2y + 2a y ( yB − y A ) = (2 cos 40°)2 + (2)(−9.81)(−1) = 21.97 m 2 /s 2 (vB ) y = 4.687 m/s tan θ = (vB ) y (vB ) x = 4.687 , 1.286 or θ = 74.6° aB = g cos 74.6° (vB ) 2x + (vB ) 2y vB 2 ρB = = (aB )n g cos 74.6° = (1.286) 2 + 21.97 9.81cos 74.6° ρ B = 9.07 m 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. 217 PROBLEM 11.148 From measurements of a photograph, it has been found that as the stream of water shown left the nozzle at A, it had a radius of curvature of 25 m. Determine (a) the initial velocity vA of the stream, (b) the radius of curvature of the stream as it reaches its maximum height at B. SOLUTION (a) We have (a A ) n = v 2A ρA or 4 v A2 = (9.81 m/s 2 ) (25 m) 5 or v A = 14.0071 m/s vA = 14.01 m/s (b) We have Where Then ( aB ) n = vB2 ρB vB = ( v A ) x = ρB 36.9° 4 vA 5 ( 4 × 14.0071 m/s ) = 5 2 9.81 m/s 2 ρ B = 12.80 m or 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. 218 PROBLEM 11.149 A child throws a ball from Point A with an initial velocity v A of 20 m/s at an angle of 25° with the horizontal. Determine the velocity of the ball at the points of the trajectory described by the ball where the radius of curvature is equal to three-quarters of its value at A. SOLUTION Assume that Points B and C are the points of interest, where yB = yC and vB = vC . Now (a A ) n = v 2A ρA or ρA = v 2A g cos 25° Then ρB = v A2 3 3 ρA = 4 4 g cos 25° vB2 We have ( aB ) n = where (aB ) n = g cos θ so that v A2 vB2 3 = 4 g cos 25° g cos θ or vB2 = ρB 3 cos θ 2 vA 4 cos 25° (1) Noting that the horizontal motion is uniform, we have ( v A ) x = ( vB ) x where Then or (v A ) x = v A cos 25° (vB ) x = vB cos θ v A cos 25° = vB cos θ cos θ = vA cos 25° vB 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. 219 PROBLEM 11.149 (Continued) Substituting for cos θ in Eq. (1), we have or vB2 = v A2 3 vA cos 25° 4 vB cos 25° vB3 = 3 3 vA 4 3 vB = 3 v A = 18.17 m/s 4 4 cos 25° 3 θ = ± 4.04° cos θ = 3 and v B = 18.17 m/s 4.04° v B = 18.17 m/s 4.04° 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. 220 PROBLEM 11.150 A projectile is fired from Point A with an initial velocity v 0 . (a) Show that the radius of curvature of the trajectory of the projectile reaches its minimum value at the highest Point B of the trajectory. (b) Denoting by θ the angle formed by the trajectory and the horizontal at a given Point C, show that the radius of curvature of the trajectory at C is ρ = ρ min /cos3 θ . SOLUTION For the arbitrary Point C, we have (aC ) n = ρC = or vC2 ρC vC2 g cos θ Noting that the horizontal motion is uniform, we have (v A ) x = (vC ) x (v A ) x = v0 cos α where v0 cos α = vC cos θ Then cos α v0 cos θ or vC = so that 1 ρC = g cos θ (a) 2 cos α v 2 cos 2 α v0 = 0 g cos3 θ cos θ In the expression for ρC , v0 , α , and g are constants, so that ρC is minimum where cos θ is maximum. By observation, this occurs at Point B where θ = 0. ρ min = ρ B = (b) (vC ) x = vC cos θ ρC = 1 cos3 θ ρC = ρ min cos3 θ v02 cos 2 α g v02 cos 2 α g Q.E.D. Q.E.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. 221 PROBLEM 11.151* Determine the radius of curvature of the path described by the particle of Problem 11.95 when t = 0. PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.) SOLUTION We have v= dr = R(cos ωn t − ωn t sin ωn t )i + cj + R (sin ωn t + ωn t cos ωn t )k dt and a= dv = R − ωn sin ωn t − ωn sin ωn t − ωn2t cos ωn t i dt ( ) ( ) + R ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t k or Now a = ωn R [−(2 sin ωn t + ωn t cos ωn t )i + (2 cos ωn t − ωn t sin ωn t ) k ] v 2 = R 2 (cos ωn t − ωn t sin ωn t )2 + c 2 + R 2 (sin ωn t + ωn t cos ωn t )2 ( ) = R 2 1 + ωn2 t 2 + c 2 Then and Now At t = 0: ( ) 1/2 v = R 2 1 + ωn2 t 2 + c 2 R 2ωn2 t dv = 1/ 2 dt 2 R 1 + ωn2 t 2 + c 2 ( 2 a = at2 + an2 ) 2 2 dv v = + dt ρ 2 dv =0 dt a = ωn R(2 k ) or a = 2ωn R v2 = R2 + c2 Then, with we have or dv = 0, dt a= 2ωn R = v2 ρ R 2 + c2 ρ= ρ R 2 + c2 2ωn R 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. 222 PROBLEM 11.152* Determine the radius of curvature of the path described by the particle of Problem 11.96 when t = 0, A = 3, and B = 1. SOLUTION With A = 3, B =1 we have r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k Now v= and 2 dv t t + 1 − a= = 3(− sin t − sin t − t cos t )i + 3 dt 2 t + 1 ) ( 3t dr = 3(cos t − t sin t )i + 2 j + (sin t + t cos t )k t + 1 dt t t2 + 1 j + (cos t + cos t − t sin t )k = − 3(2sin t + t cos t )i + 3 1 j 2 t + ( 1)1/2 + (2 cos t − t sin t )k v 2 = 9 (cos t − t sin t )2 + 9 Then t2 + (sin t + t cos t )2 t2 + 1 Expanding and simplifying yields v 2 = t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t Then v = [t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t )sin 2t ]1/ 2 and dv 4t 3 + 38t + 8(−2 cos t sin t + 4t 3sin 2 t + 2t 4 sin t cos t ) − 8[(3t 2 + 1)sin 2t + 2(t 3 + t ) cos 2t ] = dt 2[t 4 + 19t 2 + 1 + 8(cos 2 t + t 4 sin 2 t ) − 8(t 3 + t ) sin 2t ]1/ 2 Now 2 2 dv v a 2 = at2 + an2 = + dt ρ 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. 223 PROBLEM 11.152* (Continued) At t = 0: a = 3j + 2k or a = 13 ft/s 2 dv =0 dt v 2 = 9 (ft/s) 2 Then, with dv = 0, dt we have a= or ρ= v2 ρ 9 ft 2 /s 2 ρ = 2.50 ft 13 ft/s 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. 224 PROBLEM 11.153 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R /r ) 2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s 2 , determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. Earth: (umean )orbit = 107 Mm/h. SOLUTION g = 274 m/s 2 , For the sun, R= and Given that an = 1 1 D = (1.39 × 109 ) = 0.695 × 109 m 2 2 gR 2 v2 and that for a circular orbit = a n r r2 r = Eliminating an and solving for r, v = 107 × 106 m/h = 29.72 × 103 m/s For the planet Earth, Then gR 2 v2 r = (274)(0.695 × 109 ) 2 = 149.8 × 109 m (29.72 × 103 ) 2 r = 149.8 Gm 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. 225 PROBLEM 11.154 A satellite will travel indefinitely in a circular orbit around a planet if the normal component of the acceleration of the satellite is equal to g ( R /r ) 2 , where g is the acceleration of gravity at the surface of the planet, R is the radius of the planet, and r is the distance from the center of the planet to the satellite. Knowing that the diameter of the sun is 1.39 Gm and that the acceleration of gravity at its surface is 274 m/s 2 , determine the radius of the orbit of the indicated planet around the sun assuming that the orbit is circular. Saturn: (umean )orbit = 34.7 Mm/h. SOLUTION g = 274 m/s 2 For the sun, R= and Given that an = 1 1 D = (1.39 × 109 ) = 0.695 × 109 m 2 2 gR 2 v2 .and that for a circular orbit: = a n r r2 gR 2 v2 Eliminating an and solving for r, r = For the planet Saturn, v = 34.7 × 106 m/h = 9.639 × 103 m/s Then, r = (274)(0.695 × 109 ) 2 = 1.425 × 1012 m (9.639 × 103 )2 r = 1425 Gm 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. 226 PROBLEM 11.155 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Venus: g = 29.20 ft/s 2 , R = 3761 mi. SOLUTION From Problems 11.153 and 11.154, an = gR 2 r2 For a circular orbit, an = v2 r v= R Eliminating an and solving for v, For Venus, g r g = 29.20 ft/s 2 R = 3761 mi = 19.858 × 106 ft. r = 3761 + 100 = 3861 mi = 20.386 × 106 ft Then, v = 19.858 × 106 29.20 = 23.766 × 103 ft/s 20.386 × 106 v = 16200 mi/h 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. 227 PROBLEM 11.156 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Mars: g = 12.17 ft/s 2 , R = 2102 mi. SOLUTION From Problems 11.153 and 11.154, an = gR 2 r2 For a circular orbit, an = v2 r v= R Eliminating an and solving for v, For Mars, g r g = 12.17 ft/s 2 R = 2102 mi = 11.0986 × 106 ft r = 2102 + 100 = 2202 mi = 11.6266 × 106 ft Then, v = 11.0986 × 106 12.17 = 11.35 × 103 ft/s 11.6266 × 106 v = 7740 mi/h 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. 228 PROBLEM 11.157 Determine the speed of a satellite relative to the indicated planet if the satellite is to travel indefinitely in a circular orbit 100 mi above the surface of the planet. (See information given in Problems 11.153–11.154). Jupiter: g = 75.35 ft/s 2 , R = 44, 432 mi. SOLUTION From Problems 11.153 and 11.154, an = gR 2 r2 For a circular orbit, an = v2 r v= R Eliminating an and solving for v, For Jupiter, g r g = 75.35 ft/s 2 R = 44432 mi = 234.60 × 106 ft r = 44432 + 100 = 44532 mi = 235.13 × 106 ft Then, v = (234.60 × 106 ) 75.35 = 132.8 × 103 ft/s 6 235.13 × 10 v = 90600 mi/h 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. 229 PROBLEM 11.158 A satellite is traveling in a circular orbit around Mars at an altitude of 300 km. After the altitude of the satellite is adjusted, it is found that the time of one orbit has increased by 10 percent. Knowing that the radius of Mars is 3382 km, determine the new altitude of the satellite. (See information given in Problems 11.153–11.155.) SOLUTION an = g We have Then g R2 r2 and an = v2 r R2 v2 = r r2 v=R g r where r =R+h The circumference s of a circular orbit is equal to s = 2π r Assuming that the speed of the satellite in each orbit is constant, we have s = vtorbit Substituting for s and v 2π r = R torbit = = Now g torbit r 2π r 3/2 R g 2π ( R + h)3/2 R g (torbit ) 2 = 1.1(torbit )1 2π ( R + h2 )3/2 2π ( R + h1 )3/2 = 1.1 R R g g h2 = (1.1)2/3 ( R + h1 ) − R = (1.1)2/3 (3382 + 300) km − (3382 km) h2 = 542 km 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. 230 PROBLEM 11.159 Knowing that the radius of the earth is 6370 km, determine the time of one orbit of the Hubble Space Telescope, knowing that the telescope travels in a circular orbit 590 km above the surface of the earth. (See information given in Problems 11.153–11.155.) SOLUTION an = g We have Then or g R2 r2 and an = v2 r R2 v2 = r r2 v=R g r where r =R+h The circumference s of the circular orbit is equal to s = 2π r Assuming that the speed of the telescope is constant, we have s = vtorbit Substituting for s and v 2π r = R or torbit = = g torbit r 2π r 3/2 R g 2π [(6370 + 590) km]3/2 1h × −3 2 1/2 6370 km [9.81 × 10 km/s ] 3600 s torbit = 1.606 h or 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. 231 PROBLEM 11.160 Satellites A and B are traveling in the same plane in circular orbits around the earth at altitudes of 120 and 200 mi, respectively. If at t = 0 the satellites are aligned as shown and knowing that the radius of the earth is R = 3960 mi, determine when the satellites will next be radially aligned. (See information given in Problems 11.153–11.155.) SOLUTION an = g We have Then g R2 r2 R2 v2 = r r2 and an = or v2 r v=R g r r =R+h where The circumference s of a circular orbit is equal to s = 2π r Assuming that the speeds of the satellites are constant, we have s = vT Substituting for s and v 2π r = R or Now T= g T r 2π r 3/ 2 2π ( R + h)3/2 = R g R g hB > hA (T ) B > (T ) A Next let time TC be the time at which the satellites are next radially aligned. Then, if in time TC satellite B completes N orbits, satellite A must complete ( N + 1) orbits. Thus, TC = N (T ) B = ( N + 1)(T ) A or 2π ( R + hB )3/2 2π ( R + hA )3/2 ( 1) N = N + g g R R 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. 232 PROBLEM 11.160 (Continued) N= or = Then ( R + hA )3/2 ( R + hB ) 3/ 2 − ( R + hA ) 1 ( 3960 + 200 3960 +120 ) 3/2 TC = N (T ) B = N −1 3/2 = 1 ( R + hB R + hA ) 3/2 −1 = 33.835 orbits 2π ( R + hB )3/2 R g [(3960 + 200) mi] × 1 h 2π 3960 mi 32.2 ft/s 2 × 1 mi 1/2 3600s 5280 ft 3/2 = 33.835 ( ) TC = 51.2 h or Alternative solution From above, we have (T ) B > (T ) A . Thus, when the satellites are next radially aligned, the angles θ A and θ B swept out by radial lines drawn to the satellites must differ by 2π . That is, θ A = θ B + 2π For a circular orbit s = rθ From above s = vt and v = R Then θ= At time TC : or R g ( R + hA ) 3/2 TC = TC = = g r R g R g s vt 1 g = = R t t = 3/2 t = r r r r r ( R + h)3/2 R g ( R + hB )3/ 2 TC + 2π 2π 1 1 R g − ( R + hA )3/ 2 ( R + hB )3/ 2 2π ( 1 mi (3960 mi) 32.2 ft/s 2 × 5280 ft × × 1 [(3960 + 120) mi ]3/ 2 1 − ) 1/2 1 [(3960 + 200) mi ]3/ 2 1h 3600 s TC = 51.2 h or 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. 233 PROBLEM 11.161 The oscillation of rod OA about O is defined by the relation θ = (3/π )(sin π t ), where θ and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r = 6(1 − e−2t ) where r and t are expressed in inches and seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the acceleration of the collar, (c) the acceleration of the collar relative to the rod. SOLUTION Calculate the derivatives with respect to time. 3 r = 6 − 6e −2t in. θ= r = 12e −2t in/s θ = 3cosπ t rad/s r = −24e−2t in/s 2 θ = −3π sin π t rad/s2 π sin π t rad At t = 1 s, (a) 3 r = 6 − 6e −2 = 5.1880 in. θ= r = 12e −2 = 1.6240 in/s θ = 3cos π = −3 rad/s r = −24e−2 = −3.2480 in/s 2 θ = −3π sin π = 0 π sin π = 0 Velocity of the collar. v = rer + rθeθ = 1.6240 e r + (5.1880)(−3)eθ v = (1.624 in/s)er + (15.56 in/s)eθ (b) Acceleration of the collar. a = (r − rθ 2 )e r + (rθ + 2rθ)eθ = [ −3.2480 − (5.1880)( −3) 2 ]er + (5.1880)(0) + (2)(1.6240)(−3)]eθ (−49.9 in/s 2 )er + (−9.74 in/s 2 )eθ (c) Acceleration of the collar relative to the rod. a B /OA = (−3.25 in/s 2 )er a B /OA = re r 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. 234 PROBLEM 11.162 The rotation of rod OA about O is defined by the relation θ = t 3 − 4t , where θ and t are expressed in radians and seconds, respectively. Collar B slides along the rod so that its distance from O is r = 2.5t 3 − 5t 2 , where r and t are expressed in inches and seconds, respectively. When t = 1 s, determine (a) the velocity of the collar, (b) the acceleration of the collar, (c) the radius of curvature of the path of the collar. SOLUTION Calculate the derivatives with respect to time. r = 2.5t 3 − 5t 2 θ = t 3 − 4t r = 7.5t 2 − 10t θ = 3t 2 − 4 r = 15t − 10 θ = 6t At t = 1 s, (a) r = 2.5 − 5 = −2.5 in. θ = 1 − 4 = −3 rad r = 7.5 − 10 = −2.5 in./s θ = 3 − 4 = −1 rad/s r = 15 − 10 = 5 in./s 2 θ = 6 rad/s 2 Velocity of the collar. v = rer + rθeθ = −2.5er + (−2.5)( −1)eθ v = ( −2.50 in./s)er + (2.50 in./s)eθ v = (2.50) 2 + (2.50) 2 = 3.5355 in./s Unit vector tangent to the path. et = (b) v = −0.70711e r + 0.70711eθ v Acceleration of the collar. a = (r − rθ 2 )e r + (rθ + 2rθ)eθ = [5 − (−2.5)(−1)2 ]er + [(−2.5)(6) + (2)(−2.5)(−1)]eθ a = (7.50 in/s 2 )er + (−10.00 in/s 2 )eθ 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. 235 PROBLEM 11.162 (Continued) Magnitude: a = (7.50) 2 + (10.00) 2 = 12.50 in./s 2 at = aet Tangential component: at = (7.50)(−0.70711) + (−10.00)(0.70711) = −12.374 in./s 2 Normal component: (c) an = a 2 − at2 = 1.7674 in./s 2 Radius of curvature of path. an = ρ= v2 ρ v 2 (3.5355 in./s) 2 = an 1.7674 in./s 2 ρ = 7.07 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. 236 PROBLEM 11.163 The path of particle P is the ellipse defined by the relations r = 2/(2 − cos π t ) and θ = π t , where r is expressed in meters, t is in seconds, and θ is in radians. Determine the velocity and the acceleration of the particle when (a) t = 0, (b) t = 0.5 s. SOLUTION We have r= 2 2 − cos π t θ = πt Then r = −2π sin π t (2 − cos π t )2 θ = π and r = −2π π cos π t (2 − cos π t ) − sin π t (2π sin π t ) (2 − cos π t )3 = −2π 2 (a) At t = 0: Now 2cos π t − 1 − sin 2 π t (2 − cos π t )3 r=2m θ =0 r = 0 θ = π rad/s r = −2π 2 m/s 2 θ = 0 v = rer + rθeθ = (2)(π )eθ v = (2π m/s)eθ or and θ = 0 a = (r − rθ 2 )e r + (rθ + 2rθ)eθ = [ −2π 2 − (2)(π )2 ]er a = −(4π 2 m/s 2 )er or (b) At t = 0.5 s: θ= r =1 m r = −2π π = − m/s 2 2 (2) r = −2π 2 −1 − 1 π 2 = m/s 2 3 2 (2) π rad 2 θ = π rad/s θ = 0 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. 237 PROBLEM 11.163 (Continued) Now π v = rer + rθeθ = − e r + (1)(π )eθ 2 π v = − m/s er + (π m/s)eθ 2 or and a = (r − rθ 2 )e r + (rθ + 2rθ)eθ π 2 π = − (1)(π ) 2 e r + 2 − (π ) eθ 2 2 π2 a = − m/s 2 er − (π 2 m/s 2 )eθ 2 or 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. 238 PROBLEM 11.164 The two-dimensional motion of a particle is defined by the relations r = 2a cos θ and θ = bt 2 /2, where a and b are constants. Determine (a) the magnitudes of the velocity and acceleration at any instant, (b) the radius of curvature of the path. What conclusion can you draw regarding the path of the particle? SOLUTION (a) We have r = 2a cos θ 1 θ = bt 2 2 Then r = −2aθ sin θ θ = bt and r = −2a (θ sin θ + θ 2 cos θ ) θ = b Substituting for θ and θ r = −2abt sin θ r = −2ab(sin θ + bt 2 cos θ ) Now Then vθ = rθ = 2abt cos θ vr = r = −2abt sin θ v = vr2 + vθ2 = 2abt[( − sin θ ) 2 + (cos θ ) 2 ]1/2 v = 2abt or Also ar = r − rθ 2 = −2ab(sin θ + bt 2 cos θ ) − 2ab 2 t 2 cos θ = −2ab(sin θ + 2bt 2 cos θ ) and aθ = rθ + 2rθ = 2ab cos θ − 4ab 2 t 2 sin θ = −2ab(cos θ − 2bt 2 sin θ ) Then a = ar2 + aB2 = 2ab[(sin θ + 2bt 2 cos θ ) 2 + (cos θ − 2bt 2 sin θ ) 2 ]1/2 a = 2ab 1 + 4b 2 t 4 or (b) 2 at2 + an2 2 2 dv v = + dt ρ Now a = Then dv d = (2abt ) = 2ab dt dt 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. 239 PROBLEM 11.164 (Continued) so that or ( 2ab 1 + 4b 2 t 4 ) 2 = (2ab) 2 + an2 4a 2b 2 (1 + 4b 2t 4 ) = 4a 2 b 2 + an2 or an = 4ab 2 t 2 Finally an = v2 (2abt ) 2 ρ= ρ 4ab 2 t 2 ρ =a or Since the radius of curvature is a constant, the path is a circle of radius a. 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. 240 PROBLEM 11.165 As rod OA rotates, pin P moves along the parabola BCD. Knowing that the equation of this parabola is r = 2b /(1 + cos θ ) and that θ = kt , determine the velocity and acceleration of P when (a) θ = 0, (b) θ = 90°. SOLUTION 2b θ = kt 1 + cos kt 2bk sin kt θ = k θ = 0 r = (1 + cos kt ) 2 2bk [(1 + cos kt )2 k cos kt + (sin kt )2(1 + cos kt )( k sin kt )] r = (1 + cos kt ) 4 r= (a) When θ = kt = 0: 2bk 1 [(2)2 k (1) + 0] = bk 2 4 2 (2) r =b r = 0 r= θ =0 θ = k θ = 0 vθ = rθ = bk vr = r = 0 v = bk eθ 1 1 ar = r − rθ 2 = bk 2 − bk 2 = − bk 2 2 2 aθ = rθ + 2rθ = b(0) + 2(0) = 0 (b) 1 a = − bk 2 er 2 When θ = kt = 90°: r = 2b r = 2bk θ = 90° θ = k vr = r = 2bk 2bk [0 + 2k ] = 4bk 2 19 θ = 0 r = vθ = rθ = 2bk v = 2bk er + 2bk eθ ar = r − rθ 2 = 4bk 2 − 2bk 2 = 2bk 2 a = rθ + 2rθ = 2b(0) + 2(2bk )k = 4bk 2 θ a = 2bk 2 e r + 4bk 2 eθ 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. 241 PROBLEM 11.166 The pin at B is free to slide along the circular slot DE and along the rotating rod OC. Assuming that the rod OC rotates at a constant rate θ, (a) show that the acceleration of pin B is of constant magnitude, (b) determine the direction of the acceleration of pin B. SOLUTION From the sketch: r = 2b cos θ r = −2b sin θ θ Since θ = constant, θ = 0 r = −2b cos θ θ 2 ar = r − rθ 2 = −2b cos θ θ 2 − (2b cos θ )θ 2 a = −4b cos θ θ 2 r aθ = rθ + 2rθ = (2b cos θ )(0) + 2(−2b sin θ )θ 2 a = −4b sin θ θ 2 θ a = ar2 + aθ2 = 4bθ 2 (− cos θ ) 2 + (− sin θ )2 a = 4bθ 2 Since both b and θ are constant, we find that a = constant γ = tan −1 −4b sin θ θ 2 aθ = tan −1 2 ar −4b cos θ θ γ = tan −1 (tan θ ) γ =θ Thus, a is directed toward A 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. 242 PROBLEM 11.167 To study the performance of a racecar, a high-speed camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine (a) the speed of the car in terms of b, θ , and θ, (b) the magnitude of the acceleration in terms of b, θ , θ, and θ. SOLUTION (a) We have r= b cos θ Then r = bθ sin θ cos 2 θ We have v 2 = vr2 + vθ2 = (r)2 + (rθ) 2 2 or bθ sin θ bθ 2 = + 2 cos θ cos θ b2θ 2 b2θ 2 sin 2θ = + 1 = 2 2 4 cos θ cos θ cos θ bθ v=± cos 2θ For the position of the car shown, θ is decreasing; thus, the negative root is chosen. v=− bθ cos 2θ v=− bθ cos 2θ Alternative solution. From the diagram or r = −v sin θ bθ sin θ = −v sin θ cos 2θ or 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. 243 PROBLEM 11.167 (Continued) (b) For rectilinear motion a= dv dt Using the answer from Part a v=− Then a= bθ cos 2θ d bθ − dt cos 2θ = −b θ cos 2θ − θ(−2θ cos θ sin θ ) cos 4θ a=− or b (θ + 2θ 2 tan θ ) cos 2θ Alternative solution From above Then Now where and b bθ sin θ r = cos θ cos 2 θ (θ sin θ + θ 2 cos θ )(cos 2θ ) − (θ sin θ )(−2θ cos θ sin θ ) r = b cos 4θ θ sin θ θ 2 (1 + sin 2 θ ) = b + 2 cos3θ cos θ r= a 2 = ar2 + aθ2 θ sin θ θ 2 (1 + sin 2 θ ) bθ 2 + ar = r − rθ 2 = b − 2 cos 2θ cos θ cos θ b 2θ 2 sin 2 θ = + θ sin θ cos θ cos 2θ b sin θ ar = (θ + 2θ 2 tan θ ) 2 cos θ aθ = rθ + 2rθ = = Then bθ bθ 2 sin θ +2 cos θ cos 2θ b cos θ (θ + 2θ tan θ ) cos 2θ a=± b (θ + 2θ 2 tan θ )[(sin θ ) 2 + (cos θ )2 ]1/ 2 2 cos θ For the position of the car shown, θ is negative; for a to be positive, the negative root is chosen. b a=− (θ + 2θ 2 tan θ ) cos 2 θ 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. 244 PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle β . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ. SOLUTION From the diagram r d = sin (180° − β ) sin ( β − θ ) or d sin β = r (sin β cos θ − cos β sin θ ) tan β tan β cos θ − sin θ or r=d Then r = d tan β −(− tan β sin θ − cos θ ) θ (tan β cos θ − sin θ )2 tan β sin θ + cos θ = dθ tan β (tan β cos θ − sin θ ) 2 From the diagram vr = v cos ( β − θ ) where vr = r Then dθ tan β tan β sin θ + cos θ = v(cos β cos θ + sin β sin θ ) (tan β cos θ − sin θ )2 = v cos β (tan β sin θ + cos θ ) v= or dθ tan β sec β (tan β cosθ − sin θ )2 Alternative solution. We have v 2 = vr2 + vθ2 = (r) 2 + ( rθ) 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. 245 PROBLEM 11.168 (Continued) Using the expressions for r and r from above tan β sin θ + cos θ v = dθ tan β (tan β cos θ − sin θ ) 2 2 1/ 2 or (tan β sin θ + cos θ )2 dθ tan β v=± + 1 2 (tan β cos θ − sin θ ) (tan β cos θ − sin θ ) 1/ 2 tan 2 β + 1 dθ tan β =± 2 (tan β cos θ − sin θ ) (tan β cos θ − sin θ ) Note that as θ increases, the helicopter moves in the indicated direction. Thus, the positive root is chosen. v= dθ tan β sec β (tan β cos θ − sin θ )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. 246 PROBLEM 11.169 At the bottom of a loop in the vertical plane, an airplane has a horizontal velocity of 315 mi/h and is speeding up at a rate of 10 ft/s2. The radius of curvature of the loop is 1 mi. The plane is being tracked by radar at O. What are r , θ and θ for this the recorded values of r, instant? SOLUTION Geometry. The polar coordinates are r = (2400) 2 + (1800)2 = 3000 ft Velocity Analysis. 1800 = 36.87° 2400 θ = tan −1 v = 315 mi/h = 462 ft/s vr = 462 cos θ = 369.6 ft/s vθ = −462sin θ = −277.2 ft/s vr = r vθ = rθ θ = r = 370 ft/s vθ 277.2 =− r 3000 θ = −0.0924 rad/s Acceleration analysis. at = 10 ft/s 2 an = v2 ρ = (462) 2 = 40.425 ft/s 2 5280 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. 247 PROBLEM 11.169 (Continued) ar = at cos θ + an sin θ = 10 cos 36.87° + 40.425 sin 36.87° = 32.255 ft/s 2 aθ = − at sin θ + an cos θ = −10 sin 36.87° + 40.425 cos 36.87° = 26.34 ft/s 2 ar = r − rθ 2 r = ar + rθ 2 r = 32.255 + (3000)( −0.0924) 2 r = 57.9 ft/s 2 aθ = rθ + 2rθ a 2rθ θ = θ − r r 26.34 (2)(369.6)(−0.0924) = − 3000 3000 θ = 0.0315 rad/s 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. 248 PROBLEM 11.170 Pin C is attached to rod BC and slides freely in the slot of rod OA which rotates at the constant rate ω. At the instant when β = 60°, r and θ. Express your answers in terms determine (a) r and θ, (b) of d and ω. SOLUTION Looking at d and β as polar coordinates with d = 0, vβ = d β = d ω, vd = d = 0 aβ = d β + 2d β = 0, r = d 3 for angles shown. Geometry analysis: (a) Velocity analysis: ad = d − d β 2 = −d ω 2 Sketch the directions of v, er and eθ. vr = r = v ⋅ er = d ω cos120° 1 r = − d ω 2 vθ = rθ = v ⋅ eθ = d ω cos 30° θ = (b) Acceleration analysis: 3 dω cos 30° dω 2 = r d 3 θ = 1 ω 2 Sketch the directions of a, er and eθ. ar = a ⋅ e r = a cos150° = − 3 dω 2 2 3 r − rθ2 = − d ω2 2 r =− 3 3 1 dω 2 + rθ 2 = − dω 2 + d 3 ω 2 2 2 2 r = − 3 dω 2 4 1 aθ = a ⋅ eθ = d ω 2 cos120° = − d ω 2 2 a = rθ + 2rθ θ θ = 1 (aθ − 2rθ) = r 1 3d 1 1 1 2 − d ω − (2) − d ω ω 2 2 2 θ = 0 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. 249 PROBLEM 11.171 For the racecar of Problem 11.167, it was found that it took 0.5 s for the car to travel from the position θ = 60° to the position θ = 35°. Knowing that b = 25 m, determine the average speed of the car during the 0.5-s interval. PROBLEM 11.167 To study the performance of a racecar, a high-speed camera is positioned at Point A. The camera is mounted on a mechanism which permits it to record the motion of the car as the car travels on straightway BC. Determine (a) the speed of the car in terms of b, θ , and θ, (b) the magnitude of the acceleration in terms of b, θ , θ, and θ. SOLUTION From the diagram: Δr12 = 25 tan 60° − 25 tan 35° = 25.796 m Now vave = = Δr12 Δt12 25.796 m 0.5 s = 51.592 m/s vave = 185.7 km/h or 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. 250 PROBLEM 11.172 For the helicopter of Problem 11.168, it was found that when the helicopter was at B, the distance and the angle of elevation of the helicopter were r = 3000 ft and θ = 20°, respectively. Four seconds later, the radar station sighted the helicopter at r = 3320 ft and θ = 23.1°. Determine the average speed and the angle of climb β of the helicopter during the 4-s interval. PROBLEM 11.168 After taking off, a helicopter climbs in a straight line at a constant angle β . Its flight is tracked by radar from Point A. Determine the speed of the helicopter in terms of d, β , θ , and θ. SOLUTION We have r0 = 3000 ft r4 = 3320 ft θ0 = 20° θ 4 = 23.1° From the diagram: Δr 2 = 30002 + 33202 − 2(3000)(3320) cos (23.1° − 20°) or Δr = 362.70 ft Now vave = Δr Δt 362.70 ft = 4s = 90.675 ft/s vave = 61.8 mi/h or Also, or Δr cos β = r4 cos θ 4 − r0 cos θ0 cos β = 3320 cos 23.1° − 3000 cos 20° 362.70 β = 49.7° or 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. 251 PROBLEM 11.173 A particle moves along the spiral shown; determine the magnitude of the velocity of the particle in terms of b, θ , and θ. SOLUTION Hyperbolic spiral. r= b θ dr b dθ b =− 2 = − 2 θ dt θ dt θ b b vr = r = − 2 θ vθ = rθ = θ r = θ θ 2 1 1 v = vr2 + vθ2 = bθ − 2 + θ θ bθ = 2 1+θ 2 2 θ v= b θ2 1 + θ 2 θ 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. 252 PROBLEM 11.174 A particle moves along the spiral shown; determine the magnitude of the velocity of the particle in terms of b, θ , and θ. SOLUTION Logarithmic spiral. r = ebθ r = dr dθ = bebθ = bebθ θ dt dt vr = r = bebθ θ vθ = rθ = ebθ θ v= vr2 + vθ = e θ b 2 + 1 bθ 2 v = ebθ 1 + b 2 θ 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. 253 PROBLEM 11.175 A particle moves along the spiral shown. Knowing that θ is constant and denoting this constant by ω , determine the magnitude of the acceleration of the particle in terms of b, θ , and ω. SOLUTION b Hyperbolic spiral. r= From Problem 11.173 r = − θ r =− b θ2 θ b 2b 2 θ + 3θ 2 θ θ b 2b b ar = r − rθ 2 = − 2 θ + 3 θ 2 − θ 2 θ θ θ b b b b aθ = rθ + 2rθ = θ + 2 − 2 θ θ = θ − 2 2 θ 2 θ θ θ θ Since θ = ω = constant, θ = 0, and we write: b 2 bω 2 2 − ω ω = 3 (2 − θ 2 ) θ θ3 θ 2 b bω aθ = −2 2 ω 2 = − 3 (2θ ) θ θ ar = + 2b a = ar2 + aθ2 = bω 2 θ 3 (2 − θ 2 ) 2 + (2θ ) 2 = bω 2 θ 3 4 − 4θ 2 + θ 4 + 4θ 2 a= bω 2 θ3 4 +θ 4 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. 254 PROBLEM 11.176 A particle moves along the spiral shown. Knowing that θ is constant and denoting this constant by ω , determine the magnitude of the acceleration of the particle in terms of b, θ , and ω. SOLUTION Logarithmic spiral. r = ebθ dr = bebθ θ dt r = bebθ θ + b 2 ebθ θ 2 = bebθ (θ + bθ 2 ) r = ar = r − rθ 2 = bebθ (θ + bθ 2 ) − ebθ θ 2 a = rθ + 2rθ = ebθ θ + 2(bebθ θ)θ θ Since θ = ω = constant, θ = θ , and we write ar = bebθ (bω 2 ) − ebθ ω 2 = ebθ (b 2 − 1)ω 2 aθ = 2bebθ ω 2 a = ar2 + aθ2 = ebθ ω 2 (b2 − 1)2 + (2b) 2 = ebθ ω 2 b4 − 2b 2 + 1 + 4b 2 = ebθ ω 2 b 4 + 2b 2 + 1 = ebθ ω 2 (b2 + 1) 2 = ebθ ω 2 (b2 + 1) a = (1 + b 2 )ω 2 ebθ 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. 255 PROBLEM 11.177 The motion of a particle on the surface of a right circular cylinder is defined by the relations R = A, θ = 2π t , and z = B sin 2π nt , where A and B are constants and n is an integer. Determine the magnitudes of the velocity and acceleration of the particle at any time t. SOLUTION R=A R = 0 θ = 2π t z = B sin 2π nt θ = 2π z = 2π n B cos 2π nt = 0 R θ = 0 z = −4π 2 n 2 B sin 2π nt Velocity (Eq. 11.49) v = R e R + Rθeθ + zk v= + A(2π )eθ + 2π n B cos 2π nt k v = 2π A2 + n 2 B 2 cos 2 2π nt Acceleration (Eq. 11.50) - Rθ 2 )e + ( Rθ + 2 Rθ)e + a = (R zk R θ a = −4π 2 Aek − 4π 2 n 2 B sin 2π nt k a = 4π 2 A2 + n4 B 2 sin 2 2π nt 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. 256 PROBLEM 11.178 Show that r = hφ sin θ knowing that at the instant shown, step AB of the step exerciser is rotating counterclockwise at a constant rate φ. SOLUTION From the diagram r 2 = d 2 + h 2 − 2dh cos φ Then Now or 2rr = 2dhφ sin φ r d = sin φ sin θ r= d sin φ sin θ Substituting for r in the expression for r d sin φ sin θ r = dhφ sin φ or r = hφ sin θ Q.E.D. Alternative solution. First note α = 180° − (φ + θ ) Now v = v r + vθ = re r + rθeθ With B as the origin vP = dφ ( d = constant d = 0) 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. 257 PROBLEM 11.178 (Continued) With O as the origin (vP )r = r where (vP )r = vP sin α Then Now or substituting r = dφ sin α h d = sin α sin θ d sin α = h sin θ r = hφ sin θ Q.E.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. 258 PROBLEM 11.179 The three-dimensional motion of a particle is defined by the relations R = A(1 − e −t ), θ = 2π t , and z = B(1 − e− t ). Determine the magnitudes of the velocity and acceleration when (a) t = 0, (b) t = ∞. SOLUTION R = A(1 − e −t ) R = Ae −t = − Ae−t R θ = 2π t z = B (1 − e−t ) θ = 2π z = Be−t θ = 0 z = − Be−t Velocity (Eq. 11.49) v = R e R + Rθeθ + zk v = Ae −t e R + 2π A(1 − e −t )eθ + Be−t k (a) When t = 0: e−t = e0 = 1; (b) When t = ∞ : e− t = e −∞ = 0 v = A2 + B 2 v = Ae R + Bk v = 2π Aeθ v = 2π A Acceleration (Eq. 11.50) − Rθ1 )e + ( Rθ + 2 Rθ)e + a = (R zk R θ = [ − Ae−t − A(1 − e −t )4π 2 ]e R + [0 + 2 Ae−t (2π )]eθ − Be −t k (a) When t = 0: e −t = e0 = 1 a = − Ae R + 4π Aeθ − B k a = A2 + (4π A) 2 + B 2 (b) When a = (1 + 16π 2 ) A2 + B 2 t = ∞ : e −t = e−∞ = 0 a = −4π 2 Ae R a = 4π 2 A 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. 259 PROBLEM 11.180* For the conic helix of Problem 11.95, determine the angle that the osculating plane forms with the y axis. PROBLEM 11.95 The three-dimensional motion of a particle is defined by the position vector r = (Rt cos ωnt)i + ctj + (Rt sin ωnt)k. Determine the magnitudes of the velocity and acceleration of the particle. (The space curve described by the particle is a conic helix.) SOLUTION First note that the vectors v and a lie in the osculating plane. Now r = ( Rt cos ωn t )i + ctj + ( Rt sin ωn t )k Then v= dr = R(cos ωn t − ωn t sin ωn t )i + cj + R(sin ωn t + ωn t cos ωn t )k dt and a= dv dt ( ) = R −ωn sin ωn t − ωn sin ωn t − ωn2 t cos ωn t i +R ( ωn cos ωn t + ωn cos ωn t − ωn2 t sin ωn t )k = ωn R[−(2sin ωn t + ωn t cos ωn t )i + (2cos ωn t − ωn t sin ωn t )k ] It then follows that the vector ( v × a) is perpendicular to the osculating plane. i j k ( v × a) = ωn R R(cos ωn t − ωn t sin ωn t ) c R(sin ωn t + ωn t cos ωn t ) −(2sin ωn t + ωn t cos ωn t ) 0 (2cos ωn t − ωn t sin ωn t ) = ωn R{c(2 cos ωn t − ωn t sin ωn t )i + R[ −(sin ωn t + ωn t cos ωn t )(2sin ωn t + ωn t cos ωn t ) − (cos ωn t − ωn t sin ωn t )(2 cos ωn t − ωn t sin ωn t )] j + c(2sin ωn t + ωn t cos ωn t )k ( ) = ωn R c(2 cos ωn t − ωn t sin ωn t )i − R 2 + ωn2 t 2 j + c(2sin ωn t + ωn t cos ωn t )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. 260 PROBLEM 11.180* (Continued) The angle α formed by the vector ( v × a) and the y axis is found from cos α = Where ( v × a) ⋅ j | ( v × a) || j | |j| =1 ( ( v × a) ⋅ j = −ωn R 2 2 + ωn2 t 2 ) ( |( v × a) | = ωn R c 2 (2 cos ωn t − ωn t sin ωn t ) 2 + R 2 2 + ωn2 t 2 ) 2 1/ 2 + c 2 (2sin ωn t + ωn t cos ωn t )2 ( ) 2 1/2 ( ) = ωn R c 2 4 + ωn2 t 2 + R 2 2 + ωn2t 2 Then ( ) ω R c ( 4 + ω t ) + R ( 2 + ω t ) −R ( 2 + ω t ) = c 4 + ω t + R 2 + ω t ) ( ) ( −ωn R 2 2 + ωn2t 2 cos α = 2 n 2 2 n 2 2 2 n 2 1/2 2 2 n 2 2 2 n 2 2 2 n 2 1/2 The angle β that the osculating plane forms with y axis (see the above diagram) is equal to β = α − 90° Then cos α = cos ( β + 90°) = −sin β − sin β = Then tan β = ( ) ) + R ( 2 + ω t ) − R 2 + ωn2t 2 ( c 2 4 + ω 2t 2 n ( R 2 + ωn2 t 2 c 2 2 2 n 2 1/ 2 ) 4 + ωn2 t 2 ( R 2 + ωn2t 2 β = tan c 4 + ω 2t 2 n −1 or ) 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. 261 PROBLEM 11.181* Determine the direction of the binormal of the path described by the particle of Problem 11.96 when (a) t = 0, (b) t = π /2 s. SOLUTION Given: ) ( r = ( At cos t )i + A t 2 + 1 j + ( Bt sin t )k r − ft, t − s; A = 3, B − 1 First note that eb is given by eb = v×a |v ×a | ) ( Now r = (3t cos t )i + 3 t 2 + 1 j + (t sin t )k Then v= dr dt 3t = 3(cos t − t sin t )i + j + (sin t + t cos t )k ( t ) 2 dv t 2 +1 a= = 3( − sin t − sin t − t cos t )i + 3 t + 12− t j dt t +1 + (cos t + cos t − t sin t )k 3 = −3(2sin t + t cos t )i + 2 j + (2 cos t − t sin t )k (t + 1)3/2 and (a) t2 +1 At t = 0: v = (3 ft/s)i a = (3 ft/s 2 ) j + (2 ft/s 2 )k Then and Then v × a = 3i × (3j + 2k ) = 3(−2 j + 3k ) | v × a | = 3 (−2) 2 + (3) 2 = 3 13 eb = 3( −2 j + 3k ) 3 13 cos θ x = 0 or θ x = 90° = cos θ y = − 1 13 2 (−2 j + 3k ) 13 θ y = 123.7° cos θ z = 3 13 θ z = 33.7° 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. 262 PROBLEM 11.181* (Continued) (b) At t = Then π 2 s: 3π 3π v = − ft/s i + ft/s j + (1 ft/s)k 2 π2 +4 π 24 a = −(6 ft/s 2 )i + 2 ft/s 2 j − ft/s 2 k 3/2 (π + 4) 2 i 3π v×a = − 2 −6 j 3π 2 (π + 4)1/ 2 24 (π 2 + 4)3/ 2 k 1 − π 2 3π 2 24 3π 2 6 i = − + − + 2 1/2 4 (π 2 + 4)3/ 2 2(π + 4) 36π 18π k + − 2 + 2 3/2 1/2 (π + 4) (π + 4) = −4.43984i − 13.40220 j + 12.99459k and Then or j | v × a | = [(−4.43984) 2 + ( −13.40220)2 + (12.99459) 2 ]1/ 2 = 19.18829 1 (−4.43984i − 13.40220 j + 12.99459k ) 19.1829 4.43984 13.40220 12.99459 cos θ x = − cos θ y = − cos θ z = 19.18829 19.18829 19.18829 eb = θ x = 103.4° θ y = 134.3° θ z = 47.4° 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. 263 PROBLEM 11.182 The motion of a particle is defined by the relation x = 2t 3 − 15t 2 + 24t + 4, where x and t are expressed in meters and seconds, respectively. Determine (a) when the velocity is zero, (b) the position and the total distance traveled when the acceleration is zero. SOLUTION x = 2t 3 − 15t 2 + 24t + 4 dx = 6t 2 − 30t + 24 dt dv a= = 12t − 30 dt v= so (a) 0 = 6t 2 − 30t + 24 = 6 (t 2 − 5t + 4) Times when v = 0. (t − 4)(t − 1) = 0 (b) t = 1.00 s, t = 4.00 s Position and distance traveled when a = 0. a = 12t − 30 = 0 t = 2.5 s x2 = 2(2.5)3 − 15(2.5) 2 + 24(2.5) + 4 so x = 1.50 m Final position For 0 ≤ t ≤ 1 s, v > 0. For 1 s ≤ t ≤ 2.5 s, v ≤ 0. At t = 0, At t = 1 s, x1 = (2)(1)3 − (15)(1) 2 + (24)(1) + 4 = 15 m x0 = 4 m. Distance traveled over interval: x1 − x0 = 11 m For 1 s ≤ t ≤ 2.5 s, v≤0 Distance traveled over interval | x2 − x1 | = |1.5 − 15 | = 13.5 m Total distance: d = 11 + 13.5 d = 24.5 m 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. 264 PROBLEM 11.183 A particle starting from rest at x = 1 m is accelerated so that its velocity doubles in magnitude between x = 2 m and x = 8 m. Knowing that the acceleration of the particle is defined by the relation a = k[ x − (A/x)], determine the values of the constants A and k if the particle has a velocity of 29 m/s when x = 16 m. SOLUTION We have v When x = 1 ft, v = 0: v 0 dv A = a = kx− dx x vdv = x 1 A k x − dx x 1 2 1 v = k x 2 − A ln 2 2 or x x 1 1 1 = k x 2 − A ln x − 2 2 1 2 1 1 3 v2 = k (2) 2 − A ln 2 − = k − A ln 2 2 2 2 2 At x = 2 ft: 1 2 1 1 v8 = k (8) 2 − A ln 8 − = k (31.5 − A ln 8) 2 2 2 x = 8 ft: Now 1 2 v 2 8 1 2 v 2 2 v8 = 2: v2 = (2) 2 = k (31.5 − A ln 8) k ( 32 − A ln 2 ) 6 − 4 A ln 2 = 31.5 − A ln 8 1 25.5 = A(ln 8 − 4 ln 2) = A(ln 8 − ln 24 ) = A ln 2 A= When x = 16 m, v = 29 m/s: 25.5 ln 12 A = −36.8 m 2 1 1 25.5 1 (29) 2 = k (16) 2 − ln(16) − 1 2 2 2 ln ( 2 ) 1 420.5k = k 128 + 102 − = 230.5k 2 = 230.5k k = 1.832 s −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. 265 PROBLEM 11.184 A particle moves in a straight line with the acceleration shown in the figure. Knowing that the particle starts from the origin with v0 = −2 m/s, (a) construct the v −t and x −t curves for 0 < t < 18 s, (b) determine the position and the velocity of the particle and the total distance traveled when t = 18 s. SOLUTION Compute areas under a − t curve. A1 = (−0.75)(8) = −6 m/s A2 = (2)(4) = 8 m/s A3 = (6)(6) = 36 m/s v0 = −2 m/s v8 = v0 + A1 = −8 m/s v12 = v8 + A2 = 0 v18 = v12 + A3 v18 = 36 m/s Sketch v −t curve using straight line portions over the constant acceleration periods. Compute areas under the v −t curve. 1 (−2 − 8)(8) = −40 m 2 1 A5 = (−8)(4) = −16 m 2 1 A6 = (36)(6) = 108 m 2 A4 = x0 = 0 x8 = x0 + A4 = −40 m x12 = x8 + A5 = −56 m x18 = x12 + A6 Total distance traveled = 56 + 108 x18 = 52 m d = 164 m 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. 266 PROBLEM 11.185 The velocities of commuter trains A and B are as shown. Knowing that the speed of each train is constant and that B reaches the crossing 10 min after A passed through the same crossing, determine (a) the relative velocity of B with respect to A, (b) the distance between the fronts of the engines 3 min after A passed through the crossing. SOLUTION (a) We have v B = v A + v B/A The graphical representation of this equation is then as shown. Then vB2 /A = 662 + 482 − 2(66)(48) cos 155° or vB/A = 111.366 km/h and or 48 111.366 = sin α sin 155° α = 10.50° v B/A = 111.4 km/h (b) 10.50° First note that at t = 3 min, A is (66 km/h) ( 603 ) = 3.3 km west of the crossing. 7 at t = 3 min, B is (48 km/h) ( 60 ) = 5.6 km southwest of the crossing. Now rB = rA + rB/A Then at t = 3 min, we have rB2/A = 3.32 + 5.62 − 2(3.3)(5.6) cos 25° rB/A = 2.96 km or 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. 267 PROBLEM 11.186 Slider block B starts from rest and moves to the right with a constant acceleration of 1 ft/s 2. Determine (a) the relative acceleration of portion C of the cable with respect to slider block A, (b) the velocity of portion C of the cable after 2 s. SOLUTION Let d be the distance between the left and right supports. Constraint of entire cable: xB + ( xB − x A ) + 2(d − x A ) = constant 2vB − 3v A = 0 aA = and 2 2 aB = (1) = 0.667 ft/s 2 3 3 Constraint of Point C: 2aB − 3a A = 0 a A = 0.667 ft/s 2 or 2(d − x A ) + yC/ A = constant −2v A + vC/ A = 0 − 2a A + aC/ A = 0 and aC/ A = 2a A = 2(0.667) = 1.333 ft/s 2 (a) aC/ A = 1.333 ft/s 2 Velocity vectors after 2s: v A = (0.667)(2) = 1.333 ft/s vC/ A = (1.333)(2) = 2.666 ft/s v C = v A + v C/ A Sketch the vector addition. vC2 = v A2 + vC2 / A = (1.333) 2 + (2.666) 2 = 8.8889(ft/s) 2 vC = 2.981 ft/s tan θ = vC/ A vA = 2.666 = 2, 1.333 θ = 63.4° vC = 2.98 ft/s (b) 63.4° 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. 268 PROBLEM 11.187 Collar A starts from rest at t = 0 and moves downward with a constant acceleration of 7 in./s 2 . Collar B moves upward with a constant acceleration, and its initial velocity is 8 in./s. Knowing that collar B moves through 20 in. between t = 0 and t = 2 s, determine (a) the accelerations of collar B and block C, (b) the time at which the velocity of block C is zero, (c) the distance through which block C will have moved at that time. SOLUTION From the diagram − y A + ( yC − y A ) + 2 yC + ( yC − yB ) = constant Then −2v A − vB + 4vC = 0 (1) and −2a A − aB + 4aC = 0 (2) (v A )0 = 0 Given: (a A ) = 7 in./s 2 ( v B )0 = 8 in./s a B = constant At t = 2 s (a) y − (yB )0 = 20 in. y B = ( y B ) 0 + ( vB ) 0 t + We have At t = 2 s: −20 in. = (−8 in./s)(2 s) + aB = −4 in./s 2 1 aB t 2 2 1 aB (2 s) 2 2 or a B = 2 in./s 2 Then, substituting into Eq. (2) −2(7 in./s 2 ) − (−2 in./s 2 ) + 4aC = 0 aC = 3 in./s 2 or aC = 3 in./s 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. 269 PROBLEM 11.187 (Continued) (b) Substituting into Eq. (1) at t = 0 −2(0) − (−8 in./s) + 4(vC )0 = 0 or (vC )0 = −2 in./s vC = (vC )0 + aC t Now (c) When vC = 0: 0 = (−2 in./s) + (3 in./s 2 )t or t= 2 3 t = 0.667 s yC = ( yC )0 + (vC )0 t + We have At t = 2 s 3 s: 2 yC − ( yC )0 = ( −2 in./s) 3 = −0.667 in. 1 aC t 2 2 1 2 s + (3 in./s 2 ) 2 3 or s 2 y C − (y C )0 = 0.667 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. 270 PROBLEM 11.188 A golfer hits a ball with an initial velocity of magnitude v0 at an angle α with the horizontal. Knowing that the ball must clear the tops of two trees and land as close as possible to the flag, determine v0 and the distance d when the golfer uses (a) a six-iron with α = 31°, (b) a five-iron with α = 27°. SOLUTION The horizontal and vertical motions are x t cos α 1 2 1 y = (v0 sin α )t − gt = x tan α − gt 2 2 2 x = (v0 cos α )t v0 = or (1) 2( x tan α − y) g or t2 = At the landing Point C: yC = 0, And xC = (v0 cos α )t = (2) t = 2v0 sin α g 2v02 sin α cos α g (3) (a) α = 31° To clear tree A: x A = 30 m, y A = 12 m From (2), t A2 = From (1), (v0 ) A = To clear tree B: 2(30 tan 31° − 12) = 1.22851 s 2 , 9.81 t A = 1.1084 s 30 = 31.58 m/s 1.1084cos 31° xB = 100 m, yB = 14 m From (2), (t B ) 2 = 2(100 tan 31° − 14) = 9.3957 s 2 , 9.81 From (1), (v0 ) B = 100 = 38.06 m/s 3.0652 cos 31° The larger value governs, v0 = 38.06 m/s From (3), xC = t B = 3.0652 s v0 = 38.1 m/s (2)(38.06) 2 sin 31° cos 31° = 130.38 m 9.81 d = xC − 110 d = 20.4 m 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. 271 PROBLEM 11.188 (Continued) (b) α = 27° By a similar calculation, t A = 0.81846 s, t B = 2.7447 s, (v0 ) A = 41.138 m/s, (v0 ) B = 40.890 m/s, v0 = 41.138 m/s v0 = 41.1 m/s xC = 139.56 m, d = 29.6 m 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. 272 PROBLEM 11.189 As the truck shown begins to back up with a constant acceleration of 4 ft/s 2 , the outer section B of its boom starts to retract with a constant acceleration of 1.6 ft/s 2 relative to the truck. Determine (a) the acceleration of section B, (b) the velocity of section B when t = 2 s. SOLUTION For the truck, a A = 4 ft/s 2 For the boom, a B/ A = 1.6 ft/s 2 50° (a) a B = a A + a B/ A Sketch the vector addition. By law of cosines: aB2 = a A2 + aB2/ A − 2a AaB/ A cos 50° = 42 + 1.62 − 2(4)(1.6) cos 50° aB = 3.214 ft/s 2 Law of sines: sin α = aB/ A sin 50° aB α = 22.4°, = 1.6 sin 50° = 0.38131 3.214 a B = 3.21 ft/s 2 22.4° v B = (vB )0 + aBt = 0 + (3.214)(2) (b) v B = 6.43 ft/s 2 22.4° 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. 273 PROBLEM 11.190 A motorist traveling along a straight portion of a highway is decreasing the speed of his automobile at a constant rate before exiting from the highway onto a circular exit ramp with a radius of 560-ft. He continues to decelerate at the same constant rate so that 10 s after entering the ramp, his speed has decreased to 20 mi/h, a speed which he then maintains. Knowing that at this constant speed the total acceleration of the automobile is equal to one-quarter of its value prior to entering the ramp, determine the maximum value of the total acceleration of the automobile. SOLUTION First note v10 = 20 mi/h = 88 ft/s 3 While the car is on the straight portion of the highway. a = astraight = at and for the circular exit ramp a = at2 + an2 where an = v2 ρ By observation, amax occurs when v is maximum, which is at t = 0 when the car first enters the ramp. For uniformly decelerated motion v = v0 + at t and at t = 10 s: v = constant a = an = a= Then or astraight 2 v10 ρ 1 ast. 4 88 v 2 ( 3 ft/s ) 1 = at at = 10 = ρ 4 560 ft 2 at = −6.1460 ft/s 2 (The car is decelerating; hence the minus sign.) 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. 274 PROBLEM 11.190 (Continued) Then at t = 10 s: or Then at t = 0: 88 ft/s = v0 + (−6.1460 ft/s 2 )(10 s) 3 v0 = 90.793 ft/s amax = at2 v2 + 0 ρ 2 1/2 2 (90.793 ft/s) 2 2 2 = (−6.1460 ft/s ) + 560 ft amax = 15.95 ft/s 2 or 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. 275 PROBLEM 11.191 Sand is discharged at A from a conveyor belt and falls onto the top of a stockpile at B. Knowing that the conveyor belt forms an angle α = 25° with the horizontal, determine (a) the speed v0 of the belt, (b) the radius of curvature of the trajectory described by the sand at Point B. SOLUTION The motion is projectile motion. Place the origin at Point A. Then x0 = 0 and y0 = 0. The coordinates of Point B are xB = 30 ft and yB = −18 ft. Horizontal motion: Vertical motion: vx = v0 cos 25° (1) x = v0 t cos 25° (2) v y = v0 sin 25° − gt y = v0t sin 25° − (3) 1 2 gt 2 (4) At Point B, Eq. (2) gives v0 t B = xB 30 = = 33.101 ft cos 25° cos 25° Substituting into Eq. (4), 1 −18 = (33.101)(sin 25°) − (32.2)t B2 2 tB = 1.40958 s (a) Speed of the belt. v0 = v0t B 33.101 = = 23.483 1.40958 tB v0 = 23.4 ft/s Eqs. (1) and (3) give vx = 23.483cos 25° = 21.283 ft/s v y = (23.483) sin 25° − (32.2)(1.40958) = −35.464 ft/s tan θ −v y vx = 1.66632 θ = 59.03° v = 41.36 ft/s 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. 276 PROBLEM 11.191 (Continued) Components of acceleration. a = 32.2 ft/s 2 at = 32.2sin θ an = 32.2cos θ = 32.2 cos 59.03° = 16.57 ft/s 2 (b) Radius of curvature at B. an = ρ= v2 ρ v 2 (41.36) 2 = an 16.57 ρ = 103.2 ft 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. 277 PROBLEM 11.192 The end Point B of a boom is originally 5 m from fixed Point A when the driver starts to retract the boom with a constant radial acceleration of r = −1.0 m/s 2 and lower it with a constant angular acceleration θ = −0.5 rad/s 2 . At t = 2 s, determine (a) the velocity of Point B, (b) the acceleration of Point B, (c) the radius of curvature of the path. SOLUTION r0 = 5 m, r0 = 0, r = −1.0 m/s 2 Radial motion. 1 2 rt = 5 + 0 − 0.5t 2 2 r = r0 + rt = 0 − 1.0t r = r0 + r0 t + r = 5 − (0.5)(2)2 = 3 m r = (−1.0)(2) = −2 m/s At t = 2 s, θ 0 = 60° = Angular motion. π 3 rad, θ0 = 0, θ = −0.5 rad/s 2 1 π + 0 − 0.25t 2 2 3 θ = θ0 + θt = 0 − 0.5t θ = θ0 + θ0 + θt 2 = θ= At t = 2 s, π + 0 − (0.25)(2) 2 = 0.047198 rad = 2.70° 3 θ = −(0.5)(2) = −1.0 rad/s Unit vectors er and eθ . (a) Velocity of Point B at t = 2 s. v B = re r + rθeθ = −(2 m/s)e r + (3 m)(−1.0 rad/s)eθ v B = (−2.00 m/s)er + ( −3.00 m/s)eθ tan α = v= vθ −3.0 = = 1.5 vr −2.0 vr2 2 α = 56.31° 2 2 + vθ = (−2) + (−3) = 3.6055 m/s 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. 278 PROBLEM 11.192 (Continued) Direction of velocity. et = v −2er − 3eθ = = −0.55470er − 0.83205eθ 3.6055 v θ + α = 2.70 + 56.31° = 59.01° (b) v B = 3.61 m/s 59.0° Acceleration of Point B at t = 2 s. a B = ( r − rθ 2 )er + ( rθ + 2rθ)eθ = [−1.0 − (3)(−1) 2 ]e r + [(3)(−0.5) + (2)(−1.0)(−0.5)]eθ a B = ( −4.00 m/s 2 )er + (2.50 m/s 2 )eθ tan β = aθ 2.50 = = −0.625 ar −4.00 β = −32.00° a = ar2 + aθ2 = ( −4) 2 + (2.5) 2 = 4.7170 m/s 2 θ + β = 2.70° − 32.00° = −29.30° a B = 4.72 m/s 2 Tangential component: 29.3° at = (a ⋅ et )et at = (−4er + 2.5eθ ) ⋅ (−0.55470er − 0.83205eθ )et = [(−4)(−0.55470) + (2.5)( −0.83205)]et = (0.138675 m/s 2 )et = 0.1389 m/s 2 Normal component: 59.0° a n = a − at a n = −4er + 2.5eθ − (0.138675)(−0.55470e r − 0.83205eθ ) = (−3.9231 m/s 2 )er + (2.6154 m/s 2 )eθ an = (3.9231) 2 + (2.6154) 2 = 4.7149 m/s 2 (c) Radius of curvature of the path. an = ρ= v2 ρ v 2 (3.6055 m/s)2 = an 4.7149 m/s ρ = 2.76 m 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. 279 PROBLEM 11.193 A telemetry system is used to quantify kinematic values of a ski jumper immediately before she leaves the ramp. According to the system r = 500 ft, r = −105 ft/s, r = −10 ft/s 2 , θ = 25°, θ = 0.07 rad/s, θ = 0.06 rad/s 2 . Determine (a) the velocity of the skier immediately before she leaves the jump, (b) the acceleration of the skier at this instant, (c) the distance of the jump d neglecting lift and air resistance. SOLUTION (a) Velocity of the skier. (r = 500 ft, θ = 25°) v = vr e r + vθ eθ = re r + rθeθ = ( −105 ft/s)e r + (500 ft)(0.07 rad/s)eθ v = ( −105 ft/s)e r + (35 ft/s)eθ Direction of velocity: v = ( −105cos 25° − 35cos 65°)i + (35sin 65° − 105sin 25°) j = ( −109.95 ft/s)i + ( −12.654 ft/s) j v y −12.654 α = 6.565° tan α = = vx −109.95 v = (105) 2 + (35)2 = 110.68 ft/s v = 110.7 ft/s 6.57° 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. 280 PROBLEM 11.193 (Continued) (b) Acceleration of the skier. a = ar er + aθ eθ = ( r − rθ 2 )er + (rθ + 2rθ)eθ ar = −10 − (500)(0.07)2 = −12.45 ft/s 2 aθ = (500)(0.06) + (2)(−105)(0.07) = 15.30 ft/s 2 a = (−12.45 ft/s 2 )er + (15.30 ft/s 2 )eθ a = (−12.45)(i cos 25° + j sin 25°) + (15.30)(−i cos 65° + j sin 65°) = (−17.750 ft/s 2 )i + (8.6049 ft/s 2 ) j ay 8.6049 tan β = = β = −25.9° ax −17.750 a = (12.45) 2 + (15.30)2 = 19.725 ft/s 2 a = 19.73 ft/s 2 (c) 25.9° Distance of the jump d. Projectile motion. Place the origin of the xy-coordinate system at the end of the ramp with the x-coordinate horizontal and positive to the left and the y-coordinate vertical and positive downward. Horizontal motion: (Uniform motion) Vertical motion: x0 = 0 x0 = 109.95 ft/s x = x0 + x0 t = 109.95t (Uniformly accelerated motion) y0 = 0 y 0 = 12.654 ft/s y = 32.2 ft/s (from Part a ) (from Part a) 2 y = y0 + y 0 t + 1 2 yt = 12.654t − 16.1t 2 2 At the landing point, x = d cos 30° y = 10 + d sin 30° or (1) y − 10 = d sin 30° (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. 281 PROBLEM 11.193 (Continued) Multiply Eq. (1) by sin 30° and Eq. (2) by cos 30° and subtract x sin 30° − ( y − 10) cos 30° = 0 (109.95t )sin 30° − (12.654t + 16.1t 2 − 10) cos 30° = 0 −13.943t 2 + 44.016t + 8.6603 = 0 t = −0.1858 s and 3.3427 s Reject the negative root. x = (109.95 ft/s)(3.3427 s) = 367.53 ft d= x cos 30° d = 424 ft 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. 282 CHAPTER 12 PROBLEM 12.CQ1 A 1000 lb boulder B is resting on a 200 lb platform A when truck C accelerates to the left with a constant acceleration. Which of the following statements are true (more than one may be true)? (a) The tension in the cord connected to the truck is 200 lb (b) The tension in the cord connected to the truck is 1200 lb (c) The tension in the cord connected to the truck is greater than 1200 lb (d ) The normal force between A and B is 1000 lb (e) The normal force between A and B is 1200 lb ( f ) None of the above SOLUTION Answer: (c) The tension will be greater than 1200 lb and the normal force will be greater than 1000 lb. 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. 285 PROBLEM 12.CQ2 Marble A is placed in a hollow tube, and the tube is swung in a horizontal plane causing the marble to be thrown out. As viewed from the top, which of the following choices best describes the path of the marble after leaving the tube? (a) 1 (b) 2 (c) 3 (d ) 4 (e) 5 SOLUTION Answer: (d ) The particle will have velocity components along the tube and perpendicular to the tube. After it leaves, it will travel in a straight line. 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. 286 PROBLEM 12.CQ3 The two systems shown start from rest. On the left, two 40 lb weights are connected by an inextensible cord, and on the right, a constant 40 lb force pulls on the cord. Neglecting all frictional forces, which of the following statements is true? (a) Blocks A and C will have the same acceleration (b) Block C will have a larger acceleration than block A (c) Block A will have a larger acceleration than block C (d ) Block A will not move (e) None of the above SOLUTION Answer: (b) If you draw a FBD of B, you will see that since it is accelerating downward, the tension in the cable will be less than 40 lb, so the acceleration of A will be less than the acceleration of C. Also, the system on the left has more inertia, so it is harder to accelerate than the system on the right. 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. 287 PROBLEM 12.CQ4 The system shown is released from rest in the position shown. Neglecting friction, the normal force between block A and the ground is (a) less than the weight of A plus the weight of B (b) equal to the weight of A plus the weight of B (c) greater than the weight of A plus the weight of B SOLUTION Answer: (a) Since B has an acceleration component downward the normal force between A and the ground will be less than the sum of the weights. 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. 288 PROBLEM 12.CQ5 People sit on a Ferris wheel at Points A, B, C and D. The Ferris wheel travels at a constant angular velocity. At the instant shown, which person experiences the largest force from his or her chair (back and seat)? Assume you can neglect the size of the chairs, that is, the people are located the same distance from the axis of rotation. (a) A (b) B (c) C (d ) D (e) The force is the same for all the passengers. SOLUTION Answer: (c) Draw a FBD and KD at each location and it will be clear that the maximum force will be experiences by the person at Point C. 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. 289 PROBLEM 12.F1 Crate A is gently placed with zero initial velocity onto a moving conveyor belt. The coefficient of kinetic friction between the crate and the belt is μk. Draw the FBD and KD for A immediately after it contacts the belt. SOLUTION Answer: 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. 290 PROBLEM 12.F2 Two blocks weighing WA and WB are at rest on a conveyor that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Assuming the coefficient of friction between the boxes and the belt is μk, draw the FBDs and KDs for blocks A and B. How would you determine if A and B remain in contact? SOLUTION Answer: Block A Block B To see if they remain in contact assume aA = aB and then check to see if NAB is greater than zero. 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. 291 PROBLEM 12.F3 Objects A, B, and C have masses mA, mB, and mC respectively. The coefficient of kinetic friction between A and B is μk, and the friction between A and the ground is negligible and the pulleys are massless and frictionless. Assuming B slides on A draw the FBD and KD for each of the three masses A, B and C. SOLUTION Answer: Block A Block B Block C 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. 292 PROBLEM 12.F4 Blocks A and B have masses mA and mB respectively. Neglecting friction between all surfaces, draw the FBD and KD for each mass. SOLUTION Block A Block 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. 293 PROBLEM 12.F5 Blocks A and B have masses mA and mB respectively. Neglecting friction between all surfaces, draw the FBD and KD for the two systems shown. SOLUTION System 1 System 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. 294 PROBLEM 12.F6 A pilot of mass m flies a jet in a half vertical loop of radius R so that the speed of the jet, v, remains constant. Draw a FBD and KD of the pilot at Points A, B and C. SOLUTION 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. 295 PROBLEM 12.F7 Wires AC and BC are attached to a sphere which revolves at a constant speed v in the horizontal circle of radius r as shown. Draw a FBD and KD of C. SOLUTION 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. 296 PROBLEM 12.F8 A collar of mass m is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant k. Knowing that the collar has a speed v at Point B, draw the FBD and KD of the collar at this point. SOLUTION where x = 7/12 ft and r = 5/12 ft. 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. 297 PROBLEM 12.1 Astronauts who landed on the moon during the Apollo 15, 16 and 17 missions brought back a large collection of rocks to the earth. Knowing the rocks weighed 139 lb when they were on the moon, determine (a) the weight of the rocks on the earth, (b) the mass of the rocks in slugs. The acceleration due to gravity on the moon is 5.30 ft/s2. SOLUTION Since the rocks weighed 139 lb on the moon, their mass is m= (a) Wmoon 139 lb = = 26.226 lb ⋅ s 2 /ft 2 g moon 5.30 ft/s On the earth, Wearth = mg earth w = (26.226 lb ⋅ s 2 /ft)(32.2 ft/s 2 ) (b) Since 1 slug = 1 lb ⋅ s 2 /ft, w = 844 lb m = 26.2 slugs 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. 298 PROBLEM 12.2 The value of g at any latitude φ may be obtained from the formula g = 32.09(1 + 0.0053 sin 2φ ) ft/s 2 which takes into account the effect of the rotation of the earth, as well as the fact that the earth is not truly spherical. Determine to four significant figures (a) the weight in pounds, (b) the mass in pounds, (c) the mass in lb ⋅ s 2 /ft, at the latitudes of 0°, 45°, and 60°, of a silver bar, the mass of which has been officially designated as 5 lb. SOLUTION g = 32.09(1 + 0.0053 sin 2φ ) ft/s 2 (a) Weight: (b) Mass: At all latitudes: (c) or φ = 0° : g = 32.09 ft/s 2 φ = 45°: g = 32.175 ft/s 2 φ = 90°: g = 32.26 ft/s 2 W = mg φ = 0° : W = (0.1554 lb ⋅ s 2 /ft)(32.09 ft/s 2 ) = 4.987 lb φ = 45°: W = (0.1554 lb ⋅ s 2 /ft)(32.175 ft/s 2 ) = 5.000 lb φ = 90°: W = (0.1554 lb ⋅ s 2 /ft)(32.26 ft/s 2 ) = 5.013 lb m = 5.000 lb m= 5.00 lb 32.175 ft/s 2 m = 0.1554 lb ⋅ s 2 /ft 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. 299 PROBLEM 12.3 A 400-kg satellite has been placed in a circular orbit 1500 km above the surface of the earth. The acceleration of gravity at this elevation is 6.43 m/s2. Determine the linear momentum of the satellite, knowing that its orbital speed is 25.6 × 103 km/h. SOLUTION Mass of satellite is independent of gravity: m = 400 kg v = 25.6 × 103 km/h 1h 3 = (25.6 × 106 m/h) = 7.111 × 10 m/s 3600 s L = mv = (400 kg)(7.111 × 103 m/s) L = 2.84 × 106 kg ⋅ m/s 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. 300 PROBLEM 12.4 A spring scale A and a lever scale B having equal lever arms are fastened to the roof of an elevator, and identical packages are attached to the scales as shown. Knowing that when the elevator moves downward with an acceleration of 1 m/s 2 the spring scale indicates a load of 60 N, determine (a) the weight of the packages, (b) the load indicated by the spring scale and the mass needed to balance the lever scale when the elevator moves upward with an acceleration of 1 m/s2. SOLUTION Assume g = 9.81 m/s 2 m= W g ΣF = ma : Fs − W = − W a g a W 1 − = Fs g or W= Fs a 1− g = 60 1 1− 9.81 W = 66.8 N (b) ΣF = ma : Fs − W = W a g a Fs = W 1 + g 1 = 66.811 + 9.81 Fs = 73.6 N For the balance system B, ΣM 0 = 0: bFw − bFp = 0 Fw = Fp 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. 301 PROBLEM 12.4 (Continued) But a Fw = Ww 1 + g and a Fp = W p 1 + g so that Ww = W p and mw = Wp g = 66.81 9.81 mw = 6.81 kg 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. 302 PROBLEM 12.5 In anticipation of a long 7° upgrade, a bus driver accelerates at a constant rate of 3 ft/s 2 while still on a level section of the highway. Knowing that the speed of the bus is 60 mi/h as it begins to climb the grade and that the driver does not change the setting of his throttle or shift gears, determine the distance traveled by the bus up the grade when its speed has decreased to 50 mi/h. SOLUTION First consider when the bus is on the level section of the highway. alevel = 3 ft/s 2 We have ΣFx = ma: P = W alevel g Now consider when the bus is on the upgrade. We have Substituting for P ΣFx = ma: P − W sin 7° = W a′ g W W alevel − W sin 7° = a′ g g a′ = alevel − g sin 7° or = (3 − 32.2 sin 7°) ft/s 2 = −0.92419 ft/s 2 For the uniformly decelerated motion 2 v 2 = (v0 ) upgrade + 2a′( xupgrade − 0) 5 Noting that 60 mi/h = 88 ft/s, then when v = 50 mi/h = v0 , we have 6 2 5 2 2 6 × 88 ft/s = (88 ft/s) + 2(−0.92419 ft/s ) xupgrade or xupgrade = 1280.16 ft xupgrade = 0.242 mi or 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. 303 PROBLEM 12.6 A hockey player hits a puck so that it comes to rest 10 s after sliding 100 ft on the ice. Determine (a) the initial velocity of the puck, (b) the coefficient of friction between the puck and the ice. SOLUTION (a) Assume uniformly decelerated motion. Then v = v0 + at At t = 10 s: 0 = v0 + a(10) a=− v0 10 v 2 = v02 + 2a( x − 0) Also At t = 10 s: 0 = v02 + 2a(100) Substituting for a v 0 = v02 + 2 − 0 (100) = 0 10 v0 = 20.0 ft/s a=− and Alternate solution to part (a) or v0 = 20.0 ft/s 20 = −2 ft/s 2 10 1 2 at 2 1 v d = v0 t + − 0 t 2 2 t d = d0 + v0 t + 1 v0 t 2 2d v0 = t d= (b) We have + ΣFy = 0: N − W = 0 Sliding: N = W = mg F = μ k N = μ k mg − μk mg = ma ΣFx = ma : −F = ma μk = − a −2.0 ft/s 2 =− g 32.2 ft/s 2 μk = 0.0621 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. 304 PROBLEM 12.7 The acceleration of a package sliding at Point A is 3 m/s2. Assuming that the coefficient of kinetic friction is the same for each section, determine the acceleration of the package at Point B. SOLUTION For any angle θ . Use x and y coordinates as shown. ay = 0 ΣFy = ma y : N − mg cos θ = 0 N = mg cos θ ΣFx = max : mg sin θ − μ k N = max ax = g (sin θ − μ k cos θ ) At Point A. θ = 30°, ax = 3 m/s 2 μk = g sin 30° − ax g cos 30° 9.81 sin 30° − 3 9.81 cos 30° = 0.22423 = At Point B. θ = 15°, μk = 0.22423 ax = 9.81(sin 15° − 0.22423 cos 15°) = 0.414 m/s a = 0.414 m/s 2 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. 305 PROBLEM 12.8 Determine the maximum theoretical speed that may be achieved over a distance of 60 m by a car starting from rest, knowing that the coefficient of static friction is 0.80 between the tires and the pavement and that 60 percent of the weight of the car is distributed over its front wheels and 40 percent over its rear wheels. Assume (a) four-wheel drive, (b) front-wheel drive, (c) rear-wheel drive. SOLUTION (a) Four-wheel drive ΣFy = 0: N1 + N 2 − W = 0 F1 + F2 = μ N1 + μ N 2 = μ ( N1 + N 2 ) = μ w ΣFx = ma : F1 + F2 = ma μ w = ma μW mg = μ g = 0.80(9.81) m m a = 7.848 m/s 2 a= =μ v 2 = 2ax = 2(7.848 m/s 2 )(60 m) = 941.76 m 2 /s 2 v = 30.69 m/s (b) v = 110.5 km/h Front-wheel drive F2 = ma μ (0.6 W ) = ma 0.6μW 0.6μ mg = m m = 0.6 μ g = 0.6(0.80)(9.81) a= a = 4.709 m/s 2 v 2 = 2ax = 2(4.709 m/s 2 )(60 m) = 565.1 m 2 /s 2 v = 23.77 m/s v = 85.6 km/h 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. 306 PROBLEM 12.8 (Continued) (c) Rear-wheel drive F1 = ma μ (0.4 W ) = ma 0.4μW 0.4μ mg = m m = 0.4 μ g = 0.4(0.80)(9.81) a= a = 3.139 m/s 2 v 2 = 2ax = 2(3.139 m/s 2 )(60 m) = 376.7 m 2 /s 2 v = 19.41 m/s v = 69.9 km/h 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. 307 PROBLEM 12.9 If an automobile’s braking distance from 90 km/h is 45 m on level pavement, determine the automobile’s braking distance from 90 km/h when it is (a) going up a 5° incline, (b) going down a 3-percent incline. Assume the braking force is independent of grade. SOLUTION Assume uniformly decelerated motion in all cases. For braking on the level surface, v0 = 90 km/h = 25 m/s, v f = 0 x f − x0 = 45 m v 2f = v02 + 2a( x f − x0 ) a= = v 2f − v02 2( x f − x0 ) 0 − (25) 2 (2)(45) = −6.9444 m/s 2 Braking force. Fb = ma W = a g 6.944 W =− 9.81 = −0.70789W (a) Going up a 5° incline. ΣF = ma W − Fb − W sin 5° = a g F + W sin 5° a=− b g W = −(0.70789 + sin 5°)(9.81) = −7.79944 m/s 2 x f − x0 = v 2f − v02 2a 0 − (25) 2 = (2)(−7.79944) x f − x0 = 40.1 m 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. 308 PROBLEM 12.9 (Continued) (b) Going down a 3 percent incline. 3 β = 1.71835° 100 W − Fb + W sin β = a g a = −(0.70789 − sin β )(9.81) = −6.65028 m/s 0 − (25) 2 x f = x0 = (2)(−6.65028) tan β = x f − x0 = 47.0 m 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. 309 PROBLEM 12.10 A mother and her child are skiing together, and the mother is holding the end of a rope tied to the child’s waist. They are moving at a speed of 7.2 km/h on a gently sloping portion of the ski slope when the mother observes that they are approaching a steep descent. She pulls on the rope with an average force of 7 N. Knowing the coefficient of friction between the child and the ground is 0.1 and the angle of the rope does not change, determine (a) the time required for the child’s speed to be cut in half, (b) the distance traveled in this time. SOLUTION Draw free body diagram of child. ΣF = ma : x-direction: mg sin 5° − μ k N − T cos15° = ma y-direction: N − mg cos5° + T sin15° = 0 From y-direction, N = mg cos 5° − T sin15° = (20 kg)(9.81 m/s 2 ) cos 5° − (7 N) sin15° = 193.64 N From x-direction, μk N T cos15° m m (0.1)(193.64 N) (7 N) cos 15° = (9.81 m/s 2 )sin 5° − − 20 kg 20 kg a = g sin 5° − − = −0.45128 m/s 2 (in x-direction.) v0 = 7.2 km/h = 2 m/s vf = x0 = 0 1 v0 = 1 m/s 2 v f = v0 + at t= v f − v0 a = −1 m/s = 2.2159 s −0.45128 m/s 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. 310 PROBLEM 12.10 (Continued) t = 2.22 s (a) Time elapsed. (b) Corresponding distance. x = x0 + v0t + 1 2 at 2 = 0 + (2 m/s)(2.2159 s) + 1 (−0.45128 m/s 2 )(2.2159 s) 2 2 x = 3.32 m 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. 311 PROBLEM 12.11 The coefficients of friction between the load and the flat-bed trailer shown are μs = 0.40 and μk = 0.30. Knowing that the speed of the rig is 72 km/h, determine the shortest distance in which the rig can be brought to a stop if the load is not to shift. SOLUTION Load: We assume that sliding of load relative to trailer is impending: F = Fm = μs N Deceleration of load is same as deceleration of trailer, which is the maximum allowable deceleration a max . ΣFy = 0: N − W = 0 N = W Fm = μ s N = 0.40 W ΣFx = ma : Fm = mamax 0.40 W = W amax g amax = 3.924 m/s 2 a max = 3.92 m/s 2 Uniformly accelerated motion. v 2 = v02 + 2ax with v = 0 v0 = 72 km/h = 20 m/s a = − amax = 3.924 m/s 2 0 = (20)2 + 2( −3.924) x x = 51.0 m 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. 312 PROBLEM 12.12 A light train made up of two cars is traveling at 90 km/h when the brakes are applied to both cars. Knowing that car A has a mass of 25 Mg and car B a mass of 20 Mg, and that the braking force is 30 kN on each car, determine (a) the distance traveled by the train before it comes to a stop, (b) the force in the coupling between the cars while the train is showing down. SOLUTION v0 = 90 km/h = 90/3.6 = 25 m/s (a) Both cars: ΣFx = Σma: 60 × 103 N = (45 × 103 kg)a a = 1.333 m/s 2 v 2 = v01 + 2ax: 0 = (25) 2 + 2(−1.333) x x = 234 m Stopping distance: (b) Car A: ΣFx = ma: 30 × 103 + P = (25 × 103 )a P = (25 × 103 )(1.333) − 30 × 103 Coupling force: P = +3332 N P = 3.33 kN (tension) 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. 313 PROBLEM 12.13 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable. SOLUTION (a) We note that aB = 1 aA. 2 Block A ΣFx = m A a A : T − (200 lb) sin 30° = 200 aA 32.2 (1) Block B 350 1 aA 32.2 2 ΣFy = mB aB : 350 lb − 2T = (a) (2) Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T: −2(200) sin 30° + 350 = 2 150 = (b) From Eq. (1), T − (200) sin 30° = aB = 200 350 1 aA + aA 32.2 32.2 2 575 aA 32.2 1 1 a A = (8.40 ft/s 2 ), 2 2 200 (8.40) 32.2 a A = 8.40 ft/s 2 30° a B = 4.20 ft/s 2 T = 152.2 lb 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. 314 PROBLEM 12.14 Solve Problem 12.13, assuming that the coefficients of friction between block A and the incline are μs = 0.25 and μk = 0.20. PROBLEM 12.13 The two blocks shown are originally at rest. Neglecting the masses of the pulleys and the effect of friction in the pulleys and between block A and the incline, determine (a) the acceleration of each block, (b) the tension in the cable. SOLUTION We first determine whether the blocks move by computing the friction force required to maintain block A in equilibrium. T = 175 lb. When B in equilibrium, ΣFx = 0: 175 − 200sin 30° − Freq = 0 Freq = 75.0 lb ΣFy = 0: N − 200 cos 30° = 0 N = 173.2 lb FM = μs N = 0.25(173.2 lb) = 43.3 lb Since Freq > Fm , blocks will move (A up and B down). We note that aB = 1 aA. 2 Block A F = μk N = (0.20)(173.2) = 34.64 lb. ΣFx = m A 0 A : − 200sin 30° − 34.64 + T = 200 aA 32.2 (1) 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. 315 PROBLEM 12.14 (Continued) Block B ΣFy = mB aB : 350 lb − 2T = (a) (2) Multiply Eq. (1) by 2 and add Eq. (2) in order to eliminate T: −2(200) sin 30° − 2(34.64) + 350 = 2 81.32 = aB = (b) 350 1 aA 32.2 2 200 350 1 aA + aA 32.2 32.2 2 575 aA 32.2 a A = 4.55 ft/s 2 1 1 a A = (4.52 ft/s 2 ), 2 2 From Eq. (1), T − (200) sin 30° − 34.64 = 30° aB = 2.28 ft/s 2 200 (4.52) 32.2 T = 162.9 lb 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. 316 PROBLEM 12.15 Each of the systems shown is initially at rest. Neglecting axle friction and the masses of the pulleys, determine for each system (a) the acceleration of block A, (b) the velocity of block A after it has moved through 10 ft, (c) the time required for block A to reach a velocity of 20 ft/s. SOLUTION Let y be positive downward for both blocks. Constraint of cable: y A + yB = constant a A + aB = 0 For blocks A and B, aB = − a A or ΣF = ma : WA aA g Block A: WA − T = Block B: P + WB − T = WB W aB = − B a A g g P + WB − WA + WA W aA = − B aA g g Solving for aA, T = WA − or aA = WA aA g WA − WB − P g WA + WB (1) v A2 − (v A )02 = 2a A [ y A − ( y A )0 ] with (v A )0 = 0 v A = 2a A [ y A − ( y A ) 0 ] v A − (v A )0 = a A t t= (2) with (v A )0 = 0 vA aA (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. 317 PROBLEM 12.15 (Continued) (a) Acceleration of block A. System (1): By formula (1), System (2): By formula (1), System (3): By formula (1), (b) (c) WA = 200 lb, WB = 100 lb, P = 0 (a A )1 = 200 − 100 (32.2) 200 + 100 (a A )1 = 10.73 ft/s 2 WA = 200 lb, WB = 0, P = 50 lb (a A ) 2 = 200 − 100 (32.2) 200 (a A ) 2 = 16.10 ft/s 2 WA = 2200 lb, WB = 2100 lb, P = 0 ( a A )3 = 2200 − 2100 (32.2) 2200 + 2100 (a A )3 = 0.749 ft/s 2 v A at y A − ( y A )0 = 10 ft. Use formula (2). System (1): (v A )1 = (2)(10.73)(10) (v A )1 = 14.65 ft/s System (2): (v A ) 2 = (2)(16.10)(10) (v A ) 2 = 17.94 ft/s System (3): (v A )3 = (2)(0.749)(10) (v A )3 = 3.87 ft/s Time at v A = 20 ft/s. Use formula (3). System (1): t1 = 20 10.73 t1 = 1.864 s System (2): t2 = 20 16.10 t2 = 1.242 s System (3): t3 = 20 0.749 t3 = 26.7 s 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. 318 PROBLEM 12.16 Boxes A and B are at rest on a conveyor belt that is initially at rest. The belt is suddenly started in an upward direction so that slipping occurs between the belt and the boxes. Knowing that the coefficients of kinetic friction between the belt and the boxes are ( μk ) A = 0.30 and ( μk ) B = 0.32, determine the initial acceleration of each box. SOLUTION Assume that aB > a A so that the normal force NAB between the boxes is zero. A: ΣFy = 0: NA − WA cos 15° = 0 or NA = WA cos 15° Slipping: FA = ( μk ) A NA A: = 0.3WA cos 15° ΣFx = m A a A : FA − WA sin 15° = m A a A 0.3WA cos 15° − WA sin 15° = or WA aA g a A = (32.2 ft/s 2 )(0.3 cos 15° − sin 15°) or = 0.997 ft/s 2 B: B: ΣFy = 0: N B − WB cos 15° = 0 or N B = WB cos 15° Slipping: FB = ( μk ) B N B = 0.32WB cos 15° ΣFx = mB aB : FB − WB sin 15° = mB aB or or 0.32WB cos 15° − WB sin 15° = WB aB g aB = (32.2 ft/s 2 )(0.32 cos 15° − sin 15°) = 1.619 ft/s 2 aB > a A assumption is correct a A = 0.997 ft/s 2 15° a B = 1.619 ft/s 2 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. 319 PROBLEM 12.16 (Continued) Note: If it is assumed that the boxes remain in contact ( NAB ≠ 0), then assuming NAB to be compression, a A = aB and find (ΣFx = ma) for each box. A: 0.3WA cos 15° − WA sin 15° − N AB = WA a g B: 0.32WB cos 15° − WB sin 15° + N AB = WB a g Solving yields a = 1.273 ft/s 2 and NAB = −0.859 lb, which contradicts the assumption. 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. 320 PROBLEM 12.17 A 5000-lb truck is being used to lift a 1000 lb boulder B that is on a 200 lb pallet A. Knowing the acceleration of the truck is 1 ft/s2, determine (a) the horizontal force between the tires and the ground, (b) the force between the boulder and the pallet. SOLUTION aT = 1 m/s 2 Kinematics: a A = a B = 0.5 m/s 2 5000 = 155.28 slugs 32.2 200 mA = = 6.211 slugs 32.2 1000 mB = = 31.056 slugs 32.2 mT = Masses: Let T be the tension in the cable. Apply Newton’s second law to the lower pulley, pallet and boulder. Vertical components : 2T − (m A + mB ) g = (m A + mB ) a A 2T − (37.267)(32.2) = (37.267)(0.5) T = 609.32 lb Apply Newton’s second law to the truck. 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. 321 PROBLEM 12.17 (Continued) Horizontal components (a) : F − T = mT aT Horizontal force between lines and ground. F = T + mT aT = 609.32 + (155.28)(1.0) F = 765 lb Apply Newton’s second law to the boulder. Vertical components + : FAB − mB g = mB aB FAB = mB ( g + a) = 31.056(32.2 + 0.5) (b) FAB = 1016 lb Contact force: 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. 322 PROBLEM 12.18 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μs = 0.20 and μk = 0.15. If P = 0, determine (a) the acceleration of block B, (b) the tension in the cord. SOLUTION From the constraint of the cord: 2 x A + xB/A = constant Then 2v A + vB/A = 0 and 2a A + aB/A = 0 Now a B = a A + a B/A Then aB = a A + ( −2a A ) or aB = − a A (1) First we determine if the blocks will move for the given value of θ . Thus, we seek the value of θ for which the blocks are in impending motion, with the impending motion of A down the incline. B: ΣFy = 0: N AB − WB cos θ = 0 or N AB = mB g cos θ Now FAB = μ s N AB B: = 0.2mB g cos θ ΣFx = 0: − T + FAB + WB sin θ = 0 T = mB g (0.2cos θ + sin θ ) or A: A: ΣFy = 0: N A − N AB − WA cos θ = 0 or N A = (m A + mB ) g cos θ Now FA = μ s N A = 0.2( mA + mB ) g cos θ ΣFx = 0: − T − FA − FAB + WA sin θ = 0 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. 323 PROBLEM 12.18 (Continued) T = m A g sin θ − 0.2(m A + mB ) g cos θ − 0.2mB g cos θ or = g[ mA sin θ − 0.2(m A + 2mB ) cos θ ] Equating the two expressions for T mB g (0.2cos θ + sin θ ) = g[m A sin θ − 0.2(mA + 2mB ) cos θ ] 8(0.2 + tan θ ) = [40 tan θ − 0.2(40 + 2 × 8)] or tan θ = 0.4 or or θ = 21.8° for impending motion. Since θ < 25°, the blocks will move. Now consider the motion of the blocks. ΣFy = 0: N AB − WB cos 25° = 0 (a) B: or N AB = mB g cos 25° Sliding: FAB = μk N AB = 0.15mB g cos 25° ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB or T = mB [ g (0.15cos 25° + sin 25°) − aB ] = 8[9.81(0.15cos 25° + sin 25°) − aB ] = 8(5.47952 − aB ) A: (N) ΣFy = 0: N A − N AB − WA cos 25° = 0 or N A = (m A + mB ) g cos 25° Sliding: FA = μk N A = 0.15(m A + mB ) g cos 25° ΣFx = m A a A : − T − FA − FAB + WA sin 25° = m A a A Substituting and using Eq. (1) T = m A g sin 25° − 0.15( mA + mB ) g cos 25° − 0.15mB g cos 25° − m A (− aB ) = g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°] + m A aB = 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40aB = 91.15202 + 40aB (N) Equating the two expressions for T 8(5.47952 − aB ) = 91.15202 + 40aB or aB = −0.98575 m/s 2 a B = 0.986 m/s 2 (b) We have or 25° T = 8[5.47952 − (−0.98575)] T = 51.7 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. 324 PROBLEM 12.19 Block A has a mass of 40 kg, and block B has a mass of 8 kg. The coefficients of friction between all surfaces of contact are μ s = 0.20 and μk = 0.15. If P = 40 N , determine (a) the acceleration of block B, (b) the tension in the cord. SOLUTION From the constraint of the cord. 2 x A + xB/A = constant Then 2v A + vB/A = 0 and 2a A + aB/A = 0 Now a B = a A + a B/A Then aB = a A + ( −2a A ) or aB = − a A (1) First we determine if the blocks will move for the given value of P. Thus, we seek the value of P for which the blocks are in impending motion, with the impending motion of a down the incline. B: ΣFy = 0: N AB − WB cos 25° = 0 or N AB = mB g cos 25° Now FAB = μ s N AB B: = 0.2 mB g cos 25° ΣFx = 0: − T + FAB + WB sin 25° = 0 A: T = 0.2 mB g cos 25° + mB g sin 25° or = (8 kg)(9.81 m/s 2 ) (0.2 cos 25° + sin 25°) = 47.39249 N A: or ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0 N A = (m A + mB ) g cos 25° − P sin 25° 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. 325 PROBLEM 12.19 (Continued) Now FA = μ s N A or FA = 0.2[(mA + mB ) g cos 25° − P sin 25°] ΣFx = 0: − T − FA − FAB + WA sin 25° + P cos 25° = 0 −T − 0.2[(m A + mB ) g cos 25° − P sin 25°] − 0.2mB g cos 25° + mA g sin 25° + P cos 25° = 0 or or P(0.2 sin 25° + cos 25°) = T + 0.2[(m A + 2mB ) g cos 25°] − m A g sin 25° Then P(0.2 sin 25° + cos 25°) = 47.39249 N + 9.81 m/s 2 {0.2[(40 + 2 × 8) cos 25° − 40 sin 25°] kg} P = −19.04 N for impending motion. or Since P, < 40 N, the blocks will move. Now consider the motion of the blocks. ΣFy = 0: N AB − WB cos 25° = 0 (a) B: or N AB = mB g cos 25° Sliding: FAB = μk N AB = 0.15 mB g cos 25° ΣFx = mB aB : − T + FAB + WB sin 25° = mB aB T = mB [ g (0.15 cos 25° + sin 25°) − aB ] or = 8[9.81(0.15 cos 25° + sin 25°) − aB ] = 8(5.47952 − aB ) (N) A: ΣFy = 0: N A − N AB − WA cos 25° + P sin 25° = 0 or N A = (m A + mB ) g cos 25° − P sin 25° Sliding: FA = μk N A = 0.15[(m A + mB ) g cos 25° − P sin 25°] ΣFx = m A a A : − T − FA − FAB + WA sin 25° + P cos 25° = m A a A 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. 326 PROBLEM 12.19 (Continued) Substituting and using Eq. (1) T = m A g sin 25° − 0.15[( mA + mB ) g cos 25° − P sin 25°] − 0.15 mB g cos 25° + P cos 25° − m A (− aB ) = g[ mA sin 25° − 0.15(m A + 2mB ) cos 25°] + P(0.15 sin 25° + cos 25°) + m A aB = 9.81[40 sin 25° − 0.15(40 + 2 × 8) cos 25°] + 40(0.15 sin 25° + cos 25°) + 40aB = 129.94004 + 40aB Equating the two expressions for T 8(5.47952 − aB ) = 129.94004 + 40aB or aB = −1.79383 m/s 2 a B = 1.794 m/s 2 (b) We have 25° T = 8[5.47952 − (−1.79383)] T = 58.2 N or 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. 327 PROBLEM 12.20 A package is at rest on a conveyor belt which is initially at rest. The belt is started and moves to the right for 1.3 s with a constant acceleration of 2 m/s2. The belt then moves with a constant deceleration a2 and comes to a stop after a total displacement of 2.2 m. Knowing that the coefficients of friction between the package and the belt are μ s = 0.35 and μk = 0.25, determine (a) the deceleration a2 of the belt, (b) the displacement of the package relative to the belt as the belt comes to a stop. SOLUTION (a) Kinematics of the belt. vo = 0 1. Acceleration phase with a1 = 2 m/s 2 v1 = vo + a1t1 = 0 + (2)(1.3) = 2.6 m/s x1 = xo + vo t1 + 1 2 1 a1t1 = 0 + 0 + (2)(1.3) 2 = 1.69 m 2 2 2. Deceleration phase: v2 = 0 since the belt stops. v22 − v12 = 2a2 ( x2 − x1 ) a2 = t2 − t1 = (b) v22 − v12 0 − (2.6)2 = = −6.63 2( x2 − x1 ) 2(2.2 − 1.69) a 2 = 6.63 m/s 2 v2 − v1 0 − 2.6 = = 0.3923 s −6.63 a2 Motion of the package. 1. Acceleration phase. Assume no slip. (a p )1 = 2 m/s 2 ΣFy = 0: N − W = 0 or N = W = mg ΣFx = ma : F f = m(a p )1 The required friction force is Ff. The available friction force is μ s N = 0.35W = 0.35mg Ff m = (a p )1 , < μs N = μ s g = (0.35)(9.81) = 3.43 m/s 2 m Since 2.0 m/s 2 < 3.43 m/s 2, the package does not slip. (v p )1 = v1 = 2.6 m/s and (x p )1 = 1.69 m. 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. 328 PROBLEM 12.20 (Continued) 2. Deceleration phase. Assume no slip. (a p ) 2 = −11.52 m/s 2 ΣFx = ma : − F f = m( a p )2 Ff m μs N m = μs mg m = (a p ) 2 = −6.63 m/s 2 = μ s g = 3.43 m/s 2 < 6.63 m/s 2 Since the available friction force μ s N is less than the required friction force Ff for no slip, the package does slip. (a p ) 2 < 6.63 m/s 2 , F f = μk N ΣFx = m( a p )2 : − μk N = m(a p ) 2 μk N = − μk g m = −(0.25)(9.81) (a p ) 2 = − = −2.4525 m/s 2 (v p ) 2 = (v p )1 + (a p ) 2 (t2 − t1 ) = 2.6 + (−2.4525)(0.3923) = 1.638 m/s 2 1 ( a p )2 (t2 − t1 )2 2 1 = 1.69 + (2.6)(0.3923) + ( −2.4525)(0.3923) 2 2 ( x p ) 2 = ( x p )1 + (v p )1 (t2 − t1 ) + = 2.521 m Position of package relative to the belt ( x p )2 − x2 = 2.521 − 2.2 = 0.321 x p/belt = 0.321 m 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. 329 PROBLEM 12.21 A baggage conveyor is used to unload luggage from an airplane. The 10-kg duffel bag A is sitting on top of the 20-kg suitcase B. The conveyor is moving the bags down at a constant speed of 0.5 m/s when the belt suddenly stops. Knowing that the coefficient of friction between the belt and B is 0.3 and that bag A does not slip on suitcase B, determine the smallest allowable coefficient of static friction between the bags. SOLUTION Since bag A does not slide on suitcase B, both have the same acceleration. a=a 20° Apply Newton’s second law to the bag A – suitcase B combination treated as a single particle. ΣFy = ma y : − (mB + m A ) g cos 20° + N = 0 N = (m A + mB ) g cos 30° = (30)(9.81) cos 20° = 276.55 N μ B N = (0.3)(276.55) = 82.965 N ΣFx = max : μ B N + ( mA + mB ) g sin 20° = (m A + mB )a a = g sin 20° + μB N m A mB = 9.81sin 20° − a = 0.58972 m/s 2 82.965 30 a = 0.58972 m/s 2 20° Apply Newton’s second law to bag A alone. 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. 330 PROBLEM 12.21 (Continued) ΣFy = ma y : N AB − mA g cos 20° = 0 N AB = ma g sin 20° = (10)(9.81) cos 20° = 92.184 N ZFx = max : M A g sin 20° − FAB = m A a FAB = m A ( g sin 20° − a) = (10)(9.81sin 20° − 0.58972) = 27.655 N Since bag A does not slide on suitcase B, μs > FAB 27.655 = = 0.300 N AB 92.184 μs > 0.300 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. 331 PROBLEM 12.22 To unload a bound stack of plywood from a truck, the driver first tilts the bed of the truck and then accelerates from rest. Knowing that the coefficients of friction between the bottom sheet of plywood and the bed are μs = 0.40 and μk = 0.30, determine (a) the smallest acceleration of the truck which will cause the stack of plywood to slide, (b) the acceleration of the truck which causes corner A of the stack to reach the end of the bed in 0.9 s. SOLUTION Let a P be the acceleration of the plywood, aT be the acceleration of the truck, and a P/T be the acceleration of the plywood relative to the truck. (a) Find the value of aT so that the relative motion of the plywood with respect to the truck is impending. aP = aT and F1 = μs N1 = 0.40 N1 ΣFy = mP a y : N1 − WP cos 20° = − mP aT sin 20° N1 = mP ( g cos 20° − aT sin 20°) ΣFx = max : F1 − WP sin 20° = mP aT cos 20° F1 = mP ( g sin 20° + aT cos 20°) mP ( g sin 20° + aT cos 20°) = 0.40 mP ( g cos 20° − aT sin 20°) (0.40 cos 20° − sin 20°) g cos 20° + 0.40sin 20° = (0.03145)(9.81) = 0.309 aT = aT = 0.309 m/s 2 (b) xP/T = ( xP/T )o + (vP /T )t + aP /T = 2 xP /T t 2 = 1 1 aP / T t 2 = 0 + 0 + aP / T t 2 2 2 (2)(2) = 4.94 m/s 2 (0.9) 2 a P / T = 4.94 m/s 2 20° a P = aT + a P / T = (aT →) + (4.94 m/s 2 20°) Fy = mP a y : N 2 − WP cos 20° = −mP aT sin 20° N 2 = mP ( g cos 20° − aT sin 20°) 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. 332 PROBLEM 12.22 (Continued) ΣFx = Σmax : F2 − WP sin 20° = mP aT cos 20° − mP aP / T F2 = mP ( g sin 20° + aT cos 20° − aP / T ) For sliding with friction F2 = μk N 2 = 0.30 N 2 mP ( g sin 20° + aT cos 20° − aP /T ) = 0.30mP ( g cos 20° + aT sin 20°) aT = (0.30cos 20° − sin 20°) g + aP /T cos 20° + 0.30sin 20° = (−0.05767)(9.81) + (0.9594)(4.94) = 4.17 aT = 4.17 m/s 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. 333 PROBLEM 12.23 To transport a series of bundles of shingles A to a roof, a contractor uses a motor-driven lift consisting of a horizontal platform BC which rides on rails attached to the sides of a ladder. The lift starts from rest and initially moves with a constant acceleration a1 as shown. The lift then decelerates at a constant rate a2 and comes to rest at D, near the top of the ladder. Knowing that the coefficient of static friction between a bundle of shingles and the horizontal platform is 0.30, determine the largest allowable acceleration a1 and the largest allowable deceleration a2 if the bundle is not to slide on the platform. SOLUTION Acceleration a1: Impending slip. ΣFy = m A a y : F1 = μ s N1 = 0.30 N1 N1 − WA = mA a1 sin 65° N1 = WA + mA a1 sin 65° = m A ( g + a1 sin 65°) ΣFx = m A ax : F1 = m A a1 cos 65° F1 = μ s N or m A a1 cos 65° = 0.30m A ( g + a1 sin 65°) a1 = 0.30 g cos 65° − 0.30 sin 65° = (1.990)(9.81) = 19.53 m/s 2 Deceleration a 2 : Impending slip. a1 = 19.53 m/s 2 65° F2 = μ s N 2 = 0.30 N 2 ΣFy = ma y : N1 − WA = − mA a2 sin 65° N1 = WA − mA a2 sin 65° ΣFx = max : F2 = mA a2 cos 65° F2 = μ s N 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. 334 PROBLEM 12.23 (Continued) or m A a2 cos 65° = 0.30 m A ( g − a2 cos 65°) 0.30 g cos 65° + 0.30 sin 65° = (0.432)(9.81) a2 = = 4.24 m/s 2 a 2 = 4.24 m/s 2 65° 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. 335 PROBLEM 12.24 An airplane has a mass of 25 Mg and its engines develop a total thrust of 40 kN during take-off. If the drag D exerted on the plane has a magnitude D = 2.25v 2 , where v is expressed in meters per second and D in newtons, and if the plane becomes airborne at a speed of 240 km/h, determine the length of runway required for the plane to take off. SOLUTION F = ma: 40 × 103 N − 2.25v 2 = (25 × 103 kg)a a=v Substituting x1 0 dv dv : 40 × 103 − 2.25 v 2 = (25 × 103 ) v dx dx (25 × 103 )vdv 0 40 × 103 − 2.25v 2 25 × 103 x1 = − [ln(40 × 103 − 2.25v 2 )]v01 2(2.25) dx = = v1 25 × 103 40 × 103 ln 4.5 40 × 103 − 2.25v12 For v1 = 240 km/h = 66.67 m/s x1 = 25 × 103 40 × 103 ln = 5.556ln1.333 3 4.5 40 × 10 − 2.25(66.67) 2 = 1.5982 × 103 m x1 = 1.598 km 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. 336 PROBLEM 12.25 The propellers of a ship of weight W can produce a propulsive force F0; they produce a force of the same magnitude but of opposite direction when the engines are reversed. Knowing that the ship was proceeding forward at its maximum speed v0 when the engines were put into reverse, determine the distance the ship travels before coming to a stop. Assume that the frictional resistance of the water varies directly with the square of the velocity. SOLUTION At maximum speed a = 0. F0 = kv02 = 0 k= F0 v02 When the propellers are reversed, F0 is reversed. ΣFx = ma : − F0 − kv 2 = ma − F0 − F0 v2 = ma v02 dx = x 0 a− (v F0 2 0 mv02 mv02 vdv vdv = a F0 v02 + v 2 dx = − x=− =− ( mv02 F0 ) v02 mv02 1 ln v02 + v 2 F0 2 ( mv02 ) vdv + v2 0 v0 + v2 ) 0 v0 2 ln v02 − ln 2v02 = mv0 ln 2 2 F0 2 F0 ( ) x = 0.347 m0 v02 F0 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. 337 PROBLEM 12.26 A constant force P is applied to a piston and rod of total mass m to make them move in a cylinder filled with oil. As the piston moves, the oil is forced through orifices in the piston and exerts on the piston a force of magnitude kv in a direction opposite to the motion of the piston. Knowing that the piston starts from rest at t = 0 and x = 0, show that the equation relating x, v, and t, where x is the distance traveled by the piston and v is the speed of the piston, is linear in each of these variables. SOLUTION ΣF = ma : P − kv = ma dv P − kv =a= dt m t v m dv dt = 0 0 P − kv m v = − ln ( P − kv) 0 k m = − [ln ( P − kv) − ln P ] k m P − kv P − kv kt t = − ln =− or ln k P m m P − kv P v = (1 − e− kt/m ) = e− kt/m or m k x= = x= t 0 v dt = Pt k t t − 0 P k − kt/m − e k m 0 Pt P − kt/m Pt P + (e − 1) = − (1 − e− kt/m ) k m k m Pt kv − , which is linear. k m 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. 338 PROBLEM 12.27 A spring AB of constant k is attached to a support at A and to a collar of mass m. The unstretched length of the spring is . Knowing that the collar is released from rest at x = x0 and neglecting friction between the collar and the horizontal rod, determine the magnitude of the velocity of the collar as it passes through Point C. SOLUTION Choose the origin at Point C and let x be positive to the right. Then x is a position coordinate of the slider B and x0 is its initial value. Let L be the stretched length of the spring. Then, from the right triangle L = 2 + x 2 The elongation of the spring is e = L − , and the magnitude of the force exerted by the spring is Fs = ke = k ( 2 + x 2 − ) x cos θ = By geometry, + x2 2 ΣFx = max : − Fs cos θ = ma − k ( 2 + x 2 − ) a=− v 0 v dv = k x x− 2 m + x2 x + x2 2 = ma 0 0 a dx x v 1 2 k v =− m 2 0 x x− 2 x0 + x2 0 0 k 1 2 2 2 = − − + dx x x m 2 x 0 k 1 2 1 v = − 0 − 2 − x02 + 2 + x02 m 2 2 k v2 = 2 2 + x02 − 2 2 + x02 m k = 2 + x02 − 2 2 + x02 + 2 m ) ( ( ) answer: v = k m ( ) 2 + x02 − 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. 339 PROBLEM 12.28 Block A has a mass of 10 kg, and blocks B and C have masses of 5 kg each. Knowing that the blocks are initially at rest and that B moves through 3 m in 2 s, determine (a) the magnitude of the force P, (b) the tension in the cord AD. Neglect the masses of the pulleys and axle friction. SOLUTION Let the position coordinate y be positive downward. y A + yD = constant Constraint of cord AD: v A + vD = 0, a A + aD = 0 ( yB − yD ) + ( yC − yD ) = constant Constraint of cord BC: vB + vC − 2vD = 0, aB + aC − 2aD = 0 2a A + aB + aC = 0 Eliminate aD . (1) We have uniformly accelerated motion because all of the forces are constant. y B = ( y B ) 0 + ( vB ) 0 t + aB = 2[ yB − ( yB )0 ] t 2 = 1 a B t 2 , ( vB ) 0 = 0 2 (2)(3) = 1.5 m/s 2 2 (2) ΣFy = 0: 2TBC − TAD = 0 Pulley D: TAD = 2TBC ΣFy = ma y : WA − TAD = m A a A Block A: aA = or WA − TAD WA − 2TBC = mA mA (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. 340 PROBLEM 12.28 (Continued) ΣFy = ma y : WC − TBC = mC aC Block C: aC = or WC − TBC mC (3) Substituting the value for aB and Eqs. (2) and (3) into Eq. (1), and solving for TBC , WC − TBC W − 2TBC 2 A =0 + aB + mA mC m g − 2TBC 2 A mA mC g − TBC + aB + mC =0 4 1 + TBC = 3g + aB mA mC 4 1 10 + 5 TBC = 3(9.81) + 1.5 or TBC = 51.55 N ΣFy = ma y : P + WB − TBC = mB aB Block B: (a) Magnitude of P. P = TBC − WB + mB aB = 51.55 − 5(9.81) + 5(1.5) (b) P = 10.00 N Tension in cord AD. TAD = 2TBC = (2)(51.55) TAD = 103.1 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. 341 PROBLEM 12.29 A 40-lb sliding panel is supported by rollers at B and C. A 25-lb counterweight A is attached to a cable as shown and, in cases a and c, is initially in contact with a vertical edge of the panel. Neglecting friction, determine in each case shown the acceleration of the panel and the tension in the cord immediately after the system is released from rest. SOLUTION (a) F = Force exerted by counterweight Panel: T −F = ΣFx = ma : 40 a g (1) Counterweight A: Its acceleration has two components a A = a P + a A /P = a → + a ΣFx = max : F = 25 a g ΣFg = mag : 25 − T = (2) 25 a g (3) Adding (1), (2), and (3): 40 + 25 + 25 a g 25 25 a= g= (32.2) 90 90 T − F + F + 25 − T = a = 8.94 ft/s 2 Substituting for a into (3): 25 − T = 25 25 g g 90 T = 25 − 625 90 T = 18.06 lb 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. 342 PROBLEM 12.29 (Continued) (b) Panel: ΣFy = ma : T= 40 a g (1) 25 − T = 25 a g (2) Counterweight A: ΣFy = ma : Adding (1) and (2): 40 + 25 a g 25 a= g 65 T + 25 − T = Substituting for a into (1): T= (c) a = 12.38 ft/s 2 40 25 1000 g = g 65 65 T = 15.38 lb Since panel is accelerated to the left, there is no force exerted by panel on counterweight and vice versa. Panel: ΣFx = ma : T= 40 a g (1) Counterweight A: Same free body as in Part (b): ΣFy = ma : 25 − T = 25 a g (2) Since Eqs. (1) and (2) are the same as in (b), we get the same answers: a = 12.38 ft/s 2 ; T = 15.38 lb 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. 343 PROBLEM 12.30 The coefficients of friction between blocks A and C and the horizontal surfaces are μ s = 0.24 and μk = 0.20. Knowing that m A = 5 kg, mB = 10 kg, and mC = 10 kg, determine (a) the tension in the cord, (b) the acceleration of each block. SOLUTION We first check that static equilibrium is not maintained: ( FA ) m + ( FC )m = μs ( mA + mC ) g = 0.24(5 + 10) g = 3.6 g Since WB = mB g = 10g > 3.6g, equilibrium is not maintained. ΣFy : N A = m A g Block A: FA = μk N A = 0.2m A g ΣFλ = mA a A : T − 0.2mA g = mA a A (1) ΣFy : NC = mC g Block C: FC = μk NC = 0.2mC g Σ Fx = mC aC : T − 0.2mC g = mC aC ΣFy = mB aB Block B: mB g − 2T = mB aB From kinematics: (a) (2) aB = (3) 1 (a A + aC ) 2 Tension in cord. Given data: (4) m A = 5 kg mB = mC = 10 kg Eq. (1): T − 0.2(5) g = 5a A a A = 0.2T − 0.2 g (5) Eq. (2): T − 0.2(10) g = 10aC aC = 0.1T − 0.2 g (6) Eq. (3): 10 g − 2T = 10aB aB = g − 0.2T (7) 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. 344 PROBLEM 12.30 (Continued) Substitute into (4): 1 (0.2T − 0.2 g + 0.1T − 0.2 g ) 2 24 24 1.2 g = 0.35T T= g= (9.81 m/s 2 ) 7 7 g − 0.2T = (b) T = 33.6 N Substitute for T into (5), (7), and (6): 24 a A = 0.2 g − 0.2 g = 0.4857(9.81 m/s 2 ) 7 24 aB = g − 0.2 g = 0.3143(9.81 m/s 2 ) 7 24 aC = 0.1 g − 0.2 g = 0.14286(9.81 m/s 2 ) 7 a A = 4.76 m/s 2 a B = 3.08 m/s 2 aC = 1.401 m/s 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. 345 PROBLEM 12.31 A 10-lb block B rests as shown on a 20-lb bracket A. The coefficients of friction are μs = 0.30 and μk = 0.25 between block B and bracket A, and there is no friction in the pulley or between the bracket and the horizontal surface. (a) Determine the maximum weight of block C if block B is not to slide on bracket A. (b) If the weight of block C is 10% larger than the answer found in a determine the accelerations of A, B and C. SOLUTION Kinematics. Let x A and xB be horizontal coordinates of A and B measured from a fixed vertical line to the left of A and B. Let yC be the distance that block C is below the pulley. Note that yC increases when C moves downward. See figure. The cable length L is fixed. L = ( xB − x A ) + ( xP − x A ) + yC + constant Differentiating and noting that x P = 0, vB − 2v A + vC = 0 −2a A + aB + aC = 0 (1) Here, a A and aB are positive to the right, and aC is positive downward. Kinetics. Let T be the tension in the cable and FAB be the friction force between blocks A and B. The free body diagrams are: Bracket A: 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. 346 PROBLEM 12.31 (Continued) Block B: Block C: Bracket A: ΣFx = max : 2T − FAB = Block B: ΣFx = max : FAB − T = WA aA g WB aB g (2) (3) + ΣFy = ma y : N AB − WB = 0 N AB = WB or Block C: ΣFy = ma y : mC − T = WC aC g (4) Adding Eqs. (2), (3), and (4), and transposing, W WA W a A + B aB + C aC = WC g g g (5) Subtracting Eq. (4) from Eq. (3) and transposing, W WB aB − C aC = FAB − WC g g (a) No slip between A and B. aB = a A From Eq. (1), a A = aB = aC = a From Eq. (5), For impending slip, a= (6) WC g WA + WB + WC FAB = μs N AB = μsWB 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. 347 PROBLEM 12.31 (Continued) Substituting into Eq. (6), (WB − WC )(WC g ) = μ sWB − WC WA + WB + WC Solving for WC , WC = = μsWB (WA + WB ) WA + 2WB − μ sWB (0.30)(10)(20 + 10) 20 + (2)(10) − (0.30)(10) WC = 2.43 lbs (b) WC increased by 10%. WC = 2.6757 lbs Since slip is occurring, FAB = μk N AB = μkWB W WB aB − C aC = μ kWB − WC g g Eq. (6) becomes or 10aB − 2.6757 aC = [(0.25)(10) − 2.6757](32.2) (7) With numerical data, Eq. (5) becomes 20a A + 10aB + 2.6757aC = (2.6757)(32.2) (8) Solving Eqs. (1), (7), and (8) gives a A = 3.144 ft/s 2 , aB = 0.881 ft/s 2 , aC = 5.407 ft/s 2 a A = 3.14 ft/s 2 a B = 0.881 ft/s 2 aC = 5.41 ft/s 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. 348 PROBLEM 12.32 The masses of blocks A, B, C and D are 9 kg, 9 kg, 6 kg and 7 kg, respectively. Knowing that a downward force of magnitude 120 N is applied to block D, determine (a) the acceleration of each block, (b) the tension in cord ABC. Neglect the weights of the pulleys and the effect of friction. SOLUTION Note: As shown, the system is in equilibrium. From the diagram: 2 y A + 2 yB + yC = constant Cord 1: Then 2v A + 2vB + vC = 0 and 2a A + 2aB + aC = 0 ( yD − y A ) + ( yD − yB ) = constant Cord 2: A: (1) Then 2 vD − v A − vB = 0 and 2aD − a A − aB = 0 (2) ΣFy = mA a A : mA g − 2T1 + T2 = m A a A (a) 9(9.81) − 2T1 + T2 = 9a A or (3) ΣFy = mB aB : mB g − 2T1 + T2 = mB aB B: 9(9.81) − 2T1 + T2 = 9aB or (4) Note: Eqs. (3) and (4) a A = a B Then Eq. (1) aC = −4a A Eq. (2) aD = a A 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. 349 PROBLEM 12.32 (Continued) ΣFy = mC aC : mC g − T1 = mC aC C: T1 = mC ( g − aC ) = 6( g + 4a A ) or (5) ΣFy = mD aD : mD g − 2T2 + ( FD )ext = mD aD 1 1 T2 = [mD ( g − aD ) + 120] = 94.335 − (7a A ) 2 2 or (6) Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3) D: 1 9(9.81) − 2 × 6( g + 4a A ) + 94.335 − (7a A ) = 9a A 2 aA = or 9(9.81) − 2 × 6(9.81) + 94.335 = 1.0728 m/s 2 48 + 3.5 + 9 a A = a B = a D = 1.073 m/s 2 aC = −4(1.0728 m/s 2 ) and (b) or aC = 4.29 m/s 2 Substituting into Eq. (5) T1 = 6 ( 9.81 + 4(1.0728) ) or T1 = 84.6 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. 350 PROBLEM 12.33 The masses of blocks A, B, C and D are 9 kg, 9 kg, 6 kg and 7 kg, respectively. Knowing that a downward force of magnitude 50 N is applied to block B and that the system starts from rest, determine at t = 3 s the velocity (a) of D relative to A, (b) of C relative to D. Neglect the weights of the pulleys and the effect of friction. SOLUTION Note: As shown, the system is in equilibrium. From the constraint of the two cords, 2 y A + 2 yB + yC = constant Cord 1: Then 2v A + 2vB + vC = 0 and 2a A + 2aB + aC = 0 (1) ( yD − y A ) + ( yD − yB ) = constant Cord 2: Then 2 vD − v A − vB = 0 and 2aD − a A − aB = 0 (2) We determine the accelerations of blocks A, C, and D using the blocks as free bodies. WA aA g m A g − 2T1 + T2 = m A a A ΣFy = m A a A : WA − 2T1 + T2 = A: or WB aB g mB g − 2T1 + T2 + ( FB )ext = mB aB (3) ΣFy = mB aB : WB − 2T1 + T2 + ( FB )ext = B: or (4) (3) − (4) −( FB )ext = 9( a A − aB ) (F ) aB = a A + B ext mB Forming or 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. 351 PROBLEM 12.33 (Continued) Then (F ) 2a A + 2 a A + B ext mg Eq. (1): + aC = 0 aC = −4a A − or (F ) 2aD − a A − a A + B ext mB Eq. (2): =0 aD = a A + or 2( FB )ext mB C: ( FB )ext 2mB 2( FB )ext ΣFy = mC aC : WC − T1 = mC aC = mC −4a A − mB T1 = mC g + 4mC a A + or D: 2mC ( FB )ext mB (5) ΣFy = mD aD : WD − 2T2 = mD aD T2 = or (F ) 1 × mD g − a A − B ext 2 2mB (6) Substituting for T1 [Eq. (5)] and T2 [Eq. (6)] in Eq. (3) 2mC ( FB )ext 1 ( FB )ext m A g − 2 mC g + 4mC a A + + × mD g − a A − = mAa A mB 2mB 2 m A g − 2mC g − 4 mC ( FB )ext mB or aA = Then aC = −4( −2.2835 m/s 2 ) − mA + 8mC + aD = −2.2835 m/s 2 + − mD ( FB )ext 4 mB + mD 2 mD g 2 = −2.2835 m/s 2 2(50) = −1.9771 m/s 2 9 (50) = 0.4943 m/s 2 2(9) Note: We have uniformly accelerated motion, so that v = 0 + at (a) We have v D/ A = v D − v A or v D /A = aD t − a A t = [0.4943 − (−2.2835)] m/s 2 × 3 s v D /A = 8.33 m/s or (b) And v C/ D = v C = v D or vC /D = aC t − aD t = (−1.9771 − 0.4943) m/s 2 × 3 s v C /D = 7.41m/s or 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. 352 PROBLEM 12.34 The 15-kg block B is supported by the 25-kg block A and is attached to a cord to which a 225-N horizontal force is applied as shown. Neglecting friction, determine (a) the acceleration of block A, (b) the acceleration of block B relative to A. SOLUTION (a) First we note a B = a A + a B/A , where a B/A is directed along the inclined surface of A. ΣFx = mB ax : P − WB sin 25° = mB a A cos 25° + mB aB/A B: or 225 − 15 g sin 25° = 15( a A cos 25° + aB/A ) or 15 − g sin 25° = a A cos 25° + aB/A B: (1) ΣFy = mB a y : N AB − WB cos 25° = −mB a A sin 25° N AB = 15( g cos 25° − a A sin 25°) or A: ΣFx′ = mA a A : P − P cos 25° + N AB sin 25° = m A a A or N AB = [25a A − 225(1 − cos 25°)] / sin 25° A: Equating the two expressions for N AB 15( g cos 25° − a A sin 25°) = or 25a A − 225(1 − cos 25°) sin 25° 3(9.81) cos 25° sin 25° + 45(1 − cos 25°) 5 + 3sin 2 25° = 2.7979 m/s 2 aA = a A = 2.80 m/s 2 (b) From Eq. (1) aB/A = 15 − (9.81)sin 25° − 2.7979cos 25° a B/A = 8.32 m/s 2 or 25° 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. 353 PROBLEM 12.35 Block B of mass 10-kg rests as shown on the upper surface of a 22-kg wedge A. Knowing that the system is released from rest and neglecting friction, determine (a) the acceleration of B, (b) the velocity of B relative to A at t = 0.5 s. SOLUTION A: ΣFx = m A a A : WA sin 30° + N AB cos 40° = m A a A (a) NAB = or 22 ( a A − 12 g ) cos 40° Now we note: a B = a A + a B /A , where a B/A is directed along the top surface of A. B: ΣFy ′ = mB a y′ : NAB − WB cos 20° = −mB a A sin 50° NAB = 10 ( g cos 20° − a A sin 50°) or Equating the two expressions for NAB 1 22 a A − g 2 = 10( g cos 20° − a A sin 50°) cos 40° or aA = (9.81)(1.1 + cos 20° cos 40°) = 6.4061 m/s 2 2.2 + cos 40° sin 50° ΣFx′ = mB ax′ : WB sin 20° = mB aB/A − mB a A cos 50° or aB/A = g sin 20° + a A cos 50° = (9.81sin 20° + 6.4061cos 50°) m/s 2 = 7.4730 m/s 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. 354 PROBLEM 12.35 (Continued) Finally a B = a A + a B/ A We have aB2 = 6.40612 + 7.47302 − 2(6.4061 × 7.4730) cos 50° or aB = 5.9447 m/s 2 and or 7.4730 5.9447 = sin α sin 50° α = 74.4° a B = 5.94 m/s 2 (b) 75.6° Note: We have uniformly accelerated motion, so that v = 0 + at Now v B/A = v B − v A = a B t − a At = a B/At At t = 0.5 s: vB/A = 7.4730 m/s 2 × 0.5 s v B/A = 3.74 m/s or 20° 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. 355 PROBLEM 12.36 A 450-g tetherball A is moving along a horizontal circular path at a constant speed of 4 m/s. Determine (a) the angle θ that the cord forms with pole BC, (b) the tension in the cord. SOLUTION a A = an = First we note v 2A ρ ρ = l AB sin θ where ΣFy = 0: TAB cos θ − WA = 0 (a) TAB = or mA g cos θ ΣFx = m A a A : TAB sin θ = m A v A2 ρ Substituting for TAB and ρ mA g v A2 sin θ = m A cos θ l AB sin θ 1 − cos 2 θ = or sin 2 θ = 1 − cos 2 θ (4 m/s) 2 cos θ 1.8 m × 9.81 m/s 2 cos 2 θ + 0.906105cos θ − 1 = 0 cos θ = 0.64479 Solving θ = 49.9° or (b) From above TAB = m A g 0.450 kg × 9.81 m/s 2 = cos θ 0.64479 TAB = 6.85 N or 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. 356 PROBLEM 12.37 During a hammer thrower’s practice swings, the 7.1-kg head A of the hammer revolves at a constant speed v in a horizontal circle as shown. If ρ = 0.93 m and θ = 60°, determine (a) the tension in wire BC, (b) the speed of the hammer’s head. SOLUTION First we note a A = an = v 2A ρ ΣFy = 0: TBC sin 60° − WA = 0 (a) or 7.1 kg × 9.81 m/s 2 sin 60° = 80.426 N TBC = TBC = 80.4 N ΣFx = m A a A : TBC cos 60° = mA (b) or v A2 = v A2 ρ (80.426 N) cos 60° × 0.93 m 7.1 kg v A = 2.30 m/s or 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. 357 PROBLEM 12.38 A single wire ACB passes through a ring at C attached to a sphere which revolves at a constant speed v in the horizontal circle shown. Knowing that the tension is the same in both portions of the wire, determine the speed v. SOLUTION ΣFx = ma: T (sin 30° + sin 45°) = mv 2 ρ (1) ΣFy = 0: T (cos 30° + cos 45°) − mg = 0 T (cos 30° + cos 45°) = mg Divide Eq. (1) by Eq. (2): (2) sin 30° + sin 45° v 2 = cos 30° + cos 45° ρ g v 2 = 0.76733ρ g = 0.76733 (1.6 m)(9.81 m/s 2 ) = 12.044 m 2 /s 2 v = 3.47 m/s 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. 358 PROBLEM 12.39 Two wires AC and BC are tied at C to a sphere which revolves at a constant speed v in the horizontal circle shown. Determine the range of values of v for which both wires remain taut. SOLUTION ΣFx = ma: TAC sin 30° + TBC sin 45° = mv 2 (1) ρ ΣFy = 0: TAC cos 30° + TBC cos 45° − mg = 0 Divide Eq (1) by Eq. (2): TAC cos 30° + TBC cos 45° = mg (2) TAC sin 30° + TBC sin 45° v 2 = TAC cos 30° + TBC cos 45° ρ g (3) When AC is slack, TAC = 0. Eq. (3) yields v12 = ρ g tan 45° = (1.6 m) (9.81 m/s2 ) tan 45° = 15.696 m 2 /s2 Wire AC will remain taut if v ≤ v1 , that is, if v1 = 3.96 m/s v ≤ 3.96 m/s When BC is slack, TBC = 0. Eq. (3) yields v22 = ρ g tan 30° = (1.6 m)(9.81 m/s 2 ) tan 30° = 9.0621 m 2 /s 2 Wire BC will remain taut if v ≥ v2 , that is, if v2 = 3.01 m/s v ≥ 3.01 m/s Combining the results obtained, we conclude that both wires remain taut for 3.01 m/s ≤ v ≤ 3.96 m/s 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. 359 PROBLEM 12.40 Two wires AC and BC are tied at C to a sphere which revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 60 N. SOLUTION From the solution of Problem 12.39, we find that both wires remain taut for 3.01 m/s ≤ v ≤ 3.96 m/s To determine the values of v for which the tension in either wire will not exceed 60 N, we recall Eqs. (1) and (2) from Problem 12.39: TAC sin 30° + TBC sin 45° = mv 2 ρ (1) TAC cos 30° + TBC cos 45° = mg (2) Subtract Eq. (1) from Eq. (2). Since sin 45° = cos 45°, we obtain TAC (cos 30° − sin 30°) = mg − mv 2 (3) ρ Multiply Eq. (1) by cos 30°, Eq. (2) by sin 30°, and subtract: TBC (sin 45° cos 30° − cos 45° sin 30°) = TBC sin15° = mv 2 ρ mv 2 ρ cos 30° − mg sin 30° cos 30° − mg sin 30° (4) Making TAC = 60 N, m = 5 kg, ρ = 1.6 m, g = 9.81 m/s 2 in Eq. (3), we find the value v1 of v for which TAC = 60 N: 60(cos 30° − sin 30°) = 5(9.81) − 21.962 = 49.05 − 5v12 1.6 v12 0.32 v12 = 8.668, We have TAC ≤ 60 N for v ≥ v1 , that is, for v1 = 2.94 m/s v ≥ 2.94 m/s 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. 360 PROBLEM 12.40 (Continued) Making TBC = 60 N, m = 5 kg, ρ = 1.6 m, g = 9.81 m/s 2 in Eq. (4), we find the value v2 of v for which TBC = 60 N: 60sin 15° = 5v22 cos 30° − 5(9.81) sin 30° 1.6 15.529 = 2.7063v22 − 24.523 v22 = 14.80, v2 = 3.85 m/s We have TBC ≤ 60 N for v ≤ v2 , that is, for v ≤ 3.85 m/s Combining the results obtained, we conclude that the range of allowable value is 3.01 m/s ≤ v ≤ 3.85 m/s 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. 361 PROBLEM 12.41 A 100-g sphere D is at rest relative to drum ABC which rotates at a constant rate. Neglecting friction, determine the range of the allowable values of the velocity v of the sphere if neither of the normal forces exerted by the sphere on the inclined surfaces of the drum is to exceed 1.1 N. SOLUTION aD = an = First we note vD2 ρ ρ = 0.2 m where ΣFx = mD aD : N1 cos 60° + N 2 cos 20° = mD vD2 ρ (1) ΣFy = 0: N1 sin 60° + N 2 sin 20° − WD = 0 N1 sin 60° + N 2 sin 20° = mD g or (2) Case 1: N1 is maximum. N1 = 1.1 N Let Eq. (2) (1.1 N) sin 60° + N 2 sin 20° = (0.1 kg) (9.81 m/s 2 ) N 2 = 0.082954 N or ( N 2 )( N1 )max < 1.1 N OK 0.2 m (1.1cos 60° + 0.082954 cos 20°) N 0.1 kg Eq. (1) (vD2 )( N1 )max = or (vD )( N1 )max = 1.121 m/s (sin 20°) × [Eq. (1)] − (cos 20°) × [Eq. (2)] Now we form N1 cos 60° sin 20° − N1 sin 60° cos 20° = mD or − N1 sin 40° = mD vD2 ρ vD2 ρ sin 20° − mD g cos 20° sin 20° − mD g cos 20° (vD )min occurs when N1 = ( N1 ) max (vD )min = 1.121 m/s 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. 362 PROBLEM 12.41 (Continued) Case 2: N 2 is maximum. N 2 = 1.1 N Let Eq. (2) N1 sin 60° + (1.1 N)sin 20° = (0.1 kg)(9.81 m/s 2 ) N1 = 0.69834 N or ( N1 )( N 2 )max ≤ 1.1 N OK 0.2 m (0.69834 cos 60° + 1.1cos 20°) N 0.1 kg Eq. (1) (vD2 )( N 2 )max = or (vD )( N 2 )max = 1.663 m/s (sin 60°) × [Eq. (1)] − (cos 60°) × [Eq. (2)] Now we form N 2 cos 20° sin 60° − N 2 sin 20° cos 60° = mD or N 2 cos 40° = mD vD2 ρ vD2 ρ sin 60° − mD g cos 60° sin 60° − mD g cos 60° (vD ) max occurs when N 2 = ( N 2 )max (vD ) max = 1.663 m/s For N1 ≤ N 2 < 1.1 N 1.121 m/s < vD < 1.663 m/s 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. 363 PROBLEM 12.42* As part of an outdoor display, a 12-lb model C of the earth is attached to wires AC and BC and revolves at a constant speed v in the horizontal circle shown. Determine the range of the allowable values of v if both wires are to remain taut and if the tension in either of the wires is not to exceed 26 lb. SOLUTION aC = an = First note vC2 ρ ρ = 3 ft where ΣFx = mC aC : TCA sin 40° + TCB sin15° = WC vC2 g ρ ΣFy = 0: TCA cos 40° − TCB cos15° − WC = 0 (1) (2) Note that Eq. (2) implies that (a) when TCB = (TCB ) max , TCA = (TCA ) max (b) when TCB = (TCB ) min , TCA = (TCA )min Case 1: TCA is maximum. Let Eq. (2) or TCA = 26 lb (26 lb) cos 40° − TCB cos15° − (12 lb) = 0 TCB = 8.1964 lb (TCB )(TCA )max < 26 lb OK [(TCB ) max = 8.1964 lb] 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. 364 PROBLEM 12.42* (Continued) Eq. (1) (vC2 )(TCA )max = (32.2 ft/s 2 )(3 ft) (26sin 40° + 8.1964 sin15°) lb 12 lb (vC )(TCA )max = 12.31 ft/s or (cos15°)(Eq. 1) + (sin15°)(Eq. 2) Now we form TCA sin 40° cos15° + TCA cos 40° sin15° = WC vC2 cos15° + WC sin15° g ρ TCA sin 55° = WC vC2 cos15° + WC sin15° g ρ or (3) (vc ) max occurs when TCA = (TCA ) max (vC )max = 12.31 ft/s Case 2: TCA is minimum. Because (TCA ) min occurs when TCB = (TCB ) min , let TCB = 0 (note that wire BC will not be taut). Eq. (2) TCA cos 40° − (12 lb) = 0 TCA = 15.6649 lb, 26 lb OK or Note: Eq. (3) implies that when TCA = (TCA )min , vC = (vC ) min . Then (32.2 ft/s 2 )(3 ft) (15.6649 lb) sin 40° 12 lb Eq. (1) (vC2 )min = or (vC )min = 9.00 ft/s 0 < TCA ≤ TCB < 6 lb when 9.00 ft/s < vC < 12.31 ft/s 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. 365 PROBLEM 12.43* The 1.2-lb flyballs of a centrifugal governor revolve at a constant speed v in the horizontal circle of 6-in. radius shown. Neglecting the weights of links AB, BC, AD, and DE and requiring that the links support only tensile forces, determine the range of the allowable values of v so that the magnitudes of the forces in the links do not exceed 17 lb. SOLUTION v2 First note a = an = where ρ = 0.5 ft ρ W v2 g ρ (1) ΣFy = 0: TDA cos 20° − TDE cos 30° − W = 0 (2) ΣFx = ma : TDA sin 20° + TDE sin 30° = Note that Eq. (2) implies that (a) when TDE = (TDE ) max , TDA = (TDA ) max (b) when TDE = (TDE ) min , TDA = (TDA ) min Case 1: TDA is maximum. Let Eq. (2) TDA = 17 lb (17 lb) cos 20° − TDE cos 30° − (1.2 lb) = 0 or TDE = 17.06 lb unacceptable ( > 17 lb) Now let TDE = 17 lb Eq. (2) or TDA cos 20° − (17 lb) cos 30° − (1.2 lb) = 0 TDA = 16.9443 lb OK ( ≤ 17 lb) 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. 366 PROBLEM 12.43* (Continued) (TDA ) max = 16.9443 lb (TDE ) max = 17 lb (v 2 )(TDA )max = Eq. (1) (32.2 ft/s 2 )(0.5 ft) (16.9443sin 20° + 17sin 30°) lb 1.2 lb v(TDA )max = 13.85 ft/s or (cos 30°) × [Eq. (1)] + (sin 30°) × [Eq. (2)] Now form TDA sin 20° cos 30° + TDA cos 20° sin 30° = W v2 cos 30° + W sin 30° g ρ TDA sin 50° = W v2 cos 30° + W sin 30° g ρ or (3) vmax occurs when TDA = (TDA ) max vmax = 13.85 ft/s Case 2: TDA is minimum. Because (TDA )min occurs when TDE = (TDE )min , let TDE = 0. Eq. (2) TDA cos 20° − (1.2 lb) = 0 or TDA = 1.27701 lb, 17 lb OK Note: Eq. (3) implies that when TDA = (TDA ) min , v = vmin . Then Eq. (1) (v 2 )min = or (32.2 ft/s 2 ) (0.5 ft) (1.27701 lb) sin 20° 1.2 lb vmin = 2.42 ft/s 0 < TAB , TBC , TAD , TDE < 17 lb 2.42 ft/s < v < 13.85 ft/s when 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. 367 PROBLEM 12.44 A 130-lb wrecking ball B is attached to a 45-ft-long steel cable AB and swings in the vertical arc shown. Determine the tension in the cable (a) at the top C of the swing, (b) at the bottom D of the swing, where the speed of B is 13.2 ft/s. SOLUTION (a) At C, the top of the swing, vB = 0; thus an = vB2 =0 LAB ΣFn = 0: TBA − WB cos 20° = 0 TBA = (130 lb) × cos 20° or TBA = 122.2 lb or ΣFn = man : TBA − WB = mB (b) or (vB ) 2D LAB 130 lb TBA = (130 lb) + 2 32.2 ft/s (13.2 ft/s) 45 ft 2 TBA = 145.6 lb or 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. 368 PROBLEM 12.45 During a high-speed chase, a 2400-lb sports car traveling at a speed of 100 mi/h just loses contact with the road as it reaches the crest A of a hill. (a) Determine the radius of curvature ρ of the vertical profile of the road at A. (b) Using the value of ρ found in part a, determine the force exerted on a 160-lb driver by the seat of his 3100-lb car as the car, traveling at a constant speed of 50 mi/h, passes through A. SOLUTION (a) Note: 100 mi/h = 146.667 ft/s ΣFn = man : Wcar = or Wcar v A2 g ρ (146.667 ft/s) 2 32.2 ft/s 2 = 668.05 ft ρ= ρ = 668 ft or (b) Note: v is constant at = 0; 50 mi/h = 73.333 ft/s ΣFn = man : W − N = or W v 2A g ρ (73.333 ft/s) 2 N = (160 lb) 1 − 2 (32.2 ft/s )(668.05 ft) N = 120.0 lb or 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. 369 PROBLEM 12.46 A child having a mass of 22 kg sits on a swing and is held in the position shown by a second child. Neglecting the mass of the swing, determine the tension in rope AB (a) while the second child holds the swing with his arms outstretched horizontally, (b) immediately after the swing is released. SOLUTION Note: The factors of “ 12 ” are included in the following free-body diagrams because there are two ropes and only one is considered. (a) For the swing at rest 1 ΣFy = 0: TBA cos 35° − W = 0 2 TBA = or 22 kg × 9.81 m/s 2 2 cos 35° TBA = 131.7 N or (b) At t = 0, v = 0, so that an = v2 ρ =0 1 ΣFn = 0: TBA − W cos 35° = 0 2 or TBA = 1 (22 kg)(9.81 m/s 2 ) cos 35° 2 TBA = 88.4 N or 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. 370 PROBLEM 12.47 The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 km/h when the brakes are suddenly applied, causing the wheels of the car to slide on the track ( μk = 0.20). Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B. SOLUTION ΣFn = man : N − mg = m (a) v2 ρ v2 N = m g + ρ v2 F = μk N = μk m g + ρ ΣFt = mat : F = mat at = Given data: F v2 = μ k g + ρ m μk = 0.20, v = 72 km/h = 20 m/s ρ = 30 m g = 9.81 m/s 2 , (20)2 at = 0.20 9.81 + 30 (b) an = 0 at = 4.63 m/s 2 ΣFn = man = 0: N − mg = 0 N = mg F = μ k N = μk mg ΣFt = mat : F = mat at = F = μ k g = 0.20(9.81) m at = 1.962 m/s 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. 371 PROBLEM 12.47 (Continued) (c) ΣFn = man : mg − N = mv 2 ρ v2 N = m g − ρ v2 F = μk N = μk m g − ρ ΣFt = mat : F = mat at = F v2 = μk g − m ρ (20) 2 = 0.20 9.81 − 45 at = 0.1842 m/s 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. 372 PROBLEM 12.48 A 250-g block fits inside a small cavity cut in arm OA, which rotates in the vertical plane at a constant rate such that v = 3 m/s. Knowing that the spring exerts on block B a force of magnitude P = 1.5 N and neglecting the effect of friction, determine the range of values of θ for which block B is in contact with the face of the cavity closest to the axis of rotation O. SOLUTION ΣFn = man : P + mg sin θ − Q = m v2 ρ To have contact with the specified surface, we need Q ≥ 0, or Q = P + mg sin θ − sin θ > Data: mv 2 ρ >0 1 v2 P − g ρ m (1) m = 0.250 kg, v = 3 m/s, P = 1.5 N, ρ = 0.9 m Substituting into (1): 1 (3) 2 1.5 − 9.81 0.9 0.25 sin θ > 0.40775 sin θ > 24.1° < θ < 155.9° 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. 373 PROBLEM 12.49 A series of small packages, each with a mass of 0.5 kg, are discharged from a conveyor belt as shown. Knowing that the coefficient of static friction between each package and the conveyor belt is 0.4, determine (a) the force exerted by the belt on a package just after it has passed Point A, (b) the angle θ defining the Point B where the packages first slip relative to the belt. SOLUTION Assume package does not slip. at = 0, F f ≤ μ s N On the curved portion of the belt an = v2 ρ = (1 m/s) 2 = 4 m/s 2 0.250 m For any angle θ ΣFy = ma y : N − mg cos θ = − man = − mv 2 N = mg cos θ − ρ mv 2 ρ (1) ΣFx = max : −F f + mg sin θ = mat = 0 F f = mg sin θ (a) At Point A, θ = 0° N = (0.5)(9.81)(1.000) − (0.5)(4) (b) At Point B, (2) N = 2.905 N F f = μs N mg sin θ = μs ( mg cos θ − man ) a 4 sin θ = μs cos θ − n = 0.40 cos θ − g 9.81 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. 374 PROBLEM 12.49 (Continued) Squaring and using trigonometic identities, 1 − cos 2 θ = 0.16 cos 2 θ − 0.130479cos θ + 0.026601 1.16 cos 2 θ − 0.130479cos θ − 0.97340 = 0 cos θ = 0.97402 θ = 13.09° Check that package does not separate from the belt. N= Ff μs = mg sin θ μs N > 0. 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. 375 PROBLEM 12.50 A 54-kg pilot flies a jet trainer in a half vertical loop of 1200-m radius so that the speed of the trainer decreases at a constant rate. Knowing that the pilot’s apparent weights at Points A and C are 1680 N and 350 N, respectively, determine the force exerted on her by the seat of the trainer when the trainer is at Point B. SOLUTION First we note that the pilot’s apparent weight is equal to the vertical force that she exerts on the seat of the jet trainer. At A: ΣFn = man : N A − W = m v A2 ρ 1680 N v A2 = (1200 m) − 9.81 m/s 2 54 kg or = 25,561.3 m 2 /s 2 At C: ΣFn = man : N C + W = m or vC2 ρ 350 N vC2 = (1200 m) + 9.81 m/s 2 54 kg = 19,549.8 m 2 /s 2 Since at = constant, we have from A to C vC2 = v 2A + 2at Δs AC or or 19,549.8 m 2/s 2 = 25,561.3 m 2 /s 2 + 2at (π × 1200 m) at = −0.79730 m/s 2 Then from A to B vB2 = v 2A + 2at Δs AB π = 25,561.3 m 2/s 2 + 2(−0.79730 m/s 2 ) × 1200 m 2 = 22,555 m 2/s 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. 376 PROBLEM 12.50 (Continued) At B: ΣFn = man : N B = m vB2 ρ 22,555 m 2 /s 2 1200 m or N B = 54 kg or N B = 1014.98 N ΣFt = mat : W + PB = m | at | or PB = (54 kg)(0.79730 − 9.81) m/s 2 or PB = 486.69 N Finally, ( Fpilot ) B = N B2 + PB2 = (1014.98) 2 + (486.69) 2 = 1126 N (Fpilot ) B = 1126 N or 25.6° 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. 377 PROBLEM 12.51 A carnival ride is designed to allow the general public to experience high acceleration motion. The ride rotates about Point O in a horizontal circle such that the rider has a speed v0. The rider reclines on a platform A which rides on rollers such that friction is negligible. A mechanical stop prevents the platform from rolling down the incline. Determine (a) the speed v0 at which the platform A begins to roll upwards, (b) the normal force experienced by an 80-kg rider at this speed. SOLUTION Radius of circle: R = 5 + 1.5cos 70° = 5.513 m ΣF = ma: Components up the incline, 70°: − m A g cos 20° = − mv02 sin 20° R 1 (a) Components normal to the incline, N − mg sin 20° = (b) 1 g R 2 (9.81 m/s) (5.513 m 2 Speed v0 : v0 = = = 12.1898 m/s tan 20° tan 20° Normal force: v0 = 12.19 m/s 20°. mv02 cos 20°. R N = (80)(9.81)sin 20° + 80(12.1898) 2 cos 20° = 2294 N 5.513 N = 2290 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. 378 PROBLEM 12.52 A curve in a speed track has a radius of 1000 ft and a rated speed of 120 mi/h. (See Sample Problem 12.6 for the definition of rated speed). Knowing that a racing car starts skidding on the curve when traveling at a speed of 180 mi/h, determine (a) the banking angle θ , (b) the coefficient of static friction between the tires and the track under the prevailing conditions, (c) the minimum speed at which the same car could negotiate that curve. SOLUTION W = mg Weight a= Acceleration v2 ρ ΣFx = max : F + W sin θ = ma cos θ F= mv 2 ρ cos θ − mg sin θ (1) ΣFy = ma y : N − W cos θ = ma sin θ N= (a) mv 2 ρ sin θ + mg cos θ (2) Banking angle. Rated speed v = 120 mi/h = 176 ft/s. F = 0 at rated speed. 0= mv 2 ρ cos θ − mg sin θ (176)2 v2 = = 0.96199 ρ g (1000) (32.2) θ = 43.89° tan θ = (b) Slipping outward. θ = 43.9° v = 180 mi/h = 264 ft/s F = μN μ= F v 2 cos θ − ρ g sin θ = N v 2 sin θ + ρ g cos θ (264) 2 cos 43.89° − (1000) (32.2)sin 43.89° (264) 2 sin 43.89° + (1000) (32.2) cos 43.89° = 0.39009 μ= μ = 0.390 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. 379 PROBLEM 12.52 (Continued) (c) Minimum speed. F = −μ N v 2 cos θ − ρ g sin θ v 2 sin θ + ρ g cos θ ρ g (sin θ − μ cos θ ) v2 = cos θ + μ sin θ −μ = = (1000) (32.2) (sin 43.89° − 0.39009 cos 43.89°) cos 43.89° + 0.39009 sin 43.89° = 13.369 ft 2 /s 2 v = 115.62 ft/s v = 78.8 mi/h 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. 380 PROBLEM 12.53 Tilting trains, such as the American Flyer which will run from Washington to New York and Boston, are designed to travel safely at high speeds on curved sections of track which were built for slower, conventional trains. As it enters a curve, each car is tilted by hydraulic actuators mounted on its trucks. The tilting feature of the cars also increases passenger comfort by eliminating or greatly reducing the side force Fs (parallel to the floor of the car) to which passengers feel subjected. For a train traveling at 100 mi/h on a curved section of track banked through an angle θ = 6° and with a rated speed of 60 mi/h, determine (a) the magnitude of the side force felt by a passenger of weight W in a standard car with no tilt (φ = 0), (b) the required angle of tilt φ if the passenger is to feel no side force. (See Sample Problem 12.6 for the definition of rated speed.) SOLUTION Rated speed: vR = 60 mi/h = 88 ft/s, 100 mi/h = 146.67 ft/s From Sample Problem 12.6, vR2 = g ρ tan θ ρ= or vR2 (88) 2 = = 2288 ft g tan θ 32.2 tan 6° Let the x-axis be parallel to the floor of the car. ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) = (a) mv 2 ρ cos (θ + φ ) φ = 0. v2 cos (θ + φ ) − sin (θ + φ ) Fs = W gρ (146.67) 2 =W cos 6° − sin 6° (32.2)(2288) = 0.1858W Fs = 0.1858W 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. 381 PROBLEM 12.53 (Continued) (b) For Fs = 0, v2 cos (θ + φ ) − sin (θ + φ ) = 0 gρ v2 (146.67)2 = = 0.29199 g ρ (32.2)(2288) θ + φ = 16.28° φ = 16.28° − 6° tan (θ + φ ) = φ = 10.28° 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. 382 PROBLEM 12.54 Tests carried out with the tilting trains described in Problem 12.53 revealed that passengers feel queasy when they see through the car windows that the train is rounding a curve at high speed, yet do not feel any side force. Designers, therefore, prefer to reduce, but not eliminate, that force. For the train of Problem 12.53, determine the required angle of tilt φ if passengers are to feel side forces equal to 10% of their weights. SOLUTION vR = 60 mi/h = 88 ft/s, 100 mi/h = 146.67 ft/s Rated speed: From Sample Problem 12.6, vR2 = g ρ tan θ or ρ= vR2 (88) 2 = = 2288 ft g tan θ 32.2 tan 6° Let the x-axis be parallel to the floor of the car. ΣFx = max : Fs + W sin (θ + φ ) = man cos (θ + φ ) = Solving for Fs, Now So that Let Then mv 2 ρ cos (θ + φ ) v2 cos (θ + φ ) − sin (θ + φ ) Fs = W gρ v2 (146.67) 2 = = 0.29199 and Fs = 0.10W g ρ (32.2)(2288) 0.10W = W [0.29199 cos (θ + φ ) − sin (θ + φ )] u = sin (θ + φ ) cos (θ + φ ) = 1 − u 2 0.10 = 0.29199 1 − u 2 − u or 0.29199 1 − u 2 = 0.10 + u 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. 383 PROBLEM 12.54 (Continued) Squaring both sides, 0.08526(1 − u 2 ) = 0.01 + 0.2u + u 2 or 1.08526u 2 + 0.2u − 0.07526 = 0 The positive root of the quadratic equation is u = 0.18685 Then, θ + φ = sin −1 u = 10.77° φ = 10.77° − 6° φ = 4.77° 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. 384 PROBLEM 12.55 A 3-kg block is at rest relative to a parabolic dish which rotates at a constant rate about a vertical axis. Knowing that the coefficient of static friction is 0.5 and that r = 2 m, determine the maximum allowable velocity v of the block. SOLUTION Let β be the slope angle of the dish. tan β = At r = 2 m, tan β = 1 or dy 1 = r dr 2 β = 45° Draw free body sketches of the sphere. ΣFy = 0: N cos β − μS N sin β − mg = 0 N = mg cos β − μ S sin β ΣFn = man: N sin β + μ S N cos β = mv 2 ρ mg (sin β + μS N cos β ) mv 2 = cos β − μ S sin β ρ v2 = ρ g sin β + μS cos β sin 45° + 0.5cos 45° = (2)(9.81) = 58.86 m 2 /s 2 cos β − μ S sin β cos 45° − 0.5sin 45° v = 7.67 m/s 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. 385 PROBLEM 12.56 Three seconds after a polisher is started from rest, small tufts of fleece from along the circumference of the 225-mm-diameter polishing pad are observed to fly free of the pad. If the polisher is started so that the fleece along the circumference undergoes a constant tangential acceleration of 4 m/s 2 , determine (a) the speed v of a tuft as it leaves the pad, (b) the magnitude of the force required to free a tuft if the average mass of a tuft is 1.6 mg. SOLUTION (a) at = constant uniformly acceleration motion Then v = 0 + at t At t = 3 s: v = (4 m/s 2 )(3 s) v = 12.00 m/s or (b) ΣFt = mat : Ft = mat or Ft = (1.6 × 10−6 kg)(4 m/s 2 ) = 6.4 × 10−6 N ΣFn = man : Fn = m At t = 3 s: Fn = (1.6 × 10−6 kg) v2 ρ (12 m/s) 2 m) ( 0.225 2 = 2.048 × 10−3 N Finally, Ftuft = Ft 2 + Fn2 = (6.4 × 10−6 N) 2 + (2.048 × 10−3 N)2 Ftuft = 2.05 × 10−3 N or 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. 386 PROBLEM 12.57 A turntable A is built into a stage for use in a theatrical production. It is observed during a rehearsal that a trunk B starts to slide on the turntable 10 s after the turntable begins to rotate. Knowing that the trunk undergoes a constant tangential acceleration of 0.24 m/s 2 , determine the coefficient of static friction between the trunk and the turntable. SOLUTION First we note that (aB )t = constant implies uniformly accelerated motion. vB = 0 + ( a B ) t t At t = 10 s: vB = (0.24 m/s 2 )(10 s) = 2.4 m/s In the plane of the turntable ΣF = mB a B : F = mB (a B )t + mB (a B ) n Then F = mB ( aB )t2 + (aB ) n2 = mB (aB )t2 + ( ) vB2 2 ρ + ΣFy = 0: N − W = 0 or N = mB g At t = 10 s: F = μ s N = μ s mB g Then 2 2 μs mB g = mB ( aB )t2 + vρB 1/2 or 1 μs = 9.81 m/s 2 2 (2.4 m/s)2 2 2 (0.24 m/s ) + 2.5 m μs = 0.236 or 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. 387 PROBLEM 12.58 A small, 300-g collar D can slide on portion AB of a rod which is bent as shown. Knowing that α = 40° and that the rod rotates about the vertical AC at a constant rate of 5 rad/s, determine the value of r for which the collar will not slide on the rod if the effect of friction between the rod and the collar is neglected. SOLUTION First note vD = rθABC + ΣFy = 0: N sin 40° − W = 0 or N= mg sin 40° ΣFn = man : N cos 40° = m vD2 r or ( rθABC ) 2 mg cos 40° = m sin 40° r or r= g 2 θ ABC 1 tan 40° 9.81 m/s 2 1 = 2 (5 rad/s) tan 40° = 0.468 m r = 468 mm or 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. 388 PROBLEM 12.59 A small, 200-g collar D can slide on portion AB of a rod which is bent as shown. Knowing that the rod rotates about the vertical AC at a constant rate and that α = 30° and r = 600 mm, determine the range of values of the speed v for which the collar will not slide on the rod if the coefficient of static friction between the rod and the collar is 0.30. SOLUTION Case 1: v = vmin , impending motion downward ΣFx = max : N − W sin 30° = m or v2 cos 30° r v2 N = m g sin 30° + cos 30° r ΣFy = ma y : F − W cos 30° = −m v2 sin 30° r or v2 F = m g cos 30° − sin 30° r Now F = μs N Then or v2 v2 m g cos 30° − sin 30° = μs × m g sin 30° + cos 30° r r v 2 = gr 1 − μ s tan 30° μ s + tan 30° = (9.81 m/s 2 )(0.6 m) or 1 − 0.3tan 30° 0.3 + tan 30° vmin = 2.36 m/s 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. 389 PROBLEM 12.59 (Continued) Case 2: v = vmax , impending motion upward ΣFx = max : N − W sin 30° = m v2 cos 30° r v2 N = m g sin 30° + cos 30° r or ΣFy = ma y : F + W cos 30° = m v2 sin 30° r or v2 F = m − g cos 30° + sin 30° r Now F = μs N Then v2 v2 m − g cos 30° + sin 30° = μs × m g sin 30° + cos30° r r or v 2 = gr 1 + μs tan 30° tan 30° − μ s = (9.81 m/s 2 )(0.6 m) or 1 + 0.3tan 30° tan 30° − 0.3 vmax = 4.99 m/s For the collar not to slide 2.36 m/s < v < 4.99 m/s 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. 390 PROBLEM 12.60 A semicircular slot of 10-in. radius is cut in a flat plate which rotates about the vertical AD at a constant rate of 14 rad/s. A small, 0.8-lb block E is designed to slide in the slot as the plate rotates. Knowing that the coefficients of friction are μs = 0.35 and μk = 0.25, determine whether the block will slide in the slot if it is released in the position corresponding to (a) θ = 80°, (b) θ = 40°. Also determine the magnitude and the direction of the friction force exerted on the block immediately after it is released. SOLUTION First note Then 1 (26 − 10 sin θ ) ft 12 vE = ρφABCD ρ= an = vE2 ρ = ρ (φABCD ) 2 1 = (26 − 10 sin θ ) ft (14 rad/s) 2 12 98 = (13 − 5 sin θ ) ft/s 2 3 Assume that the block is at rest with respect to the plate. ΣFx = max : N + W cos θ = m or vE2 ρ v2 N = W − cos θ + E sin θ gρ ΣFy = ma y : −F + W sin θ = − m or sin θ vE2 ρ cos θ v2 F = W sin θ + E cos θ gρ 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. 391 PROBLEM 12.60 (Continued) (a) We have θ = 80° Then 1 98 N = (0.8 lb) − cos80° + × (13 − 5sin 80°) ft/s 2 × sin 80° 2 3 32.2 ft/s = 6.3159 lb 1 98 F = (0.8 lb) sin 80° + × (13 − 5sin 80°) ft/s 2 × cos80° 2 3 32.2 ft/s = 1.92601 lb Fmax = μs N = 0.35(6.3159 lb) = 2.2106 lb Now The block does not slide in the slot, and F = 1.926 lb (b) We have 80° θ = 40° Then 1 98 N = (0.8 lb) − cos 40° + × (13 − 5sin 40°) ft/s 2 × sin 40° 2 3 32.2 ft/s = 4.4924 lb 1 98 F = (0.8 lb) sin 40° + × (13 − 5sin 40°) ft/s 2 × cos 40° 2 3 32.2 ft/s = 6.5984 lb Now Fmax = μs N , from which it follows that F > Fmax Block E will slide in the slot and a E = a n + a E/plate = a n + (a E/plate )t + (a E/plate ) n At t = 0, the block is at rest relative to the plate, thus (a E/plate )n = 0 at t = 0, so that a E/plate must be directed tangentially to the slot. ΣFx = max : N + W cos 40° = m or vE2 ρ sin 40° v2 N = W − cos 40° + E sin 40° (as above) gρ = 4.4924 lb 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. 392 PROBLEM 12.60 (Continued) Sliding: F = μk N = 0.25(4.4924 lb) = 1.123 lb Noting that F and a E/plane must be directed as shown (if their directions are reversed, then ΣFx is while ma x is ), we have the block slides downward in the slot and F = 1.123 lb 40° Alternative solutions. (a) Assume that the block is at rest with respect to the plate. ΣF = ma : W + R = ma n Then tan (φ − 10°) = = or and Now so that W W g = = 2 man W vE ρ (φ ABCD ) 2 g ρ 32.2 ft/s 2 98 (13 − 5sin 80°) ft/s 2 3 (from above) φ − 10° = 6.9588° φ = 16.9588° tan φs = μs μ s = 0.35 φs = 19.29° 0 < φ < φs Block does not slide and R is directed as shown. Now F = R sin φ Then F = (0.8 lb) and R = W sin (φ − 10°) sin16.9588° sin 6.9588° = 1.926 lb F = 1.926 lb The block does not slide in the slot and 80° 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. 393 PROBLEM 12.60 (Continued) (b) Assume that the block is at rest with respect to the plate. ΣF = ma : W + R = ma n From Part a (above), it then follows that tan (φ − 50°) = g ρ (φABCD )2 = 32.2 ft/s 2 98 (13 − 5sin 40°) ft/s 2 3 φ − 50° = 5.752° or φ = 55.752° and φs = 19.29° Now φ > φs so that The block will slide in the slot and then φ = φk , where φk = 14.0362° or tan φk = μ k μk = 0.25 To determine in which direction the block will slide, consider the free-body diagrams for the two possible cases. ΣF = ma : W + R = ma n + ma E/plate Now From the diagrams it can be concluded that this equation can be satisfied only if the block is sliding downward. Then ΣFx = max : W cos 40° + R cos φk = m or ρ sin 40° F = R sin φk Now Then vE2 W cos 40° + F W vE2 = sin 40° tan φk g ρ v2 F = μ kW − cos 40° + E sin 40° gρ = 1.123 lb (see the first solution) F = 1.123 lb The block slides downward in the slot and 40° 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. 394 PROBLEM 12.61 A small block B fits inside a lot cut in arm OA which rotates in a vertical plane at a constant rate. The block remains in contact with the end of the slot closest to A and its speed is 1.4 m/s for 0 ≤ θ ≤ 150°. Knowing that the block begins to slide when θ = 150°, determine the coefficient of static friction between the block and the slot. SOLUTION Draw the free body diagrams of the block B when the arm is at θ = 150°. v = at = 0, g = 9.81 m/s 2 ΣFt = mat : − mg sin 30° + N = 0 N = mg sin 30° ΣFn = man : mg cos 30° − F = m F = mg cos30° − Form the ratio μs = v2 ρ mv 2 ρ F , and set it equal to μ s for impending slip. N F g cos 30° − v 2 /ρ 9.81 cos 30° − (1.4)2 /0.3 = = N g sin 30° 9.81 sin 30° μ s = 0.400 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. 395 PROBLEM 12.62 The parallel-link mechanism ABCD is used to transport a component I between manufacturing processes at stations E, F, and G by picking it up at a station when θ = 0 and depositing it at the next station when θ = 180°. Knowing that member BC remains horizontal throughout its motion and that links AB and CD rotate at a constant rate in a vertical plane in such a way that vB = 2.2 ft/s, determine (a) the minimum value of the coefficient of static friction between the component and BC if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending. SOLUTION ΣFx = max : F = W vB2 cos θ g ρ + ΣFy = ma y : N − W = − or Now W vB2 sin θ g ρ v2 N = W 1 − B sin θ gρ v2 Fmax = μ s N = μ s W 1 − B sin θ gρ and for the component not to slide F < Fmax or or v2 W vB2 cos θ < μ s W 1 − B sin θ g ρ gρ μs > cos θ gρ vB2 − sin θ 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. 396 PROBLEM 12.62 (Continued) We must determine the values of θ which maximize the above expression. Thus d dθ − sin θ = − sin θ cos θ gρ vB2 ( gρ vB2 ) − sin θ − (cos θ )(− cos θ ) ( gρ vB2 − sin θ ) 2 or sin θ = vB2 gρ Now sin θ = (2.2 ft/s) 2 = 0.180373 (32.2 ft/s 2 ) ( 10 ft ) 12 μs = (μ s )min for θ = 10.3915° and θ = 169.609° or (a) =0 From above, ( μs ) min = ( μs ) min = cos θ gρ vB2 where sin θ = − sin θ vB2 gρ cos θ cos θ sin θ = = tan θ − sin θ 1 − sin 2 θ 1 sin θ = tan10.3915° ( μ s )min = 0.1834 or (b) We have impending motion to the left for θ = 10.39° to the right for θ = 169.6° 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. 397 PROBLEM 12.63 Knowing that the coefficients of friction between the component I and member BC of the mechanism of Problem 12.62 are μs = 0.35 and μk = 0.25, determine (a) the maximum allowable constant speed vB if the component is not to slide on BC while being transferred, (b) the values of θ for which sliding is impending. SOLUTION ΣFx = max : F = W vB2 cos θ g ρ + ΣFy = ma y : N − W = − W vB2 sin θ g ρ v2 N = W 1 − B sin θ gρ or Fmax = μ s N Now v2 = μ s W 1 − B sin θ gρ and for the component not to slide F < Fmax or or v2 W vB2 cos θ < μ s W 1 − B sin θ g ρ gρ vB2 < μs gρ cos θ + μs sin θ (1) ( ) To ensure that this inequality is satisfied, vB2 must be less than or equal to the minimum value max of μs g ρ /(cos θ + μs sin θ ), which occurs when (cos θ + μ s sin θ ) is maximum. Thus d (cos θ + μs sin θ ) = − sin θ + μ s cos θ = 0 dθ or tan θ = μ s μ s = 0.35 or θ = 19.2900° 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. 398 PROBLEM 12.63 (Continued) (a) The maximum allowed value of vB is then (v ) 2 B max = μs gρ cos θ + μs sin θ = gρ tan θ = g ρ sin θ cos θ + (tan θ ) sin θ where tan θ = μs 10 = (32.2 ft/s 2 ) ft sin 19.2900° 12 (vB ) max = 2.98 ft/s or (b) First note that for 90° < θ < 180°, Eq. (1) becomes vB2 < μs gρ cos α + μs sin α where α = 180° − θ . It then follows that the second value of θ for which motion is impending is θ = 180° − 19.2900° = 160.7100° we have impending motion to the left for θ = 19.29° to the right for θ = 160.7° Alternative solution. ΣF = ma : W + R = man For impending motion, φ = φs . Also, as shown above, the values of θ for which motion is impending ( minimize the value of vB, and thus the value of an is an = vB2 ρ ). From the above diagram, it can be concluded that an is minimum when ma n and R are perpendicular. 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. 399 PROBLEM 12.63 (Continued) Therefore, from the diagram θ = φs = tan −1 μs and or (as above) man = W sin φs m vB2 or ρ = mg sin θ vB2 = g ρ sin θ (as above) α = 180° − θ (as above) For 90° ≤ θ ≤ 180°, we have from the diagram α = φs and or man = W sin φs vB2 = g ρ sin θ (as above) 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. 400 PROBLEM 12.64 In the cathode-ray tube shown, electrons emitted by the cathode and attracted by the anode pass through a small hole in the anode and then travel in a straight line with a speed v0 until they strike the screen at A. However, if a difference of potential V is established between the two parallel plates, the electrons will be subjected to a force F perpendicular to the plates while they travel between the plates and will strike the screen at Point B, which is at a distance δ from A. The magnitude of the force F is F = eV /d , where −e is the charge of an electron and d is the distance between the plates. Derive an expression for the deflection d in terms of V, v0 , the charge −e and the mass m of an electron, and the dimensions d, , and L. SOLUTION Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = at the right edge of the plate. At the screen, x= 2 +L Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0. Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when it impacts on the screen. For uniform horizontal motion, x = v0t , so that t1 = and t2 = v0 2v0 + L . v0 Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between plates (0 < t < t1 ), the vertical force on the electron is Fy = eV /d . After it passes the plates (t1 < t < t2 ), it is zero. For 0 < t < t1 , ΣFy = ma y : a y = Fy m = eV md v y = (v y ) 0 + a y t = 0 + y = y0 + ( v y ) 0 t + eVt md 1 eVt 2 ayt 2 = 0 + 0 + 2 2md 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. 401 PROBLEM 12.64 (Continued) At t = t1 , (v y )1 = eVt1 md and y1 = eVt12 2md For t1 < t < t2 , a y = 0 y = y1 + (v y )1 (t − t1 ) At t = t2 , y2 = δ = y1 + (v y )1 (t2 − t1 ) δ= = eVt12 eVt1 eVt 1 + ( t2 − t1 ) = 1 t2 − t1 md 2md md 2 eV L 1 + − mdv0 2v0 v0 2 v0 or δ= eV L mdv02 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. 402 PROBLEM 12.65 In Problem 12.64, determine the smallest allowable value of the ratio d / in terms of e, m, v0, and V if at x = the minimum permissible distance between the path of the electrons and the positive plate is 0.05d . Problem 12.64 In the cathode-ray tube shown, electrons emitted by the cathode and attracted by the anode pass through a small hole in the anode and then travel in a straight line with a speed v0 until they strike the screen at A. However, if a difference of potential V is established between the two parallel plates, the electrons will be subjected to a force F perpendicular to the plates while they travel between the plates and will strike the screen at point B, which is at a distance δ from A. The magnitude of the force F is F = eV /d , where −e is the charge of an electron and d is the distance between the plates. Derive an expression for the deflection d in terms of V, v0 , the charge −e and the mass m of an electron, and the dimensions d, , and L. SOLUTION Consider the motion of one electron. For the horizontal motion, let x = 0 at the left edge of the plate and x = at the right edge of the plate. At the screen, x= 2 +L Horizontal motion: There are no horizontal forces acting on the electron so that ax = 0. Let t1 = 0 when the electron passes the left edge of the plate, t = t1 when it passes the right edge, and t = t2 when it impacts on the screen. For uniform horizontal motion, x = v0t , so that t1 = and t2 = v0 2v0 + L . v0 Vertical motion: The gravity force acting on the electron is neglected since we are interested in the deflection produced by the electric force. While the electron is between the plates (0 < t < t1 ), the vertical force on the electron is Fy = eV/d . After it passes the plates (t1 < t < t2 ), it is zero. For 0 < t < t1 , ΣFy = ma y : a y = Fy m = eV md 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. 403 PROBLEM 12.65 (Continued) v y = (v y ) 0 + a y t = 0 + y = y0 + ( v y ) 0 t + At t = t1 , But so that v0 , y= y< eVt md eVt 2 1 ayt 2 = 0 + 0 + 2 2md eV 2 2mdv02 d − 0.05d = 0.450 d 2 eV 2 < 0.450 d 2mdv02 d2 1 eV eV > = 1.111 2 2 2 0.450 mv0 2mv0 d eV > 1.054 mv02 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. 404 PROBLEM 12.F9 Four pins slide in four separate slots cut in a horizontal circular plate as shown. When the plate is at rest, each pin has a velocity directed as shown and of the same constant magnitude u. Each pin has a mass m and maintains the same velocity relative to the plate when the plate rotates about O with a constant counterclockwise angular velocity ω. Draw the FBDs and KDs to determine the forces on pins P1 and P2. SOLUTION 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. 405 PROBLEM 12.F10 At the instant shown, the length of the boom AB is being decreased at the constant rate of 0.2 m/s, and the boom is being lowered at the constant rate of 0.08 rad/s. If the mass of the men and lift connected to the boom at Point B is m, draw the FBD and KD that could be used to determine the horizontal and vertical forces at B. SOLUTION Where r = 6 m, r = −0.2 m/s, r = 0, θ = −0.08 rad/s, θ = 0 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. 406 PROBLEM 12.F11 Disk A rotates in a horizontal plane about a vertical axis at the constant rate θ0 . Slider B has a mass m and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when r = 0. Knowing that the slider is released with no radial velocity in the position r = r0, draw a FBD and KD at an arbitrary distance r from O. SOLUTION 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. 407 PROBLEM 12.F12 Pin B has a mass m and slides along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate θ0 , draw a FBD and KD that can be used to determine the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively. SOLUTION 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. 408 PROBLEM 12.66 Rod OA rotates about O in a horizontal plane. The motion of the 0.5-lb collar B is defined by the relations r = 10 + 6 cosπ t and θ = π (4t 2 − 8t ), where r is expressed in inches, t in seconds, and θ in radians. Determine the radial and transverse components of the force exerted on the collar when (a) t = 0, (b) t = 0.5 s. SOLUTION Use polar coordinates and calculate the derivatives of the coordinates r and θ with respect to time. Mass of collar: (a) r = 10 + 6 cos π t in. θ = π (4t 2 − 8t ) rad r = −6π sin π t in./s θ = π (8t − 8) rad/s r = −6π 2 cos π t in./s 2 θ = 8π rad/s 2 m= 0.5 lb = 0.015528 lb s 2 /ft = 1.294 × 10−3 lb ⋅ s 2 /in. 2 32.2 ft/s t = 0: r = 16 in. θ =0 r = 0 θ = −8π = −25.1327 rad/s r = −6π 2 = −59.218 in./s 2 θ = 8π = −25.1327 rad/s 2 ar = r − rθ 2 = −59.218 − (16)( −25.1327)2 = −10165.6 in./s 2 aθ = rθ + 2rθ = (16)(25.1327) + 0 = 402.12 in./s 2 Fr = mar = (1.294 × 10−3 lb ⋅ s 2 /in.)( − 10165.6 in./s 2 ) Fθ = maθ = (1.294 × 10−3 lb ⋅ s 2 /in.)(402.12 in./s 2 ) Fr = −13.15 lb Fθ = 0.520 lb θ =0 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. 409 PROBLEM 12.66 (Continued) (b) t = 0.5 s: r = 10 + 6 cos(0.5π ) = 10 in. θ = π [(4)(0.25) − (8)(0.5)] = −9.4248 rad = −540° = 180° r = −6π sin(0.5π ) = −18.8496 in./s θ = π [(8)(0.5) − 8] = −12.5664 rad/s r = −6π 2 cos(0.5π ) = 0 θ = 8π = 25.1327 rad/s 2 ar = r − rθ 2 = 0 − (10)(−12.5664) 2 = −1579.14 in./s 2 aθ = rθ + 2rθ = (10)(25.1327) + (2)(−18.8496)(−12.5664) = 725.07 in./s 2 Fr = mar = (1.294 × 10−3 lb ⋅ s 2 /in.)( −1579.14 in./s 2 ) Fr = −2.04 lb Fθ = maθ = (1.294 × 10−3 lb ⋅ s 2 /in.)(725.07 in./s 2 ) Fθ = 0.938 lb θ = 180° 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. 410 PROBLEM 12.67 Rod OA oscillates about O in a horizontal plane. The motion of the 2-lb collar B is defined by the relations r = 6(1 − e−2t ) and θ = (3/π )(sin π t ), where r is expressed in inches, t in seconds, and θ in radians. Determine the radial and transverse components of the force exerted on the collar when (a) t = 1 s, (b) t = 1.5 s. SOLUTION Use polar coordinates and calculate the derivatives of the coordinates r and θ with respect to time. r = 6(1 − e −2t ) in. r = 12e −2 t in./s r = −24e−2t in./s 2 Mass of collar: (a) t = 1 s: m= θ = (3/π )sin π t radians θ = 3cos π t rad/s θ = −3π sin π t rad/s 2 2 lb = 0.06211 lb ⋅ s 2 /ft = 5.176 × 10−3 lb ⋅ s 2 /in. 2 32.2 ft/s e−2t = 0.13534, sin π t = 0, cos π t = −1 r = 6(1 − 0.13534) = 5.188 in. θ =0 r = (12)(0.13534) = 1.62402 in./s θ = −3.0 rad/s r = ( −24) (0.13534) = −3.2480 in./s 2 θ = 0 ar = r − rθ 2 = −3.2480 − (5.188)(−3.0)2 = −49.94 in./s 2 aθ = rθ + 2rθ = 0 + (2)(1.62402) (−3) = −9.744 in./s 2 Fr = mar = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 49.94 in./s 2 ) Fr = −0.258 lb Fθ = maθ = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 9.744 in./s 2 ) Fθ = −0.0504 lb θ =0 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. 411 PROBLEM 12.67 (Continued) (b) t = 1.5 s: e −2t = 0.049787, sin π t = −1, cos π t = 0 r = 6 (1 − 0.049787) = 5.7013 in. θ = (3/π )(−1) = −0.9549 rad = −54.7° r = (12)(0.049787) = 0.59744 in./s 2 θ = 0 r = −(24)(0.049787) = −1.19489 in./s 2 θ = −(3π )(−1) = 9.4248 rad/s 2 ar = r − rθ 2 = −1.19489 − 0 = −1.19489 in./s 2 aθ = rθ + 2rθ = (5.7013)(9.4248) + 0 = 53.733 in./s 2 Fr = mar = (5.176 × 10−3 lb ⋅ s 2 /in.)( − 1.19489 in./s 2 ) Fθ = maθ = (5.176 × 10−3 lb ⋅ s 2 /in.)(53.733 in./s 2 ) Fr = −0.00618 lb Fθ = 0.278 lb θ = −54.7° 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. 412 PROBLEM 12.68 The 3-kg collar B slides on the frictionless arm AA′. The arm is attached to drum D and rotates about O in a horizontal plane at the rate θ = 0.75t , where θ and t are expressed in rad/s and seconds, respectively. As the arm-drum assembly rotates, a mechanism within the drum releases cord so that the collar moves outward from O with a constant speed of 0.5 m/s. Knowing that at t = 0, r = 0, determine the time at which the tension in the cord is equal to the magnitude of the horizontal force exerted on B by arm AA′. SOLUTION Kinematics dr = r = 0.5 m/s dt We have At t = 0, r = 0: r 0 dr = t 0.5 dt 0 or r = (0.5t ) m Also, r =0 θ = (0.75t ) rad/s θ = 0.75 rad/s 2 Now ar = r − rθ 2 = 0 − [(0.5t ) m][(0.75t ) rad/s]2 = −(0.28125t 3 ) m/s 2 and aθ = rθ + 2rθ = [(0.5t ) m][0.75 rad/s 2 ] + 2(0.5 m/s)[(0.75t ) rad/s] = (1.125t ) m/s 2 Kinetics ΣFr = mar : − T = (3 kg)( −0.28125t 3 ) m/s 2 or T = (0.84375t 3 ) N ΣFθ = mB aθ : Q = (3 kg)(1.125t ) m/s 2 or Q = (3.375t ) N Now require that T =Q or or (0.84375t 3 ) N = (3.375t ) N t 2 = 4.000 t = 2.00 s or 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. 413 PROBLEM 12.69 The horizontal rod OA rotates about a vertical shaft according to the relation θ = 10t , where θ and t are expressed in rad/s and seconds, respectively. A 250-g collar B is held by a cord with a breaking strength of 18 N. Neglecting friction, determine, immediately after the cord breaks, (a) the relative acceleration of the collar with respect to the rod, (b) the magnitude of the horizontal force exerted on the collar by the rod. SOLUTION θ = 10t rad/s, θ = 10 rad/s 2 m = 250 g = 0.250 kg Before cable breaks: Fr = −T and r = 0. Fr = mar : − T = m( r − rθ 2 ) mrθ 2 = mr + T or θ 2 = mr + T 0 − 18 = = 144 rad 2 /s 2 mr (0.25)(0.5) θ = 12 rad/s Immediately after the cable breaks: Fr = 0, r = 0 (a) Acceleration of B relative to the rod. m( r − rθ 2 ) = 0 or r = rθ 2 = (0.5)(12)2 = 72 m/s 2 a B /rod = 72 m/s 2 radially outward (b) Transverse component of the force. Fθ = maθ : Fθ = m (rθ + 2rθ) Fθ = (0.250)[(0.5)(10) + (2)(0)(12)] = 1.25 Fθ = 1.25 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. 414 PROBLEM 12.70 Pin B weighs 4 oz and is free to slide in a horizontal plane along the rotating arm OC and along the circular slot DE of radius b = 20 in. Neglecting friction and assuming that θ = 15 rad/s and θ = 250 rad/s 2 for the position θ = 20°, determine for that position (a) the radial and transverse components of the resultant force exerted on pin B, (b) the forces P and Q exerted on pin B, respectively, by rod OC and the wall of slot DE. SOLUTION Kinematics. From the geometry of the system, we have r = 2b cos θ Then r = − (2b sin θ )θ r = −2b(θsin θ + θ 2 cos θ ) and ar = r − rθ 2 = −2b(θsin θ + θ 2 cos θ ) − (2b cos θ )θ 2 = − 2b(θsin θ + 2θ 2 cos θ ) Now 20 = − 2 ft [(250 rad/s 2 )sin 20° + 2(15 rad/s) 2 cos 20°] = −1694.56 ft/s 2 12 aθ = rθ + 2rθ = (2b cos θ )θ + 2(−2bθ sin θ )θ = 2b(θcos θ − 2θ 2 sin θ ) and 20 = 2 ft [(250 rad/s 2 ) cos 20° − 2(15 rad/s) 2 sin 20°] = 270.05 ft/s 2 12 Kinetics. (a) We have Fr = mar = and Fθ = maθ = 1 4 lb 32.2 ft/s 1 4 2 lb 32.2 ft/s 2 × (−1694.56 ft/s 2 ) = −13.1565 lb × (270.05 ft/s 2 ) = 2.0967 lb Fr = −13.16 lb Fθ = 2.10 lb ΣFr : −Fr = −Q cos 20° (b) or Q= 1 (13.1565 lb) = 14.0009 lb cos 20° ΣFθ : Fθ = P − Q sin 20° or P = (2.0967 + 14.0009sin 20°) lb = 6.89 lb P = 6.89 lb 70° Q = 14.00 lb 40° 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. 415 PROBLEM 12.71 The two blocks are released from rest when r = 0.8 m and θ = 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine (a) the initial tension in the cable, (b) the initial acceleration of block A, (c) the initial acceleration of block B. SOLUTION Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Constraint of cable: r + yB = constant, r + vB = 0, For block A, r + aB = 0 r = −aB (1) ΣFx = m Aa A : T cos θ = m Aa A or T = m Aa A sec θ (2) or + ΣFy = mB aB : mB g − T = mB aB For block B, Adding Eq. (1) to Eq. (2) to eliminate T, mB g = m Aa A secθ + mB aB (3) (4) Radial and transverse components of a A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components. r − rθ 2 = ar = a A ⋅ er = −a A cos θ (5) Noting that initially θ = 0, using Eq. (1) to eliminate r, and changing signs gives aB = a A cosθ (6) Substituting Eq. (6) into Eq. (4) and solving for a A , aA = mB g (25) (9.81) = = 5.48 m/s 2 m A secθ + mB cos θ 20sec 30° + 25cos 30° From Eq. (6), aB = 5.48cos30° = 4.75 m/s 2 (a) From Eq. (2), T = (20)(5.48)sec30° = 126.6 (b) Acceleration of block A. (c) Acceleration of block B. T = 126.6 N a A = 5.48 m/s 2 a B = 4.75 m/s 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. 416 PROBLEM 12.72 The velocity of block A is 2 m/s to the right at the instant when r = 0.8 m and θ = 30°. Neglecting the mass of the pulley and the effect of friction in the pulley and between block A and the horizontal surface, determine, at this instant, (a) the tension in the cable, (b) the acceleration of block A, (c) the acceleration of block B. SOLUTION Let r and θ be polar coordinates of block A as shown, and let yB be the position coordinate (positive downward, origin at the pulley) for the rectilinear motion of block B. Radial and transverse components of v A. Use either the scalar product of vectors or the triangle construction shown, being careful to note the positive directions of the components. r = vr = v A ⋅ er = −v A cos30° = −2cos30° = −1.73205 m/s rθ = vθ = v A ⋅ eθ = −v A sin 30° = 2sin 30° = 1.000 m/s 2 θ = vθ 1.000 = = 1.25 rad/s r 0.8 Constraint of cable: r + yB = constant, r + vB = 0, For block A, r + aB = 0 or r = −aB ΣFx = m Aa A : T cosθ = m Aa A or T = m Aa A secθ + ΣFy = mB aB : mB g − T = mB aB For block B, Adding Eq. (1) to Eq. (2) to eliminate T, mB g = m Aa A secθ + mB aB (1) (2) (3) (4) Radial and transverse components of a A. Use a method similar to that used for the components of velocity. r − rθ 2 = ar = a A ⋅ er = −a A cos θ (5) Using Eq. (1) to eliminate r and changing signs gives aB = a A cosθ − rθ 2 (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. 417 PROBLEM 12.72 (Continued) Substituting Eq. (6) into Eq. (4) and solving for a A , aA = ( mB g + rθ 2 ) m A secθ + mB cos θ = (25)[9.81 + (0.8)(1.25) 2 ] = 6.18 m/s 2 20sec30° + 25cos 30° From Eq. (6), aB = 6.18cos 30° − (0.8)(1.25) 2 = 4.10 m/s 2 (a) From Eq. (2), T = (20)(6.18) sec30° = 142.7 (b) Acceleration of block A. (c) Acceleration of block B. T = 142.7 N a A = 6.18 m/s 2 a B = 4.10 m/s 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. 418 PROBLEM 12.73* Slider C has a weight of 0.5 lb and may move in a slot cut in arm AB, which rotates at the constant rate θ0 = 10 rad/s in a horizontal plane. The slider is attached to a spring of constant k = 2.5 lb/ft, which is unstretched when r = 0. Knowing that the slider is released from rest with no radial velocity in the position r = 18 in. and neglecting friction, determine for the position r = 12 in. (a) the radial and transverse components of the velocity of the slider, (b) the radial and transverse components of its acceleration, (c) the horizontal force exerted on the slider by arm AB. SOLUTION Let l0 be the radial coordinate when the spring is unstretched. Force exerted by the spring. Fr = − k (r − l0 ) Σ Fr = mar : − k (r − l0 ) = m(r − rθ 2 ) kl k r = θ 2 − r + 0 m m (1) But d dr dr dr ( r) = = r dt dr dt dr kl k = rdr = θ 2 − r + 0 dr rdr m m r= Integrate using the condition r = r0 when r = r0 . 1 2 r 1 2 k 2 kl0 r r r = θ − r + m m r0 2 r0 2 kl 1 2 1 2 1 2 k 2 r − r0 = θ − r − r02 + 0 (r − r0 ) 2 2 2 m m 2kl0 k r 2 = r02 + θ 2 − r 2 − r02 + ( r − r0 ) m m ( ) ( Data: m= ) W 0.5 lb = = 0.01553 lb ⋅ s 2 /ft g 32.2 ft/s 2 θ = 10 rad/s, k = 2.5 lb/ft, l0 = 0 r0 = (vr )0 = 0, r0 = 18 in. = 1.5 ft, r = 12 in. = 1.0 ft 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. 419 PROBLEM 12.73* (Continued) (a) Components of velocity when r = 12 in. 2.5 2 2 r 2 = 0 + 102 − (1.0 − 1.5 ) + 0 0.01553 = 76.223 ft 2 /s 2 vr = r = ±8.7306 ft/s Since r is decreasing, vr is negative (b) r = −8.7306 ft/s vr = −8.73 ft/s vθ = rθ = (1.0)(10) vθ = 10.00 ft/s Components of acceleration. Fr = −kr + kl0 = −(2.5)(1.0) + 0 = −2.5 lb ar = Fr 2.5 =− m 0.01553 ar = 161.0 ft/s 2 aθ = rθ + 2rθ = 0 + (2)( −8.7306)(10) aθ = −174.6 ft/s 2 (c) Transverse component of force. Fθ = maθ = (0.01553)( −174.6) Fθ = −2.71 lb 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. 420 PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.27), express the radial and transverse components of the velocity v of the particle as functions of θ . SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27), h = r 2θ = h0 = r0 v0 or θ = r0 v0 r 2 = r0 v0 cos 2θ r0 2 = v0 cos 2θ r0 Radial component of velocity. vr = r = = r0 r0 dr d θ= dθ dθ cos 2θ sin 2θ v0 cos 2θ (cos 2θ )3/ 2 r sin 2θ θ θ = r0 3/2 (cos 2 ) θ vr = v0 sin 2θ cos 2θ Transverse component of velocity. vθ = h r0 v0 = cos 2θ r r0 vθ = v0 cos 2θ 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. 421 PROBLEM 12.75 For the particle of Problem 12.74, show (a) that the velocity of the particle and the central force F are proportional to the distance r from the particle to the center of force O, (b) that the radius of curvature of the path is proportional to r3. PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.27), express the radial and transverse components of the velocity v of the particle as functions of θ. SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27), h = r 2θ = h0 = r0 v0 or θ = r0 v0 r 2 = r0 v0 cos 2θ r0 2 = v0 cos 2θ r0 Differentiating the expression for r with respect to time, r = r0 dr d θ= dθ dθ cos 2θ sin 2θ sin 2θ v0 sin 2θ θ = r0 cos 2θ = v0 θ = r0 3/2 3/ 2 r (cos 2θ ) (cos 2θ ) cos 2θ 0 Differentiating again, r = (a) dr d sin 2θ 2 cos 2 2θ + sin 2 2θ v0 2 2 cos 2 2θ + sin 2 2θ θ= θ= v0 θ = v0 dθ dθ r0 (cos 2θ )3/2 cos 2θ cos 2θ vr = r = v0 sin 2θ cos 2θ = v0 r sin 2θ r0 vr vθ = rθ = 0 cos 2θ r0 v = (vr ) 2 + (vθ ) 2 = v0 r sin 2 2θ + cos 2 2θ r0 v= v0 r r0 v 2 2cos 2 2θ + sin 2 2θ r0 v0 2 ar = r − rθ 2 = 0 − cos 2 2θ r0 cos 2θ cos 2θ r0 2 = v0 2 cos 2 2θ + sin 2 2θ v0 v 2r = = 02 r0 r0 cos 2θ r0 cos 2θ Fr = mar = mv02 r r02 Fr = : mv02 r r02 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. 422 PROBLEM 12.75 (Continued) Since the particle moves under a central force, aθ = 0. Magnitude of acceleration. a = ar 2 + aθ 2 = v0 2 r r0 2 Tangential component of acceleration. at = v0 2 r dv d v0 r v0 r = = = sin 2θ dt dt r0 r0 r0 2 Normal component of acceleration. at = a 2 − at 2 = r cos 2θ = 0 r But an = Hence, (b) But an = v2 ρ or v0 2 r r0 2 1 − sin 2 2θ = v0 2 r cos 2θ r0 2 2 v0 2 r ρ= v 2 v0 2 r 2 r = 2 ⋅ 2 an r0 v0 ρ= r3 r02 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. 423 PROBLEM 12.76 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F along a semicircular path of diameter OA. Observing that r = r0 cos θ and using Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ . SOLUTION Since the particle moves under a central force, h = constant. Using Eq. (12.27), h = r 2θ = h0 = r0 v0 or θ = r0 v0 r 2 = r0 v0 r0 cos θ 2 2 = v0 r0 cos 2 θ Radial component of velocity. vr = r = d (r0 cos θ ) = −( r0 sin θ )θ dt Transverse component of velocity. vθ = rθ = ( r0 cos θ )θ Speed. v = vr 2 + vθ 2 = r0θ = r0 v0 r0 cos θ 2 v= v0 cos 2 θ 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. 424 PROBLEM 12.77 For the particle of Problem 12.76, determine the tangential component Ft of the central force F along the tangent to the path of the particle for (a) θ = 0, (b) θ = 45°. PROBLEM 12.76 A particle of mass m is projected from Point A with an initial velocity v0 perpendicular to line OA and moves under a central force F along a semicircular path of diameter OA. Observing that r = r0 cos θ and using Eq. (12.27), show that the speed of the particle is v = v0 /cos 2 θ . SOLUTION Since the particle moves under a central force, h = constant Using Eq. (12.27), h = r 2θ = h0 = r0 v0 θ = r0 v0 r 2 = r0 v0 r0 cos θ 2 2 = v0 r0 cos 2 θ Radial component of velocity. vr = r = d (r0 cos θ ) = −( r0 sin θ )θ dt Transverse component of velocity. vθ = rθ = ( r0 cos θ )θ Speed. v = vr 2 + vθ 2 = r0θ = r0 v0 r0 cos θ 2 = v0 cos 2 θ Tangential component of acceleration. at = v0 dv (−2)( − sin θ )θ 2v0 sin θ = v0 = ⋅ 3 3 dt cos θ cos θ r0 cos 2 θ = 2v0 2 sin θ r0 cos5 θ Tangential component of force. Ft = mat : Ft = (a) θ = 0, (b) θ = 45°, 2mv0 2 sin θ r0 cos5 θ Ft = 0 Ft = Ft = 0 2mv0 sin 45° Ft = cos 45° 5 8mv0 2 r0 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. 425 PROBLEM 12.78 Determine the mass of the earth knowing that the mean radius of the moon’s orbit about the earth is 238,910 mi and that the moon requires 27.32 days to complete one full revolution about the earth. SOLUTION Mm [Eq. (12.28)] r2 We have F =G and F = Fn = man = m Then or Now G v2 r Mm v2 = m r r2 M= v= r 2 v G 2π r τ 2 so that 2 r 2π r 1 2π 3 M= r = G τ G τ Noting that τ = 27.32 days = 2.3604 × 106 s and r = 238,910 mi = 1.26144 × 109 ft we have 2 2π 9 3 M= (1.26144 × 10 ft) −9 4 4 6 34.4 × 10 ft /lb ⋅ s 2.3604 × 10 s 1 M = 413 × 1021 lb ⋅ s 2 /ft or 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. 426 PROBLEM 12.79 Show that the radius r of the moon’s orbit can be determined from the radius R of the earth, the acceleration of gravity g at the surface of the earth, and the time τ required for the moon to complete one full revolution about the earth. Compute r knowing that τ = 27.3 days, giving the answer in both SI and U.S. customary units. SOLUTION Mm r2 We have F =G and F = Fn = man = m Then or Now so that G [Eq. (12.28)] v2 r Mm v2 m = r r2 v2 = GM r GM = gR 2 v2 = [Eq. (12.30)] gR 2 g or v = R r r 2π r 2π r = v R gr For one orbit, τ= or gτ 2 R 2 r = 2 4π Now τ = 27.3 days = 2.35872 × 106 s 1/ 3 Q.E.D. R = 3960 mi = 20.9088 × 106 ft 1/3 SI: 9.81 m/s 2 × (2.35872 × 106 s) 2 × (6.37 × 106 m) 2 r= 4π 2 = 382.81 × 106 m r = 383 × 103 km or U.S. customary units: 1/3 32.2 ft/s2 × (2.35872 × 106 s) 2 × (20.9088 × 106 ft) 2 r= 4π 2 = 1256.52 × 106 ft r = 238 × 103 mi or 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. 427 PROBLEM 12.80 Communication satellites are placed in a geosynchronous orbit, i.e., in a circular orbit such that they complete one full revolution about the earth in one sidereal day (23.934 h), and thus appear stationary with respect to the ground. Determine (a) the altitude of these satellites above the surface of the earth, (b) the velocity with which they describe their orbit. Give the answers in both SI and U.S. customary units. SOLUTION For gravitational force and a circular orbit, Fr = GMm mv 2 = r r2 or v= GM r Let τ be the period time to complete one orbit. But vτ = 2π r Then GM τ 2 r = 4π 2 3 or GM τ 2 = 4π 2 r 2 r GM τ 2 r = 2 4π 1/3 τ = 23.934 h = 86.1624 × 103 s Data: (a) v 2τ 2 = or In SI units: g = 9.81 m/s 2 , R = 6.37 × 106 m GM = gR 2 = (9.81)(6.37 × 106 ) 2 = 398.06 × 1012 m3 /s 2 1/3 (398.06 × 1012 )(86.1624 × 103 )2 6 r= = 42.145 × 10 m 2 4 π altitude h = r − R = 35.775 × 106 m In U.S. units: h = 35,800 km g = 32.2 ft/s 2 , R = 3960 mi = 20.909 × 106 ft GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 1/3 (14.077 × 1015 )(86.1624 × 103 ) 2 6 r= = 138.334 × 10 ft 2 4 π altitude h = r − R = 117.425 × 106 ft h = 22, 200 mi 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. 428 PROBLEM 12.80 (Continued) (b) In SI units: v= GM 398.06 × 1012 = = 3.07 × 103 m/s 6 r 42.145 × 10 v= GM 14.077 × 1015 = = 10.09 × 103 ft/s r 138.334 × 106 v = 3.07 km/s In U.S. units: v = 10.09 × 103 ft/s 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. 429 PROBLEM 12.81 Show that the radius r of the orbit of a moon of a given planet can be determined from the radius R of the planet, the acceleration of gravity at the surface of the planet, and the time τ required by the moon to complete one full revolution about the planet. Determine the acceleration of gravity at the surface of the planet Jupiter knowing that R = 71,492 km and that τ = 3.551 days and r = 670.9 × 103 km for its moon Europa. SOLUTION Mm r2 We have F =G and F = Fn = man = m Then or Now so that G [Eq. (12.28)] v2 r Mm v2 m = r r2 v2 = GM r GM = gR 2 v2 = [Eq. (12.30)] gR 2 r or v=R g r 2π r 2π r = v R gr For one orbit, τ= or gτ 2 R 2 r = 2 4π Solving for g, g = 4π 2 1/3 Q.E.D. r3 τ 2 R2 and noting that τ = 3.551 days = 306,806 s, then g Jupiter = 4π 2 = 4π 2 3 rEur 2 RJup τ Eur (670.9 × 106 m)3 (306,806 s)2 (71.492 × 106 m) 2 g Jupiter = 24.8 m/s 2 or Note: g Jupiter ≈ 2.53g Earth 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. 430 PROBLEM 12.82 The orbit of the planet Venus is nearly circular with an orbital velocity of 126.5 × 103 km/h. Knowing that the mean distance from the center of the sun to the center of Venus is 108 × 106 km and that the radius of the sun is 695 × 103 km, determine (a) the mass of the sun, (b) the acceleration of gravity at the surface of the sun. SOLUTION Let M be the mass of the sun and m the mass of Venus. For the circular orbit of Venus, GMm mv 2 = ma = n r r2 GM = rv 2 where r is radius of the orbit. r = 108 × 106 km = 108 × 109 m Data: v = 126.5 × 103 km/hr = 35.139 × 103 m/s GM = (108 × 109 )(35.139 × 103 ) 2 = 1.3335 × 1020 m3 /s 2 (a) Mass of sun. (b) At the surface of the sun, M = GM 1.3335 × 1020 m3 /s 2 = G 66.73 × 10−12 M = 1.998 × 1030 kg R = 695.5 × 103 km = 695.5 × 106 m GMm = mg R2 g = GM 1.3335 × 1020 = R2 (695.5 × 106 ) 2 g = 276 m/s 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. 431 PROBLEM 12.83 A satellite is placed into a circular orbit about the planet Saturn at an altitude of 2100 mi. The satellite describes its orbit with a velocity of 54.7 × 103 mi/h. Knowing that the radius of the orbit about Saturn and the periodic time of Atlas, one of Saturn’s moons, are 85.54 × 103 mi and 0.6017 days, respectively, determine (a) the radius of Saturn, (b) the mass of Saturn. (The periodic time of a satellite is the time it requires to complete one full revolution about the planet.) SOLUTION 2π rA Velocity of Atlas. vA = where v A = 85.54 × 103 mi = 451.651 × 106 ft and τ A = 0.6017 days = 51,987 s Gravitational force. τA vA = (2π ) (451.651 × 106 ) = 54.587 × 103 ft/s 51,987 F= GMm mv 2 = r r2 from which GM = rv 2 = constant For the satellite, rs vs2 = rAv 2A rA v A2 rs = where vs = 54.7 × 103 mi/h = 80.227 × 103 ft/s vs 2 (451.651 × 106 )(54.587 × 103 ) 2 = 209.09 × 106 ft (80.227 × 103 ) 2 rs = 39, 600 mi rs = (a) Radius of Saturn. R = rs − (altitude) = 39, 600 − 2100 (b) R = 37,500 mi Mass of Saturn. M= rAv 2A (451.651 × 106 )(54.587 × 103 ) 2 = G 34.4 × 10−9 M = 39.1 × 1024 lb ⋅ s 2 /ft 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. 432 PROBLEM 12.84 The periodic times (see Problem 12.83) of the planet Uranus’s moons Juliet and Titania have been observed to be 0.4931 days and 8.706 days, respectively. Knowing that the radius of Juliet’s orbit is 40,000 mi, determine (a) the mass of Uranus, (b) the radius of Titania’s orbit. SOLUTION 2π rJ Velocity of Juliet. vJ = where rJ = 40,000 mi = 2.112 × 108 ft and τ J = 0.4931 days = 42, 604 s Gravitational force. vJ = (2π )(2.112 × 108 ft) = 3.11476 × 10 4 ft/s 42, 604 s F= GMm mv 2 = r r2 GM = rv 2 = constant from which (a) τJ M= Mass of Uranus. rJ vJ2 G (2.112 × 108 )(3.11476 × 104 ) 2 34.4 × 10−9 = 5.95642 × 10 24 lb ⋅ s 2 /ft M= M = 5.96 × 1024 lb ⋅ s 2 /ft (b) Radius of Titania’s orbit. GM = rT vT2 = rT3 = τ rJ3 T τJ 4π 2 rT3 τT 2 = 4π 2 rJ3 τJ2 2 2 8 3 8.706 27 3 = (2.112 × 10 ) = 2.93663 × 10 ft 0.4931 rT = 1.43202 × 109 ft = 2.71216 × 105 mi rT = 2.71 × 105 mi 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. 433 PROBLEM 12.85 A 500 kg spacecraft first is placed into a circular orbit about the earth at an altitude of 4500 km and then is transferred to a circular orbit about the moon. Knowing that the mass of the moon is 0.01230 times the mass of the earth and that the radius of the moon is 1737 km, determine (a) the gravitational force exerted on the spacecraft as it was orbiting the earth, (b) the required radius of the orbit of the spacecraft about the moon if the periodic times (see Problem 12.83) of the two orbits are to be equal, (c) the acceleration of gravity at the surface of the moon. SOLUTION First note that rE = RE + hE = (6.37 × 106 + 4.5 × 106 ) m Then (a) RE = 6.37 × 106 m = 10.87 × 106 m We have and F= GMm r2 [Eq. (12.28)] GM = gR 2 [Eq. (12.29)] m R =W 2 r r 2 Then F = gR 2 For the earth orbit, 6.37 × 106 m F = (500 kg)(9.81 m/s 2 ) 6 10.87 × 10 m 2 F = 1684 N or (b) From the solution to Problem 12.78, we have 2 1 2π 3 M= r G τ Then Now τ= 2π r 3/ 2 GM τE =τM 2π rE3/ 2 GM E = 2π rM3/2 GM M (1) 1/3 or M rM = M rE = (0.01230)1/3 (10.87 × 106 m) ME or rM = 2.509 × 106 m rM = 2510 km 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. 434 PROBLEM 12.85 (Continued) (c) GM = gR 2 We have [Eq.(12.29)] Substituting into Eq. (1) 2π rE3/2 RE g E = 2π rM3/ 2 RM g M 2 or 3 2 gM R r R M = E M gE = E M gE RM rE RM M E gM 6370 km 2 = (0.01230)(9.81 m/s ) 1737 km using the results of Part (b). Then 2 g moon = 1.62 m/s 2 or Note: g moon ≈ 1 g earth 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. 435 PROBLEM 12.86 A space vehicle is in a circular orbit of 2200-km radius around the moon. To transfer it to a smaller circular orbit of 2080-km radius, the vehicle is first placed on an elliptic path AB by reducing its speed by 26.3 m/s as it passes through A. Knowing that the mass of the moon is 73.49 × 1021 kg, determine (a) the speed of the vehicle as it approaches B on the elliptic path, (b) the amount by which its speed should be reduced as it approaches B to insert it into the smaller circular orbit. SOLUTION For a circular orbit, Σ Fn = man : F = m F =G Eq. (12.28): Then G or v2 r Mm r2 Mm v2 = m r r2 GM v2 = r 66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg 2200 × 103 m Then 2 (v A )circ = or (v A )circ = 1493.0 m/s and 2 (vB )circ = or (vB )circ = 1535.5 m/s (a) We have 66.73 × 10−12 m3 /kg ⋅ s 2 × 73.49 × 1021 kg 2080 × 103 m (v A )TR = (v A )circ + Δv A = (1493.0 − 26.3) m/s = 1466.7 m/s Conservation of angular momentum requires that rA m(v A )TR = rB m(vB )TR or 2200 km × 1466.7 m/s 2080 km = 1551.3 m/s (vB )TR = (vB )TR = 1551 m/s or (b) Now or (v B )circ = (vB )TR + ΔvB ΔvB = (1535.5 − 1551.3) m/s ΔvB = −15.8 m/s or 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. 436 PROBLEM 12.87 Plans for an unmanned landing mission on the planet Mars called for the earth-return vehicle to first describe a circular orbit at an altitude dA = 2200 km above the surface of the planet with a velocity of 2771 m/s. As it passed through Point A, the vehicle was to be inserted into an elliptic transfer orbit by firing its engine and increasing its speed by Δv A = 1046 m/s. As it passed through Point B, at an altitude dB = 100,000 km, the vehicle was to be inserted into a second transfer orbit located in a slightly different plane, by changing the direction of its velocity and reducing its speed by ΔvB = −22.0 m/s. Finally, as the vehicle passed through Point C, at an altitude dC = 1000 km, its speed was to be increased by ΔvC = 660 m/s to insert it into its return trajectory. Knowing that the radius of the planet Mars is R = 3400 km, determine the velocity of the vehicle after completion of the last maneuver. SOLUTION rA = 3400 + 2200 = 5600 km = 5.60 × 106 m rB = 3400 + 100, 000 = 103, 400 km = 103.4 × 106 m rC = 3400 + 1000 = 4400 km = 4.40 × 106 m First transfer orbit. v A = 2771 m/s + 1046 m/s = 3817 m/s Conservation of angular momentum: rA m v A = rB m vB (5.60 × 10 )(3817) = (103.4 × 106 )vB 6 vB = 206.7 m/s Second transfer orbit. vB′ = vB + ΔvB = 206.7 − 22.0 = 184.7 m/s Conservation of angular momentum: rB mvB′ = rC mvC (103.4 × 106 )(184.7) = (4.40 × 106 )vC vC = 4340 m/s After last maneuver. v = vC + ΔvC = 4340 + 660 v = 5000 m/s 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. 437 PROBLEM 12.88 To place a communications satellite into a geosynchronous orbit (see Problem 12.80) at an altitude of 22,240 mi above the surface of the earth, the satellite first is released from a space shuttle, which is in a circular orbit at an altitude of 185 mi, and then is propelled by an upper-stage booster to its final altitude. As the satellite passes through A, the booster’s motor is fired to insert the satellite into an elliptic transfer orbit. The booster is again fired at B to insert the satellite into a geosynchronous orbit. Knowing that the second firing increases the speed of the satellite by 4810 ft/s, determine (a) the speed of the satellite as it approaches B on the elliptic transfer orbit, (b) the increase in speed resulting from the first firing at A. SOLUTION For earth, R = 3960 mi = 20.909 × 106 ft GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 rA = 3960 + 185 = 4145 mi = 21.8856 × 106 ft rB = 3960 + 22, 240 = 26, 200 mi = 138.336 × 106 ft Speed on circular orbit through A. (v A )circ = = GM rA 14.077 × 1015 21.8856 × 106 = 25.362 × 103 ft/s Speed on circular orbit through B. (vB )circ = = GM rB 14.077 × 1015 138.336 × 106 = 10.088 × 103 ft/s 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. 438 PROBLEM 12.88 (Continued) (a) Speed on transfer trajectory at B. (vB ) tr = 10.088 × 103 − 4810 = 5.278 × 103 Conservation of angular momentum for transfer trajectory. 5280 ft/s rA (v A ) tr = rB (vB ) tr (v A ) tr = rB (vB ) tr rA (138.336 × 106 )(5278) 21.8856 × 106 = 33.362 × 103 ft/s = (b) Change in speed at A. Δv A = (v A ) tr − (v A )circ = 33.362 × 103 − 25.362 × 103 = 8.000 × 103 Δv A = 8000 ft/s 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. 439 PROBLEM 12.89 A space shuttle S and a satellite A are in the circular orbits shown. In order for the shuttle to recover the satellite, the shuttle is first placed in an elliptic path BC by increasing its speed by ΔvB = 280 ft/s as it passes through B. As the shuttle approaches C, its speed is increased by ΔvC = 260 ft/s to insert it into a second elliptic transfer orbit CD. Knowing that the distance from O to C is 4289 mi, determine the amount by which the speed of the shuttle should be increased as it approaches D to insert it into the circular orbit of the satellite. SOLUTION R = 3960 mi = 20.9088 × 106 ft First note rA = (3960 + 380) mi = 4340 mi = 22.9152 × 106 ft rB = (3960 + 180) mi = 4140 mi = 21.8592 × 106 ft ΣFn = man : For a circular orbit, F =G Eq. (12.28): Then or G F =m Mm r2 Mm v2 =m 2 r r v2 = GM gR 2 = r r using Eq. (12.29). 32.2 ft/s 2 × (20.9088 × 106 ft) 2 22.9152 × 106 ft Then 2 (v A )circ = or (v A )circ = 24, 785 ft/s and 2 (vB )circ = or (vB )circ = 25,377 ft/s We have v2 r 32.2 ft/s 2 × (20.9088 × 106 ft) 2 21.8592 × 106 ft (vB )TRBC = (vB )circ + ΔvB = (25,377 + 280) ft/s = 25, 657 ft/s 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. 440 PROBLEM 12.89 (Continued) Conservation of angular momentum requires that From Eq. (1) BC : rB m (vB )TRBC = rC m (vC )TRBC (1) CD : rC m (vC )TRCD = rA m (vD )TRCD (2) (vC )TRBC = rB 4140 mi (vB )TRBC = × 25, 657 ft/s 4289 mi rC = 24, 766 ft/s Now (vC )TRCD = (vC )TRBC + ΔvC = (24, 766 + 260) ft/s = 25, 026 ft/s From Eq. (2) (vD )TRCD = rC 4289 mi (vC )TRCD = × 25,026 ft/s 4340 mi rA = 24, 732 ft/s Finally, or (v A )circ = (vD )TRCD + ΔvD ΔvD = (24,785 − 24, 732) ft/s ΔvD = 53 ft/s or 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. 441 PROBLEM 12.90 A 1 kg collar can slide on a horizontal rod, which is free to rotate about a vertical shaft. The collar is initially held at A by a cord attached to the shaft. A spring of constant 30 N/m is attached to the collar and to the shaft and is undeformed when the collar is at A. As the rod rotates at the rate θ = 16 rad/s, the cord is cut and the collar moves out along the rod. Neglecting friction and the mass of the rod, determine (a) the radial and transverse components of the acceleration of the collar at A, (b) the acceleration of the collar relative to the rod at A, (c) the transverse component of the velocity of the collar at B. SOLUTION Fsp = k ( r − rA ) First note (a) Fθ = 0 and at A, Fr = − Fsp = 0 (a A ) r = 0 (a A )θ = 0 ΣFr = mar : (b) Noting that −Fsp = m( r − rθ 2 ) acollar/rod = r , we have at A 0 = m[acollar/rod − (150 mm)(16 rad/s) 2 ] acollar/rod = 38400 mm/s 2 (acollar/rod ) A = 38.4 m/s 2 or (c) After the cord is cut, the only horizontal force acting on the collar is due to the spring. Thus, angular momentum about the shaft is conserved. rA m (v A )θ = rB m (vB )θ Then (vB )θ = where (v A )θ = rA θ0 150 mm [(150 mm)(16 rad/s)] = 800 mm/s 450 mm (vB )θ = 0.800 m/s or 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. 442 PROBLEM 12.91 A 1-lb ball A and a 2-lb ball B are mounted on a horizontal rod which rotates freely about a vertical shaft. The balls are held in the positions shown by pins. The pin holding B is suddenly removed and the ball moves to position C as the rod rotates. Neglecting friction and the mass of the rod and knowing that the initial speed of A is v A = 8 ft/s, determine (a) the radial and transverse components of the acceleration of ball B immediately after the pin is removed, (b) the acceleration of ball B relative to the rod at that instant, (c) the speed of ball A after ball B has reached the stop at C. SOLUTION Let r and θ be polar coordinates with the origin lying at the shaft. Constraint of rod: θ B = θ A + π radians; θB = θA = θ; θB = θA = θ. (a) Components of acceleration Sketch the free body diagrams of the balls showing the radial and transverse components of the forces acting on them. Owing to frictionless sliding of B along the rod, ( FB ) r = 0. Radial component of acceleration of B. Fr = mB (aB ) r : (a B ) r = 0 Transverse components of acceleration. (a A )θ = rAθ + 2rAθ = raθ (aB )θ = rBθ + 2rBθ (1) Since the rod is massless, it must be in equilibrium. Draw its free body diagram, applying Newton’s 3rd Law. ΣM 0 = 0: rA ( FA )θ + rB ( FB )θ = rAm A (a A )θ + rB mB (aB )θ = 0 rAmArAθ + rB mB (rBθ + 2rBθ) = 0 θ = At t = 0, rB = 0 so that −2mB rBθ m ArA2 + mB rB 2 θ = 0. (aB )θ = 0 From Eq. (1), 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. 443 PROBLEM 12.91 (Continued) (b) Acceleration of B relative to the rod. (v ) 96 At t = 0, (v A )θ = 8 ft/s = 96 in./s, θ = A θ = = 9.6 rad/s rA 10 rB − rBθ 2 = (aB ) r = 0 rB = rBθ 2 = (8) (9.6)2 = 737.28 in./s 2 rB = 61.4 ft/s 2 (c) Speed of A. Substituting d (mr 2θ) for rFθ in each term of the moment equation dt gives ( ) ( ) d d m ArA2θ + mB rB 2θ = 0 dt dt Integrating with respect to time, ( 2 2 2 m ArA θ + mB rB θ = m ArA θ ) + ( m r θ ) 2 0 B B 0 Applying to the final state with ball B moved to the stop at C, WA 2 WB 2 W W rA + rC θ f = A rA2 + B (rB )02 θ0 g g g g θ f = WArA2 + WB (rB )02 (1)(10) 2 + (2)(8) 2 = (9.6) = 3.5765 rad/s θ 0 WArA 2 + WB rC 2 (1)(10) 2 + (2)(16) 2 (v A ) f = rAθ f = (10)(3.5765) = 35.765 in./s (v A ) f = 2.98 ft/s 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. 444 PROBLEM 12.92 Two 2.6-lb collars A and B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O and a stop prevents collar B from moving. The frame is rotating at the rate θ = 12 rad/s and r = 0.6 ft when the stop is removed allowing collar A to move out along rod OE. Neglecting friction and the mass of the frame, determine, for the position r = 1.2 ft, (a) the transverse component of the velocity of collar A, (b) the tension in the cord and the acceleration of collar A relative to the rod OE. SOLUTION Masses: m A = mB = 2.6 = 0.08075 lb ⋅ s 2 /ft 32.2 (a) Conservation of angular momentum of collar A: ( H 0 ) 2 = ( H 0 )1 m Ar1(vθ )1 = m Ar2 (vθ ) 2 (vθ ) 2 = r1(vθ )1 r12θ1 (0.6) 2 (12) = = = 3.6 r2 r2 1.2 (vθ ) 2 = 3.60 ft/s θ2 = (vθ ) 2 3.6 = = 3.00 rad/s rA 1.2 (b) Let y be the position coordinate of B, positive upward with origin at O. Constraint of the cord: r − y = constant or y = r Kinematics: (aB ) y = y = r Collar B: Collar A: and (a A ) r = r − rθ 2 ΣFy = mB aB : T − WB = mB y = mB r (1) ΣFr = mA (a A )r : − T = m A ( r − rθ 2 ) (2) Adding (1) and (2) to eliminate T, −WB = (m A + mB ) r + m Arθ 2 a A/rod = r= m Arθ 2 − WB (0.08075)(1.2)(3.00) 2 − (2.6) = = −10.70 ft/s 2 m A + mB 0.08075 + 0.08075 T = mB ( r + g ) = (0.08075)(−10.70 + 32.2) T = 1.736 lb a A / rod = 10.70 ft/s 2 radially inward. 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. 445 PROBLEM 12.93 A small ball swings in a horizontal circle at the end of a cord of length l1 , which forms an angle θ1 with the vertical. The cord is then slowly drawn through the support at O until the length of the free end is l2 . (a) Derive a relation among l1 , l2 , θ1 , and θ 2 . (b) If the ball is set in motion so that initially l1 = 0.8 m and θ1 = 35°, determine the angle θ 2 when l2 = 0.6 m. SOLUTION (a) For state 1 or 2, neglecting the vertical component of acceleration, ΣFy = 0: T cos θ − W = 0 T = W cos θ ΣFx = man : T sin θ = W sin θ cos θ = But ρ = sin θ mv 2 ρ so that v2 = ρW m sin 2 θ cos θ = g sin θ tan θ v1 = 1 g sin θ1 tan θ1 and v2 = 2 g sin θ 2 tan θ 2 ΣM y = 0: H y = constant r1mv1 = r2 mv2 v11 sin θ1 = v2 2 sin θ 2 or 3/2 3/2 1 g sin θ1 sin θ1 tan θ1 = 2 sin θ 2 sin θ 2 tan θ 2 31 sin 3 θ1 tan θ1 = 32 sin 3 θ 2 tan θ 2 (b) With θ1 = 35°, 1 = 0.8 m, and 2 = 0.6 m (0.8)3 sin 3 35° tan 35° = (0.6)3 sin 3 θ 2 tan θ 2 sin 3 θ 2 tan θ 2 − 0.31320 = 0 θ 2 = 43.6° 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. 446 PROBLEM 12.CQ6 A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about Point D, that is, the upper left corner of the crate? (a) 0 (b) mv1a (c) mv1b (d ) mv1 a 2 + b 2 SOLUTION Answers: (b) The angular momentum is the moment of the momentum, so simply take the linear momentum, mv1, and multiply it by the perpendicular distance from the line of action of mv1 and Point 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. 447 PROBLEM 12.CQ7 A uniform crate C with mass mC is being transported to the left by a forklift with a constant speed v1. What is the magnitude of the angular momentum of the crate about Point A, that is, the point of contact between the front tire of the forklift and the ground? (a) 0 (b) mv1d (c) 3mv1 (d ) mv1 32 + d 2 SOLUTION Answer: (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. 448 PROBLEM 12.94 A particle of mass m is projected from Point A with an initial velocity v 0 perpendicular to OA and moves under a central force F along an elliptic path defined by the equation r = r0 /(2 − cos θ ). Using Eq. (12.37), show that F is inversely proportional to the square of the distance r from the particle to the center of force O. SOLUTION u= 1 2 − cos θ = , r r0 d 2u 2 F +u = = 2 r0 mh 2u 2 dθ Solving for F, F= du sin θ = , dθ r0 d 2 u cos θ = r0 dθ 2 by Eq. (12.37). 2mh 2 u 2 2mh 2 = r0 r0 r 2 Since m, h, and r0 are constants, F is proportional to 1 r2 , or inversely proportional to r 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. 449 PROBLEM 12.95 A particle of mass m describes the logarithmic spiral r = r0 ebθ under a central force F directed toward the center of force O. Using Eq. (12.37) show that F is inversely proportional to the cube of the distance r from the particle to O. SOLUTION u= 1 1 −bθ = e r r0 du b = − e −bθ dθ r0 d 2u b 2 −bθ e = r0 dθ 2 d 2u b 2 + 1 −bθ F +u = e = 2 r dθ mh 2u 2 0 F = = (b 2 + 1)mh 2u 2 −bθ e r0 (b 2 + 1)mh 2u 2 (b 2 + 1)mh 2 = r r3 Since b, m, and h are constants, F is proportional to 1 r3 , or inversely proportional to r 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. 450 PROBLEM 12.96 For the particle of Problem 12.74, and using Eq. (12.37), show that the central force F is proportional to the distance r from the particle to the center of force O. PROBLEM 12.74 A particle of mass m is projected from Point A with an initial velocity v 0 perpendicular to line OA and moves under a central force F directed away from the center of force O. Knowing that the particle follows a path defined by the equation r = r0 / cos 2θ and using Eq. (12.27), express the radial and transverse components of the velocity v of the particle as functions of θ . SOLUTION u = 1 = r cos 2θ , r0 sin 2θ du =− dθ r0 cos 2θ d 2u cos 2θ (2 cos 2θ ) − sin 2θ (− sin 2θ / cos 2θ ) =− dθ r0 cos 2θ =− Eq. (12.37): 2 cos 2 2θ + sin 2 2θ (1 + cos 2 2θ ) = − r0 (cos 2θ )3/2 r0 (cos 2θ )3/2 d 2u F +u = 2 dθ mh 2u 2 Solving for F, d 2u F = mh 2u 2 + u dθ = mh 2 cos 2θ r0 2 1 + cos 2 2θ + − 3/2 r0 (cos 2θ ) cos 2θ r0 = mh 2 cos 2θ r0 2 1 − − 3/ 2 r0 (cos 2θ ) cos 2θ + r0 =− mh 2 mh 2 = − 3 4 r0 cos 2θ r0 r0 cos 2θ cos 2θ r0 F =− mh 2r r0 4 The force F is proportional to r. The minus sign indicates that it is repulsive. 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. 451 PROBLEM 12.97 A particle of mass m describes the path defined by the equation r = r0 sin θ under a central force F directed toward the center of force O. Using Eq. (12.37), show that F is inversely proportional to the fifth power of the distance r from the particle to O. SOLUTION We have d 2u F +u = 2 dθ mh 2u 2 u= where 1 r Eq. (12.37) and mh 2 = constant d 2u F × u 2 2 + u dθ u= Now Then and du 1 1 1 cos θ = =− dθ dθ r0 sin θ r0 sin 2 θ d 2u 1 − sin θ (sin 2 θ ) − cos θ (2 sin θ cos θ ) = − r0 dθ 2 sin 4 θ = Then 1 1 = r r0 sin θ 1 F × 2 r 1 1 + cos 2 θ r0 sin 3 θ 2 1 1 1 + cos θ + 3 r r sin θ sin θ 0 0 = mh 2 1 1 1 + cos 2 θ sin 2 θ + r0 r 2 sin 3 θ sin 3 θ 2 1 1 = mh r0 r 2 sin 3 θ 2 = mh 2 r sin θ = r0 3 3 2r02 r5 F is proportional to 1 1 F× 5 5 r r Q.E.D. Note: F > 0 implies that F is attractive. 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. 452 PROBLEM 12.98 It was observed that during its second flyby of the earth, the Galileo spacecraft had a velocity of 14.1 km/s as it reached its minimum altitude of 303 km above the surface of the earth. Determine the eccentricity of the trajectory of the spacecraft during this portion of its flight. SOLUTION For earth, R = 6.37 × 106 m r0 = 6.37 × 106 + 303. × 103 = 6.673 × 106 m h = r0v0 = (6.673 × 106 )(14.1 × 103 ) = 94.09 × 109 m 2 /s GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 1 GM = 2 (1 + ε ) r0 h 1+ε = h2 (94.09 × 109 )2 = = 3.33 r0GM (6.673 × 106 )(398.06 × 1012 ) ε = 3.33 − 1 ε = 2.33 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. 453 PROBLEM 12.99 It was observed that during the Galileo spacecraft’s first flyby of the earth, its maximum altitude was 600 mi above the surface of the earth. Assuming that the trajectory of the spacecraft was parabolic, determine the maximum velocity of Galileo during its first flyby of the earth. SOLUTION For the earth: R = 3960 mi = 20.909 × 106 ft GM = gR 2 = (32.2) (20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 For a parabolic trajectory, ε = 1. Eq. (12.39′) : At θ = 0, 1 GM = 2 (1 + cos θ ) r h 1 2GM 2GM = = 2 2 2 r0 h r0 v0 or v0 = 2GM r0 At r0 = 3960 + 600 = 4560 mi = 24.077 × 106 ft, v0 = (2)(14.077 × 1015 ) = 34.196 × 103 ft/s 6 24.077 × 10 v0 = 6.48 mi/s 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. 454 PROBLEM 12.100 As a space probe approaching the planet Venus on a parabolic trajectory reaches Point A closest to the planet, its velocity is decreased to insert it into a circular orbit. Knowing that the mass and the radius of Venus are 4.87 × 1024 kg and 6052 km, respectively, determine (a) the velocity of the probe as it approaches A, (b) the decrease in velocity required to insert it into the circular orbit. SOLUTION First note (a) rA = (6052 + 280) km = 6332 km From the textbook, the velocity at the point of closest approach on a parabolic trajectory is given by v0 = 2GM r0 1/2 Thus, (v A )par 2 × 66.73 × 10−12 m3 /kg ⋅ s 2 × 4.87 × 1024 kg = 6332 × 103 m = 10,131.4 m/s (v A ) par = 10.13 km/s or (b) We have (v A )circ = (v A ) par + Δv A Now (v A )circ = Then GM r0 = 1 Δv A = 1 2 2 Eq. (12.44) (v A )par (v A ) par − (v A )par 1 = − 1 (10.1314 km/s) 2 = −2.97 km/s | Δv A | = 2.97 km/s 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. 455 PROBLEM 12.101 It was observed that as the Voyager I spacecraft reached the point of its trajectory closest to the planet Saturn, it was at a distance of 185 × 103 km from the center of the planet and had a velocity of 21.0 km/s. Knowing that Tethys, one of Saturn’s moons, describes a circular orbit of radius 295 × 103 km at a speed of 11.35 km/s, determine the eccentricity of the trajectory of Voyager I on its approach to Saturn. SOLUTION For a circular orbit, Eq. (12.44) v= GM r For the orbit of Tethys, GM = rT vT2 For Voyager’s trajectory, we have 1 GM = 2 (1 + ε cos θ ) r h where h = r0 v0 At O, r = r0 , θ = 0 Then 1 GM (1 + ε ) = r0 (r0 v0 ) 2 or ε= r0 v02 r v2 − 1 = 0 02 − 1 GM rT vT 2 = 185 × 103 km 21.0 km/s × −1 295 × 103 km 11.35 km/s ε = 1.147 or 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. 456 PROBLEM 12.102 A satellite describes an elliptic orbit about a planet of mass M. Denoting by r0 and r1 , respectively, the minimum and maximum values of the distance r from the satellite to the center of the planet, derive the relation 1 1 2GM + = 2 r0 r1 h where h is the angular momentum per unit mass of the satellite. SOLUTION Using Eq. (12.39), 1 GM = 2 + C cos θ A rA h and 1 GM = 2 + C cos θ B . rB h But θ B = θ A + 180°, so that Adding, cos θ A = − cos θ B . 1 1 1 1 2GM + = + = 2 rA rB r0 r1 h 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. 457 PROBLEM 12.103 A space probe is describing a circular orbit about a planet of radius R. The altitude of the probe above the surface of the planet is α R and its speed is v0. To place the probe in an elliptic orbit which will bring it closer to the planet, its speed is reduced from v0 to β v0 , where β < 1, by firing its engine for a short interval of time. Determine the smallest permissible value of β if the probe is not to crash on the surface of the planet. SOLUTION GM rA For the circular orbit, v0 = where rA = R + α R = R(1 + α ) Eq. (12.44), GM = v02 R(1 + α ) Then From the solution to Problem 12.102, we have for the elliptic orbit, 1 1 2GM + = 2 rA rB h h = hA = rA (v A ) AB Now = [ R(1 + α )]( β v0 ) Then 2v02 R(1 + α ) 1 1 + = R(1 + α ) rB [ R(1 + α ) β v0 ]2 = 2 β R(1 + α ) 2 Now β min corresponds to rB → R. Then 1 1 2 + = 2 R(1 + α ) R β min R(1 + α ) β min = or 2 2 +α 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. 458 PROBLEM 12.104 At main engine cutoff of its thirteenth flight, the space shuttle Discovery was in an elliptic orbit of minimum altitude 60 km and maximum altitude 500 km above the surface of the earth. Knowing that at Point A the shuttle had a velocity v0 parallel to the surface of the earth and that the shuttle was transferred to a circular orbit as it passed through Point B, determine (a) the speed v0 of the shuttle at A, (b) the increase in speed required at B to insert the shuttle into the circular orbit. SOLUTION For earth, R = 6370 km = 6370 × 103 m GM = gR 2 = (9.81)(6370 × 103 ) 2 = 3.9806 × 1014 m3 /s 2 rA = 6370 + 60 = 6430 km = 6430 × 103 m rB = 6370 + 500 = 6870 km = 6870 × 103 m Elliptic trajectory. Using Eq. (12.39), 1 GM = 2 + C cos θ A rA h But θ B = θ A + 180°, so that cos θ A = − cos θ B Adding, 1 GM = 2 + C cos θ B . rB h 1 1 rA + rB 2GM + = = 2 rA rB rA rB h h= (a) and 2GMrA rB (2)(3.9806 × 1014 )(6430 × 103 )(6870 × 103 ) = = 51.422 × 109 m 2 /s rA + rB 6430 × 103 + 6870 × 103 Speed v0 at A. v0 = v A = (vB )1 = h 51.422 × 109 = rA 6430 × 103 v0 = 8.00 × 103 m/s h 51.422 × 109 = = 7.48497 × 103 m/s 3 rA 6870 × 10 For a circular orbit through Point B, (vB )circ = (b) GM 3.9806 × 1014 = = 7.6119 × 103 m/s rB 6870 × 103 Increase in speed at Point B. ΔvB = (vB )circ − (vB )1 = 126.97 m/s ΔvB = 127 m/s 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. 459 PROBLEM 12.105 A space probe is to be placed in a circular orbit of 5600 mi radius about the planet Venus in a specified plane. As the probe reaches A, the point of its original trajectory closest to Venus, it is inserted in a first elliptic transfer orbit by reducing its speed by Δv A . This orbit brings it to Point B with a much reduced velocity. There the probe is inserted in a second transfer orbit located in the specified plane by changing the direction of its velocity and further reducing its speed by ΔvB . Finally, as the probe reaches Point C, it is inserted in the desired circular orbit by reducing its speed by ΔvC . Knowing that the mass of Venus is 0.82 times the mass of the earth, that rA = 9.3 × 103 mi and rB = 190 × 103 mi, and that the probe approaches A on a parabolic trajectory, determine by how much the velocity of the probe should be reduced (a) at A, (b) at B, (c) at C. SOLUTION For Earth, R = 3690 mi = 20.9088 × 106 ft, g = 32.2 ft/s 2 GM earth = gR 2 = (32.2)(20.9088 × 106 )2 = 14.077 × 1015 ft 3 /s 2 For Venus, For a parabolic trajectory with GM = 0.82GM earth = 11.543 × 1015 ft 3 /s 2 rA = 9.3 × 103 mi = 49.104 × 106 ft (v A )1 = vesc = First transfer orbit AB. 2GM (2)(11.543 × 1015 ) = = 21.683 × 103 ft/s 6 rA 49.104 × 10 rB = 190 × 103 mi = 1003.2 × 106 ft At Point A, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rA hAB hAB (1) 1 GM GM = 2 + C cos 0 = 2 + C rB hAB hAB (2) At Point B, where θ = 0° Adding, r + rA 2GM 1 1 + = B = 2 rA rB rA rB hAB 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. 460 PROBLEM 12.105 (Continued) Solving for hAB , hAB = 2GMrA rB (2)(11.543 × 1015 )(49.104 × 106 )(1003.2 × 106 ) = = 1.039575 × 1012 ft 2 /s rB + rA 1052.3 × 106 (v A ) 2 = hAB 1.039575 × 1012 = = 21.174 × 103 ft/s 6 rA 49.104 × 10 (vB )1 = hAB 1.039575 × 1012 = = 1.03626 × 103 ft/s 6 rB 1003.2 × 10 rC = 5600 mi = 29.568 × 106 ft Second transfer orbit BC. At Point B, where θ = 0 1 GM GM = 2 + C cos 0 = 2 + C rB hBC hBC At Point C, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rC hBC hBC Adding, 1 1 rB + rC 2GM + = = 2 rB rC rB rC hBC hBC = 2GMrB rC (2)(11.543 × 1015 )(1003.2 × 106 )(29.568 × 106 ) = = 814.278 × 109 ft 2 /s rB + rC 1032.768 × 106 ( vB ) 2 = hBC 814.278 × 109 = = 811.69 ft/s rB 1003.2 × 106 (vC )1 = hBC 814.278 × 109 = = 27.539 × 103 ft/s rC 29.568 × 106 Final circular orbit. rC = 29.568 × 106 ft (vC )2 = GM 11.543 × 1015 = = 19.758 × 103 ft/s rC 29.568 × 106 Speed reductions. (a) At A: (v A )1 − (v A ) 2 = 21.683 × 103 − 21.174 × 103 Δv A = 509 ft/s (b) At B: (vB )1 − (vB ) 2 = 1.036 × 103 − 811.69 ΔvB = 224 ft/s (c) At C: (vC )1 − (vC )2 = 27.539 × 103 − 19.758 × 103 ΔvC = 7.78 × 103 ft/s 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. 461 PROBLEM 12.106 For the space probe of Problem 12.105, it is known that rA = 9.3 × 103 mi and that the velocity of the probe is reduced to 20, 000 ft/s as it passes through A. Determine (a) the distance from the center of Venus to Point B, (b) the amounts by which the velocity of the probe should be reduced at B and C, respectively. SOLUTION Data from Problem 12.105: rC = 29.568 × 106 ft, M = 0.82 M earth For Earth, R = 3960 mi = 20.9088 × 106 ft, g = 32.2 ft/s 2 GM earth = gR 2 = (32.2)(20.9088 × 106 )2 = 14.077 × 1015 m3 /s 2 GM = 0.82GM earth = 11.543 × 1015 ft 3 /s 2 For Venus, v A = 20, 000 ft/s, rA = 9.3 × 103 mi = 49.104 × 106 ft Transfer orbit AB: hAB = rAv A = (49.104 × 106 )(20, 000) = 982.08 × 109 ft 2 /s At Point A, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rA hAB hAB At Point B, where θ = 0° 1 GM GM = 2 + C cos 0 = 2 + C rB hAB hAB Adding, 1 1 2GM + = 2 rA rB hAB 1 2GM 1 = 2 − rB rA hAB = (2)(11.543 × 1015 ) 1 − 9 2 (982.08 × 10 ) 49.104 × 106 = 3.57125 × 10 −9 ft −1 (a) Radial coordinate rB . rB = 280.01 × 106 ft (vB )1 = rB = 53.0 × 103 mi hAB 982.08 × 109 = = 3.5073 × 103 ft/s 6 rB 280.01 × 10 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. 462 PROBLEM 12.106 (Continued) rC = 5600 mi = 29.568 × 106 ft Second transfer orbit BC. At Point B, where θ = 0 1 GM GM = 2 + C cos 0 = 2 + C rB hBC hBC At Point C, where θ = 180° 1 GM GM = 2 + C cos 180° = 2 − C rC hBC hBC Adding, r + rC 1 1 2GM + = B = 2 rB rC rB rC hBC hBC = 2GMrB rC = rB + rC (2)(11.543 × 1015 )(280.01 × 106 )(29.568 × 106 ) 309.578 × 106 = 785.755 × 109 ft 2 /s ( vB ) 2 = hBC 785.755 × 109 = = 2.8062 × 103 ft/s 6 rB 280.01 × 10 (vC )1 = hBC 785.755 × 109 = = 26.575 × 103 ft/s 6 rC 29.568 × 10 rC = 29.568 × 106 ft Circular orbit with (vC )2 = (b) GM 11.543 × 1015 = = 19.758 × 103 ft/s 6 rC 29.568 × 10 Speed reductions at B and C. At B: (vB )1 − (vB ) 2 = 3.5073 × 103 − 2.8062 × 103 ΔvB = 701 ft/s At C: (vC )1 − (vC )2 = 26.575 × 103 − 19.758 × 103 ΔvC = 6.82 × 103 ft/s 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. 463 PROBLEM 12.107 As it describes an elliptic orbit about the sun, a spacecraft reaches a maximum distance of 202 × 106 mi from the center of the sun at Point A (called the aphelion) and a minimum distance of 92 × 106 mi at Point B (called the perihelion). To place the spacecraft in a smaller elliptic orbit with aphelion at A′ and perihelion at B ′, where A′ and B ′ are located 164.5 × 106 mi and 85.5 × 106 mi, respectively, from the center of the sun, the speed of the spacecraft is first reduced as it passes through A and then is further reduced as it passes through B ′. Knowing that the mass of the sun is 332.8 × 103 times the mass of the earth, determine (a) the speed of the spacecraft at A, (b) the amounts by which the speed of the spacecraft should be reduced at A and B ′ to insert it into the desired elliptic orbit. SOLUTION First note Rearth = 3960 mi = 20.9088 × 106 ft rA = 202 × 106 mi = 1066.56 × 109 ft rB = 92 × 106 mi = 485.76 × 109 ft From the solution to Problem 12.102, we have for any elliptic orbit about the sun 1 1 2GM sun + = r1 r2 h2 (a) For the elliptic orbit AB, we have r1 = rA , r2 = rB , h = hA = rA v A Also, GM sun = G[(332.8 × 103 ) M earth ] 2 = gRearth (332.8 × 103 ) Then using Eq. (12.30). 2 (332.8 × 103 ) 1 1 2 gRearth + = rA rB (rAv A )2 1/2 or R v A = earth rA 665.6 g × 103 1 + r1 r A B 3960 mi 665.6 × 103 × 32.2 ft/s 2 = 202 × 106 mi 1066.561× 109 ft + 485.761× 109 ft = 52, 431 ft/s 1/ 2 v A = 52.4 × 103 ft/s or 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. 464 PROBLEM 12.107 (Continued) (b) From Part (a), we have 1 1 2GM sun = (rA v A ) 2 + rA rB Then, for any other elliptic orbit about the sun, we have ( 2 1 1 1 (rA v A ) rA + + = r1 r2 h2 1 rB ) For the elliptic transfer orbit AB′, we have r1 = rA , r2 = rB′ , h = htr = rA (v A ) tr Then or ( 2 1 1 1 1 ( rAv A ) rA + rB + = rA rB′ [rA (v A ) tr ]2 (v A ) tr = v A 1 rA + 1 rB 1 rA + 1 rB′ 1/2 ) 1 + rA rB = vA 1 + rA rB′ 1/ 2 1/2 1 + 202 92 = (52, 431 ft/s) 1 + 202 85.5 = 51,113 ft/s htr = ( hA ) tr = (hB′ ) tr : rA (v A ) tr = rB′ (vB′ ) tr Now Then (vB′ ) tr = 202 × 106 mi × 51,113 ft/s = 120,758 ft/s 85.5 × 106 mi For the elliptic orbit A′B′, we have r1 = rA′ , r2 = rB′ , h = rB′ vB′ Then or ( 2 1 1 1 (rA v A ) rA + + = rA′ rB′ (rB′vB′ )2 r vB ′ = v A A rB′ 1 rB 1 rA + 1 rB 1 rA′ + 1 rB′ ) 1/ 2 1 1 202 × 106 mi 202 × 106 + 92 × 106 = (52, 431 ft/s) 85.5 × 106 mi 164.51× 106 + 85.51× 106 = 116,862 ft/s 1/ 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. 465 PROBLEM 12.107 (Continued) Finally, or (v A ) tr = v A + Δv A Δv A = (51,113 − 52, 431) ft/s |Δv A | = 1318 ft/s or and or vB′ = (vB′ ) tr + ΔvB ΔvB′ = (116,862 − 120, 758) ft/s = −3896 ft/s |ΔvB | = 3900 ft/s or 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. 466 PROBLEM 12.108 Halley’s comet travels in an elongated elliptic orbit for which the minimum distance from the sun is approximately 12 rE , where rE = 150 × 106 km is the mean distance from the sun to the earth. Knowing that the periodic time of Halley’s comet is about 76 years, determine the maximum distance from the sun reached by the comet. SOLUTION We apply Kepler’s Third Law to the orbits and periodic times of earth and Halley’s comet: 2 τH aH = τE aE Thus 3 τ aH = aE H τE 2/3 76 years = rE 1 year = 17.94rE But 2/3 1 ( rmin + rmax ) 2 11 17.94rE = rE + rmax 22 aH = 1 rE 2 = (35.88 − 0.5) rE rmax = 2(17.94 rE ) − = 35.38 rE rmax = (35.38)(150 × 106 km) rmax = 5.31 × 109 km 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. 467 PROBLEM 12.109 Based on observations made during the 1996 sighting of comet Hyakutake, it was concluded that the trajectory of the comet is a highly elongated ellipse for which the eccentricity is approximately ε = 0.999887. Knowing that for the 1996 sighting the minimum distance between the comet and the sun was 0.230 RE , where RE is the mean distance from the sun to the earth, determine the periodic time of the comet. SOLUTION For Earth’s orbit about the sun, v0 = 2π RE 2π RE 3/ 2 GM = , τ0 = RE v0 GM or GM = 2π RE3/ 2 τ0 (1) For the comet Hyakutake, 1 GM = 2 = (1 + ε ), r0 h a= 1 GM 1+ ε = 2 (1 + ε ), r1 = r0 r1 1− ε h r 1 1+ ε r0 (r0 + r1 ) = 0 , b = r0 r1 = 2 1− ε 1−ε h = GMr0 (1 + ε ) τ= = 2π r02 (1 + ε )1/ 2 2π ab = h (1 − ε )3/ 2 GMr0 (1 + ε ) 2π r03/ 2 GM (1 − ε )3/2 r = 0 RE 3/2 1 (1 − ε )3/2 = (0.230)3/2 Since = 2π r03/ 2τ 0 2π RE3 (1 − ε )3/2 τ0 1 τ 0 = 91.8 × 103τ 0 (1 − 0.999887)3/ 2 τ 0 = 1 yr, τ = (91.8 × 103 )(1.000) τ = 91.8 × 103 yr 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. 468 PROBLEM 12.110 A space probe is to be placed in a circular orbit of radius 4000 km about the planet Mars. As the probe reaches A, the point of its original trajectory closest to Mars, it is inserted into a first elliptic transfer orbit by reducing its speed. This orbit brings it to Point B with a much reduced velocity. There the probe is inserted into a second transfer orbit by further reducing its speed. Knowing that the mass of Mars is 0.1074 times the mass of the earth, that rA = 9000 km and rB = 180,000 km, and that the probe approaches A on a parabolic trajectory, determine the time needed for the space probe to travel from A to B on its first transfer orbit. SOLUTION For earth, R = 6373 km = 6.373 × 106 m GM = gR 2 = (9.81) (6.373 × 106 ) 2 = 398.43 × 1012 m3 /s 2 For Mars, GM = (0.1074)(398.43 × 1012 ) = 42.792 × 1012 m3 /s 2 rA = 9000 km = 9.0 × 106 m rB = 180000 km = 180 × 106 m For the parabolic approach trajectory at A, (v A )1 = 2GM = rA (2)(42.792 × 1012 ) = 3.0837 × 103 m/s 9.0 × 106 First elliptic transfer orbit AB. Using Eq. (12.39), But 1 GM = 2 + C cosθ A rA hAB θ B = θ A + 180°, Adding, so that and 1 GM = 2 + C cosθ B . rB hAB cosθ A = − cosθ B . 1 1 r + rB 2GM + = A = 2 rA rB rArB hAB hAB = 2GMrArB = rA + rB (2)(42.792 × 1012 )(9.0 × 106 )(180 × 106 ) 189.0 × 106 hAB = 27.085 × 109 m 2 /s a = b= 1 1 (rA + rB ) = (9.0 × 106 + 180 × 106 ) = 94.5 × 106 m 2 2 rArB = (9.0 × 106 )(180 × 106 ) = 40.249 × 106 m 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. 469 PROBLEM 12.110 (Continued) Periodic time for full ellipse: For half ellipse AB, τ AB = τ AB = τ = 2π ab h 1 π ab τ = 2 h π (94.5 × 106 )(40.249 × 106 ) 27.085 × 10 9 = 444.81 × 103 s τ AB = 122.6 h 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. 470 PROBLEM 12.111 A space shuttle is in an elliptic orbit of eccentricity 0.0356 and a minimum altitude of 300 km above the surface of the earth. Knowing that the radius of the earth is 6370 km, determine the periodic time for the orbit. SOLUTION For earth, g = 9.81 m/s 2 , R = 6370 km = 6.370 × 106 m GM = gR 2 = (9.81)(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2 For the orbit, r0 = 6370 + 300 = 6670 km = 6.670 × 106 m 1 GM = 2 (1 − ε ) r1 h 1 GM = 2 (1 + ε ) r0 h r1 = r0 a = 1+ε 1.0356 − (6.670 × 106 ) = 7.1624 × 106 m 1−ε 0.9644 1 (r0 + r1) = 6.9162 × 106 m 2 b= r0r1 = 6.9118 × 106 m h= (1 + ε )GMr0 = (1.0356)(398.06 × 1012 )(6.670 × 106 ) = 52.436400 × 109 m 2 /s τ = 2π ab 2π (6.9118 × 106 )(6.9162 × 106 ) = h 52.436400 × 109 m 2 /s τ = 95.5 min = 5.7281 × 103 s 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. 471 PROBLEM 12.112 The Clementine spacecraft described an elliptic orbit of minimum altitude hA = 400 km and a maximum altitude of hB = 2940 km above the surface of the moon. Knowing that the radius of the moon is 1737 km and that the mass of the moon is 0.01230 times the mass of the earth, determine the periodic time of the spacecraft. SOLUTION For earth, R = 6370 km = 6.370 × 106 m GM = gR 2 = (9.81)(6.370 × 106 )2 = 398.06 × 1012 m3 /s 2 For moon, GM = (0.01230)(398.06 × 1012 ) = 4.896 × 1012 m3 /s 2 rA = 1737 + 400 = 2137 km = 2.137 × 106 m rB = 1737 + 2940 = 4677 km = 4.677 × 106 m Using Eq. (12.39), 1 GM = 2 + C cos θ A rA h But θ B = θ A + 180°, so that cos θ A = − cos θ B . Adding, and 1 GM = 2 + C cos θ B . rB h 1 1 r +r 2GM + = A B = 2 rA rB rA rB hAB hAB = 2GMrA rB (2)(4.896 × 1012 )(2.137 × 106 )(4.677 × 106 ) = rA + rB 6.814 × 106 = 3.78983 × 109 m 2 /s 1 a = ( rA + rB ) = 3.402 × 106 m 2 b = rA rB = 3.16145 × 106 m Periodic time. τ= 2π ab 2π (3.402 × 106 )(3.16145 × 106 ) = = 17.831 × 103 s hAB 3.78983 × 109 τ = 4.95 h 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. 472 PROBLEM 12.113 Determine the time needed for the space probe of Problem 12.100 to travel from B to C. SOLUTION From the solution to Problem 12.100, we have (v A )par = 10,131.4 m/s (v A )circ = and 1 2 (v A ) par = 7164.0 m/s rA = (6052 + 280) km = 6332 km Also, For the parabolic trajectory BA, we have 1 GM v = 2 (1 + ε cos θ ) r hBA [Eq. (12.39′)] where ε = 1. Now at A, θ = 0: 1 GM v = 2 (1 + 1) rA hBA or rA = at B, θ = −90°: 1 GM v = 2 (1 + 0) rB hBA or rB = 2 hBA 2GM v 2 hBA GM v rB = 2rA As the probe travels from B to A, the area swept out is the semiparabolic area defined by Vertex A and Point B. Thus, (Area swept out) BA = ABA = Now 2 4 rA rB = rA2 3 3 dA 1 = h dt 2 where h = constant 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. 473 PROBLEM 12.113 (Continued) Then A= 2A 1 ht or tBA = BA 2 hBA tBA = = 2 × 43 rA2 rAv A hBA = rAv A = 8 rA 3 vA 8 6332 × 103 m 3 10,131.4 m/s = 1666.63 s For the circular trajectory AC, t AC = Finally, π r 2 A (v A )circ = π 6332 × 103 m 2 7164.0 m/s = 1388.37 s t BC = t BA + t AC = (1666.63 + 1388.37) s =3055.0 s t BC = 50 min 55 s or 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. 474 PROBLEM 12.114 A space probe is describing a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O. As the probe passes through Point A, its velocity is reduced from v0 to β v0, where β < 1, to place the probe on a crash trajectory. Express in terms of n and β the angle AOB, where B denotes the point of impact of the probe on the planet. SOLUTION r0 = rA = nR For the circular orbit, v0 = GM GM = r0 nR The crash trajectory is elliptic. v A = β v0 = β 2 GM nR h = rAv A = nRv A = β 2 nGMR GM 1 = 2 h2 β nR 1 GM 1 + ε cos θ = 2 (1 + ε cos θ ) = r β 2 nR h At Point A, θ = 180° 1 1 1− ε = = 2 rA nR β nR or β 2 = 1 − ε or ε = 1 − β 2 At impact Point B, θ = π − φ 1 1 = rB R 1 1 + ε cos (π − φ ) 1 − ε cos φ = = R β 2 nR β 2 nR ε cos φ = 1 − nβ 2 or cos φ = 1 − nβ 2 ε = 1 − nβ 2 1− β 2 φ = cos −1[(1 − nβ 2 ) /(1 − β 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. 475 PROBLEM 12.115 A long-range ballistic trajectory between Points A and B on the earth’s surface consists of a portion of an ellipse with the apogee at Point C. Knowing that Point C is 1500 km above the surface of the earth and the range Rφ of the trajectory is 6000 km, determine (a) the velocity of the projectile at C, (b) the eccentricity ε of the trajectory. SOLUTION For earth, R = 6370 km = 6.37 × 106 m GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 For the trajectory, rC = 6370 + 1500 = 7870 km = 7.87 × 106 m rA = rB = R = 6.37 × 106 m, rC 7870 = = 1.23548 rA 6370 Range A to B: s AB = 6000 km = 6.00 × 106 m s AB 6.00 × 106 = = 0.94192 rad = 53.968° R 6.37 × 106 ϕ = For an elliptic trajectory, At A, θ = 180° − At C, θ = 180° , ϕ 2 1 GM = 2 (1 + ε cos θ ) r h 1 GM = 2 (1 + ε cos153.016°) rA h = 153.016°, 1 GM = 2 (1 − ε ) rC h (1) (2) Dividing Eq. (1) by Eq. (2), rC 1 + ε cos153.016° = = 1.23548 rA 1−ε ε = 1.23548 − 1 = 0.68384 1.23548 + cos153.016° 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. 476 PROBLEM 12.115 (Continued) From Eq. (2), h = h= GM (1 − ε )rC (398.06 × 1012 )(0.31616)(7.87 × 106 ) = 31.471 × 109 m 2 /s (a) Velocity at C. (b) vC = h 31.471 × 109 = = 4.00 × 103 m/s 6 rC 7.87 × 10 vC = 4 km/s ε = 0.684 Eccentricity of trajectory. 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. 477 PROBLEM 12.116 A space shuttle is describing a circular orbit at an altitude of 563 km above the surface of the earth. As it passes through Point A, it fires its engine for a short interval of time to reduce its speed by 152 m/s and begin its descent toward the earth. Determine the angle AOB so that the altitude of the shuttle at Point B is 121 km. (Hint: Point A is the apogee of the elliptic descent trajectory.) SOLUTION GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 rA = 6370 + 563 = 6933 km = 6.933 × 106 m rB = 6370 + 121 = 6491 km = 6.491 × 106 m For the circular orbit through Point A, vcirc = GM = rA 398.06 × 1012 = 7.5773 × 103 m/s 6.933 × 106 For the descent trajectory, v A = vcirc + Δv = 7.5773 × 103 − 152 = 7.4253 × 103 m/s h = rAv A = (6.933 × 106 )(7.4253 × 103 ) = 51.4795 × 109 m 2 /s 1 GM = 2 (1 + ε cos θ ) r h At Point A, θ = 180°, r = rA 1 GM = 2 (1 − ε ) rA h 1−ε = h2 (51.4795 × 109 )2 = = 0.96028 GM rA (398.06 × 1012 )(6.933 × 106 ) ε = 0.03972 1 GM = 2 (1 + ε cos θ B ) rB h 1 + ε cos θ B = h2 (51.4795 × 109 ) 2 = = 1.02567 GM rB (398.06 × 1012 )(6.491 × 106 ) cosθ B = θ B = 49.7° 1.02567 − 1 ε = 0.6463 AOB = 180° − θ B = 130.3° AOB = 130.3° 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. 478 PROBLEM 12.117 As a spacecraft approaches the planet Jupiter, it releases a probe which is to enter the planet’s atmosphere at Point B at an altitude of 280 mi above the surface of the planet. The trajectory of the probe is a hyperbola of eccentricity ε = 1.031. Knowing that the radius and the mass of Jupiter are 44423 mi and 1.30 × 1026 slug, respectively, and that the velocity vB of the probe at B forms an angle of 82.9° with the direction of OA, determine (a) the angle AOB, (b) the speed vB of the probe at B. SOLUTION First we note (a) rB = (44.423 × 103 + 280) mi = 44.703 × 103 mi We have 1 GM j = 2 (1 + ε cos θ ) r h At A, θ = 0: 1 GM j = 2 (1 + ε ) rA h or At B, θ = θ B = AOB : or Then or [Eq. (12.39′)] h2 = rA (1 + ε ) GM j 1 GM j = 2 (1 + ε cos θ B ) rB h h2 = rB (1 + ε cos θ B ) GM j rA (1 + ε ) = rB (1 + ε cos θ B ) cos θ B = = 1 rA (1 + ε ) − 1 ε rB 1 44.0 × 103 mi (1 + 1.031) − 1 3 1.031 44.703 × 10 mi = 0.96902 or θ B = 14.2988° AOB = 14.30° 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. 479 PROBLEM 12.117 (Continued) (b) From above where Then h 2 = GM j rB (1 + ε cos θ B ) 1 | rB × mv B | = rB vB sin φ m φ = (θ B + 82.9°) = 97.1988° h= (rB vB sin φ )2 = GM j rB (1 + ε cos θ B ) or 1/2 vB = 1 GM j (1 + ε cos θ B ) sin φ rB 1/2 −9 4 4 26 1 34.4 × 10 ft /lb ⋅ s × (1.30 × 10 slug) = × [1 + (1.031)(0.96902)] 6 sin 97.1988° 236.03 × 10 ft vB = 196.2 ft/s or 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. 480 PROBLEM 12.118 A satellite describes an elliptic orbit about a planet. Denoting by r0 and r1 the distances corresponding, respectively, to the perigee and apogee of the orbit, show that the curvature of the orbit at each of these two points can be expressed as 1 ρ = 1 1 1 + 2 r0 r1 SOLUTION Using Eq. (12.39), 1 GM = 2 + C cos θ A rA h and 1 GM = 2 + C cos θ B . rB h But θ B = θ A + 180°, so that Adding, cos θ A = − cos θ B 1 1 2GM + = 2 rA rB h At Points A and B, the radial direction is normal to the path. an = But Fn = 1 ρ = v2 ρ = h2 r 2ρ GMm mh 2 ma = = n r2 r 2ρ GM 1 1 1 = + 2 rA rB h2 1 ρ = 1 1 1 + 2 r0 r1 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. 481 PROBLEM 12.119 (a) Express the eccentricity ε of the elliptic orbit described by a satellite about a planet in terms of the distances r0 and r1 corresponding, respectively, to the perigee and apogee of the orbit. (b) Use the result obtained in Part a and the data given in Problem 12.109, where RE = 149.6 × 106 km, to determine the approximate maximum distance from the sun reached by comet Hyakutake. SOLUTION (a) We have 1 GM = 2 (1 + ε cos θ ) r h At A, θ = 0: 1 GM = 2 (1 + ε ) r0 h or At B, θ = 180°: or Then Eq. (12.39′) h2 = r0 (1 + ε ) GM 1 GM = 2 (1 − ε ) r1 h h2 = r1 (1 − ε ) GM r0 (1 + ε ) = r1 (1 − ε ) ε= or (b) 1+ ε r0 1−ε From above, r1 = where r0 = 0.230 RE Then r1 = 1 + 0.999887 × 0.230(149.6 × 109 m) 1 − 0.999887 r1 = 609 × 1012 m or Note: r1 = 4070 RE r1 − r0 r1 + r0 or r1 = 0.064 lightyears. 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. 482 PROBLEM 12.120 Derive Kepler’s third law of planetary motion from Eqs. (12.39) and (12.45). SOLUTION For an ellipse, 2a = rA + rB Using Eq. (12.39), 1 GM = 2 + C cos θ A rA h and 1 GM = 2 + C cos θ B . rB h But θ B = θ A + 180°, so that Adding, and b = rA rB cos θ A = − cos θ B . 1 1 rA + rB 2a 2GM + = = 2 = 2 rA rB rA rB b h h=b GM a By Eq. (12.45), τ= τ2 = 2π ab 2π ab a 2π a3/ 2 = = h b GM GM 4π 2 a3 GM For Orbits 1 and 2 about the same large mass, and τ12 = 4π 2 a13 GM τ 22 = 4π 2 a23 GM 2 3 τ1 a1 = τ 2 a2 Forming the ratio, 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. 483 PROBLEM 12.121 Show that the angular momentum per unit mass h of a satellite describing an elliptic orbit of semimajor axis a and eccentricity ε about a planet of mass M can be expressed as h = GMa(1 − ε 2 ) SOLUTION By Eq. (12.39′), 1 GM = 2 (1 + ε cos θ ) r h At A, θ = 0°: 1 GM = 2 = (1 + ε ) rA h or rA = h2 GM (1 + ε ) At B, θ = 180°: 1 GM = 2 = (1 − ε ) rB h or rB = h2 GM (1 − ε ) h2 1 2h 2 1 = + = GM 1 + ε 1 − ε GM (1 − ε 2 ) Adding, rA + rB = But for an ellipse, rA + rB = 2a 2a = 2h2 GM (1 − ε 2 ) h = GMa(1 − ε 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. 484 PROBLEM 12.122 In the braking test of a sports car its velocity is reduced from 70 mi/h to zero in a distance of 170 ft with slipping impending. Knowing that the coefficient of kinetic friction is 80 percent of the coefficient of static friction, determine (a) the coefficient of static friction, (b) the stopping distance for the same initial velocity if the car skids. Ignore air resistance and rolling resistance. SOLUTION (a) Coefficient of static friction. ΣFy = 0: N −W = 0 N =W v0 = 70 mi/h = 102.667 ft/s v 2 v02 − = at (s − s0 ) 2 2 at = v 2 − v02 0 − (102.667) 2 = = − 31.001 ft/s 2 2(s − s0 ) (2)(170) For braking without skidding μ = μ s , so that μ s N = m | at | ΣFt = mat : − μ s N = mat μs = − (b) mat a 31.001 = − t = W g 32.2 μ s = 0.963 Stopping distance with skidding. Use μ = μk = (0.80)(0.963) = 0.770 ΣF = mat : μk N = −mat at = − μk N m = − μ k g = − 24.801 ft/s 2 Since acceleration is constant, (s − s0 ) = v 2 − v02 0 − (102.667) 2 = 2at (2)(− 24.801) s − s0 = 212 ft 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. 485 PROBLEM 12.123 A bucket is attached to a rope of length L = 1.2 m and is made to revolve in a horizontal circle. Drops of water leaking from the bucket fall and strike the floor along the perimeter of a circle of radius a. Determine the radius a when θ = 30°. SOLUTION Initial velocity of drop = velocity of bucket ΣFy = 0: T cos 30° = mg ΣFx = max : Divide (2) by (1): (1) T sin 30° = man tan 30° = (2) an v2 = g pg Thus v 2 = ρ g tan 30° But ρ = L sin 30° = (1.2 m) sin 30° = 0.6 m Thus v 2 = 0.6(9.81) tan 30° = 3.398 m 2 /s 2 v = 1.843 m/s Assuming the bucket to rotate clockwise (when viewed from above), and using the axes shown, we find that the components of the initial velocity of the drop are (v0 ) x = 0, (v0 ) y = 0, (v0 ) 2 = 1.843 m/s Free fall of drop y = y0 + (v0 ) y t − 1 2 gt 2 y = y0 − 1 2 gt 2 When drop strikes floor: y =0 y0 − 1 2 gt = 0 2 But y0 = 2L − L cos 30° = 2(1.2) − 1.2 cos 30° = 1.361 m Thus 1.361 − 1 (9.81) t 2 = 0 2 t = 0.5275 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. 486 PROBLEM 12.123 (Continued) Projection on horizontal floor (uniform motion) x = x0 + (v0 ) z t = L sin 30° + 0, x = 0.6 m z = z0 + (v0 ) z t = 0 + 1.843(0.527) = 0.971 m Radius of circle: a = a = x2 + z 2 (0.6)2 + (0.971)2 a = 1.141 m Note: The drop travels in a vertical plane parallel to the yz plane. 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. 487 PROBLEM 12.124 A 12-lb block B rests as shown on the upper surface of a 30-lb wedge A. Neglecting friction, determine immediately after the system is released from rest (a) the acceleration of A, (b) the acceleration of B relative to A. SOLUTION Acceleration vectors: a A = aA 30°, a B/A = aB/A a B = a A + a B/ A Block B: ΣFx = max : mB aB /A − mB a A cos 30° = 0 aB/A = a A cos 30° (1) ΣFy = ma y : N AB − WB = − mB a A sin 30° N AB = WB − (WB sin 30°) Block A: aA g (2) ΣF = ma : WA sin 30° + N AB sin 30° = WA WA sin 30° + WB sin 30° − (WB sin 2 30°) aA = aA g aA a = WA A g g (WA + WB ) sin 30° (30 + 12)sin 30° g= (32.2) = 20.49 ft/s 2 2 2 WA + WB sin 30° 30 + 12sin 30° a A = 20.49 ft/s 2 (a) 30° aB/ A = (20.49) cos 30° = 17.75 ft/s 2 a B/A = 17.75 ft/s 2 (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. 488 PROBLEM 12.125 A 500-lb crate B is suspended from a cable attached to a 40-lb trolley A which rides on an inclined I-beam as shown. Knowing that at the instant shown the trolley has an acceleration of 1.2 ft/s2 up and to the right, determine (a) the acceleration of B relative to A, (b) the tension in cable CD. SOLUTION (a) First we note: a B = a A + a B/A , where a B/A is directed perpendicular to cable AB. ΣFx = mB ax : 0 = −mB ax + mB a A cos 25° B: aB/A = (1.2 ft/s 2 ) cos 25° or a B/A = 1.088 ft/s 2 or (b) For crate B ΣFy = mB a y : TAB − WB = or WB a A sin 25° g A: (1.2 ft/s 2 )sin 25° TAB = (500 lb) 1 + 32.2 ft/s 2 = 507.87 lb For trolley A ΣFx = m A a A : TCD − TAB sin 25° − WA sin 25° = or WA aA g 1.2 ft/s 2 TCD = (507.87 lb)sin 25° + (40 lb) sin 25° + 32.2 ft/s 2 TCD = 233 lb or 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. 489 PROBLEM 12.126 The roller-coaster track shown is contained in a vertical plane. The portion of track between A and B is straight and horizontal, while the portions to the left of A and to the right of B have radii of curvature as indicated. A car is traveling at a speed of 72 km/h when the brakes are suddenly applied, causing the wheels of the car to slide on the track (μ k = 0.25). Determine the initial deceleration of the car if the brakes are applied as the car (a) has almost reached A, (b) is traveling between A and B, (c) has just passed B. SOLUTION v = 72 km/h = 20 m/s (a) Almost reached Point A. ρ = 30 m an = v2 ρ = (20) 2 = 13.333 m/s 2 30 ΣFy = ma y : N R + N F − mg = man N R + N F = m( g + an ) F = μk ( N R + N F ) = μ k m( g + an ) ΣFx = max : − F = mat at = − F = − μ k ( g + an ) m | at | = μk ( g + an ) = 0.25(9.81 + 13.33) (b) Between A and B. | at | = 5.79 m/s 2 ρ =∞ an = 0 | at | = μk g = (0.25)(9.81) (c) Just passed Point B. | at | = 2.45 m/s 2 ρ = 45 m an = v2 ρ = (20)2 = 8.8889 m/s 2 45 ΣFy = ma y : N R + N F − mg = − man 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. 490 PROBLEM 12.126 (Continued) or N R + N F = m ( g − an ) F = μ k ( N R + N F ) = μ k m ( g − an ) ΣFx = max : − F = mat at = − F = − μ k ( g − an ) m | at | = μk ( g − an ) = (0.25)(9.81 − 8.8889) | at | = 0.230 m/s 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. 491 PROBLEM 12.127 The 100-g pin B slides along the slot in the rotating arm OC and along the slot DE which is cut in a fixed horizontal plate. Neglecting friction and knowing that rod OC rotates at the constant rate θ0 = 12 rad/s, determine for any given value of θ (a) the radial and transverse components of the resultant force F exerted on pin B, (b) the forces P and Q exerted on pin B by rod OC and the wall of slot DE, respectively. SOLUTION Kinematics From the drawing of the system, we have r= Then 0.2 m cos θ sin θ r = 0.2 θ m/s 2 cos θ θ = 12 rad/s θ = 0 and cos θ (cos θ ) − sin θ (−2cos θ sin θ ) 2 θ cos 4 θ 1 + sin 2θ 2 = 0.2 θ m/s 2 3 cos θ r = 0.2 2 Substituting for θ sin θ sin θ (12) = 2.4 m/s 2 cos θ cos 2 θ 1 + sin 2 θ 1 + sin 2 θ (12)2 = 28.8 r = 0.2 3 cos θ cos3 θ r = 0.2 Now 2 m/s ar = r − rθ 2 1 + sin 2 θ 0.2 = 28.8 − cos3 θ cos θ sin 2 θ 2 = 57.6 3 m/s θ cos 2 (12) 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. 492 PROBLEM 12.127 (Continued) aθ = rθ + 2rθ and sin θ = 0 + 2 2.4 (12) cos 2 θ sin θ 2 = 57.6 2 m/s θ cos Kinetics (a) We have sin 2 θ Fr = mB ar = (0.1 kg) 57.6 cos3 θ 2 m/s Fr = (5.76 N) tan 2 θ sec θ or and sin θ Fθ = mB aθ = (0.1 kg) 57.6 cos 2 θ 2 m/s Fθ = (5.76 N) tan θ sec θ or (b) Now or ΣFy : Fθ cos θ + Fr sin θ = P cos θ P = 5.76 tan θ sec θ + (5.76 tan 2 θ sec θ ) tan θ P = (5.76 N) tan θ sec2 θ or θ ΣFr : Fr = Q cos θ or Q = (5.76 tan 2 θ sec θ ) 1 cos θ Q = (5.76 N) tan 2 θ sec2 θ or 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. 493 PROBLEM 12.128 A small 200-g collar C can slide on a semicircular rod which is made to rotate about the vertical AB at the constant rate of 6 rad/s. Determine the minimum required value of the coefficient of static friction between the collar and the rod if the collar is not to slide when (a) θ = 90°, (b) θ = 75°, (c) θ = 45°. Indicate in each case the direction of the impending motion. SOLUTION vC = (r sin θ )φAB First note = (0.6 m)(6 rad/s)sin θ = (3.6 m/s)sin θ (a) With θ = 90°, vC = 3.6 m/s ΣFy = 0: F − WC = 0 or F = mC g Now F = μs N or N= 1 μs ΣFn = mC an : N = mC 1 or or μs mC g = mC μs = mC g vC2 r vC2 r gr (9.81 m/s 2 )(0.6 m) = vC2 (3.6 m/s) 2 ( μ s ) min = 0.454 or The direction of the impending motion is downward. 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. 494 PROBLEM 12.128 (Continued) (b) and (c) First observe that for an arbitrary value of θ, it is not known whether the impending motion will be upward or downward. To consider both possibilities for each value of θ, let Fdown correspond to impending motion downward, Fup correspond to impending motion upward, then with the “top sign” corresponding to Fdown, we have ΣFy = 0: N cos θ ± F sin θ − WC = 0 F = μs N Now N cos θ ± μ s N sin θ − mC g = 0 Then or N= mC g cos θ ± μs sin θ and F= μ s mC g cos θ ± μ s sin θ ΣFn = mC an : N sin θ F cos θ = mC vC2 ρ ρ = r sin θ Substituting for N and F mC g μs mC g v2 sin θ cos θ = mC C cos θ ± μs sin θ cos θ ± μ s sin θ r sin θ μs vC2 tan θ = 1 ± μs tan θ 1 ± μs tan θ gr sin θ or μs = ± or 1+ vC2 gr sin θ vC2 gr sin θ tan θ vC2 [(3.6 m/s) sin θ ]2 = = 2.2018sin θ gr sin θ (9.81 m/s 2 )(0.6 m)sin θ Now Then (b) tan θ − μs = ± tan θ − 2.2018 sin θ 1 + 2.2018sin θ tan θ μs = ± tan 75° − 2.2018sin 75° = ± 0.1796 1 + 2.2018sin 75° tan 75° θ = 75° 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. 495 PROBLEM 12.128 (Continued) Then downward: μs = + 0.1796 upward: μs < 0 not possible ( μ s )min = 0.1796 The direction of the impending motion is downward. (c) θ = 45° μs = ± tan 45° − 2.2018sin 45° = ± (− 0.218) 1 + 2.2018sin 45° tan 45° Then downward: μs < 0 upward: μs = 0.218 not possible ( μ s ) min = 0.218 The direction of the impending motion is upward. Note: When or tan θ − 2.2018sin θ = 0 θ = 62.988°, μs = 0. Thus, for this value of θ , friction is not necessary to prevent the collar from sliding on the rod. 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. 496 PROBLEM 12.129 Telemetry technology is used to quantify kinematic values of a 200-kg roller coaster cart as it passes overhead. According to the system, r = 25 m, r = −10 m/s, r = −2 m/s 2 , θ = 90°, θ = −0.4 rad/s, θ = −0.32 rad/s 2. At this instant, determine (a) the normal force between the cart and the track, (b) the radius of curvature of the track. SOLUTION Find the acceleration and velocity using polar coordinates. vr = r = −10 m/s vθ = rθ = (25 m)( − 0.4 rad/s) = −10 m/s So the tangential direction is 45° and v = 10 2 m/s. ar = r − rθ 2 = −2 m/s − (25 m)( − 0.4 rad/s)2 = −6 m/s 2 aθ = rθ + 2rθ = (25 m)(−0.32) rad/s 2 | +2(−10 m/s)( − 0.4 rad/s) =0 So the acceleration is vertical and downward. (a) To find the normal force use Newton’s second law. y-direction N − mg sin 45° = −ma cos 45° N = m( g sin 45° − a cos 45°) = (200 kg)(9.81) m/s 2 − 6 m/s 2 )(0.70711) = 538.815 N N = 539 N (b) Radius l curvature of the track. an = ρ = v2 ρ v2 (10 2)2 = an 6 cos 45° ρ = 47.1 m 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. 497 PROBLEM 12.130 The radius of the orbit of a moon of a given planet is equal to twice the radius of that planet. Denoting by ρ the mean density of the planet, show that the time required by the moon to complete one full revolution about the planet is (24 π /G ρ )1/2 , where G is the constant of gravitation. SOLUTION For gravitational force and a circular orbit, Fr = GMm mv 2 = r r2 or v= GM r Let τ be the periodic time to complete one orbit. vτ = 2π r Solving for τ, But τ= τ= GM = 2π r r hence, GM = 2 2π r 3/2 GM 4 M = π R3 ρ , 3 Then τ or 3π r G ρ R π 3 G ρ R3/2 3/2 Using r = 2R as a given leads to τ = 23/2 3π = Gρ 24π Gρ τ = (24π /G ρ )1/2 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. 498 PROBLEM 12.131 At engine burnout on a mission, a shuttle had reached Point A at an altitude of 40 mi above the surface of the earth and had a horizontal velocity v0. Knowing that its first orbit was elliptic and that the shuttle was transferred to a circular orbit as it passed through Point B at an altitude of 170 mi, determine (a) the time needed for the shuttle to travel from A to B on its original elliptic orbit, (b) the periodic time of the shuttle on its final circular orbit. SOLUTION For Earth, R = 3960 mi = 20.909 × 106 ft, g = 32.2 ft/s 2 GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 (a) For the elliptic orbit, rA = 3960 + 40 = 4000 mi = 21.12 × 106 ft rB = 3960 + 170 = 4130 mi = 21.8064 × 106 ft 1 (rA + rB ) = 21.5032 × 106 ft 2 b = rA rB = 21.4605 × 106 ft a= Using Eq. 12.39, 1 GM = 2 + C cos θ A rA h and 1 GM = 2 + C cos θ B rB rB But θ B = θ A + 180°, so that cos θ A = − cos θ B Adding, 1 1 rA + rB 2a 2GM + = = 2 = 2 rA rB rA rB b h or h= Periodic time. τ= τ= GMb 2 a 2π ab 2π ab a 2π a3/ 2 = = h GM GMb2 2π (21.5032 × 106 )3/ 2 14.077 × 1015 = 5280.6 s = 1.4668 h The time to travel from A to B is one half the periodic time τ AB = 0.7334 h τ AB = 44.0 min 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. 499 PROBLEM 12.131 (Continued) (b) For the circular orbit, a = b = rB = 21.8064 × 106 ft τ circ = 2π a3/2 GM = 2π (21.8064 × 106 )3/2 14.077 × 1015 τ circ = 1.498 h = 5393 s τ circ = 89.9 min 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. 500 PROBLEM 12.132 It was observed that as the Galileo spacecraft reached the point on its trajectory closest to Io, a moon of the planet Jupiter, it was at a distance of 1750 mi from the center of Io and had a velocity of 49.4 × 103 ft/s. Knowing that the mass of Io is 0.01496 times the mass of the earth, determine the eccentricity of the trajectory of the spacecraft as it approached Io. SOLUTION First note r0 = 1750 mi = 9.24 × 106 ft Rearth = 3960 mi = 20.9088 × 106 ft We have 1 GM = 2 (1 + ε cos θ ) r h At Point O, r = r0 , θ = 0, h = h0 = r0 v0 Also, Eq. (12.39) GM Io = G (0.01496 M earth ) 2 = 0.01496gRearth using Eq. (12.30). Then 2 1 0.01496 gRearth = (1 + ε ) r0 (r0 v0 ) 2 or ε= = r0 v02 2 0.01496 gRearth −1 (9.24 × 106 ft)(49.4 × 103 ft/s) 2 −1 0.01496(32.2 ft/s 2 )(20.9088 × 106 ft) 2 ε = 106.1 or 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. 501 PROBLEM 12.133* Disk A rotates in a horizontal plane about a vertical axis at the constant rate θ0 = 10 rad/s. Slider B has mass 1 kg and moves in a frictionless slot cut in the disk. The slider is attached to a spring of constant k, which is undeformed when r = 0. Knowing that the slider is released with no radial velocity in the position r = 500 mm, determine the position of the slider and the horizontal force exerted on it by the disk at t = 0.1 s for (a) k = 100 N/m, (b) k = 200 N/m. SOLUTION First we note r = 0, xsp = 0 Fsp = kr when r0 = 500 mm = 0.5 m and θ = θ0 = 12 rad/s then θ = 0 ΣFr = mB ar : − Fsp = mB ( r − rθ02 ) k − θ02 r = 0 r + mB ΣFθ = mB aθ : FA = mB (0 + 2rθ0 ) (a) (1) (2) k = 100 N/m Substituting the given values into Eq. (1) 100 N/m r + − (10 rad/s)2 r = 0 1 kg r =0 Then dr r = 0 and at t = 0, r = 0 : = dt r 0 dr = 0.1 (0) dt 0 r = 0 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. 502 PROBLEM 12.133* (Continued) and dr = r = 0 and at t = 0, r0 = 0.5 m dt r r0 dr = 0.1 (0) dt 0 r = r0 r = 0.5 m Note: r = 0 implies that the slider remains at its initial radial position. With r = 0, Eq. (2) implies FH = 0 (b) k = 200 N/m Substituting the given values into Eq. (1) 200 r+ − (10 rad/s)2 r = 0 1 kg r + 100 r = 0 d (r) dt Now r= Then r = vr so that d dr d d = = vr dt dt dr dr dvr dr dvr + 100r = 0 dr vr At t = 0, vr = 0, r = r0 : r = vr vr 0 vr d vr = −100 r r dr r0 vr2 = −100(r 2 − r02 ) vr = 10 r02 − r 2 vr = Now At t = 0, r = r0 : r r0 dr r02 −r 2 = dr = 10 r02 − r 2 dt t 10 dt = 10t 0 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. 503 PROBLEM 12.133* (Continued) r = r0 sin φ , Let sin −1 ( r/r0 ) π Then /2 dr = r0 cos φ dφ r0 cos φ dφ r02 − r02 sin 2 φ sin −1 ( r/r0 ) π /2 r sin −1 r0 = 10t dφ = 10t π − = 10 t 2 π r = r0 sin 10 t + = r0 cos10 t = (0.5 ft) cos10 t 2 r = −(5 m/s) sin10 t Then Finally, at t = 0.1 s: r = (0.5 ft) cos (10 × 0.1) r = 0.270 m Eq. (2) FH = 1 kg × 2 × [ −(5 ft/s)sin (10 × 0.1)] (10 rad/s) FH = −84.1 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. 504 CHAPTER 13 PROBLEM 13.CQ1 Block A is traveling with a speed v0 on a smooth surface when the surface suddenly becomes rough with a coefficient of friction of μ causing the block to stop after a distance d. If block A were traveling twice as fast, that is, at a speed 2v0, how far will it travel on the rough surface before stopping? (a) d/2 (b) d (c) 2d (d) 2d (e) 4d SOLUTION Answer: (e) 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. 507 PROBLEM 13.1 A 400-kg satellite was placed in a circular orbit 1500 km above the surface of the earth. At this elevation the acceleration of gravity is 6.43 m/s2. Determine the kinetic energy of the satellite, knowing that its orbital speed is 25.6 × 103 km/h. SOLUTION Mass of satellite: m = 400 kg Velocity: v = 25.6 × 103 km/h = 7.111 × 103 m/s Kinetic energy: T= 1 2 1 mv = (400 kg)(7.111 × 103 m/s) 2 2 2 T = 10.113 × 109 J T = 10.11 GJ Note: Acceleration of gravity has no effect on the mass of the satellite. 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. 508 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 W lb 1 lb = = 0.031056 lb ⋅ s 2 /ft g 32.2 ft/s 2 Mass of stone: m= Initial kinetic energy: T1 = 0 (a) (rest) Kinetic energy at ground strike: T2 = 1 2 1 mv2 = (0.031056)(95)2 = 140.14 ft ⋅ lb 2 2 T2 = 140.1 ft ⋅ lb Use work and energy: where T1 + U1→ 2 = T2 U1→2 = wh = mgh 0 + mgh = h= (b) On the moon: 1 2 mv2 2 v22 (95) 2 = 2 g (2)(32.2) h = 140.1 ft g = 5.31 ft/s 2 T2 = 140.1 ft ⋅ lb T1 and T2 will be the same, hence h= v22 (95) 2 = 2 g (2)(5.31) h = 850 ft 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 13.3 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 1 lb W = (5.1 oz) = 0.31875 lb 16 oz Mass of baseball: m= (a) W 0.31875 lb = = 0.009899 lb ⋅ s 2 /ft 2 g 32.2 ft/s Kinetic energy immediately after hit. v = v0 = 140 ft/s T1 = (b) 1 2 1 mv = (0.009899)(140) 2 2 2 T1 = 97.0 ft ⋅ lb Kinetic energy at maximum height: v = v0 cos 40° = 140cos 40° = 107.246 ft/s T2 = Principle of work and energy: 1 2 1 mv = (0.009899)(107.246) 2 2 2 T2 = 56.9 ft ⋅ lb T1 + U1→ 2 = T2 U1→2 = T2 − T1 = −40.082 ft ⋅ lb Work of weight: U1→2 = −Wd Maximum height above impact point. d= (c) T2 − T1 −40.082 ft ⋅ lb = = 125.7 ft −W −0.31875 lb 125.7 ft Maximum height above ground: h = 125.7 ft + 2 ft h = 127.7 ft 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 13.4 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: R = 6370 km Radius of orbit: r = R + h = 6370 + 35800 = 42170 km = 42.170 × 106 m Time one revolution: t = 23 h + 56 min t = (23 h)(3600 s/h) + (56 min)(60 s/min) = 86.160 × 103 s Speed: v= 2π r 2π (42.170 × 106 ) = = 3075.2 m/s t 86.160 × 103 Kinetic energy: T= 1 2 mv 2 T= 1 (500 kg)(3075.2 m/s) 2 = 2.3643 × 109 J 2 T = 2.36 GJ 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 13.5 In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The bucket is to swing no more than 10 ft horizontally when the crane is brought to a sudden stop. Determine the maximum allowable speed v of the crane. SOLUTION Let position be the position with bucket B directly below A, and position be that of maximum swing where d = 10 ft. Let L be the length AB. T1 = Kinetic energies: 1 2 mv , T2 = 0 2 U1→2 = −Wh = − mgh Work of the weight: where h is the vertical projection of position above position From geometry (see figure), y = L2 − d 2 h= L− y = L − L2 − d 2 = 30 − (30) 2 − (10)2 = 1.7157 ft Principle of work and energy: T1 + U1→ 2 = T2 1 2 mv − mgh = 0 2 v 2 = 2 gh = (2)(32.2 ft/s 2 )(1.7157 ft) = 110.49 ft 2 /s 2 v = 10.51 ft/s 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. 512 PROBLEM 13.6 In an ore-mixing operation, a bucket full of ore is suspended from a traveling crane which moves along a stationary bridge. The crane is traveling at a speed of 10 ft/s when it is brought to a sudden stop. Determine the maximum horizontal distance through which the bucket will swing. SOLUTION Let position be the position with bucket B directly below A, and position be that of maximum swing where the horizontal distance is d. Let L be the length AB. T1 = Kinetic energies: 1 2 mv , T2 = 0 2 U1→2 = −Wh = − mgh Work of the weight: where h is the vertical projection of position above position . Principle of work and energy: T1 + U1→ 2 = T2 1 2 mv − mgh = 0 2 v2 (10 ft/s) 2 h= = = 1.5528 ft 2 g (2)(32.2 ft/s 2 ) From geometry (see figure), d = L2 − y 2 = L2 − ( L − h) 2 = (30) 2 − (30 − 1.5528) 2 = 9.53 ft d = 9.53 ft 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. 513 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) N = 0.60W = 0.60 mg μs = 0.75 Front wheel drive: Maximum friction force without slipping: F = μ s N = (0.75)(0.60W ) = 0.45mg U1→2 = Fd = 0.45 mgd T1 = 0, Principle of work and energy: T2 = 1 2 mv2 2 T1 + U1→ 2 = T2 1 2 mv2 2 v22 = (2)(0.45 gd ) = (2)(0.45)(9.81 m/s 2 )(110 m) = 971.19 m 2 /s 2 0 + 0.45mgd = v2 = 31.164 m/s (b) v2 = 112.2 km/h N = 0.40W = 0.40mg μs = 0.75 Rear wheel drive: Maximum friction force without slipping: F = μ s N = (0.75)(0.40W ) = 0.30mg U1→2 = Fd = 0.30mgd T1 = 0, Principle of work and energy: T2 = 1 2 mv2 2 T1 + U1→ 2 = T2 1 mv22 2 v22 = (2)(0.30) gd = (2)(0.30)(9.81 m/s 2 )(110 m) = 647.46 m 2 /s 2 0 + 0.30 mgd = v2 = 25.445 m/s v2 = 91.6 km/h 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. 514 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 F = μ k N = μk W = μ k mg Principle of work and energy: T1 + U1→ 2 = T2 1 2 mv2 2 1 0 + μk mgd = mv22 2 2 v2 = 2 μk gd = (2)(0.6)(9.81 m/s2 )(20 m) = 235.44 m 2 /s 2 0 + Fd = (b) v2 = 15.34 m/s Assume that for the remainder of the race, sliding is impending and N = 0.6 W F = μ s N = μ s (0.6W ) = (0.75)(0.6 mg ) = 0.45 mg Principle of work and energy: T2 + U 2→3 = T3 1 2 1 mv2 + (0.45mg )d ′ = mv32 2 2 v32 = v22 (2)(0.45) gd ′ = 235.44 m 2 /s 2 + (2)(0.45)(9.81 m/s 2 )(400 m − 20 m) = 3590.5 m 2 /s 2 v3 = 59.9 m/s 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 13.9 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: 1 2 1W W mv A = (8 m/s) 2 = 32 TB = 0 2 2 g g = ( −W sin15° − F )d F = μk N = 0.12 N TA = U A−B ΣF = 0 N − W cos15° = 0 N = W cos15° U A−B = −W (sin15° + 0.12 cos15°) d = −Wd (0.3747) TA + U A−B = TB : 32 W − Wd (0.3743) = 0 g d= (b) 32 (9.81)(0.3747) d = 8.71 m Down the plane from B to A: (F reverses direction) U B−A 1W 2 vA TB = 0 2 g = (W sin15° − F )d U B−A = W (sin15° − 0.12cos15°)(8.70 m/s) = 1.245W TA = TB + U B−A = TA 0 + 1.245W = d = 8.71 m/s 1W 2 vA 2 g v A2 = (2)(9.81)(1.245) = 24.43 v A = 4.94 m/s vA = 4.94 m/s 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. 516 PROBLEM 13.10 A 1.4 kg model rocket is launched vertically from rest with a constant thrust of 25 N until the rocket reaches an altitude of 15 m and the thrust ends. Neglecting air resistance, determine (a) the speed of the rocket when the thrust ends, (b) the maximum height reached by the rocket, (c) the speed of the rocket when it returns to the ground. SOLUTION Weight: W = mg = (1.4)(9.81) = 13.734 N (a) T1 = 0 First stage: U1→2 = (25 − 13.734)(15) = 169.0 N ⋅ m T1 + U1→ 2 = T2 (b) T2 = 1 2 mv = U1→2 = 169.0 N ⋅ m 2 v2 = 2U1→2 (2) (169.0) = m 1.4 v2 = 15.54 m/s Unpowered flight to maximum height h: T2 = 169.0 N ⋅ m T3 = 0 U 2→3 = −W (h − 15) T2 + U 2→3 = T3 W (h − 15) = T2 h − 15 = (c) T2 169.0 = W 13.734 h = 27.3 m Falling from maximum height: T3 = 0 T4 = 1 2 mv4 2 U 3→4 = Wh = mgh T3 = U 3→4 = T4 : 0 + mgh = 1 2 mv4 2 v42 = 2 gh = (2)(9.81 m/s 2 )(27.3 m) = 535.6 m 2 /s 2 v4 = 23.1 m/s 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. 517 PROBLEM 13.11 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 μk = 0.25 between the packages and the surface ABC, determine the distance d if the packages are to arrive at C with a velocity of 2 m/s. SOLUTION N AB = mg cos 30° On incline AB: FAB = μk N AB = 0.25 mg cos 30° U A→ B = mgd sin 30° − FAB d = mgd (sin 30° − μk cos 30°) N BC = mg On level surface BC: xBC = 7 m FBC = μk mg U B →C = − μk mg xBC At A, TA = 1 2 mv A 2 and v A = 1 m/s At C, TC = 1 2 mvC 2 and vC = 2 m/s Assume that no energy is lost at the corner B. TA + U A→ B + U B →C = TC Work and energy. 1 2 1 mv A + mgd (sin 30° − μ k cos 30°) − μ k mg xBC = mv02 2 2 Dividing by m and solving for d, vC2 /2 g + μ k xBC − v A2 /2 g d= (sin 30° − μk cos 30°) = (2) 2/(2)(9.81) + (0.25)(7) − (1)2/(2)(9.81) sin 30° − 0.25cos 30° d = 6.71 m 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. 518 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 d = 7.5 m and μk = 0.25 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: N AB = mg cos 30° FAB = μk N AB = 0.25 mg cos 30° UA→ B = mgd sin 30° − FAB d = mgd (sin 30° − μ k cos 30°) On level surface BC: N BC = mg xBC = 7 m FBC = μ k mg U B →C = − μk mg xBC At A, TA = 1 2 mv A 2 and v A = 1 m/s At C, TC = 1 2 mvC 2 and vC = 2 m/s Assume that no energy is lost at the corner B. Work and energy. TA + U A→ B + U B →C = TC 1 2 1 mv A + mgd (sin 30° − μ k cos 30°) − μ k mg xBC = mv02 2 2 Solving for vC2 , vC2 = v 2A + 2 gd (sin 30° − μk cos 30°) − 2 μk g xBC = (1) 2 + (2)(9.81)(7.5)(sin 30° − 0.25cos 30°) − (2)(0.25)(9.81)(7) = 8.3811 m 2/s 2 (b) vC = 2.90 m/s Box on belt: Let xbelt be the distance moves by a package as it slides on the belt. ΣFy = ma y N − mg = 0 N = mg Fx = μk N = μk mg At the end of sliding, v = vbelt = 2 m/s 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 (Continued) Principle of work and energy: 1 2 1 2 mvC − μ k mg xbelt = mvbelt 2 2 2 vC2 − vbelt xbelt = 2 μk g = 8.3811 − (2) 2 (2)(0.25)(9.81) xbelt = 0.893 m 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 13.13 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 μk = 0.40, 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. ΣFy = 0: N − W cos15° = 0 N = W cos15° Box is sliding on AB. F f = μk N = μkW cos15° Distance AB = d = 20 ft Work of gravity force: (U A− B ) g = Wd sin15° − F f d = − μkWd cos15° Work of friction force: Total work U A→ B = Wd (sin15° − μ k sin15°) Kinetic energy: TA = 1W 2 v0 2 g 1W 2 TB = vB 2 g Principle of work and energy: TA + U A→ B = TB 1W 2 1W 2 v0 + Wd (sin15° − μk cos15°) = vB 2 g 2 g v02 = vB2 − 2 gd (sin15° − μk cos15°) = (8) 2 − (2)(32.2)(20)[sin15° − (0.40)(cos15°)] = 228.29 ft 2 /s 2 v 0 = 15.11 ft/s 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. 521 PROBLEM 13.14 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 μk = 0.40, determine the velocity of the conveyor belt if the boxes are to have zero velocity at B. SOLUTION Forces when box is on AB. ΣFy = 0: N − W cos15° = 0 N = W cos15° Box is sliding on AB. F f = μk N = μkW cos15° Distance AB = d = 20 ft Work of gravity force: (U A− B ) g = Wd sin15° − F f d = − μkWd cos15° Work of friction force: Total work U A− B = Wd (sin15° − μk cos15°) Kinetic energy: TA = 1W 2 v0 2 g 1W 2 TB = vB 2 g Principle of work and energy: TA + U A− B = TB 1W 2 1W 2 v0 + Wd (sin15° − μk cos15°) = vB 2 g 2 g v02 = vB2 − 2 gd (sin15° − μk cos15°) = 0 − (2)(32.2)(20)[sin15° − (0.40)(cos15°)] = 164.29 ft 2 /s 2 v 0 = 12.81 ft/s 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. 522 PROBLEM 13.15 A 1200-kg trailer is hitched to a 1400-kg car. The car and trailer are traveling at 72 km/h when the driver applies the brakes on both the car and the trailer. Knowing that the braking forces exerted on the car and the trailer are 5000 N and 4000 N, respectively, determine (a) the distance traveled by the car and trailer before they come to a stop, (b) the horizontal component of the force exerted by the trailer hitch on the car. SOLUTION Let position 1 be the initial state at velocity v1 = 72 km/h = 20 m/s and position 2 be at the end of braking (v2 = 0). The braking forces and FC = 5000 N for the car and 4000 N for the trailer. (a) Car and trailer system. (d = braking distance) 1 (mC + mT )v12 2 = −( FC + FT )d T1 = U1→ 2 T2 = 0 T1 + U1→ 2 = T2 1 (mC + mT )v12 − ( FC + FT )d = 0 2 d= (b) (mC + mT )v12 (2600)(20) 2 = = 57.778 2( FC + FT ) (2)(9000) d = 57.8 m Car considered separately. Let H be the horizontal pushing force that the trailer exerts on the car through the hitch. 1 mC v12 2 = ( H − FC )d T1 = U1→ 2 T2 = 0 T1 + U1→ 2 = T2 1 mC v12 + ( H − FC ) d = 0 2 H = FC − mC v12 (1400)(20) 2 = 5000 − 2d (2)(57.778) H = 154 N Trailer hitch force on car: 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. 523 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: v1 = 72 km/h = 20 m/s Final speed: v2 = 108 km/h = 30 m/s Vertical rise: h = (0.02)(300) = 6.00 m Distance traveled: d = 300 m (a) Traction force. Use cab and trailer as a free body. m = 1800 + 5400 = 7200 kg T1 + U1→ 2 = T2 Work and energy: Ft = 1 d W = mg = (7200)(9.81) = 70.632 × 103 N 1 2 1 mv1 − Wh + Ft d = mv22 2 2 1 1 1 1 2 2 2 3 2 2 mv1 + Wh − mv1 = 300 2 (7200)(30) + (70.632 × 10 )(6.00) − 2 (7200)(20) = 7.4126 × 103 N (b) Ft = 7.41 kN Coupling force Fc . Use the trailer alone as a free body. m = 5400 kg W = mg = (5400)(9.81) = 52.974 × 103 N Assume that the tangential force at the trailer wheels is zero. Work and energy: T1 + U1→ 2 = T2 1 2 1 mv1 − Wh + Fc d = mv22 2 2 The plus sign before Fc means that we have assumed that the coupling is in tension. Fc = 1 1 2 1 1 1 1 mv2 + Wh − mv12 = (5400)(30) 2 + (52.974 × 103 )(6.00) − (5400)(20) 2 2 d 2 2 300 2 = 5.5595 × 103 N Fc = 5.56 kN (tension) 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 13.17 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 μk = 0.35 FB = (0.35)(100 kips) = 35 kips FC = (0.35)(80 kips) = 28 kips v1 = 30 mi/h = 44 ft/s (a) Entire train: v2 = 0 T2 = 0 T1 + U1− 2 = T2 1 (80 kips + 100 kips + 80 kips) (44 ft/s) 2 − (28 kips +35 kips) x = 0 2 32.2 ft/s 2 x = 124.07 ft (b) Force in each coupling: Recall that x = 124.07 ft x = 124.1 ft Car A: Assume FAB to be in tension T1 + V1− 2 = T2 1 80 kips (44) 2 − FAB (124.07 ft) = 0 2 32.2 FAB = +19.38 kips FAB = 19.38 kips (tension) Car C: T1 + U1− 2 = T2 1 80 kips (44)2 + ( FBC − 28 kips)(124.07 ft) = 0 2 32.2 FBC − 28 kips = −19.38 kips FBC = +8.62 kips FBC = 8.62 kips (tension) 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 13.18 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: FA = μ N A = (0.35)(80 kips) = 28 kips v2 = 0 T2 = 0 v1 = 30 mi/h = 44 ft/s ← T1 + V1− 2 = T2 1 (80 kips + 100 kips + 80 kips) (44 ft/s) 2 − (28 kips) x = 0 2 2 32.2 ft/s x = 279.1 ft (b) x = 279 ft Force in each coupling: Car A: Assume FAB to be in tension T1 + V1− 2 = T2 1 80 kips (44 ft/s) 2 − (28 kips + FAB )(279.1 ft) = 0 2 32.2 ft/s 2 28 kips + FAB = +8.62 kips FAB = −19.38 kips Car C: FAB = 19.38 kips (compression) T1 + V1− 2 = T2 1 80 kips (44 ft/s) 2 + FBC (279.1 ft) = 0 2 32.2 ft/s 2 FBC = −8.617 kips FBC = 8.62 kips (compression) 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 13.19 Blocks A and B weigh 25 lbs and 10 lbs, respectively, and they are both at a height 6 ft above the ground when the system is released from rest. Just before hitting the ground block A is moving at a speed of 9 ft/s. Determine (a) the amount of energy dissipated in friction by the pulley, (b) the tension in each portion of the cord during the motion. SOLUTION By constraint of the cable block B moves up a distance h when block A moves down a distance h. (h = 6 ft) Their speeds are equal. Let FA and FB be the tensions on the A and B sides, respectively, of the pulley. Masses: WA 25 = = 0.7764 lb ⋅ s 2 /ft 32.2 g W 10 MB = B = = 0.31056 lb ⋅ s 2 /ft 32.2 g MA = Let position 1 be the initial position with both blocks a distance h above the ground and position 2 be just before block A hits the ground. Kinetic energies: (T1 ) A = 0, (T1 ) B = 0 1 1 mA v 2 = (0.7764)(9)2 = 31.444 ft ⋅ lb 2 2 1 1 (T2 ) B = mB v 2 = (0.31056)(9) 2 = 12.578 ft ⋅ lb 2 2 (T2 ) A = Principle of work and energy: T1 + U1→ 2 = T2 Block A: U1→2 = (WA − FA ) h 0 + (25 − FA )(6) = 31.444 Block B: FA = 19.759 lb U1→2 = ( FB − WB ) h 0 + ( FB − 10)(6) = 12.578 FB = 12.096 lb At the pulley FA moves a distance h down, and FB moves a distance h up. The work done is U1→2 = ( FA − FB ) h = (19.759 − 12.096)(6) = 46.0 ft ⋅ lb 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. 527 PROBLEM 13.19 (Continued) Since the pulley is assumed to be massless, it cannot acquire kinetic energy; hence, (a) Energy dissipated by the pulley: (b) Tension in each portion of the cord: E p = 46.0 ft ⋅ lb A :19.76 lb B :12.10 lb 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. 528 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. (T1 ) B + (U1→ 2 ) B = (T2 ) B 0 + 30d − (2 F )(2) = 1 18 2 vB 2 32.2 30d − 4 F = 0.27950vB2 (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, v A = 2vB . (T1 ) A + (U1− 2 ) A = (T2 ) B 1 WA 2 vA 2 g 1 6 4 F − (6)(4) = (2vB ) 2 2 32.2 0 + ( F − WA )(4) = 4 F − 24 = 0.37267 vB2 (2) 30d − 24 = 0.65217vB2 (3) Add Eqs. (1) and (2) to eliminate F. (a) Case a: d = 2 ft, vB = ? (30)(2) − (24) = 0.65217vB2 vB2 = 55.2 ft 2 /s 2 (b) Case b: vB = 7.43 ft/s d = ?, vB = 0. 30d − 24 = 0 d = 0.800 ft 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 13.21 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 Given: Car B tows car A at 10 m/s uphill. μk = 0.8 Car A brakes so 4 wheels skid. Car B continues in same gear and throttle setting. Find: (a) Distance d, traveled to stop (b) Tension in cable (a) F1 = traction force (from equilibrium) F1 = (1400 g ) sin 5° + (1200 g )sin 5° F = 0.8 N A = 2600(9.81)sin 5° For system: A + B U1− 2 = [( F1 − 1400 g sin 5° − 1200 g sin 5°) − F ]d = T2 − T1 = 0 − Since 1 1 mA + Bv 2 = − (2600)(10) 2 2 2 ( F1 − 1400 g sin 5° − 1200 g sin 5°) = 0 − Fd = −0.8[1400(9.81) cos 5°]d = −130, 000 N ⋅ m d = 11.88 m (b) Cable tension, T U1− 2 = [T − 0.8N A ](11.88) = T2 − T1 (T − 0.8(1400)(9.81) cos 5°)11.88 = − 1400 (10) 2 2 (T − 10945) = −5892 = 5.053 kN T = 5.05 kN 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 13.22 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: x A + 3 yB = constant Δx A + 3ΔyB = 0 v A + 3vB = 0 Let F be the tension in the cable. Block A: m A = 30 kg, P = 250 N, (T1 ) A = 0 (T1 ) A + (U1→2 ) A = (T2 ) A 1 mA v A2 2 1 0 + (250 − F )(2) = (30)(3vB ) 2 2 0 + ( P − F )(Δx A ) = 500 − 2 F = 135vB2 Block B: (1) mB = 25 kg, WB = mB g = 245.25 N (T1 ) B + (U1→ 2 ) B = (T2 ) B 1 mB vB2 2 2 1 (3F ) − 245.25) = (25) vB2 3 2 0 + (3F − WB )(−ΔyB ) = 2 F − 163.5 = 12.5 vB2 (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. 531 PROBLEM 13.22 (Continued) Add Eqs. (1) and (2) to eliminate F. 500 − 163.5 = 147.5vB2 vB2 = 2.2814 m 2 /s 2 (a) Velocity of B. (b) Tension in the cable. From Eq. (2), v B = 1.510 m/s 2 F − 163.5 = (12.5)(2.2814) F = 96.0 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. 532 PROBLEM 13.23 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 assuming that the coefficients of friction between block A and the horizontal surface are μ s = 0.25 and μ k = 0.20, determine (a) the velocity of block B after block A has moved 2 m, (b) the tension in the cable. SOLUTION Check the equilibrium position to see if the blocks move. Let F be the tension in the cable. Block B: 3F − mB g = 0 F= Block A: mB g (25)(9.81) = = 81.75 N 3 3 ΣFy = 0: N A − mA g = 0 N A = m A g = (30)(9.81) = 294.3 N ΣFx = 0: 250 − FA − F = 0 FA = 250 − 81.75 = 168.25 N μs N A = (0.25)(294.3) = 73.57 N Available static friction force: Since FA > μs N A , the blocks move. The friction force, FA, during sliding is FA = μk N A = (0.20)(294.3) = 58.86 N Constraint of cable: x A + 3 yB = constant Δ x A + 3ΔyB = 0 v A + 3vB = 0 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. 533 PROBLEM 13.23 (Continued) m A = 30 kg, P = 250 N, (T1 ) A = 0. Block A: (T1 ) A + (U1→2 ) A = (T2 ) A 1 mA v 2A 2 1 0 + (250 − 58.86 − F )(2) = (30)(3vB ) 2 2 0 + ( P − FA − F )(Δx A ) = 382.28 − 2 F = 135vB2 (1) M B = 25 kg, WB = mB g = 245.25 N Block B: (T1 ) B + (U1→2 ) B = (T2 ) B 1 mB vB2 2 2 1 (3F − 245.25) = (25)vB2 3 2 0 + (3F − WB )(−ΔyB ) = 2 F − 163.5 = 12.5vB2 (2) Add Eqs. (1) and (2) to eliminate F. 382.28 − 163.5 = 147.5vB2 vB2 = 1.48325 m 2 /s 2 v B = 1.218 m/s (a) Velocity of B: (b) Tension in the cable: From Eq. (2), 2 F − 163.5 = (12.5)(1.48325) F = 91.0 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. 534 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 . v1 = 0 T1 = 0 At before C is removed from the system 1 1 ( mA + mB + mC )v22 = (12 kg)v22 = 6v22 2 2 = (m A + mC − mB ) g (0.9 m) T2 = U1− 2 U1− 2 = (4 + 3 − 5)( g )(0.9 m) = (2 kg)(9.81 m/s 2 )(0.9 m) U1− 2 = 17.658 J T1 + U1− 2 = T2 : 0 + 17.658 = 6v22 v22 = 2.943 At Position , collar C is removed from the system. Position to Position . T3 = T2′ = 1 9 (m A + mB )v22 = kg (2.943) = 13.244 J 2 2 1 9 ( mA + mB )(v3 ) 2 = v32 2 2 U 2′−3 = ( mA − mB )( g )(0.7 m) = (−1 kg)(9.81 m/s 2 )(0.7 m) = −6.867 J T2′ + U 2 −3 = T3 13.244 − 6.867 = 4.5v32 v32 = 1.417 v A = v3 = 1.190 m/s v A = 1.190 m/s 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 13.25 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: (a) sin β = 6 ft 15 ft β = 23.6° Package falls off the belt and 2, 3, 4 move down 6 = 2 ft. 3 1 3 6 lb T2 = 3 mv22 = 2 2 2 32.2 ft/s 2 2 v2 = 0.2795v2 U1− 2 = (3)(W )( R) = (3)(6 lb)(2 ft) = 36 lb ⋅ ft T1 + U1− 2 = T2 0 + 36 = 0.2795 v22 v22 = 128.8 v 2 = 11.35 ft/s (b) 23.6° Package 2 falls off the belt and its energy is lost to the system and 3 and 4 move down 2 ft. 1 6 lb (128.8) T2′ = (2) m v22 = 2 2 32 ft/s T2′ = 24 lb ⋅ ft 1 6 lb 2 (v3 ) = 0.18634v32 T3 = (2) m v32 = 2 2 32.2 ft/s U 2−3 = (2)(W )(2) = (2)(6 lb)(2 ft) = 24 lb ⋅ ft T2 + U 2−3 = T3 24 + 24 = 0.18634v32 v32 = 257.6 v3 = 16.05 ft/s 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. 536 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. (a) m A = 2 kg, mB = 3 kg Position 1: Block B has just been removed. Spring force: FS = −(m A + mB ) g = −k x Spring stretch: x1 = − (m A + mB ) g (5 kg)(9.81 m/s 2 ) =− = −1.22625 m k 40 N/m Let position 2 be a later position while the spring still contacts block A. Work of the force exerted by the spring: (U1→2 )e = − x2 k x dx x1 1 = − k x2 2 = Work of the gravitational force: x2 x1 = 1 1 k x12 − k x22 2 2 1 1 (40)(−1.22625) 2 − (40) x22 = 30.074 − 20 x22 2 2 (U1→2 ) g = − mA g ( x2 − x1 ) = −(2)(9.81)( x2 + 1.22625) = −19.62 x2 − 24.059 Total work: Kinetic energies: U1→2 = −20 x22 + 19.62 x2 + 6.015 T1 = 0 T2 = Principle of work and energy: 1 1 m Av22 = (2)v22 = v22 2 2 T1 + U1→ 2 = T2 0 + 20 x22 − 19.62 x2 + 6.015 = v22 Speed squared: v22 = −20 x22 − 19.62 x2 + 6.015 At maximum speed, dv2 =0 dx2 (1) 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. 537 PROBLEM 13.26 (Continued) Differentiating Eq. (1), and setting equal to zero, 2v2 Substituting into Eq. (1), dv2 = −40 x2 = −19.62 = 0 dx 19.62 x2 = − = −0.4905 m 40 v22 = −(20)(−0.4905) 2 − (19.62)( −0.4905) + 6.015 = 10.827 m 2 /s 2 v 2 = 3.29 m/s Maximum speed: (b) Position 3: Block A reaches maximum height. Assume that the block has separated from the spring. Spring force is zero at separation. Work of the force exerted by the spring: (U1→3 )e = − 0 x1 kxdx = 1 2 1 kx1 = (40)(1.22625)2 = 30.074 J 2 2 Work of the gravitational force: (U1→3 ) g = − m A gh = −(2)(9.81) h = −19.62 h Total work: U1→3 = 30.074 − 19.62 h At maximum height, v3 = 0, T3 = 0 Principle of work and energy: T1 + U1→3 = T3 0 + 30.074 − 19.62 h = 0 h = 1.533 m Maximum height: 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. 538 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. (a) m A = 2 kg, mB = 3 kg Position 1: Block B has just been removed. Spring force: FS = −(m A + mB ) g = − kx1 Spring stretch: x1 = − (m A + mB ) g (5 kg)(9.81 m/s 2 ) =− = −1.22625 m k 40 N/m Let position 2 be a later position. Note that the spring remains attached to block A. Work of the force exerted by the spring: (U1→2 )e = − x2 kxdx x1 1 = − kx 2 2 = Work of the gravitational force: x2 x1 = 1 2 1 2 kx1 − kx2 2 2 1 1 (40)(−1.22625) 2 − (40) x22 = 30.074 − 20 x22 2 2 (U1→2 ) g = − mA g ( x2 − x1 ) = −(2)(9.81)( x2 + 1.22625) = −19.62 x2 − 24.059 Total work: Kinetic energies: U1→2 = −20 x22 − 19.62 x2 + 6.015 T1 = 0 T2 = Principle of work and energy: 1 1 m Av22 = (2)v22 = v22 2 2 T1 + U1→ 2 = T2 0 + 20 x22 − 19.62 x2 + 6.015 = v22 Speed squared: v22 = −20 x22 − 19.62 x2 + 6.015 At maximum speed, dv2 =0 dx2 (1) 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 (Continued) Differentiating Eq. (1) and setting equal to zero, 2v2 dv2 = −40 x2 = −19.62 = 0 dx2 x2 = − Substituting into Eq. (1), 19.62 = −0.4905 m 40 v22 = −(20)( −0.4905)2 − (19.62)(−0.4905) + 6.015 = 10.827 m1 /s 2 v2 = 3.29 m/s Maximum speed: (b) Maximum height occurs when v2 = 0. Substituting into Eq. (1), 0 = −20 x22 − 19.62 x2 + 6.015 Solving the quadratic equation x2 = −1.22625 m and 0.24525 m Using the larger value, x2 = 0.24525 m Maximum height: h = x2 − x1 = 0.24525 + 1.22625 h = 1.472 m 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 13.28 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 μk = 0.35. SOLUTION k B = 144 lb/ft (a) k A = 216 lb/ft Since the collar C leaves the spring at B and there is no friction, it must engage the spring at A. TA = 0 U A− B = TB = 0 2/12 0 k B xdx − y 0 k A xdx 2 U A− B 144 lb/ft 2 216 lb/ft 2 = ft − ( y) 2 2 12 TA + U A− B = TB : 0 + 2 − 108 y 2 = 0 y = 0.1361 ft = 1.633 in. Total distance d = 2 + 16 − (6 − 1.633) d = 13.63 in. (b) Assume that C does not reach the spring at B because of friction. N = W = 6 lb F f = (0.35)(8 lb) = 2.80 lb TA = TD = 0 U A− D = 2/12 0 TA + U A− D = TD 144 × dx − F f ( y ) = 2 − 2.80 y 0 + 2 − 2.80 y = 0 y = 0.714 ft = 8.57 in. The collar must travel 16 − 6 + 2 = 12 in. before it engages the spring at B. Since y = 8.57 in., it stops before engaging the spring at B. d = 8.57 in. Total distance 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 13.29 A 6-lb block is attached to a cable and to a spring as shown. The constant of the spring is k = 8 lb/in. and the tension in the cable is 3 lb. If the cable is cut, determine (a) the maximum displacement of the block, (b) the maximum speed of the block. SOLUTION k = 8 lb/in. = 96 lb/ft ΣFy = 0: ( Fs )1 = 6 − 3 = 3 lb C v1 = 0 T1 = 0: T2 = For weight: U1− 2 = (6 lb)x = 6 x For spring: U1− 2 = − x 0 1 6 lb 2 2 v2 = 0.09317v2 2 32.2 (3 + 96 x)dx = −3x − 48 x 2 T1 + U1− 2 = T2 : 0 + 6 x − 3x − 48 x 2 = 0.09317v22 3x − 48 x 2 = 0.09317v22 (a) For xm , v2 = 0: (1) 3x − 48 x 2 = 0 x = 0, xm = 3 1 = ft 48 16 x m = 0.75 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. 542 PROBLEM 13.29 (Continued) (b) For vm we see maximum of U1− 2 = 3x − 48 x 2 dU1− 2 = 3 − 96 x = 0 dx x= 3 1 ft = ft 96 32 2 Eq. (1): 1 1 3 ft − 48 ft = 0.09317vm2 32 32 vm2 = 0.5031 vm = 0.7093 ft/s vm = 8.51 in./s Note: U1− 2 for the spring may be computed using F6 − x curve U1− 2 = area 1 = 3x + 96x 2 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. 543 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 xB = 1 xA 2 1 1 k A ( x A ) 2 − k B ( xB ) 2 2 2 (Gravity) (Spring A) (Spring B) U1− 2 = W ( x A ) − U1 − 2 = (98.1 N)(0.05 m) − − 1 (2000 N/m)(0.05 m) 2 2 1 (2000 N/m)(0.025 m) 2 2 U1− 2 = 1 1 (m)v 2 = (10 kg) v 2 2 2 4.905 − 2.5 − 0.625 = 1 (10)v 2 2 v = 0.597 m/s (b) Let x = distance moved down by the 10 kg block 2 U1− 2 1 1 x 1 = W ( x) − k A ( x) 2 − k B = (m)v 2 2 2 2 2 d 1 k (m)v 2 = 0 = W − k A ( x) − B (2 x) dx 2 8 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 (Continued) 0 = 98.1 − 2000 ( x) − 2000 (2 x) = 98.1 − (2000 + 250) x 8 x = 0.0436 m (43.6 mm) For x = 0.0436, U = 4.2772 − 1.9010 − 0.4752 = 1 (10)v 2 2 vmax = 0.6166 m/s vmax = 0.617 m/s 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 13.31 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. T1 + U1→ 2 = T2 Since collar is initially at rest, T1 = 0. In position 2, where the upper spring is compressed 75 mm and v2 = 1.00 m/s, the kinetic energy is T2 = 1 2 1 mv2 = (5 kg)(1.00 m/s) 2 = 2.5 J 2 2 As the collar is raised from level A to level B, the work of the weight force is (U1→2 ) g = − mgh where m = 5 kg, g = 9.81 m/s 2 and h = 450 mm = 0.450 m Thus, (U1→2 ) g = −(5)(9.81)(0.450) = −22.0725 J In position 1, the force exerted by the lower spring is equal to the weight of collar A. F1 = mg = −(5 kg)(9.81 m/s) = −49.05 N As the collar moves up a distance x1, the spring force is F = F1 − k1 x2 until the collar separates from the spring at xf = F1 49.05 N = = 0.122625 m = 122.625 mm k1 400 N/m 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 (Continued) Work of the force exerted by the lower spring: (U1→2 )1 = xf 0 ( F1 − k1 x)dx = F1 x f − = 1 2 1 1 kx f = k1 x 2f − k1 x 2f = k1 x 2f 2 2 2 1 (400 N/m)(0.122625)2 = 3.0074 J 2 In position 2, the upper spring is compressed by y = 75 mm = 0.075 m. The work of the force exerted by this spring is 1 1 (U1→2 ) 2 = − k2 y 2 = − k2 (0.075) 2 = −0.0028125 k 2 2 2 Finally, we must calculate the work of the 150 N force applied to the cable. In position 1, the length AB is (l AB )1 = (450) 2 + (400) 2 = 602.08 mm In position 2, the length AB is (l AB ) 2 = 400 mm. The displacement d of the 150 N force is d = (l AB )1 − (l AB )2 = 202.08 mm = 0.20208 m The work of the 150 N force P is (U1→2 ) P = Pd = (150 N)(0.20208 m) = 30.312 J Total work: U1→2 = −22.0725 + 3.0074 − 0.0028125k2 + 30.312 = 11.247 − 0.0028125k2 Principle of work and energy: T1 + U1→ 2 = T2 0 + 11.247 − 0.0028125k2 = 2.5 k2 = 3110 N/m k2 = 3110 N/m 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. 547 PROBLEM 13.32 A piston of mass m and cross-sectional area A is equilibrium under the pressure p at the center of a cylinder closed at both ends. Assuming that the piston is moved to the left a distance a/2 and released, and knowing that the pressure on each side of the piston varies inversely with the volume, determine the velocity of the piston as it again reaches the center of the cylinder. Neglect friction between the piston and the cylinder and express your answer in terms of m, a, p, and A. SOLUTION Pressures vary inversely as the volume pL Aa = P Ax pR Aa = P A(2a − x) Initially at , v=0 x= pL = pa x pR = pa (2a − x) a 2 T1 = 0 At , x = a, T2 = U1− 2 = a a/2 1 2 mv 2 ( pL − pR ) Adx = 1 1 paA − dx a/2 x 2a − x a U1− 2 = paA[ln x + ln (2a − x)]aa/2 a 3a U1− 2 = paA ln a + ln a − ln − ln 2 2 3a 2 4 U1− 2 = paA ln a 2 − ln = paA ln 4 3 4 1 T1 + U1− 2 = T2 0 + paA ln = mv 2 3 2 v2 = 2 paA ln ( 43 ) m = 0.5754 paA m v = 0.759 paA m 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. 548 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 ≤ d ≤ 14 ft. v1 = 95.33 ft/s 1 2 1 2250 lb mv1 2 2 32.2 ft/s 2 T1 = 317,530 lb ⋅ ft T1 = 2 (95.3 ft/s) = 317.63 k ⋅ ft v2 = 0 T2 = 0 U1− 2 = (18 k)(5 ft) + (27 k)( d − 5) T1 + U1− 2 = 90 + 27d − 135 = 27d − 45 k ⋅ ft = T2 317.53 = 27d − 45 d = 13.43 ft Assumption that d ≤ 14 ft is ok. (b) Maximum deceleration occurs when F is largest. For d = 13.43 ft, F = 27 k. Thus, F = maD 2250 lb (27, 000 lb) = 2 32.2 ft/s ( aD ) aD = 386 ft/s 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. 549 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 W1 = (90 ton)(2000 lb/ton) = 180 × 103 lb Weight: W 180 × 103 = = 5590 lb ⋅ s 2 /ft g 32.2 Mass: m= Speed: v1 = 1 mi/h = Kinetic energy: T1 = 5280 ft = 1.4667 ft/s 3600 s 1 2 1 mv1 = (5590)(1.4667) 2 2 2 = 6012 ft ⋅ lb T2 = 0 Principle of work and energy: (rest) T1 + U1→ 2 = T2 6012 + U1→2 = 0 U1→2 = −6012 ft ⋅ lb = −72.15 kip ⋅ in. 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 F = kx Area = x 0 fdx = k= x 0 Fmax 60 = = 5 kip/in. xmax 12 kx dx = 1 2 kx 2 1 (5) x 2 = 72.51 2 x 2 = 28.86 in.2 Fender B: We divide area under curve B into trapezoids Partial area x = 5.37 in. Total Area From x = 0 to x = 2 in.: 1 (2 in.)(4 kips) = 4 kip ⋅ in. 2 4 kip ⋅ in. From x = 2 in. to x = 4 in.: 1 (2 in.)(4 + 10) = 14 kip ⋅ in. 2 18 kip ⋅ in. From x = 4 in. to x = 6 in.: 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. 550 PROBLEM 13.34 (Continued) We still need ΔU = 72.15 − 46 = 26.15 kip ⋅ in. Equation of straight line approximating curve B from x = 6 in. to x = 8 in. is Δx F − 18 = F = 18 + 6Δx 2 30 − 18 1 ΔU = 18Δx + (6Δx)Δx = 26.15 kip ⋅ in. 2 2 (Δx) + 6Δx − 8.716 = 0 Δx = 1.209 in. Thus: x = 6 in. + 1.209 in. = 7.209 in. x = 7.21 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 13.35 Nonlinear springs are classified as hard or soft, depending upon the curvature of their force-deflection curve (see figure). If a delicate instrument having a mass of 5 kg is placed on a spring of length l so that its base is just touching the undeformed spring and then inadvertently released from that position, determine the maximum deflection xm of the spring and the maximum force Fm exerted by the spring, assuming (a) a linear spring of constant k = 3 kN/m, (b) a hard, nonlinear spring, for which F = (3 kN/m)( x + 160 x 2 ) . SOLUTION W = mg = (5 kg) g W = 49.05 N T1 = T2 = 0, T1 + U1− 2 = T2 yields U1− 2 = 0 Since U1− 2 = Wxm − (a) xm 0 Fdx = 49.05 xm − xm 0 Fdx = 0 For F = kx = (300 N/m) x Eq. (1): 49.05 xm − xm 0 3000 x dx = 0 xm = 32.7 × 10−3 m = 32.7 mm 49.05 xm − 1500 xm2 = 0 Fm = 3000 xm = 3000(32.7 × 10−3 ) (b) (1) For Eq. (1) F = (3000 N/m) x(1 + 160 x 2 ) 49.05 xm − xm 0 3000( x + 160 x3 )dx = 0 1 49.05 xm − 3000 xm2 + 40 xm4 = 0 2 Solve by trial: Fm = 98.1 N (2) xm = 30.44 × 10−3 m xm = 30.4 mm Fm = (3000)(30.44 × 10−3 )[1 + 160(30.44 × 10−3 ) 2 ] Fm = 104.9 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. 552 PROBLEM 13.36 A rocket is fired vertically from the surface of the moon with a speed v0 . Derive a formula for the ratio hn /hu of heights reached with a speed v, if Newton’s law of gravitation is used to calculate hn and a uniform gravitational field is used to calculate hu . Express your answer in terms of the acceleration of gravity g m on the surface of the moon, the radius Rm of the moon, and the speeds v and v0 . SOLUTION Newton’s law of gravitation T1 = U1− 2 = 1 2 mv0 2 Rm + hn Rm T2 = 1 2 mv 2 ( − Fn )dr U1− 2 = − mg m Rm2 Rm + hn Rm Fn = mg m Rm2 r2 dr r2 1 1 U1− 2 = mg m Rm2 − Rm Rm + hn T1 + U1− 2 = T2 Rm 1 2 1 2 mv0 + mg m Rm − = mv Rm + hn 2 2 hn = (v ) Rm − v2 2 2 (v −v ) 2 g m Rm − 0 2 gm 2 0 (1) Uniform gravitational field T1 = U1− 2 = 1 2 mv0 T2 = mv 2 2 Rm + hn Rm 1 2 1 mv0 − mg m hu = mv 2 2 2 T1 + U1− 2 = T2 hu ( − Fu )dr = −mg m ( Rm + hu − Rm ) = −mghu (v = 2 0 − v2 ) (2) 2 gm hn = hu Dividing (1) by (2) 1 ( v −v ) 1 − (2 g R ) 2 0 2 m m 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. 553 PROBLEM 13.37 Express the acceleration of gravity g h at an altitude h above the surface of the earth in terms of the acceleration of gravity g0 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 F= At earth’s surface, (h = 0) = (h + R ) 2 GM E m / R 2 ( Rh + 1) 2 mg h GM E m = mg0 R2 GM E R 2 GM E = g0 gh = Thus, GM E m gh = R2 h +1 R 2 g0 h +1 R 2 R = 6370 km At altitude h, “true” weight Assumed weight F = mg h = WT W0 = mg0 Error = E = W0 − WT mg0 − mg h g0 − g h = = W0 mg0 g0 g0 − gh = g0 ( Rh + 1) 2 E= g0 (1 + ) h R 2 g0 1 = 1 − 2 h (1 + R ) (a) h = 1 km: 1 P = 100E = 100 1 − 2 1 (1 + 6370 ) P = 0.0314% (b) h = 1000 km: 1 P = 100E = 100 1 − 1000 2 (1 + 6370 ) 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. 554 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 hm . At maximum height, the total velocity is the horizontal component of the velocity, which is constant and the same in both cases. 1 2 mv 2 = −mge he T1 = U1− 2 T2 = Earth U1− 2 = −mg m hm Earth 1 2 1 mv − mge he = mvH2 2 2 Moon 1 2 1 mv − mg m hm = mvH2 2 2 Moon hm g e = he g m − ge he + g m hm = 0 Subtracting 1 2 mvH 2 ge hm = (60 m) 0.165 ge hm = 364 m 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 13.39 The sphere at A is given a downward velocity v 0 of magnitude 5 m/s and swings in a vertical plane at the end of a rope of length l = 2 m 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 1 2 1 mv0 = m (5) 2 2 2 T1 = 12.5 m T1 = 1 2 mv 2 = mg (l ) sin θ T2 = U1− 2 T1 + U1− 2 = T2 12.5m + 2mg sin θ = 25 + 4 g sin θ = v 2 1 2 mv 2 (1) Newton’s law at . v2 v2 =m 2 2 v = 4 g − 2 g sin θ 2mg − mg sin θ = m (2) Substitute for v 2 from Eq. (2) into Eq. (1) 25 + 4 g sin θ = 4 g − 2 g sin θ (4)(9.81) − 25 sin θ = = 0.2419 (6)(9.81) θ = 14.00° 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 13.40 The sphere at A is given a downward velocity v 0 and swings in a vertical circle of radius l and center O. Determine the smallest velocity v 0 for which the sphere will reach Point B as it swings about Point O (a) if AO is a rope, (b) if AO is a slender rod of negligible mass. SOLUTION 1 2 mv0 2 1 T2 = mv 2 2 U1− 2 = − mgl T1 = 1 2 1 mv0 − mgl = mv 2 2 2 v02 = v 2 + 2 gl T1 + U1− 2 = T2 Newton’s law at (a) For minimum v, tension in the cord must be zero. Thus, v 2 = gl v02 = v 2 + 2 gl = 3gl (b) v0 = 3gl Force in the rod can support the weight so that v can be zero. v02 = 0 + 2 gl Thus, v0 = 2 gl 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. 557 PROBLEM 13.41 A small sphere B of weight W is released from rest in the position shown and swings freely in a vertical plane, first about O and then about the peg A after the cord comes in contact with the peg. Determine the tension in the cord (a) just before the sphere comes in contact with the peg, (b) just after it comes in contact with the peg. SOLUTION Velocity of the sphere as the cord contacts A vB = 0 TB = 0 1 2 mvC 2 = ( mg )(1) TC = U B −C TB + U B −C = TC 0 + 1mg = 1 2 mvC 2 vC2 = (2)( g ) Newton’s law (a) Cord rotates about Point O ( R = L) T − mg (cos 60°) = m vC2 L T = mg (cos 60°) + T= 3 mg 2 m(2) g 2 T = 1.5 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. 558 PROBLEM 13.41 (Continued) (b) L Cord rotates about A R = 2 T − mg (cos 60°) = T= mvC2 L 2 mg m(2)( g ) + 2 1 5 1 T = + 2 mg = mg 2 2 T = 2.5W 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 13.42 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. T1 = 0 T2 = 1 2 mvP 2 U1→2 = mgyP T1 + U1→ 2 = T2 : 0 + mgyP = 1 2 mvP 2 vP2 = 2 gyP Magnitude of normal acceleration at P: ( aP ) n = (a) vP2 ρP = 2 gyP ρP Rider at Point B. yB = h = 60 ft ρ B = r = 20 ft an = (2 g )(60) = 6g 20 ΣF = ma: N B − mg = m(6 g ) N B = 7 mg = 7W = (7)(160 lb) N B = 1120 lb 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 (Continued) Rider at Point D. yD = h − 2r = 20 ft ρ D = 20 ft an = (2 g )(20) = 2g 20 ΣF = ma: N D + mg = m(2 g ) N D = mg = W = 160 lb N D = 160 lb (b) Car at Point E. yE = h − r = 40 ft NE = 0 ΣF = man : mg = m ⋅ 2 gyE ρE ρ E = 2 yE ρ = 80.0 ft 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 13.43 In Problem 13.42, determine the range of values of h for which the roller coaster will not leave the track at D or E, knowing that the radius of curvature at E is ρ = 75 ft. Assume no energy loss due to friction. 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. T2 = T1 = 0 1 2 mvP 2 U1→2 = mg yP T1 + U1→ 2 = T2 : 0 + mgy p = 1 2 mvP 2 vP2 = 2 g yP Magnitude of normal acceleration of P: ( aP ) n = vP2 ρP = 2 g yP ρP The condition of loss of contact with the track at P is that the curvature of the path is equal to ρp and the normal contact force N P = 0. Car at Point D. ρ D = r = 20 ft y D = h − 2r ( aD ) n = 2 g (h − 2r ) r ΣF = ma 2 g ( h − 2r ) r 2 h − 5r N D = mg r N D + mg = m 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. 562 PROBLEM 13.43 (Continued) For N D > 0 2 h − 5r > 0 5 h > r = 50 ft 2 Car at Point E. ρ E = ρ = 75 ft yE = h − r = h − 20 ft ( aE ) n = 2 g (h − 20) 75 ΣF = ma 2mg ( h − 20) 75 115 − 2h N E = mg 75 N E − mg = − For N E > 0, 115 − 2h > 0 h < 57.5 ft 50.0 ft ≤ h ≤ 57.5 ft Range of values for h: 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. 563 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: N − mg cos θ = −man = − mvC2 h vc2 = gh cos θ With N = 0, we get 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. T1 = 1 2 mvB 2 T2 = 1 1 mvC2 = mgh cos θ 2 2 U1→2 = weight × change in vertical distance = mgh (1 − cos θ ) 1 2 1 mvB + mg (1 − cos θ ) = mgh cos θ 2 2 2 vB = gh cos θ − 2 gh(1 − cos θ ) = gh(3cos θ − 2) T1 + U1→ 2 = T2 : Data: g = 9.81 m/s 2 , h = 0.9 m, θ = 30°. vB2 = (9.81)(0.9)(3cos 30° − 2) = 5.2804 m 2 /s 2 vB = 2.30 m/s 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 13.45 A small block slides at a speed v = 8 ft/s on a horizontal surface at a height h = 3 ft above the ground. Determine (a) the angle θ 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 N = 0. mg cos θ = man g cos θ = vC2 h vC2 = gh cos θ = gy (1) Work-energy principle. TB = (a) 1 2 1 mv = m(8) 2 = 32m 2 2 1 2 mvC 2 = TC TC = TB + U B −C Use Eq. (1) 32m + mg (h − y ) = 1 2 mvC 2 32 + g (h − yC ) = 1 g yC 2 U B −C = W (h − g ) = mg ( h − yC ) (2) 3 g yC 2 (32 + gh) yC = ( 32 g ) 32 + gh = yC = (32 + (32.2)(3)) 3 (32.2) 2 yC = 2.6625 ft yC = h cos θ cos θ = yC 2.6625 = = 0.8875 3 h (3) θ = 27.4° 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 (Continued) (b) From (1) and (3) vC = gy vC = (32.2)(2.6625) vC = 9.259 ft/s At C: (vC ) x = vC cos θ = (9.259)(cos 27.4°) = 8.220 ft/s (vC ) y = −vC sin θ = −(9.259)(sin 27.4°) = 4.261 ft/s y = yC + (vC ) y t − At E: yE = 0: 1 2 gt = 2.6625 − 4.261t − 16.1t 2 2 t 2 + 0.2647t − 0.1654 = 0 t = 0.2953 s At E: x = h(sin θ ) + (vC ) x t = (3)(sin 27.4°) + (8.220)(0.2953) x = 1.381 + 2.427 = 3.808 ft x = 3.81 ft 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 13.46 A chair-lift is designed to transport 1000 skiers per hour from the base A to the summit B. The average mass of a skier is 70 kg and the average speed of the lift is 75 m/min. Determine (a) the average power required, (b) the required capacity of the motor if the mechanical efficiency is 85 percent and if a 300 percent overload is to be allowed. SOLUTION Note: Solution is independent of speed. (a) Average power = ΔU (1000)(70 kg)(9.81 m/s 2 )(300 m) N⋅m = = 57, 225 Δt 3600 s s Average power = 57.2 kW (b) Maximum power required with 300% over load = 100 + 300 (57.225 kW) = 229 kW 100 Required motor capacity (85% efficient) Motor capacity = 229 kW = 269 kW 0.85 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. 567 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 (a) ( PP ) A = ( F )(v A ) = (mC + mL )( g )(v A ) v A = s/t = (2.8 m)/(15 s) = 0.18667 m/s ( PP ) A = [(1200 kg) + (300 kg)](9.81 m/s 2 )(0.18667 m/s)3 (b) ( PP ) A = 2.747 kJ/s ( PP ) A = 2.75 kW ( PE ) A = ( PP )/η = (2.75 kW)/(0.82) ( PE ) A = 3.35 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. 568 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 Mg = ( M C + M L ) g = (1200 + 300) g Mg = 1500 g ΣF = F − 1500 g = 1500a (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. a= vmax 7.5 s P = (6000 W) = ( F )(vmax ) vmax = (6000)/F a = (6000)/(7.5)( F ) Thus, (2) Substitute (2) into (1) F − 1500 g = (1500)(6000)/(7.5)( F ) F 2 − (1500 kg)(9.81 m/s 2 ) F − (1500 kg)(6000 N ⋅ m/s) =0 (7.5 s) F 2 − 14, 715F − 1.2 × 106 = 0 F = 14,800 N F = 14.8 kN 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 13.49 (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 tan θ = 3 100 θ = 1.718° W = WB + WW = 15 + 120 (a) W = 135 lb PW = W ⋅ v = (W sin θ ) (v) PW = (135)(sin 1.718°)(5) (a) PW = 20.24 ft ⋅ lb/s PW = 20.2 ft ⋅ lb/s W = WB + Wm = 18 + 180 (b) W = 198 lb Brakes must dissipate the power generated by the bike and the man going down the slope at 20 ft/s. PB = W ⋅ v = (W sin θ )(v) PB = (198)(sin 1.718)(20) PB = 118.7 ft ⋅ lb/s (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. 570 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, (m kg/h)( g m/s 2 ) = [mg (N/h)] Thus, ΔU [mg (N/h)][b(m)] mgb = = N ⋅ m/s Δt (3600 s/h) 3600 1000 N ⋅ m/s = 1 kW Thus, including motor efficiency, η P (kw) = mgb (N ⋅ m/s) 1000 N ⋅ m/s (3600) (η ) kW P(kW) = 0.278 × 10−6 mgb η ΔU [W (tons/h)(2000 lb/ton)][b(ft)] = Δt 3600 s/h (b) = With η , Wb ft ⋅ lb/s; 1hp = 550 ft ⋅ lb/s 1.8 Wb 1 hp 1 hp = (ft ⋅ lb/s) 1.8 550 ft ⋅ lb/s η hp = 1.010 × 10−3Wb 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. 571 PROBLEM 13.51 In an automobile drag race, the rear (drive) wheels of a 1000 kg car skid for the first 20 m and roll with sliding impending during the remaining 380 m. The front wheels of the car are just off the ground for the first 20 m, and for the remainder of the race 80 percent of the weight is on the rear wheels. Knowing that the coefficients of friction are μs = 0.90 and μk = 0.68, determine the power developed by the car at the drive wheels (a) at the end of the 20-m portion of the race, (b) at the end of the race: Give your answer in kW and in hp. Ignore the effect of air resistance and rolling friction. SOLUTION (a) First 20 m. (Calculate velocity at 20 m.) Force generated by rear wheels = μkW , since car skids. Thus, Fs = (0.68)(1000)( g ) Fs = (0.68)(1000 kg)(9.81 m/s 2 ) = 6670.8 N Work and energy. T1 = 0, T2 = 1 2 2 mv20 = 500v20 2 T1 + U1− 2 = T2 U1− 2 = (20 m)(Fs ) = (20 m)(6670.8 N) U1− 2 = 133, 420 J 2 0 + 133,420 = 500v20 2 v20 = 133,420 = 266.83 500 v20 = 16.335 m/s Power = ( Fs )(v20 ) = (6670.8 N)(16.335 m/s) Power = 108,970 J/s =108.97 kJ/s 1 kJ/s = 1 kW 1 hp = 0.7457 kW Power = 109.0 kJ/s = 109.0 kW Power = (109.0 kW) = 146.2 hp (0.7457 kW/hp) 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. 572 PROBLEM 13.51 (Continued) (b) End of race. (Calculate velocity at 400 m.) For remaining 380 m, with 80% of weight on rear wheels, the force generated at impending sliding is ( μ s )(0.80)(mg ) FI = (0.90)(0.80)(1000 kg)(9.81 m/s 2 ) FI = 7063.2 N Work and energy, from 20 m to 28 m . v2 = 16.335 m/s [from part (a)] T2 = 1 (1000 kg)(16.335 m/s) 2 2 T2 = 133, 420 J T3 = 1 2 2 mv380 = 500v380 2 U 2 −3 = ( FI )(380 m) = (7063.2 N)(380 m) U 2 −3 = 2, 684,000 J T2 + U 2 −3 = T3 2 (133, 420 J) + (2, 684, 000 J) = 500v30 v30 = 75.066 m/s Power = ( FI )(v30 ) = (7063.2 N)(75.066 m/s) = 530, 200 J/s kW Power = 530, 200 J = 530 kW hp Power = 530 kW = 711 hp (0.7457 kW/hp) 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. 573 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. v0 = 4.5 km/h = 1.25 m/s F0 = 300 kN P0 = F0 v0 = (300 kN)(1.25 m/s) (b) P0 = 375 kW Maximum speed. Power required to tow ship at speed v: 1.75 v F = F0 v0 1.75 v P = Fv = F0 v v0 v = F0 v0 v0 2.75 (1) Since we have two tugboats, the available power is twice maximum power F0 v0 developed by one tugboat. v 2 F0 v0 = F0 v0 v0 v v0 Recalling that 2.75 2.75 =2 v = v0 (2)1/ 2.75 = v0 (1.2867) v0 = 4.5 km/h v = (4.5 km/h)(1.2867) = 5.7902 km/h v = 5.79 km/h 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 13.53 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. (t < 50 s): v = v0 + at v0 = 0 t = 50 s, At v = 90 km/h = 25 m/s 25 m/s = 0 + a(50) a = 0.5 m/s 2 v = (0.5 m/s 2 )t FP − FR = ma Newton’s second law: FR = 15 kN = 15 × 103 N where m = 500 Mg = 500 × 103 kg a = 0.5 m/s 2 FP = FR + ma = 15 × 103 + (500 × 103 )(0.5) = 265 × 103 N = 265 kN FP v = (265 × 103 )(0.5t ) Power: (0 < 50s) Uniform motion. (t > 50 s): Power = (132.5 kW/s)t a=0 FP = FR = 15 × 103 N; v = 25 m/s Power: FP v = (15 × 103 )(25 m/s) = 375 × 103W (t > 50 s) 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. 575 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 0.18 ft/s 2 . SOLUTION (a) Acceleration = 0 Elevator Counterweight Motor ΣFy = 0: TW − WW = 0 ΣF = 0: 2TC + TW − 6600 = 0 TW = 2200 lb Kinematics: TC = 2200 lb 2 xE = xC , 2 x E = xC , vC = 2vE = 2 ft/s P = TC ⋅ vC = (2200 lb)(2 ft/s) = 4400 lb ⋅ ft/s = 8.00 hp P = 8.00 hp aE = 0.18 ft/s 2 , vE = 1 ft/s (b) Counterweight Elevator Counterweight: ΣF = Ma : TW − W = W (aW ) g 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 (Continued) TW = (2200 lb) + (2200 lb)(0.18 ft/s 2 ) (32.2 ft/s 2 ) TW = 2212 lb Elevator ΣF = ma 2TC + TW − WE = 2TC = (−2212 lb) + (6600 lb) − −WE (a E ) g (6600 lb)(0.18 ft/s 2 ) (32.2 ft/s 2 ) 2TC = 4351 lb TC = 2175.6 lb vC = 2 ft/s (see part(a)) P = TC ⋅ vC = (2175.6 lb)(2 ft/s) = 4351.2 lb ⋅ ft/s = 7.911 hp P = 7.91 hp 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. 577 PROBLEM 13.CQ2 Two small balls A and B with masses 2m and m respectively are released from rest at a height h above the ground. Neglecting air resistance, which of the following statements are true when the two balls hit the ground? (a) The kinetic energy of A is the same as the kinetic energy of B. (b) The kinetic energy of A is half the kinetic energy of B. (c) The kinetic energy of A is twice the kinetic energy of B. (d) The kinetic energy of A is four times the kinetic energy of B. SOLUTION Answer: (c) 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. 578 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. 579 PROBLEM 13.55 A force P is slowly applied to a plate that is attached to two springs and causes a deflection x0 . In each of the two cases shown, derive an expression for the constant ke , in terms of k1 and k2 , of the single spring equivalent to the given system, that is, of the single spring which will undergo the same deflection x0 when subjected to the same force P. SOLUTION System is in equilibrium in deflected x0 position. Case (a) Force in both springs is the same = P x0 = x1 + x2 Thus, x0 = P ke x1 = P k1 x2 = P k2 P P P = + ke k1 k2 1 1 1 = + ke k1 k2 Case (b) ke = k1k2 k1 + k2 Deflection in both springs is the same = x0 P = k1 x0 + k2 x0 P = (k1 + k2 ) x0 P = ke x0 Equating the two expressions for P = (k1 + k2 ) x0 = ke x0 ke = k1 + k2 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 13.56 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. 1 2 mv0 2 VA = 0 (zero force in springs) TB = 0 (v = 0 at maximum deflection) TA = 1 2 1 k1 x1 + k2 x22 2 2 where x1 is deflection of spring k1 and x2 is that of spring k2. VB = Conservation of energy: TA + VA = TB + VB 1 2 1 1 mv0 + 0 = 0 + k1 x12 + k2 x22 2 2 2 k1 x12 + k2 x22 = mv02 (a) (1) Springs are in series. Let F be the force carried by the two springs. F k1 Then, x1 = Eq. (1) becomes 1 1 F 2 + = mv02 k1 k2 so that The maximum deflection is and x2 = F k2 1 1 F = v0 m / + k1 k2 1 1 + F k1 k2 δ = x1 + x2 = 1 1 1 1 = + v0 m / + k1 k2 k1 k2 1 1 = v0 m + k1 k2 δ = v0 m(k1 + k2 )/k1k2 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 (Continued) (b) Springs are in parallel. x1 = x2 = δ Eq. (1) becomes (k1 + k2 )δ 2 = mv02 δ = v0 m k1 + k2 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. 582 PROBLEM 13.57 A 600-g collar C may slide along a horizontal, semicircular rod ABD. The spring CE has an undeformed length of 250 mm and a spring constant of 135 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 D. SOLUTION First calculate the lengths of the spring when the collar is at positions A, B, and D. l A = 4402 + 3002 + 1802 = 562.14 mm lB = 2402 + 3002 + 202 = 384.71 mm lD = 402 + 3002 + 1802 = 352.14 mm The elongations of springs are given by e = l − l0 . eA = 562.14 − 250 = 312.14 mm = 0.31214 m eB = 384.71 − 250 = 134.71 mm = 0.13471 m eD = 352.14 − 250 = 102.14 mm = 0.10214 m V= Potential energies: 1 2 ke 2 1 (135 N/m)(0.31214 m) 2 = 6.5767 J 2 1 VB = (135 N/m)(0.13471 m) 2 = 1.2249 J 2 1 VD = (135 N/m)(0.10214 m)2 = 0.7042 J 2 VA = Since the semicircular rod ABC is horizontal, there is no change in gravitational potential energy. m = 600 g = 0.600 kg Mass of collar: Kinetic energies: 1 2 mv A = 0.300 v A2 = 0 2 1 TB = mvB2 = 0.300 vB2 2 1 2 TD = mvD = 0.300 vD2 2 TA = 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. 583 PROBLEM 13.57 (Continued) (a) Speed of collar at B. Conservation of energy: TA + VA = TB + VB 0 + 6.5767 = 0.300vB2 + 1.2249 vB2 = 17.839 m 2 /s 2 (b) vB = 4.22 m/s Speed of collar at D. Conservation of energy: TA + VA = TD + VD 0 + 6.5767 = 0.300 v02 + 0.7042 vD2 = 19.575 m 2 /s 2 vD = 4.42 m/s 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 13.58 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 k = 1.5 lb/in. 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 L0 = 7 in., LDA = 20 in. LDB = (8 + 6)2 + 62 = 15.23 in. LDC = 8 in. Δ LDA = 20 − 7 = 13 in. Δ LDB = 15.23 − 7 = 8.23 in. Δ LDC = 8 − 7 = 1 in. (a) TA = 0, VA = 1 1 k (ΔLDA )2 = (1.5)(13) 2 = 126.75 lb ⋅ in. 2 2 = 10.5625 lb ⋅ ft TB = VB = 1 2 1.5 2 mvB = vB 2 g 1 (1.5)(8.23) 2 = 50.8 lb ⋅ in. = 4.233 lb ⋅ ft 2 TA + VA = TB + VB : 0 + 10.5625 = 1.5vB2 + 4.233 32.2 vB = 11.66 ft/s 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 (Continued) (b) TA = 0, VA = 10.5625 lb ⋅ ft, TC = VC = 1.5 2 vC 32.2 1 (1.5)(1) 2 = 0.75 lb ⋅ in. = 0.0625 lb ⋅ ft 2 TA + VA = TC + VC : 0 + 10.5625 = 1.5 2 vc + 0.0625 32.2 vC = 15.01 ft/s 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 13.59 A 3-lb collar C may slide without friction along a horizontal rod. It is attached to three springs, each of constant k = 2 lb/in. and 6 in. undeformed length. Knowing that the collar is released from rest in the position shown, determine the maximum speed it will reach in the ensuing motion. SOLUTION Maximum velocity occurs at E where collar is passing through position of equilibrium. Position T1 = 0 Note: Undeformed length of springs is 6 in. = 0.5 ft. Spring AC: L = (1 ft) 2 + (0.5 ft) 2 = 1.1180 ft Δ = 1.1180 − 0.50 = 0.6180 ft Spring CD: L = (0.5 ft)2 + (0.5 ft)2 = 0.70711 ft Δ = 0.70711 − 0.50 = 0.20711 ft Spring BD: L = 0.50 ft, Δ = 0 Potential energy. (k = 2 lb/in. = 24 lb/ft for each spring) 1 1 1 V1 = Σ k Δ 2 = k ΣΔ 2 = (24 lb/ft)[(0.6180 ft) 2 + (0.20711 ft) 2 + 0] 2 2 2 V1 = 5.0983 lb ⋅ ft Position 3.0 lb 1 1 m= = 0.093168 slug; T2 = mv22 = (0.093168 slug) v22 2 2 2 32.2 ft/s Spring AC: L = (0.5 ft)2 + (0.5 ft)2 = 0.7071067 ft Δ = 0.70711 − 0.50 = 0.20711 ft Spring CD: L = 0.50 ft Δ=0 Spring BC: L = (0.5 ft) 2 + (0.5 ft)2 = 0.7071067 ft Δ = 0.70711 − 0.50 = 0.20711 ft 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. 587 PROBLEM 13.59 (Continued) Potential energy. 1 1 V2 = Σ k Δ 2 = k ΣΔ 2 2 2 1 V2 = (24 lb/ft)[(0.20711 ft) 2 + 0 + (0.20711 ft) 2 ] = 1.0294 lb ⋅ ft 2 Conservation of energy. T1 + V1 = T2 + V2 : 0 + 5.0983 lb ⋅ ft = 1 (0.093168 slug)v22 + 1.0294 lb ⋅ ft 2 v22 = 87.345 v 2 = 9.35 ft/s ↔ 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. 588 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 k = 400 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: TA = 1 2 0.5 mv A = kg v A2 2 2 TA = (0.25)v A2 ΔLA = 0.150 m − 0.080 m ΔLA = 0.070 m 1 k ( ΔLA )2 2 1 VA = (400 × 103 N/m)(0.070 m)2 2 VA = 980 J VA = vB = 0 TB = 0 ΔLB = 0.200 m − 0.080 m = 0.120 m 1 1 k ( ΔLB )2 = (400 × 103 N/m)(0.120 m)2 2 2 VB = 2880 J VB = Substitute into conservation of energy. TA + VA = TB + VB v A2 = 0.25v 2A + 980 = 0 + 2880 (2880 − 980) (0.25) v A2 = 7600 m 2 /s 2 vA = 87.2 m/s 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 (Continued) (b) Velocity at C: Since slope at B is positive, the component of the spring force FP , parallel to the rod, causes the block to move back toward A. TB = 0, VB = 2880 J [from part (a)] 1 2 (0.5 kg) 2 mvC = vC = 0.25vC2 2 2 ΔLC = 0.100 m − 0.080 m = 0.020 m TC = VC = 1 1 k (Δ LC ) 2 = (400 × 103 N/m)(0.020 m) 2 = 80.0 J 2 2 Substitute into conservation of energy. TB + VB = TC + VC 0 + 2880 = 0.25vC2 + 80.0 vC2 = 11, 200 m 2 /s 2 vC = 105.8 m/s 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 13.61 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. Length AC ′B = 1.0 m; Elongation = x1 = 1.0 − 3 ΣFx = 0: 2 F1 − 240 N = 0 F1 = 200 N 5 Position 2. Length ACB = 0.8 m; Elongation = x2 = 0.8 − Given F2 = 40 N F1 = kx, F2 = k x2 F1 − F2 = k ( x1 − x2 ) 200 − 40 = k [(1.0 − ) − (0.8 − )] = 0.2k k= 160 = 800 N/m 0.2 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. 591 PROBLEM 13.61 (Continued) Position : Position: T1 = 0 x1 = F1 200 N = = 0.25 m k 800 N/m x2 = F2 40 N = = 0.05 m k 800 N/m V1 = 1 2 1 kx1 = (800 N/m)(0.25 m) 2 = 25.0 N ⋅ m 2 2 m = 0.10 kg 1 2 1 mv2 = (0.1 kg) v22 = 0.05 v22 2 2 1 1 V2 = k x22 = (800 N/m)(0.05 m) 2 = 1 N ⋅ m 2 2 T2 = Conservation of energy: T1 + V1 = T2 + V2 0 + 25.0 N ⋅ m = 0.05v22 + 1.0 N ⋅ m 24.0 = 0.05v22 v2 = 21.909 m/s v2 = 21.9 m/s Note: The horizontal force applied at the midpoint of the cord is not proportional to the horizontal distance C ′C. A solution based on the work of the horizontal force would be rather involved. 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. 592 PROBLEM 13.62 An elastic cable is to be designed for bungee jumping from a tower 130 ft high. The specifications call for the cable to be 85 ft long when unstretched, and to stretch to a total length of 100 ft when a 600-lb weight is attached to it and dropped from the tower. Determine (a) the required spring constant k of the cable, (b) how close to the ground a 186-lb man will come if he uses this cable to jump from the tower. SOLUTION (a) Conservation of energy: V1 = 0 T1 = 0 V1 = 100 W V1 = (100 ft)(600 lb) Datum at : = 6 × 104 ft ⋅ lb V2 = 0 T2 = 0 V2 = Vg + Ve = 0 + 1 k (15 ft) 2 2 T1 + V1 = T2 + V2 0 + 6 × 104 = 0 + (112.5)k k = 533 lb/ft (b) From (a), k = 533 lb/ft T1 = 0 W = 186 lb V1 = (186)(130 − d ) T2 = 0 Datum: 1 (533)(130 − 85 − d ) 2 2 V2 = (266.67)(45 − d )2 V2 = Vg + Ve = 0 + 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. 593 PROBLEM 13.62 (Continued) d = distance from the ground T1 + V1 = T2 + V2 0 + (186)(130 − d ) = 0 + (266.67)(45 − d ) 2 266.7 d 2 − 23815d + 515827 = 0 d= 23815 (23815) 2 − 4(266.7)(515827) 36.99 ft = 52.3 ft (2)(266.7) Discard 52.3 ft (since the cord acts in compression when rebound occurs). d = 37.0 ft 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 13.63 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 m= Mass of machinery: W 4000 = = 124.22 lb ⋅ s 2 /ft g 32.2 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, A= π 4 (diameter)2 = π 4 (0.4 in.)2 = 0.12566 in 2 AE = (0.12556 in.2 )(29 × 106 lb/in 2 ) = 3.6442 × 106 lb AE 3.6442 × 106 lb = = 121.47 × 103 lb/ft L 30 ft For L = 30 ft, Initial force in cable (equilibrium): F1 = W = 4000 lb. Elongation in position 1: x1 = F1 4000 = = 0.03293 ft k 121.47 × 103 Potential energy: V1 = 1 2 F12 kx1 = 2 2k V1 = (4000 lb) 2 = 65.860 ft ⋅ lb (2)(121.47 × 103 lb/ft) T1 = 1 2 mv1 2 T1 = 1 (124.22 lb ⋅ s 2 /ft)(3 ft/s2 ) = 558.99 ft ⋅ lb 2 Kinetic energy: 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. 595 PROBLEM 13.63 (Continued) Let position 2 be the position of maximum downward displacement. Let x2 be the elongation in this position. Potential energy: V2 = 1 2 kx2 − W ( x2 − x1 ) 2 1 (121.47 × 103 ) x22 − (4000)( x2 − 0.03293) 2 = 60.735 × 103 x 2 − 4000 x2 + 131.72 V2 = Kinetic energy: T2 = 0 (since v2 = 0) Principle of work and energy: T1 + V1 = T2 + V2 558.99 + 65.860 = 60.735 × 103 x22 − 4000 x2 + 131.72 60.735 × 103 x22 − 4000 x2 − 493.13 = 0 x2 = 0.12887 ft Maximum displacement: δ = x2 − x1 = 0.09594 ft δ = 1.151 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. 596 PROBLEM 13.64 A 2-kg collar is attached to a spring and slides without friction in a vertical plane along the curved rod ABC. The spring is undeformed when the collar is at C and its constant is 600 N/m. If the collar is released at A with no initial velocity, determine its velocity (a) as it passes through B, (b) as it reaches C. SOLUTION Spring elongations: At A, x A = 250 mm − 150 mm = 100 mm = 0.100 m At B, xB = 200 mm − 150 mm = 50 mm = 0.050 m At C , xC = 0 Potential energies for springs. 1 2 1 kx A = (600)(0.100) 2 = 3.00 J 2 2 1 2 1 (VB )e = kxB = (600)(0.050) 2 = 0.75 J 2 2 (VC )e = 0 (VA )e = Gravitational potential energies: Choose the datum at level AOC. (VA ) g = (VC ) g = 0 (VB ) g = −mg y = −(2)(9.81)(0.200) = −3.924 J Kinetic energies: TA = 0 1 2 mvB = 1.00 vB2 2 1 TC = mvC2 = 1.00 vC2 2 TB = (a) Velocity as the collar passes through B. Conservation of energy: TA + VA = TB + VB 0 + 3.00 + 0 = 1.00 vB2 + 0.75 − 3.924 vB2 = 6.174 m 2 /s 2 v B = 2.48 m/s 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. 597 PROBLEM 13.64 (Continued) (b) Velocity as the collar reaches C. Conservation of energy: TA + VA = TC + VC 0 + 3.00 + 0 = 1.00vC2 + 0 + 0 vC2 = 3.00 m 2 /s 2 vC = 1.732 m/s 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. 598 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 LAF = (0.5) 2 + (0.4) 2 + (0.3) 2 LAF = 0.70711 m LBF = (0.4) 2 + (0.3) 2 LBF = 0.5 m LFE = (0.5) 2 + (0.3) 2 LFE = 0.58309 m V = Ve + Vg (a) Speed at B: v A = 0, TA = 0 Point A: (VA )e = 1 k (Δ LAF ) 2 2 Δ LAF = LAF − L0 = 0.70711 − 0.25 Δ LAF = 0.45711 m 1 (75 N/m)(0.45711 m) 2 2 (VA )e = 7.8355 N ⋅ m (VA )e = (VA ) g = ( mg )(0.4) = (1.0 kg)(9.81 m/s 2 )(0.4 m) = 3.9240 N ⋅ m VA = (VA )e + (VA ) g = 7.8355 + 3.9240 = 11.7595 N ⋅ m 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 (Continued) Point B: 1 2 1 mvB = (1.0 kg)vB2 2 2 2 TB = 0.5 vB TB = (VB )e = 1 k (Δ LBF ) 2 2 Δ LBF = LBF − L0 = 0.5 − 0.25 Δ LBF = 0.25 m 1 (75 N/m)(0.25 m) 2 = 2.3438 N ⋅ m 2 (VB ) g = ( mg )(0.4) = (1.0 kg)(9.81 m/s2 )(0.4 m) = 3.9240 N ⋅ m (VB )e = VB = (vB )e + (VB ) g = 2.3438 + 3.9240 = 6.2678 N ⋅ m TA + VA = TB + VB 0 + 11.7595 = 0.5 vB2 + 6.2678 vB2 = (5.49169)/(0.5) vB2 = 10.983 m 2 /s 2 (b) vB = 3.31 m/s Speed at E: Point A: TA = 0 VA = 11.7595 N ⋅ m (from part (a)) Point E: 1 2 1 mvE = (1.0 kg)vE2 = 0.5vE2 2 2 1 (VE )e = k ( Δ LFE ) 2 Δ LFE = LFE − L0 = 0.5831 − 0.25 2 TE = ΔLFE = 0.3331 m 1 (75 N/m)(0.3331 m)2 = 4.1607 N ⋅ m 2 (VE ) g = 0 VE = 4.1607 N ⋅ m (VE )e = TA + VA = TE + VE 0 + 11.7595 = 0.5 vE2 + 4.1607 vE2 = 7.5988/0.5 vE2 = 15.1976 m 2 /s 2 vE = 3.90 m/s 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 13.66 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 k = 3 lb/ft 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. vC = 0 vD = 0 TC = 0 TD = 0 V = Ve + Vg Point C: Δ LBC = (1 ft)(θ ) = θ ft 1 k (ΔLBC ) 2 2 3 (VC )e = θ 2 2 (VC )e = R = 12 in. = 1 ft (VC ) g = WR(1 − cos θ ) 8 oz (VC ) g = (1 ft)(1 − cosθ ) 16 oz/lb (VC ) g = 1 (1 − cos θ ) 2 3 1 VC = (VC )e + (VC ) g = θ 2 + (1 − cos θ ) 2 2 Point D: (VD )e = 0 (spring is unattached) (VD ) g = W (2 R ) = (2)(0.5 lb)(1 ft) = 1 lb ⋅ ft TC + VC = TD + VD By trial, 3 1 0 + θ 2 + (1 − cos θ ) = 1 2 2 2 (1.5)θ − (0.5) cos θ = 0.5 θ = 0.7592 rad θ = 43.5° 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 (Continued) (b) Velocity at A: Point D: VD = 0 TD = 0 VD = 1 lb ⋅ ft[see Part (a)] Point A: TA = 1 2 1 (0.5 lb) 2 mv A = vA 2 2 (32.2 ft/s 2 ) TA = 0.0077640v A2 VA = (VA ) g = W ( R) = (0.5 lb)(1 ft) = 0.5 lb ⋅ ft TA + VA = TD + VD 0.0077640v A2 + 0.5 = 0 + 1 v A2 = 64.4 ft 2 /s 2 vA = 8.02 ft/s 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. 602 PROBLEM 13.67 The system shown is in equilibrium when φ = 0. Knowing that initially φ = 90° and that block C is given a slight nudge when the system is in that position, determine the velocity of the block as it passes through the equilibrium position φ = 0. Neglect the weight of the rod. SOLUTION Find the unstretched length of the spring. 1.1 0.3 = 1.3045 rad θ = 74.745° θ = tan −1 LBD = (1.1) 2 + .32 LBD = 1.140 ft Equilibrium ΣM A = (0.3)( Fs sin θ ) − (25)(2.1) = 0 (25 lb)(2.1 ft) (0.3 ft)(sin 74.745°) = 181.39 lb Fs = k ΔLBD Fs = 181.39 lb = (600 lb/ft)(ΔLBD ) ΔLBD = 0.30232 ft Unstretched length L0 = LBD − ΔLBD L0 = 1.140 − 0.3023 = 0.83768 ft ′ , when φ = 90°. Spring elongation, ΔLBD ′ = (1.1 ft + 0.3 ft) − L0 ΔLBD ′ = 1.4 ft − 0.8377 ft ΔLBD = 0.56232 ft 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. 603 PROBLEM 13.67 (Continued) At , (φ = 90°) v1 = 0, T1 = 0 V1 = (V1 )e + (V1 ) g 1 ′ )2 k (ΔLBD 2 1 (V1 )e = (600 lb/ft)(0.5623 ft) 2 2 (V1 )e = 94.86 lb ⋅ ft (V1 )e = (V1 ) g = −(25 lb)(2.1 ft) = −52.5 ft ⋅ lb V1 = 94.86 − 52.5 = 42.36 ft ⋅ lb At , (φ = 0°) 1 1 k ( ΔLBD ) 2 = (600 lb/ft)(0.3023 ft) 2 2 2 (V2 )e = 27.42 lb ⋅ ft (V2 )e = (V2 ) g = 0 V2 = 27.42 ft ⋅ lb 1 2 1 25 lb mv2 = 2 2 32.2 ft/s 2 T1 + V1 = T2 + V2 T2 = 2 2 v2 = 0.3882 v2 0 + 42.36 = 0.3882 v22 + 27.42 v22 = (14.941)/(0.3882) v22 = 38.48 ft 2 /s 2 v2 = 6.20 ft/s 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. 604 PROBLEM 13.68 A spring is used to stop a 50-kg package which is moving down a 20º incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest. SOLUTION Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the spring. Use the principle of conservation of energy. T1 + V1 = T2 + V2 . Position 1: 1 2 1 mv1 = (50)(2)2 = 100 J 2 2 V1g = mgh1 = (50)(9.81)(8 sin 20°) = 1342.09 J T1 = V1e = 1 2 1 ke1 = (30 × 103 )(0.050) 2 = 37.5 J 2 2 1 2 mv2 = 0 since v2 = 0. 2 = mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x T2 = Position 2: V2 g V2e = 1 2 1 ke2 = (30 × 103 )(0.05 + x)2 = 37.5 + 1500 x + 15000 x 2 2 2 Principle of conservation of energy: 100 + 1342.09 + 37.5 = −167.61x + 37.5 + 1500 x + 15000 x 2 15, 000 x 2 + 1332.24 x − 1442.09 = 0 Solving for x, x = 0.26882 and − 0.357 64 x = 0.269 m 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. 605 PROBLEM 13.69 Solve Problem 13.68 assuming the kinetic coefficient of friction between the package and the incline is 0.2. PROBLEM 13.68 A spring is used to stop a 50-kg package which is moving down a 20° incline. The spring has a constant k = 30 kN/m and is held by cables so that it is initially compressed 50 mm. Knowing that the velocity of the package is 2 m/s when it is 8 m from the spring and neglecting friction, determine the maximum additional deformation of the spring in bringing the package to rest. SOLUTION Let position 1 be the starting position 8 m from the end of the spring when it is compressed 50 mm by the cable. Let position 2 be the position of maximum compression. Let x be the additional compression of the spring. Use the principle of work and energy. T1 + V1 + U1→2 = T2 + V2 Position 1. 1 2 1 mv1 = (50)(2) 2 = 100 J 2 2 V1g = mgh1 = (50)(9.81)(8sin 20°) = 1342.09 J T1 = V1e = Position 2. 1 2 1 ke1 = (30 × 103 )(0.05) 2 = 37.5 J 2 2 1 2 mv2 = 0 since v2 = 0. 2 = mgh2 = (50)(9.81)(− x sin 20°) = −167.76 x T2 = V2 g V2e = 1 2 1 ke2 = (30 × 103 )(0.05 + x) 2 = 37.5 + 1500 x + 15,000 x 2 2 2 Work of the friction force. ΣFn = 0 N − mg cos 20° = 0 N = mg cos 20° = (50)(9.81) cos 20° = 460.92 N F f = μk N U1→2 = (0.2)(460.92) = 92.184 = − Ff d = −92.184(8 + x) = −737.47 − 92.184 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. 606 PROBLEM 13.69 (Continued) Principle of work and energy: T1 + V1 + U1− 2 = T2 + V2 100 + 1342.09 + 37.5 − 737.47 − 92.184 x = −167.76 x + 37.5 + 1500 x + 15, 000 x 2 15, 000 x 2 + 1424.42 x − 704.62 = 0 Solving for x, x = 0.17440 and −0.26936 x = 0.1744 m 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. 607 PROBLEM 13.70 A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach Point A with practically no velocity and then drop freely along the track. Determine the normal force exerted by the track on the car as the car reaches point B. Ignore air resistance and rolling resistance. SOLUTION Calculate the speed of the car as it reaches Point B using the principle of conservation of energy as the car travels from position A to position B. Position A: v A = 0, TA = Position B: VB = −mgh 1 2 mv A = 0, VA = 0 (datum) 2 where h is the decrease in elevation between A and B. TB = Conservation of energy: 1 2 mvB 2 TA + VA = TB + VB : 1 2 mvB − mgh 2 vB2 = 2 gh 0+0= = (2)(9.81 m/s 2 )(27 m)(1 − cos 40°) = 123.94 m 2 /s 2 Normal acceleration at B: ( aB ) n = vB2 ρ = 123.94 m 2 /s 2 = 4.59 m/s 2 27 m (a B )n = 4.59 m/s 2 50° Apply Newton’s second law to the car at 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. 608 PROBLEM 13.70 (Continued) 50°ΣFn = man : N − mg cos 40° = − man N = mg cos 40° − man = m( g cos 40° − an ) = (250 kg)[(9.81 m/s 2 ) cos 40° − 4.59 m/s 2 ] = 1878.7 − 1147.5 N = 731 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. 609 PROBLEM 13.71 A section of track for a roller coaster consists of two circular arcs AB and CD joined by a straight portion BC. The radius of AB is 27 m and the radius of CD is 72 m. The car and its occupants, of total mass 250 kg, reach Point A with practically no velocity and then drop freely along the track. Determine the maximum and minimum values of the normal force exerted by the track on the car as the car travels from A to D. Ignore air resistance and rolling resistance. SOLUTION Calculate the speed of the car as it reaches Point P, any point on the roller coaster track. Apply the principle of conservation of energy. Position A: v A = 0, TA = Position P: VP = −mgh 1 2 mv A = 0, VA = 0 (datum) 2 where h is the decrease in elevation along the track. TP = Conservation of energy: 1 mv 2 2 TA + VA = TP + VP 0+0= 1 2 mv − mgh 2 v 2 = 2 gh (1) Calculate the normal force using Newton’s second law. Let θ be the slope angle of the track. ΣFn = man : N − mg cos θ = man N = mg cos θ + man Over portion AB of the track, and (2) h = ρ (1 − cos θ ) an = − mv 2 ρ 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. 610 PROBLEM 13.71 (Continued) where ρ is the radius of curvature. ( ρ = 27 m) N = mg cos θ − 2mg ρ (1 − cos θ ) ρ = mg (3cos θ − 2) At Point A (θ = 0) N A = mg = (250)(9.81) = 2452.5 N At Point B (θ = 40°) N B = (2452.5)(3cos 40° − 2) N B = 731 N Over portion BC, θ = 40°, an = 0 (straight track) N BC = mg cos 40° = 2452.5cos 40° N BC = 1879 N Over portion CD, h = hmax − r (1 − cos θ ) an = and mv 2 r where r is the radius of curvature. (r = 72 m) 2mgh r h = mg cos θ + 2mg max − 1 − cos θ r N = mg cos θ + 2h = mg 3cos θ − 2 + max r which is maximum at Point D, where 2h N D = mg 1 + max r Data: hmax = 27 + 18 = 45 m, r = 72 m (2) (45) N D = (2452.5) 1 + = 5520 N 72 Summary: minimum (just above B): 731 N 5520 N maximum (at 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. 611 PROBLEM 13.72 A 1-lb collar is attached to a spring and slides without friction along a circular rod in a vertical plane. The spring has an undeformed length of 5 in. and a constant k = 10 lb/ft. Knowing that the collar is released from being held at A determine the speed of the collar and the normal force between the collar and the rod as the collar passes through B. SOLUTION W 1 = = 0.031056 lb ⋅ s 2 /ft g 32.2 For the collar, m= For the spring, k = 10 lb/ft l0 = 5 in. A = 7 + 5 + 5 = 17 in. At A: Δ − 0 = 12 in. = 1 ft B = (7 + 5) 2 + 52 = 13 in. At B: B − 0 = 1.8 in. = 2 ft 3 Velocity of the collar at B. Use the principle of conservation of energy. TA + VA = TB + VB Where TA = 1 2 mv A = 0 2 1 k ( A − 0 ) 2 + W (0) 2 1 = (10)(1) 2 + 0 = 5 ft ⋅ lb 2 1 2 1 TB = mvB = (0.031056)vB2 = 0.015528vB2 2 2 1 VB = k ( B − 0 ) 2 + Wh 2 VA = 2 1 2 5 (10) + (1) − 2 3 12 = 1.80556 ft ⋅ lb = 0 + 5 = 0.015528vB2 = 1.80556 vB = 14.34 ft/s vB2 = 205.72 ft 2 /s 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. 612 PROBLEM 13.72 (Continued) Forces at B. 2 Fs = k ( B − 0 ) = (10) = 6.6667 lb. 3 5 sinα = 13 5 ρ = 5 in. = ft 12 mvB2 man = ρ (0.031056)(205.72) 5/12 = 15.3332 lb = ΣFy = ma y : Fs sin α − W + N = man N = man + W − Fs sin α 5 = 15.3332 + 1 − (6.6667) 13 N = 13.769 lb N = 13.77 lb 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. 613 PROBLEM 13.73 A 10-lb collar is attached to a spring and slides without friction along a fixed rod in a vertical plane. The spring has an undeformed length of 14 in. and a constant k = 4 lb/in. Knowing that the collar is released from rest in the position shown, determine the force exerted by the rod on the collar at (a) Point A, (b) Point B. Both these points are on the curved portion of the rod. SOLUTION Mass of collar: m= W 10 = = 0.31056 lb ⋅ s 2 /ft g 32.2 Let position 1 be the initial position shown, and calculate the potential energies of the spring for positions 1, A, and B. l0 = 14 in. l1 = (14 + 14)2 + (14) 2 = 31.305 in. x1 = l1 − l0 = 31.305 − 14 = 17.305 in. (V1 )e = 1 2 1 kx1 = (4)(17.305) 2 = 598.92 in ⋅ lb = 49.910 ft ⋅ lb 2 2 l A = (14) 2 + (14) 2 = 19.799 in. x A = l A − l0 = 19.799 − 14 = 5.799 in. (VA )e = 1 2 1 kx A = (4)(5.799)2 = 67.257 in ⋅ lb = 5.605 ft ⋅ lb 2 2 lB = 14 + 14 = 28 in. xB = lB − l0 = 28 − 14 = 14 in. (VB )e = Gravitational potential energies: 1 2 1 kxB = (4)(14) 2 = 392 in ⋅ lb = 32.667 ft ⋅ lb 2 2 Datum at level A. (V1 ) g = 0 (VA ) g = 0 (VB ) g = Wy = (10 lb)(−14 in.) = −140 in ⋅ lb = −11.667 ft ⋅ lb 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. 614 PROBLEM 13.73 (Continued) Total potential energies: V = Ve + Vg V1 = 49.910 ft ⋅ lb, VA = 5.605 ft ⋅ lb, VB = 21.0 ft ⋅ lb Kinetic energies: T1 = 0 1 2 1 mv A = (0.31056)v A2 = 0.15528v A2 2 2 1 2 1 TB = mvB = (0.31056)vB2 = 0.15528vB2 2 2 TA = Conservation of energy: T1 + V1 = TA + VA : 0 + 49.910 = 0.15528v 2A + 5.605 Conservation of energy: T1 + V1 = TB + VB 0 + 49.910 = 0.15528vB2 + 21.0 Normal accelerations at A and B. v 2A = 285.32 ft 2 /s 2 vB2 = 186.18 ft 2 /s2 an = v 2 /ρ ρ = 14 in. = 1.16667 ft Spring forces at A and B: (a A )n = 285.32 ft 2 /s 2 1.16667 ft (a A )n = 244.56 ft/s 2 ( aB ) n = 189.10 ft 2 /s 2 1.16667 ft (a B ) n = 159.58 ft/s 2 F = kx FA = (4 lb/in.)(5.799 in.) FB = (4 lb/in.)(14 in.) FA = 23.196 lb 45° FB = 56.0 lb To determine the forces (N A and N B ) exerted by the rod on the collar, apply Newton’s second law. (a) At Point A: ΣF = m ( a A ) n : W + N A − FA sin 45° = m( a A ) n 10 + N A − 23.196sin 45° = (0.31056)(244.56) N A = 82.4 lb (b) At Point B: ΣF = m ( a B ) n : N B = (0.31056)(159.58) N B = 49.6 lb 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. 615 PROBLEM 13.74 An 8-oz package is projected upward with a velocity v0 by a spring at A; it moves around a frictionless loop and is deposited at C. For each of the two loops shown, determine (a) the smallest velocity v0 for which the package will reach C, (b) the corresponding force exerted by the package on the loop just before the package leaves the loop at C. SOLUTION Loop 1 (a) The smallest velocity at B will occur when the force exerted by the tube on the package is zero. ΣF = 0 + mg = mvB2 r vB2 = rg = 1.5 ft(32.2 ft/s 2 ) vB2 = 48.30 TA = At A 1 2 mv0 2 0.5 VA = 0 8 oz = 0.5 lb = = 0.01553 32.2 TB = At B 1 2 1 mvB = m (48.30) = 24.15 m 2 2 VB = mg (7.5 + 1.5) = 9 mg = 9(0.5) = 4.5 lb ⋅ ft TA + VA = TB + VB : 1 (0.01553)v02 = 24.15(0.01553) + 4.5 2 v02 = 627.82 v0 = 25.056 v0 = 25.1 ft/s At C TC = 1 2 mvC = 0.007765vC2 2 VC = 7.5 mg = 7.5(0.5) = 3.75 TA + VA = TC + VC : 0.007765v02 = 0.007765vC2 + 3.75 0.007765(25.056) 2 − 3.75 = 0.007765vC2 vC2 = 144.87 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. 616 PROBLEM 13.74 (Continued) (b) ΣF = man : N = 0.01553 (144.87) 1.5 N = 1.49989 Loop 2 {Package in tube} NC = 1.500 lb (a) At B, tube supports the package so, vB ≈ 0 vB = 0, TB = 0 VB = mg (7.5 + 1.5) = 4.5 lb ⋅ ft TA + VA = TB + VB 1 (0.01553)v A2 = 4.5 v A = 24.073 2 v A = 24.1 ft/s TC = 0.007765vC2 , VC = 7.5 mg = 3.75 (b) At C TA + VA = TC + VC : 0.007765(24.073) 2 = 0.007765vC2 + 3.75 vC2 = 96.573 96.573 NC = 0.01553 = 0.99985 1.5 {Package on tube} NC = 1.000 lb 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. 617 PROBLEM 13.75 If the package of Problem 13.74 is not to hit the horizontal surface at C with a speed greater than 10 ft/s, (a) show that this requirement can be satisfied only by the second loop, (b) determine the largest allowable initial velocity v 0 when the second loop is used. SOLUTION (a) Loop 1 From Problem 13.74, at B vB2 = gr = 48.3 ft 2 /s 2 vB = 6.9498 ft/s TB = 1 2 1 mvB = (0.01553)(48.3) = 0.37505 2 2 VB = mg (7.5 + 1.5) = (0.5)(9) = 4.5 lb ⋅ ft TC = 1 2 1 mvC = (0.01553)vC2 = 0.007765vC2 2 2 VC = 7.5(0.5) = 3.75 lb ⋅ ft TB + VB = TC + VC : 0.37505 + 4.5 = 0.007765vC2 + 3.75 vC2 = 144.887 vC = 12.039 ft/s 12.04 ft/s > 10 ft/s Loop (1) does not work 1 TA = mv02 = 0.007765v02 2 VA = 0 (b) Loop 2 at A vC = 10 ft/s At C assume TC = 1 2 mvC = 0.007765(10) 2 = 0.7765 2 vC = 7.5(0.5) = 3.75 TA + VA = TC + VC : 0.007765v02 = 0.7765 + 3.75 v0 = 24.144 v0 = 24.1 ft/s 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. 618 PROBLEM 13.76 A small package of weight W is projected into a vertical return loop at A with a velocity v0. The package travels without friction along a circle of radius r and is deposited on a horizontal surface at C. For each of the two loops shown, determine (a) the smallest velocity v0 for which the package will reach the horizontal surface at C, (b) the corresponding force exerted by the loop on the package as it passes Point B. SOLUTION Loop 1: (a) Newton’s second law at position C: ΣF = ma: mg = m vc2 r vc2 = gr Conservation of energy between position A and B. 1 2 mv0 2 VA = 0 TA = 1 2 1 mvC = mgr 2 2 VC = mg (2r ) = 2mgr TC = TA + VA = TC + VC : 1 2 1 mv0 + 0 = mgr + 2mgr 2 2 v02 = 5gr v0 = Smallest velocity v 0: 5 gr (b) Conservation of energy between positions A and B. (b) TB = 1 2 mvB ; 2 VB = mg (r ) TA + VA = TB + VB: 1 2 1 mv A + 0 = mvB2 + mgr 2 2 1 1 m(5 gr ) + 0 = mvB2 + mgr 2 2 vB2 = 3gr 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. 619 PROBLEM 13.76 (Continued) Newton’s second law at position B. man = m vB2 3gr =m = 3 mg r r ΣF = ΣFeff : N B = 3 mg N B = 3W Force exerted by the loop: Loop 2: (a) At point C, vc = 0 Conservation of energy between positions A and C. 1 2 mvC = 0 2 VC = mg (2r ) = 2mgr TC = TA + VA = TC + VC : 1 2 mv0 + 0 = 0 + 2mgr 2 v02 = 4 gr v0 = Smallest velocity v 0: (b) 4 gr Conservation of energy between positions A and B. TA + VA − TB + VB : 1 2 1 mv0 + 0 = mvB2 + mgr 2 2 1 1 m(4 gr ) = mvB2 + mgr vB2 = 2 gr 2 2 Newton’s second law at position B. man = m vB2 2 gr =m = 2 mg r r ΣF = Σeff : N = 2 mg Force exerted by loop: N = 2W 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. 620 PROBLEM 13.77 The 1 kg ball at A is suspended by an inextensible cord and given an initial horizontal velocity of 5 m/s. If l = 0.6 m and xB = 0, determine yB so that the ball will enter the basket. SOLUTION Let position 1 be at A. v1 = v0 Let position 2 be the point described by the angle where the path of the ball changes from circular to parabolic. At position 2, the tension Q in the cord is zero. Relationship between v2 and θ based on Q = 0. Draw the free body diagram. ΣF = 0: Q + mg sin θ = man = With Q = 0, v22 = gl sin θ mv22 l or v2 = gl sin θ (1) Relationship among v0 , v2 and θ based on conservation of energy. T1 + V1 = T2 + V2 1 2 1 mv0 − mgl = mv22 + mgl sin θ 2 2 v02 − v22 = 2 gl (1 + sin θ ) (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. 621 PROBLEM 13.77 (Continued) Eliminating v2 from Eqs. (1) and (2), v02 − gl sin θ = 2 gl (1 + sin θ ) 1 (5) 2 1 v02 − 2 = 0.74912 − 2 = 3 gl 3 (9.81)(0.6) θ = 48.514° sin θ = From Eq. (1), v22 = (9.81)(0.6)sin 48.514° = 4.4093 m 2/s 2 v2 = 2.0998 m/s x and y coordinates at position 2. x2 = l cos θ = 0.6cos 48.514° = 0.39746 m y2 = l sin θ = 0.6sin 48.514° = 0.44947 m Let t2 be the time when the ball is a position 2. Motion on the parabolic path. The horizontal motion is x = −v2 sin θ = −2.0998 sin 48.514° = −1.5730 m/s x = x2 − 1.5730(t − t2 ) At Point B, xB = 0 0 = 0.39746 − 1.5730 (t B − t2 ) t B − t2 = 0.25267 s The vertical motion is y = y2 + v2 cos θ (t − t2 ) − At Point B, 1 g (t − t2 ) 2 2 yB = y2 + v2 cos θ (t B − t2 ) − 1 g (t B − t2 ) 2 2 yB = 0.44947 + (2.0998 cos 48.514°)(0.25267) 1 − (9.81)(0.25267) 2 2 = 0.48779 m yB = 0.448 m 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. 622 PROBLEM 13.78* Packages are moved from Point A on the upper floor of a warehouse to Point B on the lower floor, 12 ft directly below A, by means of a chute, the centerline of which is in the shape of a helix of vertical axis y and radius R = 8 ft. The cross section of the chute is to be banked in such a way that each package, after being released at A with no velocity, will slide along the centerline of the chute without ever touching its edges. Neglecting friction, (a) express as a function of the elevation y of a given Point P of the centerline the angle φ formed by the normal to the surface of the chute at P and the principal normal of the centerline at that point, (b) determine the magnitude and direction of the force exerted by the chute on a 20-lb package as it reaches Point B. Hint: The principal normal to the helix at any Point P is horizontal and directed toward the y axis, and the radius of curvature of the helix is ρ = R[1 – (h/2πR)2]. SOLUTION (a) v A = 0 TA = 0 At Point A: VA = mgh TP = At any Point P: 1 2 mv 2 VP = Wy = mgy TA + VA = TP + VP 1 2 mv + mgy 2 v 2 = 2 g (h − y ) 0 + mgh = en along principal normal, horizontal and directed toward y axis et tangent to centerline of the chute eD along binormal β = tan −1 h (12 ft) = tan −1 = 13.427° 2π R 2π (8 ft) mab = 0 since ab = 0 Note: Friction is zero, ΣFt = mat : mg sin β = mat at = g sin β ΣFb = mab : N b − W cos β = 0 N B = W cos β ΣFn = man : N n = mv 2 m 2 g (h − y ) (h − y) = = 2W e e e 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. 623 PROBLEM 13.78* (Continued) The total normal force is the resultant of Nb and Nm, it lies in the b–m plane, and forms angle φ with m axis. tan φ = Nb /N n 2( w(h − y ) e tan φ = (e /2(h − y )) cos β tan φ = W cos β h 2 R e = R 1 + = R(1 + tan 2 β ) = cos 2 β 2π R Given: e R cos β = 2( h − y ) 2( h − y ) cos β 8 ft 4.112 tan φ = = 2(12 − y ) cos13.427° 12 − y tan φ = Thus, cot φ = 0.243(12 − y ) or (b) At Point B: y=0 for x, y, z axes, we write, with W = 20 lb, N x = Nb sin β = W cos β sin β = (20 lb) cos 13.427° sin13.427° N x = 4.517 lb N y = Nb cos β = W cos 2 β = (20 lb) cos 2 13.427° N y = 18.922 lb N z = − N n = −2w N z = −2(20 lb) h− y h− y = −2W e R/cos 2 β (12 ft-0) cos 2 13.427° N z = −56.765 lb 8 ft N = (4.517) 2 + (18.922) 2 + (−56.765) 2 N = 60.0 lb cos θ x = N x 4.517 = N 60 θ x = 85.7° cos θ y = Ny θ y = 71.6° cos θ z = Nz 56.742 =− N 60 N = 18.922 60 θ z = 161.1° 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. 624 PROBLEM 13.79* Prove that a force F(x, y, z) is conservative if, and only if, the following relations are satisfied: ∂ Fx ∂ Fy = ∂y ∂x ∂ Fy ∂z ∂ Fz ∂y = ∂ Fz ∂ Fx = ∂x ∂z SOLUTION For a conservative force, Equation (13.22) must be satisfied. Fx = − We now write Since ∂V ∂x ∂ Fx ∂ 2V =− ∂y ∂ x∂ y Fy = − ∂ Fy ∂x ∂V ∂y =− Fz = − ∂V ∂z ∂ 2V ∂ y∂ x ∂ Fx ∂ Fy = ∂y ∂x ∂ 2V ∂ 2V = : ∂ x∂ y ∂ y∂ x We obtain in a similar way ∂ Fy ∂z = ∂ Fz ∂y ∂ Fz ∂ Fx = ∂x ∂z 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. 625 PROBLEM 13.80 The force F = ( yzi + zxj + xyk )/xyz acts on the particle P( x, y, z ) which moves in space. (a) Using the relation derived in Problem 13.79, show that this force is a conservative force. (b) Determine the potential function associated with F. SOLUTION Fx = (a) yz xyz Fy = zx xyz Fz = xy xyz () ∂ Fy ∂ 1y ∂ Fx ∂ ( 1x ) = =0 = =0 ∂y ∂y ∂x ∂x Thus, ∂ Fx ∂ Fy = ∂y ∂x The other two equations derived in Problem 13.79 are checked in a similar way. (b) Recall that Fx = − ∂V , ∂x Fy = − ∂V , ∂y Fz = − ∂V ∂z Fx = 1 ∂V =− x ∂x V = − ln x + f ( y, z ) (1) Fy = 1 ∂V =− y ∂y V = − ln y + g ( z , x) (2) Fz = 1 ∂V =− z ∂z V = − ln z + h( x, y ) (3) Equating (1) and (2) − ln x + f ( y , z ) = − ln y + g ( z , x) Thus, f ( y, z ) = − ln y + k ( z ) (4) g ( z , x) = − ln x + k ( z ) (5) Equating (2) and (3) − ln z + h( x, y ) = − ln y + g ( z , x) g ( z , x) = − ln z + l ( x) From (5), g ( z , x) = − ln x + k ( z ) 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. 626 PROBLEM 13.80 (Continued) Thus, k ( z ) = − ln z l ( x) = − ln x From (4), f ( y, z ) = − ln y − ln z Substitute for f ( y, z ) in (1) V = − ln x − ln y − ln z V = − ln xyz 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. 627 PROBLEM 13.81* A force F acts on a particle P(x, y) which moves in the xy plane. Determine whether F is a conservative force and compute the work of F when P describes in a clockwise sense the path A, B, C, A including the quarter circle x 2 + y 2 = a 2, if (a) F = kyi, (b) F = k ( yi + xj). SOLUTION (a) Fx = ky ∂ Fy ∂ Fx =k ∂y Fy = 0 =0 ∂x ∂ Fx ∂ Fy ≠ ∂y ∂x Thus, F is not conservative. UABCA = F ⋅ dr = ABCA C B B A B A kyi ⋅ dyj + C B kyi ⋅ (dxi + dyj) + A B kyi ⋅ dxj = 0, F is perpendicular to the path. kyi ⋅ (dxi + dyj) = C ky dx B From B to C, the path is a quarter circle with origin at A. x2 + y2 = a2 Thus, y = a2 − x2 Along BC, A C C B kydx = Fx = k y Fy = kx 0 k a 2 − x 2 dx = π ka 2 4 kyi ⋅ dx j = 0 ( y = 0 on CA) UABCA = (b) a B C A + B + A C =0+ π ka 2 4 +0 U ABCA = π ka 2 4 ∂ Fy ∂ Fx =k =k ∂y ∂x ∂ Fx ∂ Fy = , F is conservative. ∂y ∂x Since ABCA is a closed loop and F is conservative, U ABCA = 0 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. 628 PROBLEM 13.82* The potential function associated with a force P in space is known to be V ( x, y, z ) = −( x 2 + y 2 + z 2 )1/2. (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D. SOLUTION (a) (b) Px = − ∂V ∂ [− ( x 2 + y 2 + z 2 )1/2 ] =− = x( x 2 + y 2 + z 2 ) −1/ 2 ∂x ∂x Py = − ∂V ∂ [− ( x 2 + y 2 + z 2 )1/ 2 ] =− = y ( x 2 + y 2 + z 2 ) −1/2 ∂y ∂y Pz = − ∂V ∂ [−( x 2 + y 2 + z 2 )1/2 ] =− = z ( x 2 + y 2 + z 2 ) −1/ 2 ∂z ∂z U OABD = U OA + U AB + U BD O–A: Py and Px are perpendicular to O–A and do no work. x = y = 0 and Pz = 1 Also, on O–A UO− A = Thus, a 0 Pz dz = a 0 dz = a A–B: Pz and Py are perpendicular to A–B and do no work. y = 0, z = a and Px = Also, on A–B U A− B = Thus, a 0 x (x + a 2 )1/ 2 2 xdx ( x + a 2 )1/2 2 = a ( 2 − 1) B–D: Px and Pz are perpendicular to B–D and do no work. On B−D, k =a z=a Py = y ( y + 2a 2 )1/ 2 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. 629 PROBLEM 13.82* (Continued) Thus, U BD = a 0 a y dy = ( y 2 + 2a 2 )1/ 2 2 1/ 2 0 ( y + 2a ) 2 U BD = (a 2 + 2a 2 )1/ 2 − (2a 2 )1/ 2 = a ( 3 − 2 ) U OABD = U O − A + U A− B + U B − D = a + a ( 2 − 1) + a ( 3 − 2) U OABD = a 3 ΔVOD = V (a, a, a ) − V (0, 0, 0) = −(a 2 + a 2 + a 2 )1/ 2 − 0 Thus, ΔVOD = − a 3 U OABD = −ΔVOD 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. 630 PROBLEM 13.83* (a) Calculate the work done from D to O by the force P of Problem 13.82 by integrating along the diagonal of the cube. (b) Using the result obtained and the answer to part b of Problem 13.82, verify that the work done by a conservative force around the closed path OABDO is zero. PROBLEM 13.82 The potential function associated with a force P in space is known to be V(x, y, z) = −( x 2 + y 2 + z 2 )1/ 2. (a) Determine the x, y, and z components of P. (b) Calculate the work done by P from O to D by integrating along the path OABD, and show that it is equal to the negative of the change in potential from O to D. SOLUTION From solution to (a) of Problem 13.82 P= (a) U OD = D O xi + yj + zk ( x + y 2 + z 2 )1/ 2 2 P ⋅ dr r = xi + yj + zk dr = dxi + dyj + dzk xi + yj + zk P= 2 ( x + y 2 + z 2 )1/ 2 Along the diagonal. Thus, x= y=z P ⋅ dr = UO−D = 3x = 3 (3x 2 )1/2 a 0 3 dx = 3a U OD = 3a U OABDO = U OABD + U DO (b) From Problem 13.82 U OABD = 3a at left The work done from D to O along the diagonal is the negative of the work done from O to D. U DO = −U OD = − 3a [see part (a)] Thus, U OABDO = 3a − 3a = 0 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. 631 PROBLEM 13.84* The force F = ( xi + yj + zk )/(x 2 + y 2 + z 2 )3/ 2 acts on the particle P( x, y , z ) which moves in space. (a) Using the relations derived in Problem 13.79, prove that F is a conservative force. (b) Determine the potential function V(x, y, z) associated with F. SOLUTION Fx = (a) x ( x + y + z 2 )3/2 y Fy = 2 2 ( x + y + z 2 )1/2 2 2 x ( − 32 ) (2 y ) ∂ Fx = 2 ∂ y ( x + y 2 + z 2 )5/2 y ( − 32 ) 2 y ∂ Fy = 2 ∂ x ( x + y 2 + z 2 )5/2 Thus, ∂ Fx ∂ Fy = ∂y ∂x The other two equations derived in Problem 13.79 are checked in a similar fashion. (b) Recalling that ∂V ∂V ∂V , Fy = − , Fz = − ∂x ∂y ∂z x ∂V Fx = − V =− dx 2 2 ∂x ( x + y + z 2 )3/2 Fx = − V = ( x 2 + y 2 + z 2 )−1/2 + f ( y , z ) Similarly integrating ∂ V/∂ y and ∂ V/∂ z shows that the unknown function f ( x, y ) is a constant. V= 1 ( x + y + z 2 )1/2 2 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. 632 PROBLEM 13.85 (a) Determine the kinetic energy per unit mass which a missile must have after being fired from the surface of the earth if it is to to reach an infinite distance from the earth. (b) What is the initial velocity of the missile (called the escape velocity)? Give your answers in SI units and show that the answer to part b is independent of the firing angle. SOLUTION g = 9.81 m/s 2 At the surface of the earth, r1 = R = 6370 km = 6.37 × 106 m Centric force at the surface of the earth, GMm R2 GM = gR 2 = (9.81)(6.37 × 106 )2 = 398.06 × 1012 m3 /s 2 F = mg = Let position 1 be on the surface of the earth (r1 = R) and position 2 be at r2 = OD. Apply the conservation of energy principle. T1 + V1 = T2 + V2 1 2 GMm 1 2 GMm = mv2 + mv1 − r1 r2 2 2 GMm GMm − R ∞ T1 T2 GM T2 = + = + gR m m R m T1 = T2 + For the escape condition set T2 =0 m T1 = gR = (9.81 m/s 2 )(6.37 × 106 m) = 62.49 × 106 m 2 /s 2 m T1 = 62.5 MJ/kg m (a) 1 2 mvesc = gr 2 vesc = 2 gR (b) vesc = (2)(9.81)(6.37 × 106 ) = 11.18 × 103 m/s vesc = 11.18 km/s Note that the escape condition depends only on the speed in position 1 and is independent of the direction of the velocity (firing angle). 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. 633 PROBLEM 13.86 A satellite describes an elliptic orbit of minimum altitude 606 km above the surface of the earth. The semimajor and semiminor axes are 17,440 km and 13,950 km, respectively. Knowing that the speed of the satellite at Point C is 4.78 km/s, determine (a) the speed at Point A, the perigee, (b) the speed at Point B, the apogee. SOLUTION rA = 6370 + 606 = 6976 km = 6.976 × 106 m rC = (17440 − 6976) 2 + (13950) 2 = 17438.4 km = 17.4384 × 106 m rB = (2)(17440) − 6976 = 27904 km = 27.904 × 106 m For earth, R = 6370 km = 6.37 × 106 m GM = gR 2 = (9.81 m/s 2 )(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2 vC = 4.78 km/s = 4780 m/s (a) Speed at Point A: Use conservation of energy. TA + VA = TC + VC 1 2 GMm 1 2 GMm mv A − = mvC − rA rC 2 2 1 1 v A2 = vC2 + 2GM − rA rC 1 1 = (4780) 2 + (2)(398.06 × 1012 ) − 6 6 6.976 × 10 17.4384 × 10 = 91.318 × 106 m 2 /s 2 VA = 9.556 × 103 m/s v A = 9.56 km/s 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. 634 PROBLEM 13.86 (Continued) (b) Speed at Point B: Use conservation of energy. TB + VB = TC + VC 1 2 GMm 1 2 GMm mvB − = mvC − rB rC 2 2 1 1 vB2 = vC2 + 2GM − rB rC 1 1 = (4780) 2 + (2)(398.06 × 1012 ) − 6 6 × × 27.904 10 17.4384 10 = 5.7258 × 106 m 2 /s 2 vB = 2.39 × 103 m/s vB = 2.39 km/s 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. 635 PROBLEM 13.87 While describing a circular orbit 200 mi above the earth a space vehicle launches a 6000-lb communications satellite. Determine (a) the additional energy required to place the satellite in a geosynchronous orbit at an altitude of 22,000 mi above the surface of the earth, (b) the energy required to place the satellite in the same orbit by launching it from the surface of the earth, excluding the energy needed to overcome air resistance. (A geosynchronous orbit is a circular orbit in which the satellite appears stationary with respect to the ground). SOLUTION Geosynchronous orbit r1 = 3960 + 200 = 4160 mi = 21.965 × 106 ft r2 = 3960 + 22,000 = 25,960 mi = 137.07 × 106 ft E = T +V = Total energy 1 2 GMm mv − 2 r M = mass of earth m = mass of satellite Newton’s second law T = F = man : 1 2 GM mv = m 2 2r E = T +V = GM = gRE2 E =− GMm mv 2 GM = v2 = 2 r r r V =− GMm r 1 GMm GMm 1 GMm − =− 2 r r 2 r E =− 1 gRE2 m 1 RE2W =− where (W = mg ) 2 r 2 r 1 (6000)(20.9088 × 106 ft) 2 1.3115 × 1018 =− ft ⋅ lb 2 r r Geosynchronous orbit at r2 = 137.07 × 106 ft EGs = −1.3115 × 1018 = −9.5681 × 109 ft ⋅ lb 6 137.07 × 10 (a) At 200 mi, E200 = − r1 = 21.965 × 106 ft 1.3115 × 1018 = −5.9709 × 1010 21.965 × 106 ΔE300 = EGs − E200 = 5.0141 × 1010 Δ E300 = 50.1 × 109 ft ⋅ lb 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. 636 PROBLEM 13.87 (Continued) (b) Launch from earth At launch pad EE = − − gRE2 m GMm = = −WRE RE RE EE = −6000(3960 × 5280) = −1.25453 × 1011 Δ EE = EGs − EE = −9.5681 × 109 + 125.453 × 109 Δ E E = 115.9 × 109 ft ⋅ lb 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. 637 PROBLEM 13.88 A lunar excursion module (LEM) was used in the Apollo moon-landing missions to save fuel by making it unnecessary to launch the entire Apollo spacecraft from the moon’s surface on its return trip to earth. Check the effectiveness of this approach by computing the energy per pound required for a spacecraft (as weighed on the earth) to escape the moon’s gravitational field if the spacecraft starts from (a) the moon’s surface, (b) a circular orbit 50 mi above the moon’s surface. Neglect the effect of the earth’s gravitational field. (The radius of the moon is 1081 mi and its mass is 0.0123 times the mass of the earth.) SOLUTION Note: GM moon = 0.0123 GM earth By Eq. 12.30, GM moon = 0.0123 gRE2 At ∞ distance from moon: r2 = ∞, Assume v2 = 0 E2 = T2 + V2 GM M m ∞ =0−0 =0− =0 (a) On surface of moon: RM = 1081 mi = 5.7077 × 106 ft v1 = 0 T1 = 0 RE = 3960 mi = 20.909 × 106 ft V1 = − GM M m RM E1 = T1 + V1 = 0 − E1 = − 0.0123 gRE2 m RM (0.0123)(32.2 ft/s2 )(20.909 × 106 ft)2 m (5.7077 × 106 ft) WE = Weight of LEM on the earth E1 = ( −30.336 × 106 ft 2 /s 2 )m m= WE g 30.336 × 106 2 2 E1 = − ft /s WE 2 32.2 ft/s ΔE = E2 − E1 = 0 + (942.1 × 103 ft ⋅ lb/lb)WE ΔE = 942 × 103 ft ⋅ lb/lb WE Energy per pound: 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. 638 PROBLEM 13.88 (Continued) (b) r1 = RM + 50 mi r1 = (1081 mi + 50 mi) = 1131 mi = 5.9717 × 106 ft Newton’s second law: F = man : GM M m r12 =m v12 = v12 r1 GMm r1 V1 = − 1 2 1 GM M mv1 = m r1 2 2 GM M m r1 E1 = T1 + V1 = E1 = − T1 = 1 GM M m GM M m − r1 r1 2 1 GM M m 1 0.0123 gRE2 m =− r1 r1 2 2 1 (0.0123)(32.2 ft/s 2 )(20.909 × 106 ft)2 m 2 5.9717 × 106 ft 6 2 2 (14.498 × 10 ft /s )WE E1 = = 450.2 × 103 ft ⋅ lb/lb WE 2 (32.2 ft/s ) E1 = − ΔE = E2 − E1 = 0 + 450.2 × 103 ft ⋅ lb/lb WE ΔE = 450 × 103 ft ⋅ lb/lb WE Energy per pound: 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. 639 PROBLEM 13.89 Knowing that the velocity of an experimental space probe fired from the earth has a magnitude v A = 32.5 Mm/h at Point A, determine the speed of the probe as it passes through Point B. SOLUTION rA = R + hA = 6370 + 4300 = 10740 km = 10.670 × 106 m rB = 6370 + 12700 = 19070 km = 19.070 × 106 m GM = gR 2 = (9.81)(6.370 × 106 ) 2 = 398.06 × 1012 m3 /s 2 v A = 32.5 Mm/h = 9.0278 × 103 m/s Use conservation of energy. TB + VB = TA + VA 1 2 GMm 1 2 GMm mvB − = mv A − 2 rB 2 rA 1 1 vB2 = v 2A + 2GM − rB rA 1 1 = (9.0278 × 103 ) 2 + (2)(398.06 × 1012 ) − 6 6 19.070 × 10 10.670 × 10 = 48.635 × 106 m 2 /s 2 vB = 6.97 × 103 m/s vB = 25.1 Mm/h 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. 640 PROBLEM 13.90 A spacecraft is describing a circular orbit at an altitude of 1500 km above the surface of the earth. As it passes through Point A, its speed is reduced by 40 percent and it enters an elliptic crash trajectory with the apogee at Point A. Neglecting air resistance, determine the speed of the spacecraft when it reaches the earth’s surface at Point B. SOLUTION Circular orbit velocity vC2 GM = 2 , GM = gR 2 r r vC2 = GM gR 2 (9.81 m/s 2 )(6.370 × 106 m) 2 = = r r (6.370 × 106 m + 1.500 × 106 m) vC2 = 50.579 × 106 m 2 /s 2 vC = 7112 m/s Velocity reduced to 60% of vC gives v A = 4267 m/s. Conservation of energy: TA + VA = TB + VB 1 GM m 1 GM m = m vB2 − m v A2 − rB rA 2 2 1 9.81(6.370 × 106 )2 vB2 9.81(6.370 × 106 ) 2 (4.267 × 103 ) 2 − = − 2 2 (7.870 × 106 ) (6.370 × 106 ) vB = 6.48 × 103 m/s vB = 6.48 km/s 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. 641 PROBLEM 13.91 Observations show that a celestial body traveling at 1.2 × 106 mi/h appears to be describing about Point B a circle of radius equal to 60 light years. Point B is suspected of being a very dense concentration of mass called a black hole. Determine the ratio MB/MS of the mass at B to the mass of the sun. (The mass of the sun is 330,000 times the mass of the earth, and a light year is the distance traveled by light in one year at a velocity of 186,300 mi/s.) SOLUTION One light year is the distance traveled by light in one year. Speed of light = 186,300 mi/s r = (60 yr)(186,300 mi/s)(5280 ft/mi)(365 days/yr)(24 h/day)(3600 s/h) r = 1.8612 × 1018 ft Newton’s second law GM B m v2 = m r r2 2 rv MB = G 2 GM earth = gRearth F= = (32.2 ft/s 2 )(3960 mi × 5280 ft/mi) 2 = 14.077 × 1015 (ft 3 /s 2 ) M sun = 330, 000 M E : GM sun = 330, 000 GM earth GM sun = (330,000)(14.077 × 1015 ) = 4.645 × 1021 ft 3 /s 2 G= MB = 4.645 × 1021 M sun rv 2 M sun rv 2 = G 4.645 × 1021 MB (1.8612 × 1018 )(1.76 × 106 ) 2 = M sun 4.645 × 1021 MB = 1.241 × 109 M sun 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. 642 PROBLEM 13.92 (a) Show that, by setting r = R + y in the right-hand member of Eq. (13.17′) and expanding that member in a power series in y/R, the expression in Eq. (13.16) for the potential energy Vg due to gravity is a first-order approximation for the expression given in Eq. (13.17′). (b) Using the same expansion, derive a second-order approximation for Vg. SOLUTION Vg = − WR 2 WR 2 WR setting r = R + y : Vg = − =− r R+ y 1 + Ry y Vg = −WR 1 + R −1 (−1) y (−1)(−2) y 2 = −WR 1 + + + 1 R 1⋅ 2 R We add the constant WR, which is equivalent to changing the datum from r = ∞ to r = R : y y 2 Vg = WR − + R R (a) First order approximation: y Vg = WR = Wy R [Equation 13.16] (b) Second order approximation: y y 2 Vg = WR − R R Vg = Wy − Wy 2 R 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. 643 PROBLEM 13.93 Collar A has a mass of 3 kg and is attached to a spring of constant 1200 N/m and of undeformed length equal to 0.5 m. The system is set in motion with r = 0.3 m, vθ = 2 m/s, and vr = 0. Neglecting the mass of the rod and the effect of friction, determine the radial and transverse components of the velocity of the collar when r = 0.6 m. SOLUTION Let position 1 be the initial position. r1 = 0.3 m (vr )1 = 0, (vθ )1 = 2 m/s, v1 = 2 m/s x1 = r1 − l0 = (0.3 − 0.5) = −0.2 m Let position 2 be when r = 0.6 m. r2 = 0.6 m (vr )2 = ?, (vθ ) 2 = ?, v2 = ? x2 = r2 − l0 = (0.6 − 0.5) = 0.1 m Conservation of angular momentum: r1m(vθ )1 = v2 m(vθ ) 2 (vθ ) 2 = Conservation of energy: r1 (vθ )1 (0.3)(2) = = 1.000 m/s 0.6 r2 T1 + V1 = T2 + V2 1 2 1 2 1 2 1 2 mv1 + kx1 = mv2 + kx2 2 2 2 2 k 2 2 2 2 v2 = v1 + x1 − x2 m 1200 (0.2)2 − (0.1)2 = 16 m 2 /s 2 = (2) 2 + 3 2 2 2 2 2 (vr ) 2 = v2 − (vθ ) = 16 − 1 = 15 m /s ( ) vr = ±3.87 m/s vr = ±3.87 m/s vθ = 1.000 m/s 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. 644 PROBLEM 13.94 Collar A has a mass of 3 kg and is attached to a spring of constant 1200 N/m and of undeformed length equal to 0.5 m. The system is set in motion with r = 0.3 m, vθ = 2 m/s, and vr = 0. Neglecting the mass of the rod and the effect of friction, determine (a) the maximum distance between the origin and the collar, (b) the corresponding speed. (Hint: Solve the equation obtained for r by trial and error.) SOLUTION Let position 1 be the initial position. r1 = 0.3 m (vr )1 = 0, (vθ )1 = 2 m/s, v1 = 2 m/s x1 = r1 − l0 = 0.3 − 0.5 = −0.2 m 1 2 1 mv = (3)(2) 2 = 6 J 2 2 1 2 1 V1 = kx1 = (1200)(−0.2) 2 = 24 J 2 2 T1 = Let position 2 be when r is maximum. (vr ) 2 = 0 r2 = rm x2 = (rm − 0.5) 1 2 1 mv2 = (3)(vθ ) 22 = 1.5(vθ ) 22 2 2 1 1 V2 = kx22 = (1200)(rm − 0.5) 2 2 2 T2 = = 600(rm − 0.5)2 Conservation of angular momentum: r1m(vθ )1 = v2 m(vθ ) 2 (vθ ) 2 = Conservation of energy: r1 (0.3) 0.6 (vθ ), = (2) = r2 rm rm T1 + V1 = T2 + V2 6 + 24 = 1.5(vθ ) 22 + 600( rm − 0.5)2 2 0.6 2 30 = (1.5) + 600( rm − 0.5) r m 0.54 f (rm ) = 2 + 600( rm − 0.5) 2 − 30 = 0 rm 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. 645 PROBLEM 13.94 (Continued) Solve for rm by trial and error. rm (m) 0.5 1.0 0.8 0.7 0.72 0.71 f (rm ) –27.8 120.5 24.8 –4.9 0.080 –2.469 rm = 0.72 − (a) Maximum distance. (b) Corresponding speed. (0.01)(0.08) = 0.7197 m 2.467 + 0.08 rm = 0.720 m (vθ ) 2 = 0.6 = 0.8337 m/s 0.7197 (vr ) 2 = 0 v2 = 0.834 m/s 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. 646 PROBLEM 13.95 A 4-lb collar A and a 1.5-lb collar B can slide without friction on a frame, consisting of the horizontal rod OE and the vertical rod CD, which is free to rotate about CD. The two collars are connected by a cord running over a pulley that is attached to the frame at O. At the instant shown, the velocity vA of collar A has a magnitude of 6 ft/s and a stop prevents collar B from moving. If the stop is suddenly removed, determine (a) the velocity of collar A when it is 8 in. from O, (b) the velocity of collar A when collar B comes to rest. (Assume that collar B does not hit O, that collar A does not come off rod OE, and that the mass of the frame is negligible.) SOLUTION 4 = 0.12422 lb ⋅ s 2 /ft 32.2 1.5 mB = = 0.04658 lb ⋅ s 2 /ft 32.2 mA = Masses: Constraint of the cord. Let r be the radial distance to the center of collar A and y be the distance that collar B moves up from its initial level. y = Δr ; y = vr (a) Let position 1 be the initial position just after the stop at B is removed and position 2 be when the collar is 8 in. (0.66667 ft) from O. r1 = 4 in. = 0.33333 ft (vr )1 = 0 r2 = 8 in. = 0.66667 ft Δr = y2 = 8 − 4 = 4 in. = 0.33333 ft V1 = 0, Potential energy: V2 = WB y2 = (1.5)(0.33333) = 0.5 ft ⋅ lb Conservation of angular momentum of collar A: m A r1 (vθ )1 = m A r2 (vθ ) 2 (vθ ) 2 = Conservation of energy: r1 (vθ )1 (0.33333)(6) = = 3 ft/s 0.66667 r2 T1 + V1 = T2 + V2 1 1 1 1 m A [(vr )12 + (vθ )12 ] + mB y12 = m A [(vr ) 22 + (vθ ) 22 ] + mB y 2 + V2 2 2 2 2 1 1 1 m A [0 + (vθ )12 ] + 0 + 0 = m A [(vr ) 22 + (vθ ) 22 ] + mB (vr ) 22 + 0.5 2 2 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. 647 PROBLEM 13.95 (Continued) 1 1 1 (0.12422)(6) 2 = (0.12422)[(vr ) 22 + (3) 2 ] + (0.04658)(vr ) 22 + 0.5 2 2 2 2.236 = 0.06211(vr )22 + 0.559 + 0.02329(vr ) 22 + 0.5 0.0854(vr )22 = 1.177 (vr ) 2 = 13.78 ft 2 /s 2 (vr ) 2 = 3.71 ft/s (vθ ) 2 = 3.00 ft/s v = 4.77 ft/s (b) Let position 3 be when collar B comes to rest. y3 = r3 − 0.33333, (vr )3 = 0, y3 = 0 Conservation of angular momentum of collar A. m A r1 (vθ )1 = m A r3 (vθ )3 (vθ )3 = r1 (vθ )1 (0.33333)(6) 2 = = r3 r3 r3 T1 + V1 = T3 + V3 Conservation of energy: 1 1 1 1 m A [(vr )12 + (vθ )12 ] + mB y12 = m A [(vr )32 + (vθ )32 ] + mB y32 + wB y3 2 2 2 2 2 2 1 1 (0.12422)[0 + (6)2 ] + 0 = (0.12422) 0 + + 0 + (1.5)(r3 − 0.33333) 2 2 r3 2.236 = 0.24844 + 1.5r3 − 0.5 r32 1.5r33 − 2.736r32 + 0.24844 = 0 Solving the cubic equation for r3, r3 = 1.7712 ft, − 0.2805 ft, 0.33333 ft Since r3 > r1 = 0.33333 ft, the required root is r3 = 1.7712 ft Corresponding velocity of collar A: (vr )3 = 0 (vθ )3 = 2 2 = r3 1.7712 (vθ )3 = 1.129 ft/s v3 = 1.129 ft/s 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. 648 PROBLEM 13.96 A 1.5-lb ball that can slide on a horizontal frictionless surface is attached to a fixed Point O by means of an elastic cord of constant k = 1 lb/in. and undeformed length 2 ft. The ball is placed at Point A, 3 ft from O, and given an initial velocity v 0 perpendicular to OA. Determine (a) the smallest allowable value of the initial speed v0 if the cord is not to become slack, (b) the closest distance d that the ball will come to Point O if it is given half the initial speed found in part a. SOLUTION Let L1 be the initial stretched length of the cord and L2 the length of the closest approach to Point O if the cord does not become slack. Let position 1 be the initial state and position 2 be that of closest approach to Point O. The only horizontal force acting on the ball is the conservative central force due to the elastic cord. At the point of closest approach the velocity of the ball is perpendicular to the cord. Conservation of angular momentum: r1m v1 = r2m v2 L1m v0 = L2m v2 or v2 = L1v0 L2 T1 + V1 = T2 + V2 Conservation of energy: 1 2 1 1 1 mv1 + k ( L1 − L0 ) 2 = mv22 + k ( L2 − L0 ) 2 2 2 2 2 k v12 − v22 = − [( L1 − L0 ) 2 + ( L2 − L0 ) 2 ] m v02 − L12 2 k v = − [( L1 − L0 ) 2 + ( L2 − L0 ) 2 ] 2 0 m L2 L0 = 2 ft, L1 = 3 ft Data: L2 = L0 = 2 ft for zero tension in the cord at the point of closest approach. k = 1 lb/in. = 12 lb/ft m = W /g = 1.5/32.2 = 0.04658 lb ⋅ s 2 /ft v02 − (3) 2 12 v0 = − [(3 − 2) 2 + (2 − 2) 2 ] 0.04658 (2) 2 −1.25 v02 = −257.6 (a) v02 = 206.1 ft 2 /s 2 v0 = 14.36 ft/s 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. 649 PROBLEM 13.96 (Continued) (b) Let v0 = 1 2 (14.36 ft/s) = 7.18 ft/s 2 so that the cord is slack in the position of closest approach to Point O. Let position 1 be the initial position and position 2 be position of closest approach with the cord being slack. Conservation of energy: T1 + V1 = T2 + V2 1 2 1 1 mv0 + k ( L1 − L0 )2 = mv22 2 2 2 k ( L − L0 ) 2 m 12 (3 − 2) 2 = 309.17 ft 2 /s 2 = (7.18) 2 + 0.04658 v2 = 17.583 ft/s v22 = v02 + Conservation of angular momentum: r1m v1 = r2m v2 sin φ r2 sin φ = d = d = r1v1 Lv = 10 v2 v2 (3)(7.18) 17.583 d = 1.225 ft 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. 650 PROBLEM 13.97 A 1.5-lb ball that can slide on a horizontal frictionless surface is attached to a fixed Point O by means of an elastic cord of constant k = 1 lb/in. and undeformed length 2 ft. The ball is placed at Point A, 3 ft from O, and given an initial velocity v 0 perpendicular to OA, allowing the ball to come within a distance d = 9 in. of Point O after the cord has become slack. Determine (a) the initial speed v0 of the ball, (b) its maximum speed. SOLUTION Let L1 be the initial stretched length of the cord. Let position 1 be the initial position. Let position 2 be the position of closest approach to point after the cord has become slack. While the cord is slack there are no horizontal forces acting on the ball, so the velocity remains constant. While the cord is stretched, the only horizontal force acting on the ball is the conservative central force due to the elastic cord. At the point of closest approach the velocity of the ball is perpendicular to the radius vector. Conservation of angular momentum: r1m v1 = r2 mv2 L1v0 = d v2 Conservation of energy: v2 = or L1 v0 d T1 + V1 = T2 + V2 1 2 1 1 mv0 + k ( L1 − L0 )2 = mv22 + 0 2 2 2 k v02 − v22 = − ( L1 − L0 ) 2 m 2 k L v02 − 12 v0 = − ( L1 − L0 ) m d L0 = 2 ft, L1 = 3 ft, d = 9 in. = 0.75 ft Data: k = 1 lb/in. = 12 lb/ft m = W /g = 1.5/32.2 = 0.04658 lb ⋅ s 2 /ft 2 12 3v v02 − 0 = − (3 − 2) 2 0.04658 0.75 − 15 v02 = −257.6 v02 = 17.17 ft 2 /s 2 (a) (b) Maximum speed. vm = v2 = 3v0 0.75 v0 = 4.14 ft/s vm = 16.58 ft/s 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. 651 PROBLEM 13.98 Using the principles of conservation of energy and conservation of angular momentum, solve part a of Sample Problem 12.9. SOLUTION R = 6370 km r0 = 500 km + 6370 km r0 = 6870 km = 6.87 × 106 m v0 = 36,900 km/h = 36.9 × 106 m 3.6 × 103 s = 10.25 × 103 m/s Conservation of angular momentum: r0 mv0 = r1mv A, r0 = rmin , r1 = rmax 6.870 × 106 r v A′ = 0 v0 = r1 r1 v A′ = 3 (10.25 × 10 ) 70.418 × 109 r1 (1) Conservation of energy: Point A: v0 = 10.25 × 103 m/s 1 2 1 mv0 = m(10.25 × 103 ) 2 2 2 TA = (m)(52.53 × 106 )(J) TA = VA = − GMm r0 GM = gR 2 = (9.81 m/s 2 )(6.37 × 106 m) 2 GM = 398 × 1012 m3 /s 2 r0 = 6.87 × 106 m VA = − (398 × 1012 m3 /s 2 )m (6.87 × 106 m) = −57.93 × 106 m (J) 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. 652 PROBLEM 13.98 (Continued) Point A′: 1 2 mv A′ 2 GMm VA′ = − r1 TA′ = =− 398 × 1012 m (J) r1 TA + VA = TA′ + VA′ 52.53 × 106 m − 57.93 × 106 m = 1 398 × 1012 m m v A2 ′ − r1 2 Substituting for v A′ from (1) −5.402 × 106 = (70.418 × 109 ) 2 398 × 1012 − r1 (2)(r1 ) 2 −5.402 × 106 = (2.4793 × 1021 ) 398 × 1012 − r1 r12 (5.402 × 106 )r12 − (398 × 1012 )r1 + 2.4793 × 1021 = 0 r1 = 66.7 × 106 m, 6.87 × 106 m rmax = 66,700 km 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. 653 PROBLEM 13.99 Solve sample Problem 13.8, assuming that the elastic cord is replaced by a central force F of magnitude (80/r2) N directed toward O. 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) The force exerted on the sphere passes through O. Angular momentum about O is conserved. Minimum velocity is at B, where the distance from O is maximum. Maximum velocity is at C, where distance from O is minimum. rA mv A sin 60° = rm mvm (0.5 m)(0.6 kg)(20 m/s)sin 60° = rm (0.6 kg)vm vm = 8.66 rm (1) Conservation of energy: At Point A, 1 2 1 mv A = (0.6 kg)(20 m/s)2 = 120 J 2 2 80 −80 , V = Fdr = 2 dr = r r −80 VA = = −160 J 0.5 TA = At Point B, TB = 1 2 1 mvm = (0.6 kg)vm2 = 0.3vm2 2 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. 654 PROBLEM 13.99 (Continued) VB = and Point C : −80 rm TA + VA = TB + VB 120 − 160 = 0.3vm2 − 80 rm (2) Substitute (1) into (2) 2 8.66 80 −40 = (0.3) − rm rm rm2 − 2 rm + 0.5625 = 0 rm′ = 0.339 m and rm = 1.661 m rmax = 1.661 m rmin = 0.339 m (b) Substitute rm′ and rm from results of part (a) into (1) to get corresponding maximum and minimum values of the speed. vm′ = 8.66 = 25.6 m/s 0.339 vmax = 25.6 m/s vm = 8.66 = 5.21 m/s 1.661 vmin = 5.21 m/s 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. 655 PROBLEM 13.100 A spacecraft is describing an elliptic orbit of minimum altitude hA = 2400 km and maximum altitude hB = 9600 km above the surface of the earth. Determine the speed of the spacecraft at A. SOLUTION rA = 6370 km + 2400 km rA = 8770 km rB = 6370 km + 9600 km = 15,970 km rA mv A = rB mvB Conservation of momentum: vB = Conservation of energy: TA = 1 2 mv A 2 rA 8770 vA = v A = 0.5492v A 15,970 rB VA = −GMm rA TB = 1 2 mvB 2 (1) VB = −GMm rB GM = gR 2 = (9.81 m/s 2 )(6370 × 103 m)2 = 398.1 × 1012 m3/s 2 −(398.1 × 1012 ) m = −45.39 × 106 m 3 8770 × 10 −(398.1 × 1012 ) m VB = = −24.93 m (15,970 × 103 ) VA = TA + VA = TB + VB : 1 1 m v 2A − 45.39 × 106 m = m vB2 − 24.93 × 106 m 2 2 (2) Substituting for vB in (2) from (1) v A2 [1 − (0.5492) 2 ] = 40.92 × 106 v A2 = 58.59 × 106 m 2/s 2 v A = 7.65 × 103 m/s v A = 27.6 × 103 km/h 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. 656 PROBLEM 13.101 While describing a circular orbit, 185 mi above the surface of the earth, a space shuttle ejects at Point A an inertial upper stage (IUS) carrying a communication satellite to be placed in a geosynchronous orbit (see Problem 13.87) at an altitude of 22,230 mi above the surface of the earth. Determine (a) the velocity of the IUS relative to the shuttle after its engine has been fired at A, (b) the increase in velocity required at B to plae the satellite in its final orbit. SOLUTION For earth, R = 3960 mi = 20.909 × 106 ft g = 32.2 ft 2 GM = gR 2 = (32.2)(20.909 × 106 ) 2 = 14.077 × 1015 ft 3 /s 2 Speed on a circular orbits of radius r, rA, and rB. F = man GMm mv 2 = r r2 GM v2 = r v= GM r rA = 3960 + 185 = 4145 mi = 21.886 × 106 ft (v A )circ = 14.077 × 1015 = 25.362 × 103 ft/s 21.886 × 106 rB = 3960 + 22230 = 26190 mi = 138.283 × 106 ft (vB )circ = 14.077 × 1015 = 10.089 × 103 ft/s 138.283 × 106 Calculate speeds at A and B for path AB. Conservation of angular momentum: mrA v A sin φ A = mrB vB sin φ A vB = rA v A sin 90° 21.886 × 106 v A = = 0.15816 v A rB sin 90° 138.283 × 106 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. 657 PROBLEM 13.101 (Continued) TA + VA = TB + VB Conservation of energy: 1 2 GMm 1 2 GMm mv A + = mvB − 2 rA 2 rB 1 1 2GM (rB − rA ) v A2 − vB2 = 2GM − = rA rB rA rB v A2 − (0.15816v A ) 2 = (2)(14.077 × 1015 )(116.397 × 106 ) (21.886 × 106 )(138.283 × 106 ) 0.97499v A2 = 1.082796 × 109 v A = 33.325 × 103 ft/s vB = (0.15816)(33.325 × 106 ) = 5.271 × 103 ft/s (a) (b) Increase in speed at A: Δv A = 33.325 × 103 − 25.362 × 103 = 7.963 × 103 ft/s Δv A = 7960 ft/s ΔvB = 10.089 × 103 − 5.271 × 103 = 4.818 × 103 ft/s ΔvB = 4820 ft/s Increase in speed at 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. 658 PROBLEM 13.102 A spacecraft approaching the planet Saturn reaches Point A with a velocity v A of magnitude 68.8 × 103 ft/s. It is to be placed in an elliptic orbit about Saturn so that it will be able to periodically examine Tethys, one of Saturn’s moons. Tethys is in a circular orbit of radius 183 × 103 mi about the center of Saturn, traveling at a speed of 37.2 × 103 ft/s. Determine (a) the decrease in speed required by the spacecraft at A to achieve the desired orbit, (b) the speed of the spacecraft when it reaches the orbit of Tethys at B. SOLUTION (a) rA = 607.2 × 106 ft rB = 966.2 × 106 ft v′A = speed of spacecraft in the elliptical orbit after its speed has been decreased. Elliptical orbit between A and B. Conservation of energy 1 mv′A2 2 −GM sat m VA = rA TA = Point A: M sa = Mass of Saturn, determine GM sa from the speed of Tethys in its circular orbit. vcirc = (Eq. 12.44) GM sat r 2 GM sat = rB vcirc GM sat = (966.2 × 106 ft 2 )(37.2 × 103 ft/s) 2 = 1.337 × 1018 ft 3/s 2 VA = − (1.337 × 1018 ft 3/s 2 ) m (607.2 × 106 ft) = −2.202 × 109 m 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. 659 PROBLEM 13.102 (Continued) TB = Point B: 1 2 mvB 2 VB = −GM sat m (1.337 × 1018 ft 3 /s 2 ) m =− rB (966.2 × 106 ft) VB = 1.384 × 109 TA + VA = TB + VB ; 1 1 m v′A2 − 2.202 × 109 m = m vB2 − 1.384 × 109 m 2 2 v′A2 − vB2 = 1.636 × 109 Conservation of angular momentum: rA mv′A = rB mvB vB = rA 607.2 × 106 v′A = v′A = 0.6284v′A rB 966.2 × 106 v′A2 [1 − (0.6284) 2 ] = 1.636 × 109 v′A = 52, 005 ft/s (a) Δv A = v A − v′A = 68,800 − 52, 005 (b) vB = Δv A = 16,795 ft/s rA v′A = (0.6284)(52, 005) rB vB = 32, 700 ft/s 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. 660 PROBLEM 13.103 A spacecraft traveling along a parabolic path toward the planet Jupiter is expected to reach Point A with a velocity vA of magnitude 26.9 km/s. Its engines will then be fired to slow it down, placing it into an elliptic orbit which will bring it to within 100 × 103 km of Jupiter. Determine the decrease in speed Δv at Point A which will place the spacecraft into the required orbit. The mass of Jupiter is 319 times the mass of the earth. SOLUTION Conservation of energy. Point A: 1 m(v A − Δv A ) 2 2 −GM J m VA = rA TA = GM J = 319GM E = 319 gRE2 RE = 6.37 × 106 m GM J = (319)(9.81 m/s 2 )(6.37 × 106 m) 2 GM J = 126.98 × 1015 m3 /s 2 rA = 350 × 106 m VA = −(126.98 × 1015 m3 /s 2 )m (350 × 106 m) VA = −(362.8 × 106 )m Point B: 1 2 mvB 2 −GM J m −(126.98 × 1015 m3 /s 2 )m VB = = rB (100 × 106 m) TB = VB = −(1269.8 × 106 ) m TA + VA = TB + VB 1 1 m(v A − Δv A ) 2 − 362.8 × 106 m = mvB2 − 1269.8 × 106 m 2 2 (v A − Δv A ) 2 − vB2 = −1814 × 106 (1) 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. 661 PROBLEM 13.103 (Continued) Conservation of angular momentum. rA = 350 × 106 m rB = 100 × 106 m rA m(v A − Δv A ) = rB mvB r vB = A (v A − Δv A ) rB 350 = (v A − Δv A ) 100 (2) Substitute vB in (2) into (1) (v A − Δv A ) 2 [1 − (3.5) 2 ] = −1814 × 106 (v A − Δv A ) 2 = 161.24 × 106 (v A − Δv A ) = 12.698 × 103 m/s (Take positive root; negative root reverses flight direction.) v A = 26.9 × 103 m/s (given) Δv A = (26.9 × 103 m/s − 12.698 × 103 m/s) Δv A = 14.20 km/s 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. 662 PROBLEM 13.104 As a first approximation to the analysis of a space flight from the earth to Mars, it is assumed that the orbits of the earth and Mars are circular and coplanar. The mean distances from the sun to the earth and to Mars are 149.6 × 106 km and 227.8 × 106 km, respectively. To place the spacecraft into an elliptical transfer orbit at Point A, its speed is increased over a short interval of time to v A which is faster than the earth’s orbital speed. When the spacecraft reaches Point B on the elliptical transfer orbit, its speed vB is increased to the orbital speed of Mars. Knowing that the mass of the sun is 332.8 × 103 times the mass of the earth, determine the increase in velocity required (a) at A, (b) at B. SOLUTION M = mass of the sun GM = 332.8(10)3 (9.81 m/s 2 )(6.37 × 106 m)2 = 1.3247(10)20 m3 /s 2 Circular orbits Earth vE = GM = 29.758 m/s 149.6(10)9 Mars vM = GM = 24.115 m/s 227.8(10)9 Conservation of angular momentum Elliptical orbit v A (149.6) = vB (227.8) Conservation of energy 1 2 GM 1 GM vA − = vB2 − 9 2 2 149.6(10) 227.8(10)9 v A = vB (227.8) = 1.52273 vB (149.6) 1 1.3247(10) 20 1 1.3247(10)20 = vB2 − (1.52273) 2 vB2 − 9 2 2 149.6(10) 227.8(10)9 0.65935vB2 = 3.0398(10)8 vB2 = 4.6102(10)8 vB = 21, 471 m/s, v A = 32, 695 m/s (a) Increase at A, v A − vE = 32.695 − 29.758 = 2.94 km/s (b) Increase at B, vB − vM = 24.115 − 21.471 = 2.64 km/s 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. 663 PROBLEM 13.105 The optimal way of transferring a space vehicle from an inner circular orbit to an outer coplanar circular orbit is to fire its engines as it passes through A to increase its speed and place it in an elliptic transfer orbit. Another increase in speed as it passes through B will place it in the desired circular orbit. For a vehicle in a circular orbit about the earth at an altitude h1 = 200 mi, which is to be transferred to a circular orbit at an altitude h2 = 500 mi, determine (a) the required increases in speed at A and at B, (b) the total energy per unit mass required to execute the transfer. SOLUTION Elliptical orbit between A and B Conservation of angular momentum mrAv A = mrBvB vA = rB 7.170 vB = vB 6.690 rA rA = 6370 km + 320 km = 6690 km, rA = 6.690 × 106 m v A = 1.0718vB rB = 6370 km + 800 km = 7170 km, (1) rB = 7.170 × 106 m R = (6370 km) = 6.37 × 106 m Conservation of energy GM = gR 2 = (9.81 m/s 2 )(6.37 × 106 m)2 = 398.060 × 1012 m3 /s 2 Point A: TA = 1 2 mv A 2 VA = − GMm (398.060 × 1012 )m =− rA (6.690 × 106 ) VA = 59.501 × 106 m Point B: TB = 1 2 mvB 2 VB = − GMm (398.060 × 1012 )m =− rB (7.170 × 106 ) VB = 55.5 × 106 m 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. 664 PROBLEM 13.105 (Continued) TA + VA = TB + VB 1 2 1 mv A − 59.501 × 106 m = mvB2 − 55.5 × 106 m 2 2 v A2 − vB2 = 8.002 × 106 From (1) v A = 1.0718vB vB2 [(1.0718) 2 − 1] = 8.002 × 106 vB2 = 53.79 × 106 m 2 /s 2 , vB = 7334 m/s v A = (1.0718)(7334 m/s) = 7861 m/s Circular orbit at A and B (Equation 12.44) (v A )C = GM = rA 398.060 × 1012 = 7714 m/s 6.690 × 106 (vB )C = GM = rB 398.060 × 1012 = 7451 m/s 7.170 × 106 (a) Increases in speed at A and B Δv A = v A − (v A )C = 7861 − 7714 = 147 m/s ΔvB = (vB )C − vB = 7451 − 7334 = 117 m/s (b) Total energy per unit mass E/m = E/m = 1 [(v A ) 2 − (v A )C2 + (vB )C2 − (vB ) 2 ] 2 1 [(7861) 2 − (7714) 2 + (7451) 2 − (7334)2 ] 2 E/m = 2.01 × 106 J/kg 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. 665 PROBLEM 13.106 During a flyby of the earth, the velocity of a spacecraft is 10.4 km/s as it reaches its minimum altitude of 990 km above the surface at Point A. At Point B the spacecraft is observed to have an altitude of 8350 km. Determine (a) the magnitude of the velocity at Point B, (b) the angle φB . SOLUTION At A: hA = vr = [1.04(10) 4 m/s][6.37(10)6 m + 0.990(10)6 m] hA = 76.544(10)9 m 2 /s 1 1 GM (TA + VA ) = v 2 − m 2 r = 1 (9.81)[6.37(10)6 ]2 ≅0 [1.04(10) 4 ]2 − 2 [6.37(10)6 + 0.990(10)6 ] (Parabolic orbit) At B: 1 1 GM (TB + VB ) = vB2 − =0 m 2 rB 1 2 (9.81)[6.37(106 )]2 vB = 2 [6.37(10)6 + 8.35(10)6 ] vB2 = 54.084(10)6 vB = 7.35 km/s (a) hB = vB sin φB rB = 76.544(10)9 sin φB = 76.544(10)9 7.35(106 )[6.37(10)6 + 8.35(10)6 ] = 0.707483 φ B = 45.0° (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. 666 PROBLEM 13.107 A space platform is in a circular orbit about the earth at an altitude of 300 km. As the platform passes through A, a rocket carrying a communications satellite is launched from the platform with a relative velocity of magnitude 3.44 km/s in a direction tangent to the orbit of the platform. This was intended to place the rocket in an elliptic transfer orbit bringing it to Point B, where the rocket would again be fired to place the satellite in a geosynchronous orbit of radius 42,140 km. After launching, it was discovered that the relative velocity imparted to the rocket was too large. Determine the angle γ at which the rocket will cross the intended orbit at Point C. SOLUTION R = 6370 km rA = 6370 km + 300 km rA = 6.67 × 106 m rC = 42.14 × 106 m GM = gR 2 GM = (9.81 m/s)(6.37 × 106 m) 2 GM = 398.1 × 1012 m3 /s 2 For any circular orbit: Fn = man = 2 mvcirc r 2 mv GMm = m circ 2 r r GM = r Fn = vcirc Velocity at A: (v A )circ = GM (398.1 × 1012 m3/s3 ) = = 7.726 × 103 m/s rA (6.67 × 106 m) v A = (v A )circ + (v A ) R = 7.726 × 103 + 3.44 × 103 = 11.165 × 103 m/s 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. 667 PROBLEM 13.107 (Continued) Velocity at C: Conservation of energy: TA + VA = TC + VC 1 GM m 1 GM m m v A2 − = m vC2 − 2 2 rA rC 1 1 vC2 = v 2A + 2GM − rC rA 1 1 = (11.165 × 103 ) 2 + 2(398.1 × 1012 ) − 6 6 42.14 10 6.67 10 × × vC2 = 124.67 × 106 − 100.48 × 106 = 24.19 × 106 m 2 /s 2 vC = 4.919 × 103 m/s Conservation of angular momentum: rA mv A = rC mvC cos γ cos γ = rA v A rC vC (6.67 × 106 )(11.165 × 103 ) (42.14 × 106 )(4.919 × 103 ) cos γ = 0.35926 = γ = 68.9° 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. 668 PROBLEM 13.108 A satellite is projected into space with a velocity v0 at a distance r0 from the center of the earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle α with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite. SOLUTION For circular orbit of radius r0 v02 GMm = m r0 r02 F = man v02 = GM r0 But v0 forms an angle α with the intended circular path. For elliptic orbit. Conservation of angular momentum: r0 mv0 cos α = rA mv A r v A = 0 cos α v0 rA (1) Conservation of energy: 1 2 GMm 1 2 GMm mv0 − = mv A − 2 r0 2 rA r0 2GM v02 − v 2A = 1 − r0 rA Substitute for v A from (1) r 2 2GM v02 1 − 0 cos 2 α = r0 rA But v02 = GM , r0 r0 1 − rA 2 thus r r 1 − 0 cos 2 α = 2 1 − 0 rA rA 2 r r cos 2 α 0 − 2 0 + 1 = 0 rA rA 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. 669 PROBLEM 13.108 (Continued) Solving for r0 rA r0 + 2 ± 4 − 4cos 2 α 1 ± sin α = = 2 rA 2cos α 1 − sin 2 α (1 + sin α )(1 − sin α ) rA = r0 = (1 sin α )r0 1 ± sin α also valid for Point A′ Thus, rmax = (1 + sin α )r0 rmin = (1 − sin α )r0 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. 670 PROBLEM 13.109 Upon the LEM’s return to the command module, the Apollo spacecraft of Problem 13.88 was turned around so that the LEM faced to the rear. The LEM was then cast adrift with a velocity of 200 m/s relative to the command module. Determine the magnitude and direction (angle φ formed with the vertical OC) of the velocity vC of the LEM just before it crashed at C on the moon’s surface. SOLUTION Command module in circular orbit rB = 1740 + 140 = 1880 km = 1.88 × 106 m GM moon = 0.0123 GM earth = 0.0123 gR 2 = 0.0123(9.81)(6.37 × 106 )2 = 4.896 × 1012 m3 /s 2 R = 1740 km ΣF = man v0 = GM m m = rB2 GMm = rB mv02 rB 4.896 × 1012 1.88 × 106 v0 = 1614 m/s vB = 1614 − 200 = 1414 m/s Conservation of energy between B and C: 1 2 GM m m 1 2 GM m m mvB − = mvC − rB rC 2 2 vC2 = vB2 + rC = R 2GMm rB − 1 rB R vC2 = (1414 m/s)2 + 2 (4.896 × 1012 m3/s 2 ) 1.88 × 106 − 1 6 6 (1.88 × 10 m) 1.74 × 10 vC2 = 1.999 × 106 + 0.4191 × 106 = 2.418 × 106 m 2 /s 2 vC = 1555 m/s Conservation of angular momentum: rB mvB = RmvC sin φ rB vB (1.88 × 106 m)(1414 m/s) = = 0.98249 rC vC (1.74 × 106 m)(1555 m/s) φ = 79.26° sin φ = φ = 79.3° 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. 671 PROBLEM 13.110 A space vehicle is in a circular orbit at an altitude of 225 mi above the earth. To return to earth, it decreases its speed as it passes through A by firing its engine for a short interval of time in a direction opposite to the direction of its motion. Knowing that the velocity of the space vehicle should form an angle φB = 60° with the vertical as it reaches Point B at an altitude of 40 mi, determine (a) the required speed of the vehicle as it leaves its circular orbit at A, (b) its speed at Point B. SOLUTION (a) rA = 3960 mi + 225 mi = 4185 mi rA = 4185 mi × 5280 ft/mi = 22,097 × 103 ft rB = 3960 mi + 40 mi = 4000 mi rB = 4000 × 5280 = 21,120 × 103 ft R = 3960 mi = 20,909 × 103 ft GM = gR 2 = (32.2 ft/s 2 )(20,909 × 103 ft) 2 GM = 14.077 × 1015 ft 3 /s 2 Conservation of energy: 1 2 mv A 2 −GMm VA = rA TA = = −14.077 × 1015 m 22, 097 × 103 = −637.1 × 106 m 1 2 mvB 2 −GMm VB = rB TB = = −14.077 × 1015 m 21,120 × 103 = −666.5 × 106 m TA + VA = TB + VB 1 1 m v A2 − 637.1 × 106 m = m vB2 − 666.5 × 106 m 2 2 2 v A = vB2 − 58.94 × 106 (1) 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. 672 PROBLEM 13.110 (Continued) Conservation of angular momentum: rA mv A = rB mvB sin φB vB = (rA )v A 4185 1 = vA (rB )(sin φB ) 4000 sin 60° vB = 1.208 v A (2) Substitute vB from (2) in (1) v 2A = (1.208v A ) 2 − 58.94 × 106 v A2 [(1.208) 2 − 1] = 58.94 × 106 v 2A = 128.27 × 106 ft 2 /s 2 v A = 11.32 × 103 ft/s (a) (b) From (2) vB = 1.208v A = 1.208(11.32 × 106 ) = 13.68 × 103 ft/s vB = 13.68 × 103 ft/s 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. 673 PROBLEM 13.111* In Problem 13.110, the speed of the space vehicle was decreased as it passed through A by firing its engine in a direction opposite to the direction of motion. An alternative strategy for taking the space vehicle out of its circular orbit would be to turn it around so that its engine would point away from the earth and then give it an incremental velocity Δv A toward the center O of the earth. This would likely require a smaller expenditure of energy when firing the engine at A, but might result in too fast a descent at B. Assuming this strategy is used with only 50 percent of the energy expenditure used in Problem 13.109, determine the resulting values of φB and vB . SOLUTION rA = 3960 mi + 225 mi rA = 4185 mi = 22.097 × 106 ft rB = 3960 mi + 40 mi = 4000 mi rB = 21.120 × 106 ft GM = gR 2 = (32.2 ft/s 2 )[(3960)(5280) ft 2 ] GM = 14.077 × 1015 ft 3 /s 2 Velocity in circular orbit at 225 m altitude: Newton’s second law F = man : (v A )circ = 2 GMm m(v A )circ = rA rA2 GM 14.077 × 1015 = rA 22.097 × 106 = 25.24 × 103 ft/s Energy expenditure: From Problem 13.110, Energy, v A = 11.32 × 103 ft/s 1 1 2 m(v A )circ − mv 2A 2 2 1 1 = m(25.24 × 103 )2 − m(11.32 × 103 ) 2 2 2 6 = 254.46 × 10 m ft ⋅ lb ΔE109 = ΔE109 ΔE109 ΔE110 = (0.50)ΔE109 = (254.46 × 106 m) ft ⋅ lb 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. 674 PROBLEM 13.111* (Continued) Thus, additional kinetic energy at A is 1 (254.46 × 106 m) m( Δv A ) 2 = ΔE110 = ft ⋅ lb 2 2 (1) Conservation of energy between A and B: TA = 1 2 m[(v A )circ + ( Δv A ) 2 ] 2 TB = 1 2 mvB 2 VB = VA = −GMm rA −GMm rA TA + VA = TB + VB 1 254.46 × 106 m 14.077 × 1015 m 1 2 14.077 × 1015 m m(25.24 × 103 ) 2 + − = mvB − 2 2 2 22.097 × 106 21.120 × 106 vB2 = 637.06 × 106 + 254.46 × 106 − 1274.1 × 106 + 1333 × 106 vB2 = 950.4 × 103 vB = 30.88 × 103 ft/s Conservation of angular momentum between A and B: rA m(v A )circ = rB mvB sin φB r (v ) (4185) (25.24 × 103 ) = 0.8565 sin φB = A A circ = (4000) (30.88 × 103 ) rB (vB ) φB = 58.9° 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. 675 PROBLEM 13.112 Show that the values vA and vP of the speed of an earth satellite at the apogee A and the perigee P of an elliptic orbit are defined by the relations v A2 = 2GM rP rA + rP rA vP2 = 2GM rA rA + rP rP where M is the mass of the earth, and rA and rP represent, respectively, the maximum and minimum distances of the orbit to the center of the earth. SOLUTION Conservation of angular momentum: rA mv A = rP mvP vA = rP vP rA (1) Conservation of energy: 1 2 GMm 1 2 GMm mvP − = mv A − 2 rP 2 rA (2) Substituting for v A from (1) into (2) 2 vP2 2GM rP 2 2GM − = vP − rP rA rA r 2 1 − P vP2 = 2GM 1 − 1 rA rP rA rA2 − rP2 rA2 with vP2 = 2GM rA − rP rA rP rA2 − rP2 = (rA − rP )( rA + rP ) vP2 = 2GM rA + rP rA (3) rP Exchanging subscripts P and A v A2 = 2GM rA + rP rP rA Q.E.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. 676 PROBLEM 13.113 Show that the total energy E of an earth satellite of mass m describing an elliptic orbit is E = −GMm /(rA + rP ), where M is the mass of the earth, and rA and rP represent, respectively, the maximum and minimum distances of the orbit to the center of the earth. (Recall that the gravitational potential energy of a satellite was defined as being zero at an infinite distance from the earth.) SOLUTION See solution to Problem 13.112 (above) for derivation of Equation (3). vP2 = 2GM rA (rA + rP ) rP Total energy at Point P is E = TP + VP = = 1 2 GMm mvP − 2 rP 1 2GMm rA GMm − rP 2 ( rA + r0 ) rP rA 1 = GMm − rP ( rA + rP ) rP (r − r − r ) = GMm A A P rP (rA + rP ) E=− GMm rA + rP Note: Recall that gravitational potential of a satellite is defined as being zero at an infinite distance from the earth. 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. 677 PROBLEM 13.114* A space probe describes a circular orbit of radius nR with a velocity v0 about a planet of radius R and center O. Show that (a) in order for the probe to leave its orbit and hit the planet at an angle θ with the vertical, its velocity must be reduced to α v0, where α = sin θ 2(n − 1) n − sin 2 θ (b) the probe will not hit the planet if α is larger than 2 /(1 + n). 2 SOLUTION (a) Conservation of energy: 1 m (α v0 ) 2 2 GMm VA = − nR TA = At A: 1 mv 2 2 GMm VB = − R TB = At B: M = mass of planet m = mass of probe TA + VA = TB + VB 1 GMm 1 GMm = mv 2 − m (α v0 ) 2 − 2 nR 2 R (1) Conservation of angular momentum: nR mα v0 = Rmv sin θ v= nα v0 sin θ (2) Replacing v in (1) by (2) 2 (α v0 )2 − 2GM nα v0 2GM = − nR R sin θ (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. 678 PROBLEM 13.114* (Continued) For any circular orbit. an = v2 r Newton’s second law 2 −GMm m(v) circ = r r2 GM vcirc = r v0 = vcirc = For r = nR, GM nR Substitute for v0 in (3) α2 GM 2GM n 2α 2 GM 2GM − = − nR nR R sin 2 θ nR α 2 1 − n2 = 2(1 − n) sin 2 θ α2 = 2(1 − n)(sin 2 θ ) 2(n − 1)sin 2 θ = 2 (sin 2 θ − n 2 ) (n − sin 2 θ ) α = sin θ (b) 2(n − 1) n − sin 2 θ 2 Q.E.D. Probe will just miss the planet if θ > 90°, α = sin 90° Note: 2( n − 1) = n − sin 2 90° 2 2 n +1 n 2 − 1 = ( n − 1)(n + 1) 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. 679 PROBLEM 13.115 A missile is fired from the ground with an initial velocity v 0 forming an angle φ 0 with the vertical. If the missile is to reach a maximum altitude equal to α R, where R is the radius of the earth, (a) show that the required angle φ 0 is defined by the relation sin φ0 = (1 + α ) 1 − α vesc 1 + α v0 2 where vesc is the escape velocity, (b) determine the range of allowable values of v0 . SOLUTION rA = R (a) Conservation of angular momentum: Rmv0 sin φ0 = rB mvB rB = R + α R = (1 + α ) R vB = Rv0 sin φ0 v0 sin φ0 = (1 + α ) R (1 + α ) (1) Conservation of energy: 1 2 GMm 1 2 GMm = mvB − mv0 − R 2 2 (1 + α ) R TA + VA = TB + VB v02 − vB2 = 2 GMm 1 1− R 1+ α 2 GMm α = R 1+α Substitute for vB from (1) sin 2 φ0 v02 1 − (1 + α ) 2 From Equation (12.43): 2 GMm α = R 1+α 2 = vesc sin 2 φ0 v02 1 − (1 + α ) 2 2GM R 2 α = vesc 1+α 2 v α = 1 − esc 2 (1 + α ) v0 1 + α sin 2 φ0 α vesc sin φ0 = (1 + α ) 1 − 1 + α v0 (2) 2 Q.E.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. 680 PROBLEM 13.115 (Continued) (b) Allowable values of v0 (for which maximum altitude = α R) 0 < sin 2 φ0 < 1 For sin φ0 = 0, from (2) 2 v α 0 = 1 − esc v0 1 + α v0 = vesc α 1+ α For sin φ0 = 1, from (2) 2 v α 1 = 1 − esc 2 (1 + α ) v0 1 + α 2 vesc 1 1 = 1 + α − α 1+ α v0 v0 = vesc 2 1 + 2α + α − 1 2 + α = = α (1 + α ) 1+ α 1+α 2 +α vesc α 1+α < v0 < vesc 1+α 2 +α 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. 681 PROBLEM 13.116 A spacecraft of mass m describes a circular orbit of radius r1 around the earth. (a) Show that the additional energy ΔE which must be imparted to the spacecraft to transfer it to a circular orbit of larger radius r2 is ΔE = GMm(r2 − r1 ) 2r1r2 where M is the mass of the earth. (b) Further show that if the transfer from one circular orbit to the other is executed by placing the spacecraft on a transitional semielliptic path AB, the amounts of energy Δ E A and Δ EB which must be imparted at A and B are, respectively, proportional to r2 and r1 : Δ EA = r2 r1 + r2 ΔE Δ EB = r1 r1 + r2 ΔE SOLUTION (a) For a circular orbit of radius r F = man : GMm v2 = m r r2 GM v2 = r E = T +V = 1 2 GMm 1 GMm mv − =− r 2 2 r (1) Thus ΔE required to pass from circular orbit of radius r1 to circular orbit of radius r2 is 1 GMm 1 GMm + 2 r1 2 r2 GMm(r2 − r1 ) ΔE = Q.E.D. 2r1r2 Δ E = E1 − E2 = − (b) (2) For an elliptic orbit, we recall Equation (3) derived in Problem 13.113 (with vP = v1 ) v12 = 2Gm r2 (r1 + r2 ) r1 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. 682 PROBLEM 13.116 (Continued) At Point A: Initially spacecraft is in a circular orbit of radius r1 . GM 2 = vcirc r1 1 2 1 GM Tcirc = mvcirc = m r1 2 2 After the spacecraft engines are fired and it is placed on a semi-elliptic path AB, we recall and v12 = 2GM r2 ⋅ (r1 + r2 ) r1 T1 = 2GMr2 1 2 1 mv1 = m 2 2 r1 (r1 + r2 ) At Point A, the increase in energy is ΔE A = T1 − Tcirc = Recall Equation (2): 2GMr2 1 1 GM − m m 2 r1 (r1 + r2 ) 2 r1 ΔE A = GMm(2r2 − r1 − r2 ) GMm( r2 − r1 ) = 2r1 (r1 + r2 ) 2r1 (r1 + r2 ) ΔE A = r2 r1 + r2 ΔE A = r2 ΔE (r1 + r2 ) Q.E.D. r1 ΔE (r1 + r2 ) Q.E.D. GMm(r2 − r1 ) 2r1r2 A similar derivation at Point B yields, ΔEB = 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. 683 PROBLEM 13.117* Using the answers obtained in Problem 13.108, show that the intended circular orbit and the resulting elliptic orbit intersect at the ends of the minor axis of the elliptic orbit. PROBLEM 13.108 A satellite is projected into space with a velocity v0 at a distance r0 from the center of the earth by the last stage of its launching rocket. The velocity v0 was designed to send the satellite into a circular orbit of radius r0. However, owing to a malfunction of control, the satellite is not projected horizontally but at an angle α with the horizontal and, as a result, is propelled into an elliptic orbit. Determine the maximum and minimum values of the distance from the center of the earth to the satellite. SOLUTION If the point of intersection P0 of the circular and elliptic orbits is at an end of the minor axis, then v0 is parallel to the major axis. This will be the case only if α + 90° = θ0 , that is if cos θ0 = − sin α . We must therefore prove that cos θ0 = − sin α (1) We recall from Equation (12.39): 1 GM = 2 + C cos θ r h When θ = 0, r = rmin (2) and rmin = r0 (1 − sin α ) 1 GM = 2 +C r0 (1 − sin α ) h For θ = 180°, (3) r = rmax = r0 (1 + sin α ) 1 GM = 2 −C r0 (1 + sin α ) h (4) Adding (3) and (4) and dividing by 2: GM 1 1 1 = + 2 2r0 1 − sin α 1 + sin α h 1 = r0 cos 2 α Subtracting (4) from (3) and dividing by 2: 1 1 1 − 2r0 1 − sin α 1 + sin α sin α C= r0 cos 2 α C= 1 = 2r0 2sin α 2 1 − sin α 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. 684 PROBLEM 13.117* (Continued) Substituting for GM h2 and C into Equation (2) 1 1 (1 + sin α cos θ ) = r r0 cos 2 α (5) Letting r = r0 and θ = θ0 in Equation (5), we have cos 2 α = 1 + sin α cos θ0 cos 2 α − 1 sin α sin 2 α =− sin α = − sin α cos θ0 = This proves the validity of Equation (1) and thus P0 is an end of the minor axis of the elliptic orbit. 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. 685 PROBLEM 13.118* (a) Express in terms of rmin and vmax the angular momentum per unit mass, h, and the total energy per unit mass, E/m, of a space vehicle moving under the gravitational attraction of a planet of mass M (Figure 13.15). (b) Eliminating vmax between the equations obtained, derive the formula 1 rmin GM = 2 h 2 1 + 1 + 2 E h m GM (c) Show that the eccentricity ε of the trajectory of the vehicle can be expressed as ε = 1+ 2E h m GM 2 (d ) Further show that the trajectory of the vehicle is a hyperbola, an ellipse, or a parabola, depending on whether E is positive, negative, or zero. SOLUTION (a) Point A: Angular momentum per unit mass. H0 m r mv = min max m h= h = rmin vmax (1) Energy per unit mass E 1 = (T + V ) m m E 1 1 2 GMm 1 2 GM = mvmax − = vmax − m m2 rmin 2 rmin (b) (2) From Eq. (1): vmax = h/rmin substituting into (2) E 1 h 2 GM = − 2 m 2 rmin rmin E 2 2 1 2GM 1 m − 2 =0 − 2 ⋅ rmin h h rmin 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. 686 PROBLEM 13.118* (Continued) Solving the quadratic: 1 rmin = 2 2 ( mE ) GM GM + + 2 h2 h2 h = GM h2 Rearranging 1 rmin (c) 2 1 + 1 + 2 E h m GM (3) Eccentricity of the trajectory: Eq. (12.39′) 1 GM = 2 (1 + ε cos θ ) r h When θ = 0, cos θ = 1 and r = rmin Thus, 1 rmin Comparing (3) and (4), (d ) = GM (1 + ε ) h2 ε = 1+ (4) 2E h m GM 2 (5) Recalling discussion in section 12.12 and in view of Eq. (5) 1. Hyperbola if ε > 1, that is, if E > 0 2. Parabola if ε = 1, that is, if E = 0 3. Ellipse if ε < 1, that is, if E < 0 Note: For circular orbit ε = 0 and 2 2E h 1+ =0 m GM GM r 2 GM m , E = − h 2 or but for circular orbit v2 = thus 1 (GM ) 2 1 GMm E=− m =− 2 GMr 2 r and h2 = v 2 r 2 = GMr , 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. 687 PROBLEM 13.CQ4 A large insect impacts the front windshield of a sports car traveling down a road. Which of the following statements is true during the collision? (a) The car exerts a greater force on the insect than the insect exerts on the car. (b) The insect exerts a greater force on the car than the car exerts on the insect. (c) The car exerts a force on the insect, but the insect does not exert a force on the car. (d) The car exerts the same force on the insect as the insect exerts on the car. (e) Neither exerts a force on the other; the insect gets smashed simply because it gets in the way of the car. SOLUTION Answer: (d) This is Newton’s 3rd Law. 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. 688 PROBLEM 13.CQ5 The expected damages associated with two types of perfectly plastic collisions are to be compared. In the first case, two identical cars traveling at the same speed impact each other head on. In the second case, the car impacts a massive concrete wall. In which case would you expect the car to be more damaged? (a) Case 1 (b) Case 2 (c) The same damage in each case SOLUTION Answer: (c) In both cases the car will come to a complete stop, so the applied impulse will be the same. 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. 689 PROBLEM 13.F1 The initial velocity of the block in position A is 30 ft/s. The coefficient of kinetic friction between the block and the plane is μk = 0.30. Draw impulse-momentum diagrams that could be used to determine the time it takes for the block to reach B with zero velocity, if θ = 20°. SOLUTION Answer: 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. 690 PROBLEM 13.F2 A 4-lb collar which can slide on a frictionless vertical rod is acted upon by a force P which varies in magnitude as shown. Knowing that the collar is initially at rest, draw impulse-momentum diagrams that could be used to determine its velocity at t = 3 s. SOLUTION Answer: Where ò t2 = 3 t1 = 0 Pdt is the area under the curve. 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. 691 PROBLEM 13.F3 The 15-kg suitcase A has been propped up against one end of a 40-kg luggage carrier B and is prevented from sliding down by other luggage. When the luggage is unloaded and the last heavy trunk is removed from the carrier, the suitcase is free to slide down, causing the 40-kg carrier to move to the left with a velocity vB of magnitude 0.8 m/s. Neglecting friction, draw impulse-momentum diagrams that could be used to determine (a) the velocity of A as it rolls on the carrier and (b) the velocity of the carrier after the suitcase hits the right side of the carrier without bouncing back. SOLUTION Answer: (a) (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. 692 PROBLEM 13.F4 Car A was traveling west at a speed of 15 m/s and car B was traveling north at an unknown speed when they slammed into each other at an intersection. Upon investigation it was found that after the crash the two cars got stuck and skidded off at an angle of 50° north of east. Knowing the masses of A and B are mA and mB respectively, draw impulse-momentum diagrams that could be used to determine the velocity of B before impact. SOLUTION Answer: 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. 693 PROBLEM 13.F5 Two identical spheres A and B, each of mass m, are attached to an inextensible inelastic cord of length L and are resting at a distance a from each other on a frictionless horizontal surface. Sphere B is given a velocity v0 in a direction perpendicular to line AB and moves it without friction until it reaches B' where the cord becomes taut. Draw impulse-momentum diagrams that could be used to determine the magnitude of the velocity of each sphere immediately after the cord has become taut. SOLUTION Answer: Where v A¢ y = vB¢ y since the cord is inextensible. 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. 694 PROBLEM 13.119 A 35,000 Mg ocean liner has an initial velocity of 4 km/h. Neglecting the frictional resistance of the water, determine the time required to bring the liner to rest by using a single tugboat which exerts a constant force of 150 kN. SOLUTION m = 35, 000 Mg = 35 × 106 kg F = 150 × 103 N v1 = 4 km/hr = 1.1111 m/s mv1 − Ft = 0 (35 × 10 kg)(1.1111 m/s) − (150 × 103 N)t = 0 t = 259.26 s 6 t = 4 min19 s 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. 695 PROBLEM 13.120 A 2500-lb automobile is moving at a speed of 60 mi/h when the brakes are fully applied, causing all four wheels to skid. Determine the time required to stop the automobile (a) on dry pavement ( μ k = 0.75), (b) on an icy road ( μ k = 0.10). SOLUTION v1 = 60 mph = 88 ft/s mv1 − μkWt = 0 t= (a) (b) mv1 mv1 v = = 1 μkW μk mg μk g For μk = 0.75 t= 88 ft/s (0.75)(32.2 ft/s 2 ) t = 3.64 s t= 88 ft/s (0.10)(32.2 ft/s 2 ) t = 27.3 s For μk = 0.10 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. 696 PROBLEM 13.121 A sailboat weighing 980 lb with its occupants is running down wind at 8 mi/h when its spinnaker is raised to increase its speed. Determine the net force provided by the spinnaker over the 10-s interval that it takes for the boat to reach a speed of 12 mi/h. SOLUTION v1 = 8 mi/h = 11.73 ft/s t1− 2 = 10 sec v2 = 12 mi/h = 17.60 ft/s m ⋅ v1 + imp1− 2 = mv2 m(11.73 ft/s) + Fn (10 s) = m(17.60 ft/s) Fn = (980 lb)(17.60 ft/s − 11.73 ft/s) (32.2 ft/s 2 )(10 s) Fn = 178.6 lb Note: Fn is the net force provided by the sails. The force on the sails is actually greater and includes the force needed to overcome the water resistance on the hull. 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. 697 PROBLEM 13.122 A truck is hauling a 300-kg log out of a ditch using a winch attached to the back of the truck. Knowing the winch applies a constant force of 2500 N and the coefficient of kinetic friction between the ground and the log is 0.45, determine the time for the log to reach a speed of 0.5m/s. SOLUTION Apply the principle of impulse and momentum to the log. mv1 + ΣImp1 → 2 = mv 2 Components in y-direction: 0 + Nt − mgt cos 20° = 0 N = mg cos 20° Components in x-direction: 0 + Tt − mgt sin 20° − μk Nt = mv2 (T − mg sin 20° − μk mg cos 20°)t = mv2 [T − mg (sin 20° + μk cos 20°)]t = mv2 Data: T = 2500 N, m = 300 kg, g = 9.81 m/s 2 , μk = 0.45, v2 = 0.5 m/s [2500 − (300)(9.81)(sin 20° + 0.45cos 20°)] t = (300)(0.5) 248.95 t = 150 t = 0.603 s 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. 698 PROBLEM 13.123 A truck is traveling down a road with a 3-percent grade at a speed of 55 mi/h when the brakes are applied. Knowing the coefficients of friction between the load and the flatbed trailer shown are μs = 0.40 and μk = 0.35, determine the shortest time in which the rig can be brought to a stop if the load is not to shift. SOLUTION Apply the principle impulse-momentum to the crate, knowing that, if the crate does not shift, the velocity of the crate matches that of the truck. For impending slip the friction and normal components of the contact force between the crate and the flatbed trailer satisfy the following equation: F f = μs N mv1 + ΣImp1→2 = mv 2 Components in y-direction: 0 + Nt − mgt cos θ = 0 N = mg cos θ mv1 + mgt sin θ − μ s Nt = mv2 Components in x-direction: mv1 + mgt (sin θ − μ s cos θ ) = 0 t= Data: v1 g ( μ s cos θ − sin θ ) v1 = 55 mi/h = 80.667 ft/s, v2 = 0, g = 32.2 ft/s 2 , μ s = 0.40, tan θ = 3/100 θ = 1.71835° μs cos θ − sin θ = 0.36983 t= 80.667 (32.2)(0.36983) t = 6.77 s 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. 699 PROBLEM 13.124 Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop. A 10-ton truck enters a 15° ramp at a high speed v0 = 108 ft/s and travels for 6 s before its speed is reduced to 36 ft/s. Assuming constant deceleration, determine (a) the magnitude of the braking force, (b) the additional time required for the truck to stop. Neglect air resistance and rolling resistance. SOLUTION W = 20, 000 lb m= 20,000 = 621.118 lb ⋅ s 2 /ft 32.2 Momentum in the x direction x : mv0 − ( F + mg sin15°)t = mv1 621.118(108) − ( F + mg sin15°)6 = (621.118)(36) F + mg sin15° = 7453.4 (a) F = 7453.4 − 20, 000 sin15° = 2277 lb F = 2280 lb (b) mv0 − ( F + mg sin15°)t = 0 t = total time 621.118(108) − 7453.4 t = 0; t = 9.00 s Additional time = 9 – 6 t = 3.00 s 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. 700 PROBLEM 13.125 Baggage on the floor of the baggage car of a high-speed train is not prevented from moving other than by friction. The train is travelling down a 5 percent grade when it decreases its speed at a constant rate from 120 mi/h to 60 mi/h in a time interval of 12 s. Determine the smallest allowable value of the coefficient of static friction between a trunk and the floor of the baggage car if the trunk is not to slide. SOLUTION v1 = 120 mi/h = 176 ft/s v2 = 60 mi/h = 88 ft/s t1− 2 = 12 s Nt1− 2 = Wt1− 2 cos θ m v1 − μs m gt1− 2 cos θ + m gt1− 2 sin θ = m v2 (176 ft/s) − μs (32.2 ft/s 2 )(12 s)(cos 2.86°) + (32.2 ft/s 2 )(12 s)(sin 2.86°) = 88 ft/s μs = 176 − 88 + (32.2)(12)(sin 2.86°) (32.2)(12)(cos 2.86°) μs = 0.278 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. 701 PROBLEM 13.126 A 2-kg particle is acted upon by the force, expressed in newtons, F = (8 − 6t )i + (4 − t 2 ) j + (4 + t )k. Knowing that the velocity of the particle is v = (150 m/s)i + (100 m/s)j − (250 m/s)k at t = 0, determine (a) the time at which the velocity of the particle is parallel to the yz plane, (b) the corresponding velocity of the particle. SOLUTION mv 0 + Fdt = mv Where Fdt = (1) t [(8 − 6t )i + (4 − t 0 2 ) j + (4 + t )k ]dt 1 1 = (8t − 3t 2 ) i + 4t − t 3 j + 4t + t 2 k 3 2 Substituting m = 2 kg, v0 = 150i + 100 j − 250k into (1): 1 1 (2 kg)(150i + 100 j − 250k ) + (8t − 3t 2 )i + 4t − t 3 j + 4t + t 2 k = (2 kg) v 3 2 3 1 1 v = 150 + 4t − t 2 i + 100 + 2t − t 3 j + −250 + 2t + t 2 k 2 6 4 (a) v is parallel to yz plane when vx = 0, that is, when 3 150 + 4t − t 2 = 0 t = 11.422 s 2 (b) t = 11.42 s 1 v = 100 + 2(11.422) − (11.422)3 j 6 1 + −250 + 2(11.422) + (11.422) 2 k 4 v = −(125.5 m/s) j − (194.5 m/s)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. 702 PROBLEM 13.127 A truck is traveling down a road with a 4-percent grade at a speed of 60 mi/h when its brakes are applied to slow it down to 20 mi/h. An antiskid braking system limits the braking force to a value at which the wheels of the truck are just about to slide. Knowing that the coefficient of static friction between the road and the wheels is 0.60, determine the shortest time needed for the truck to slow down. SOLUTION θ = tan −1 4 = 2.29° 100 mv1 + Σ imp1− 2 = mv 2 mv1 + Wt sin θ − Ft = mv2 v1 = 60 mi/h = 88 ft/s N = W cos θ W = mg v2 = 20 mi/h = 29.33 ft/s F = μ s N = μ sW cos θ ( m )(88 ft/s) + ( m )(32.2 ft/s 2 )(t )(sin 2.29°) − (0.60)( m )(32.2 ft/s 2 )(cos 2.29°)(t ) = ( m )(29.33 ft/s) t= 88 − 29.33 32.2[(0.60) cos 2.29° − sin 2.29°] t = 3.26 s 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. 703 PROBLEM 13.128 Skid marks on a drag race track 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 shortest possible time for the car to travel the initial 20-m portion of the track if it starts from rest with its front wheels just off the ground. (b) Determine the minimum time for the car to run the whole race if, after skidding for 20 m, the wheels roll without sliding for the remainder of the race. Assume for the rolling portion of the race that 65 percent of the weight is on the rear wheels and that the coefficient of static friction is 0.85. Ignore air resistance and rolling resistance. SOLUTION (a) First 20 m Velocity at 20 m. Rear wheels skid to generate the maximum force resulting in maximum velocity and minimum time since all the weight is on the rear wheel: This force is F = μ k N = 0.60W . T0 + U 0− 20 = T20 Work and energy. T0 = 0 U 0 − 20 = ( F )(20) T20 = 1 2 mv20 2 1 2 m v20 2 = (2)(0.60)(20 m)(9.81 m/s 2 ) 0 + μk mg (20) = 2 v20 v20 = 15.344 m/s Impulse-momentum. 0 + μk mgt0 − 20 = mv20 t0− 20 = v20 = 15.344 m/s 15.344 m/s (0.60)(9.81 m/s 2 ) t0− 20 = 2.61 s 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. 704 PROBLEM 13.128 (Continued) (b) For the whole race: The maximum force on the wheels for the first 20 m is F = μk mg = 0.60 mg. For remaining 360 m, the maximum force, if there is no sliding and 65 percent of the weight is on the rear (drive) wheels, is F = μ s (0.65) mg = (0.85)(0.65) mg = 0.5525 mg Velocity at 400 m. T0 + U 0− 20 + U 20 − 400 = T400 Work and energy. T0 = 0 T400 = U 0 − 20 = (0.60mg )(20 m), U 60 − 400 = (0.5525mg )(380 m) 1 2 mv400 2 0 + 12 mg + (0.5525)(380)mg = 1 2 mv400 2 v400 = 65.990 m/s Impulse–momentum. From 20 m to 400 m F = μs N = 0.510mg v20 = 15.344 m/s v400 = 65.990 m/s m (15.344) + 0.5525 mgt20 − 400 = m (65.990); t20 − 400 = 9.3442 s t0− 400 = t0 − 20 + t20 − 400 = 2.61 + 9.34 t0− 400 = 11.95 s 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. 705 PROBLEM 13.129 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 time required to bring the train to a stop, (b) the force in each coupling. SOLUTION Weights of cars: WA = WC = 80,000 lb, WB = 100, 000 lb Masses of cars: m A = mC = 2484 lb ⋅ s 2 /ft, mB = 3106 lb ⋅ s 2 /ft For each car the normal force (upward) is equal in magnitude to the weight of the car. N A = NC = 80,000 lb Friction forces: FA = 0 N B = 100, 000 lb (brakes not applied) FB = (0.35)(100, 000) = 35000 lb FC = (0.35)(80, 000) = 28, 000 lb Stopping data: (a) v1 = 30 mi/h = 44 ft/s, v2 = 0. Apply the principle of impulse-momentum to the entire train. m = m A + mB + mC = 8074 lb ⋅ s 2 /ft F = FA + FB + FC = 63, 000 lb − mv1 + Ft = mv2 t = (b) m(v1 − v2 ) (8074)(44) = = 5.639 s F 63, 000 t = 5.64 s Coupling force FAB: Apply the principle of impulse-momentum to car A alone. −m Av1 + FAt + FABt = 0 −(2484)(44) + 0 + FAB (5.639) = 0 FAB = 19,390 lb FAB = 19,390 lb (tension) 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. 706 PROBLEM 13.129 (Continued) Coupling force FBC : Apply the principle of impulse-momentum to car C alone. −mC v1 + FC t − FBC t = 0 −(2484)(44) + (28000)(5.639) − FBC (5.639) = 0 FBC = 8620 lb FBC = 8620 lb (tension) 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. 707 PROBLEM 13.130 Solve Problem 13.129 assuming that the brakes are applied only on the wheels of car A. PROBLEM 13.129 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 time required to bring the train to a stop, (b) the force in each coupling. SOLUTION Weights of cars: WA = WC = 80,000 lb, WB = 100, 000 lb Masses of cars: m A = mC = 2484 lb ⋅ s 2 /ft, mB = 3106 lb ⋅ s 2 /ft For each car the normal force (upward) is equal in magnitude to the weight of the car. N A = N C = 80,000 lb N B = 100, 000 lb FA = (0.35)(80, 000) = 28, 000 lb Friction forces: FB = 0 (brakes not applied) FC = 0 Stopping data: (a) v1 = 30 mi/h = 44 ft/s, v2 = 0. Apply the principle of impulse-momentum to the entire train. m = m A + mB + mC = 8074 lb ⋅ s 2 /ft F = FA + FB + FC = 28, 000 lb − mv1 + Ft = mv2 t = (b) m(v1 − v2 ) (8074)(44) = = 12.688 s F 28, 000 t = 12.69 s Coupling force FAB: Apply the principle of impulse-momentum to car A alone. −mAv1 + FAt + FABt = 0 −(2484)(44) + (28, 000)(12.688) + FAB (12.688) = 0 FAB = −19,390 lb FAB = 19,390 lb (compression) 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. 708 PROBLEM 13.130 (Continued) Coupling force FBC : Apply the principle of impulse-momentum to car C alone. −mC v1 + FC t − FBC t = 0 −(2484)(44) + (0) − FBC (12.688) = 0 FBC = −8620 lb FBC = 8620 lb (compression) 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. 709 PROBLEM 13.131 A trailer truck with a 2000-kg cab and an 8000-kg trailer is traveling on a level road at 90 km/h. The brakes on the trailer fail and the antiskid system of the cab provides the largest possible force which will not cause the wheels of the cab to slide. Knowing that the coefficient of static friction is 0.65, determine (a) the shortest time for the rig to come to a stop, (b) the force in the coupling during that time. SOLUTION v = 90 km/h = 25 m/s (a) The shortest time for the rig to come to a stop will be when the friction force on the wheels is maximum. The downward force exerted by the trailer on the cab is assumed to be zero. Since the trailer brakes fail, all of the braking force is supplied by the wheels of the cab, which is maximum when the wheels of the cab are at impending sliding. Ft1− 2 = μs N C t1− 2 N C = mC g = (2000) g Ft1− 2 = (0.65)(2000) gt [( mC + mT )v]2 = − Ft + [(mC + mT )v]1 0 = −(0.65)(2000 kg)(9.81 m/s 2 )(t1− 2 ) = 10, 000 kg(25 m/s) t1− 2 = 19.60 s (b) For the trailer: [mT v]2 = −Qt1− 2 + [mT v]1 From (a), t1− 2 = 19.60 s 0 = −Q(19.60 s) + (8000 kg)(25 m/s) Q = 10, 204 N Q = 10.20 kN (compression) 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. 710 PROBLEM 13.132 The system shown is at rest when a constant 150-N force is applied to collar B. Neglecting the effect of friction, determine (a) the time at which the velocity of collar B will be 2.5 m/s to the left, (b) the corresponding tension in the cable. SOLUTION Constraint of cord. When the collar B moves 1 unit to the left, the weight A moves up 2 units. Thus v A = 2 vB Masses and weights: vB = m A = 3 kg 1 vA 2 WA = 29.43 N mB = 8 kg Let T be the tension in the cable. Principle of impulse and momentum applied to collar B. : 0 + 150t − 2Tt = mB (vB ) 2 For (vB ) 2 = 2.5 m/s 150t − 2Tt = (8 kg)(2.5 m/s) 150t − 2Tt = 20 (1) 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. 711 PROBLEM 13.132 (Continued) Principle of impulse and momentum applied to weight A. : 0 + Tt − WAt = mA (v A ) 2 Tt + WAt = m A (2VB 2 ) Tt − 29.43t = (3 kg)(2)(2.5 m/s) Tt − 29.43t = 15 (2) To eliminate T multiply Eq. (2) by 2 and add to Eq. (1). (a) (b) Time: 91.14t = 50 From Eq. (2), T= t = 0.549 s 15 + 29.43 t T = 56.8 N Tension in the cable. 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. 712 PROBLEM 13.133 An 8-kg cylinder C rests on a 4-kg platform A supported by a cord which passes over the pulleys D and E and is attached to a 4-kg block B. Knowing that the system is released from rest, determine (a) the velocity of block B after 0.8 s, (b) the force exerted by the cylinder on the platform. SOLUTION (a) Blocks A and C: [( mA + mC )v]1 − T (t1− 2 ) + ( mA + mC ) gt1− 2 = [(m A + mC )v]2 0 + (12 g − T )(0.8) = 12v (1) Block B: [mB v]1 + (T )t1− 2 − mB gt1− 2 = (mB v) 2 0 + (T − 4 g )(0.8) = 4v (2) Adding (1) and (2), (eliminating T) (12 g − 4 g )(0.8) = (12 + 4) v= (b) (8 kg)(9.81 m/s 2 )(0.8 s) 16 kg v = 3.92 m/s Collar A: (m Av)1 = 0 0 + ( FC + m A g ) (3) From Eq. (2) with v = 3.92 m/s 4v T= + 4g 0.8 (4 kg)(3.92 m/s) T= + (4 kg)(9.81 m/s 2 ) (0.8 s) T = 58.84 N Solving for FC in (3) FC = (4 kg)(3.92 m/s) − (4 kg)(9.81 m/s 2 ) + 58.84 N (0.8 s) FC = 39.2 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. 713 PROBLEM 13.134 An estimate of the expected load on over-the-shoulder seat belts is to be made before designing prototype belts that will be evaluated in automobile crash tests. Assuming that an automobile traveling at 45 mi/h is brought to a stop in 110 ms, determine (a) the average impulsive force exerted by a 200-lb man on the belt, (b) the maximum force Fm exerted on the belt if the force-time diagram has the shape shown. SOLUTION (a) Force on the belt is opposite to the direction shown. v1 = 45 mi/h = 66 ft/s, W = 200 lb mv1 − Fdt = mv 2 Fdt = F ave Δt Δt = 0.110 s (200 lb)(66 ft/s) − Fave (0.110 s) = 0 (32.2 ft/s 2 ) Fave = (b) (200)(66) = 3727 lb (32.2)(0.110) Fave = 3730 lb 1 Fm (0.110 s) 2 From (a), impulse = Fave Δt = (3727 lb)(0.110 s) Impulse = area under F −t diagram = 1 Fm (0.110) = (3727)(0.110) 2 Fm = 7450 lb 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. 714 PROBLEM 13.135 A 60-g model rocket is fired vertically. The engine applies a thrust P which varies in magnitude as shown. Neglecting air resistance and the change in mass of the rocket, determine (a) the maximum speed of the rocket as it goes up, (b) the time for the rocket to reach its maximum elevation. SOLUTION m = 0.060 kg Mass: mg = (0.060)(9.81) = 0.5886 N Weight: Forces acting on the model rocket: Thrust: P(t )(given function of t ) Weight: W (constant) Support: S (acts until P > W ) Over 0 < t < 0.2 s: 13 t = 65t 0.2 W = 0.5886 N P= S = W − P = 0.5886 − 65t Before the rocket lifts off, S become zero when t = t1. 0 = 0.5886 − 65t1 t1 = 0.009055 s. Impulse due to S: (t > t1 ) t 0 Sdt = t1 Sdt 0 1 mgt1 2 = (0.5)(0.5886)(0.009055) = = 0.00266 N ⋅ s The maximum speed occurs when dv = a = 0. dt At this time, W − P = 0, which occurs at t2 = 0.8 s. 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. 715 PROBLEM 13.135 (Continued) (a) Maximum speed (upward motion): Apply the principle of impulse-momentum to the rocket over 0 ≤ t ≤ t2 . 0.8 0 Pdt = area under the given thrust-time plot. 1 1 (0.2)(13) + (0.1)(13 + 5) + (0.8 − 0.3)(5) 2 2 = 4.7 N ⋅ s = 0.8 0 m1v1 + Wdt = (0.5886)(0.8) = 0.47088 N ⋅ S 0.8 0 Pdt + 0.8 0 Sdt − 0.8 0 Wdt = mv2 0 + 4.7 + 0.00266 − 0.47088 = 0.060 v2 v2 = 70.5 m/s (b) Time t3 to reach maximum height: (v3 = 0) mv1 + t3 0 Pdt + t3 0 Sdt − Wt3 = mv3 0 + 4.7 + 0.00266 − 0.5886t3 = 0 t3 = 7.99 s 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. 716 PROBLEM 13.136 A simplified model consisting of a single straight line is to be obtained for the variation of pressure inside the 10-mm-diameter barrel of a rifle as a 20-g bullet is fired. Knowing that it takes 1.6 ms for the bullet to travel the length of the barrel and that the velocity of the bullet upon exit is 700 m/s, determine the value of p0. SOLUTION At t = 0, p = p0 = c1 − c2t c1 = p0 At t = 1.6 × 10−3 s, p=0 0 = c1 − c2 (1.6 × 10−3 s) c2 = p0 1.6 × 10−3 s m = 20 × 10−3 kg 0+ A 1.6 × 10 −3 s 0 pdt = mv2 π (10 × 10−3 )2 A= 4 A = 78.54 × 10−6 m 2 0+ A 1.6 ×10−3 s 0 (c1 − c2t ) dt = 20 × 10−3 g (c )(1.6 × 10−3 s) 2 −3 (78.54 × 10−6 m 2 ) (c1 )(1.6 × 10−3 s) − 2 = (20 × 10 kg)(700 m/s) 2 1.6 × 10−3 c1 − 1.280 × 10−6 c2 = 178.25 × 103 (1.6 × 10−3 m 2 ⋅ s) p0 − (1.280 × 10−6 m 2 ⋅ s 2 ) p0 = 178.25 × 103 kg ⋅ m/s −3 (1.6 × 10 s) p0 = 222.8 × 106 N/m 2 p0 = 223 MPa 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. 717 PROBLEM 13.137 A 125-lb block initially at rest is acted upon by a force P which varies as shown. Knowing that the coefficients of friction between the block and the horizontal surface are μs = 0.50 and μk = 0.40, determine (a) the time at which the block will start moving, (b) the maximum speed reached by the block, (c) the time at which the block will stop moving. SOLUTION 0 + Pdt − Fdt = mv v= At any time: (a) 1 Pdt − Fdt m (1) Block starts moving at t. P = Fs = μsW = (0.50)(125 lb) = 62.5 lb t1 t1 8s 8s ; = = Fs 100 lb 62.5 lb 100 lb (b) t1 = 5.00 s Maximum velocity: At t = tm P = Fk = μkW = 0.4(125) = 50 lb where Block moves at t = 5 s. Shaded area is maximum net impulse Pdt − F dt R when t = tm1 v = vm Eq. (1): vm = vm = 1 shaded 1 1 1 1 (12.5 + 50)(3) + (50)(4) = (193.75) = 2 m area m 2 m 1 125 lb 32.2 [193.75] = 49.91 ft/s v m = 49.9 ft/s 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. 718 PROBLEM 13.137 (Continued) (c) Block stops moving when Pdt − Fdt = 0; or Qdt = Fdt Assume tm > 16 s. 1 Pdt = 2 (100)(16) = 800 lb ⋅ s 1 Fdt = 2 (62.5)(5) + (50)(t m Pdt − Fdt = 800 − [156.25 + 50(t tm > 16 s OK m − 5) − 5)] = 0 tm = 17.875 s tm = 17.88 s 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. 719 PROBLEM 13.138 Solve Problem 13.137, assuming that the weight of the block is 175 lb. PROBLEM 13.137 A 125-lb block initially at rest is acted upon by a force P which varies as shown. Knowing that the coefficients of friction between the block and the horizontal surface are μs = 0.50 and μk = 0.40, determine (a) the time at which the block will start moving, (b) the maximum speed reached by the block, (c) the time at which the block will stop moving. SOLUTION See solution of Problem 13.137. W = 175 lb (a) Block starts moving: v= 1 Pdt − Fdt m (1) P = Fs = μsW = (0.50)(175) = 87.5 lb See first figure of Problem 13.137. t1 t1 8s 8s = = ; Fs 100 lb 87.5 lb 100 s (b) Maximum velocity: t1 = 7.00 s P = Fk = μkW = 0.4(175) = 70 lb 16 − tm 8s = 70 lb 100 lb 8 16 − tm = 70 = 5.6 100 tm = 10.40 s Eq. (1): vm = 1 shaded m area 1 1 1 (17.5 + 30)(1.0) + (30)(10.4 − 8) m 2 2 1 = (59.75) m = vm = 1 175 lb 32.2 [59.75] = 10.994 ft/s v m = 10.99 ft/s 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. 720 PROBLEM 13.138 (Continued) (c) Block stops moving when net impulse ( P − F )dt = 0 Assume ts , < 16 s. ts 0 Pdt = = ts 0 Fdt = (16 − t s ) 1 1 (100)(8) + 100 + 100 (t s − 8) 2 2 8 1 1 100 (100)(16) − (16 − t s ) 2 2 2 8 1 (87.5)(7) + (70)(t s − 7) 2 100 Pdt − Fdt = 800 − 16 (16 − t ) 2 s Solving for ts , ts = 13.492 s − 306.25 − 70(ts − 7) = 0 ts = 13.49 s 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. 721 PROBLEM 13.139 A baseball player catching a ball can soften the impact by pulling his hand back. Assuming that a 5-oz ball reaches his glove at 90 mi/h and that the player pulls his hand back during the impact at an average speed of 30 ft/s over a distance of 6 in., bringing the ball to a stop, determine the average impulsive force exerted on the player’s hand. SOLUTION v = 90 mi/h = 132 ft/s m= 5 /g 16 ( ) d = 12 = 1 s 60 vav 30 6 t= 0 = Fav t + mv Fav = ( ) Wv gt Fav = = mv t ( 165 lb ) (132 ft/s) 1 s (32.2 ft/s 2 ) ( 60 ) Fav = 76.9 lb 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. 722 PROBLEM 13.140 A 1.62 ounce golf ball is hit with a golf club and leaves it with a velocity of 100 mi/h. We assume that for 0 ≤ t ≤ t0, where t0 is the duration of the impact, the magnitude F of the force exerted on the ball can be expressed as F = Fm sin (π t/t0 ). Knowing that t0 = 0.5 ms, determine the maximum value Fm of the force exerted on the ball. SOLUTION W = 1.62 ounces = 0.10125 lb m = 3.1444 × 10−3 slug t = 0.5 ms = 0.5 × 10−3 s v = 100 mi/h = 146.67 ft/s The impulse applied to the ball is t0 0 Fdt = t0 0 =− Fm sin Fm t0 π πt t0 dt = − Fm t0 π (cos π − cos 0) = cos πt t0 t0 0 2 Fm t0 π Principle of impulse and momentum. mv1 + t0 0 F dt = mv 2 with v1 = 0, 0+ Solving for Fm, Fm = 2 Fm t0 π π mv2 2t0 = mv2 = π (3.1444 × 10−3 )(146.67) (2)(0.5 × 10−3 ) = 1.4488 × 103 lb Fm = 1.45 kip 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. 723 PROBLEM 13.141 The triple jump is a track-and-field event in which an athlete gets a running start and tries to leap as far as he can with a hop, step, and jump. Shown in the figure is the initial hop of the athlete. Assuming that he approaches the takeoff line from the left with a horizontal velocity of 10 m/s, remains in contact with the ground for 0.18 s, and takes off at a 50° angle with a velocity of 12 m/s, determine the vertical component of the average impulsive force exerted by the ground on his foot. Give your answer in terms of the weight W of the athlete. SOLUTION mv1 + (P − W)Δt = mv 2 Δt = 0.18 s Vertical components W (12)(sin 50°) g (12)(sin 50°) Pv = W + W (9.81)(0.18) 0 + ( Pv − W )(0.18) = Pv = 6.21W 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. 724 PROBLEM 13.142 The last segment of the triple jump track-and-field event is the jump, in which the athlete makes a final leap, landing in a sand-filled pit. Assuming that the velocity of a 80-kg athlete just before landing is 9 m/s at an angle of 35° with the horizontal and that the athlete comes to a complete stop in 0.22 s after landing, determine the horizontal component of the average impulsive force exerted on his feet during landing. SOLUTION m = 80 kg Δt = 0.22 s mv1 + (P − W)Δt = mv 2 Horizontal components m(9)(cos 35°) − PH (0.22) = 0 PH = (80 kg)(9 m/s)(cos 35°) = 2.6809 kN (0.22 s) PH = 2.68 kN 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. 725 PROBLEM 13.143 The design for a new cementless hip implant is to be studied using an instrumented implant and a fixed simulated femur. Assuming the punch applies an average force of 2 kN over a time of 2 ms to the 200 g implant determine (a) the velocity of the implant immediately after impact, (b) the average resistance of the implant to penetration if the implant moves 1 mm before coming to rest. SOLUTION m = 200 g = 0.200 kg Fave = 2 kN = 2000 N Δt = 2 ms = 0.002 s (a) Velocity immediately after impact: Use principle of impulse and momentum: v1 = 0 v2 = ? Imp1→2 = Fave (Δt ) mv1 + Imp1→2 = mv 2 0 + Fave (Δt ) = mv2 v2 = (b) Fave (Δt ) (2000)(0.002) = m 0.200 v2 = 20.0 m/s Average resistance to penetration: Δx = 1 mm = 0.001 m v2 = 20.0 ft/s v3 = 0 Use principle of work and energy. T2 + U 2→3 = T3 Rave = or 1 2 mv2 − Rave ( Δx) = 0 2 mv22 (0.200)(20.0) 2 = = 40 × 103 N 2(Δx) (2)(0.001) Rave = 40.0 kN 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. 726 PROBLEM 13.144 A 25-g steel-jacketed bullet is fired horizontally with a velocity of 600 m/s and ricochets off a steel plate along the path CD with a velocity of 400 m/s. Knowing that the bullet leaves a 10-mm scratch on the plate and assuming that its average speed is 500 m/s while it is in contact with the plate, determine the magnitude and direction of the average impulsive force exerted by the bullet on the plate. SOLUTION Impulse and momentum. Bullet alone: mv1 + F Δ t = mv 2 t direction: mv1 cos15° − Ft Δt = mv2 cos 20° Ft Δt = (0.025 kg)[600 m/s cos 15° − 400 m/s cos 20°] = 5.092 kg ⋅ m/s Δt = S BC 0.010 m = = 20 × 10−6 s v AV 500 m/s Ft = (5.092 kg ⋅ m/s)/(20 × 10−6 s) = 254.6 × 103 kg ⋅ m/s 2 = 254.6 kN n direction: − mv1 sin15° + Fn Δt = mv2 sin 20° Fn Δt = (0.025 kg)[600 m/s sin 15° + 400 m/s sin 20°] = 7.3025 kg ⋅ m/s Fn = (43025 kg ⋅ m/s)/(20 × 10−6 ) = 365.1 × 103 kg ⋅ m/s 2 = 365.1 kN Force on bullet: F = Fn2 + Ft 2 = 365.12 + 254.62 = 445 kN tanθ = Fn 365.1 = Ft 254.6 θ = 55.1° θ − 15° = 40.1° F = 445 kN Force on plate: 40.1° F ′ = −F F ′ = 445 kN 40.1° 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. 727 PROBLEM 13.145 A 25-ton railroad car moving at 2.5 mi/h is to be coupled to a 50 ton car which is at rest with locked wheels ( μk = 0.30). Determine (a) the velocity of both cars after the coupling is completed, (b) the time it takes for both cars to come to rest. SOLUTION Weight and mass: (Label cars A and B.) Car A : WA = 50 tons = 100, 000 lb, mA = 3106 lb ⋅ s 2 /ft Car B: WB = 25 tons = 50,000 lb, mB = 1553 lb ⋅ s 2 /ft Initital velocities: vA = 0 vB = 2.5 mi/h = 3.6667 ft/s (a) v B = 3.6667 ft/s The momentum of the system consisting of the two cars is conserved immediately before and after coupling. Let v′ be the common velocity of that cars immediately after coupling. Apply conservation of momentum. : mB vB = m A v′ + mB v′ v′ = (b) mB vB (3106)(3.6667) = = 2.444 ft/s 4569 mA + mB v′ = 1.667 mi/h After coupling: The friction force acts only on car A. + ΣF = 0 A : N A − W A = 0 N A = WA FA = μk N A = μkWA (sliding) FB = 0 (Car B is rolling.) Apply impuslse-momentum to the coupled cars. 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. 728 PROBLEM 13.145 (Continued) : −(m A + mB )v′ + FAt = 0 t= (m A + mB )v11 mB vB = μk W A FA t= (1553)(3.6667) = 0.1898 (0.30)(100, 000) t = 0.190 s 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. 729 PROBLEM 13.146 At an intersection car B was traveling south and car A was traveling 30° north of east when they slammed into each other. Upon investigation it was found that after the crash the two cars got stuck and skidded off at an angle of 10° north of east. Each driver claimed that he was going at the speed limit of 50 km/h and that he tried to slow down but couldn’t avoid the crash because the other driver was going a lot faster. Knowing that the masses of cars A and B were 1500 kg and 1200 kg, respectively, determine (a) which car was going faster, (b) the speed of the faster of the two cars if the slower car was traveling at the speed limit. SOLUTION (a) Total momentum of the two cars is conserved. Σmv, x : m A v A cos 30° = (m A + mB )v cos 10° (1) Σmv, y : m A v A sin 30° − mB vB = (mA + mB )v sin 10° (2) Dividing (1) into (2), mB vB sin 30° sin 10° − = cos 30° m Av A cos 30° cos 10° vB (tan 30° − tan 10°)( mA cos 30°) = vA mB vB (1500) = (0.4010) cos 30° vA (1200) vB = 0.434 vA v A = 2.30 vB A was going faster. Thus, (b) Since vB was the slower car. vB = 50 km/h v A = (2.30)(50) v A = 115.2 km/h 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. 730 PROBLEM 13.147 The 650-kg hammer of a drop-hammer pile driver falls from a height of 1.2 m onto the top of a 140-kg pile, driving it 110 mm into the ground. Assuming perfectly plastic impact (e = 0), determine the average resistance of the ground to penetration. SOLUTION Velocity of the hammer at impact: Conservation of energy. T1 = 0 VH = mg (1.2 m) VH = (650 kg)(9.81 m/s 2 )(1.2 m) V1 = 7652 J 1 m 2 650 2 VH2 = v = 325 vH2 2 V2 = 0 T2 = T1 + V1 = T2 + V2 0 + 7652 = 325 v 2 v 2 = 23.54 m 2 /s 2 v = 4.852 m/s Velocity of pile after impact: Since the impact is plastic (e = 0), the velocity of the pile and hammer are the same after impact. Conservation of momentum: The ground reaction and the weights are non-impulsive. 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. 731 PROBLEM 13.147 (Continued) Thus, mH vH = (mH + m p )v′ v′ = Work and energy: mH vH (650) = (4.852 m/s) = 3.992 m/s (mH + m p ) (650 + 140) d = 0.110 m T2 + U 2 −3 = T3 1 (mH + mH )(v′)2 2 T3 = 0 T2 = 1 (650 + 140)(3.992) 2 2 T2 = 6.295 × 103 J T2 = U 2 −3 = ( mH + m p ) gd − FAV d = (650 + 140)(9.81)(0.110) − FAV (0.110) U 2 −3 = 852.49 − (0.110) FAV T2 + U 2 −3 = T3 6.295 × 103 + 852.49 − (0.110) FAV = 0 FAV = (7147.5)/(0.110) = 64.98 × 103 N FAV = 65.0 kN 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. 732 PROBLEM 13.148 A small rivet connecting two pieces of sheet metal is being clinched by hammering. Determine the impulse exerted on the rivet and the energy absorbed by the rivet under each blow, knowing that the head of the hammer has a weight of 1.5 lbs and that it strikes the rivet with a velocity of 20 ft/s. Assume that the hammer does not rebound and that the anvil is supported by springs and (a) has an infinite mass (rigid support), (b) has a weight of 9 lb. SOLUTION Weight and mass: Hammer: WH = 1.5 lb mH = 0.04658 lb ⋅ s 2 /ft Anvil: Part a: WA = ∞ mA = ∞ Part b: WA = 9 lb mA = 0.2795 lb ⋅ s 2 /ft Kinetic energy before impact: 1 1 mH vH2 = (0.04658)(20) 2 = 9.316 ft ⋅ lb 2 2 Let v2 be the velocity common to the hammer and anvil immediately after impact. Apply the principle of conservation of momentum to the hammer and anvil over the duration of the impact. T1 = : Σmv1 = Σ mv2 mH vH = ( mH + mA )v2 v2 = mH vH mH + m A TA = 1 1 mH2 vH2 (mH + m A )v22 = 2 2 mH + mA T2 = mH T1 mH + m A (1) Kinetic energy after impact: (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. 733 PROBLEM 13.148 (Continued) Impulse exerted on the hammer: : mH vH − F (Δt ) = mH v2 F Δt = mH (vH − v2 ) (a) (3) WA = ∞ : By Eq. (1), v2 = 0 By Eq. (2), T2 = 0 T1 − T2 = 9.32 ft ⋅ lb Energy absorbed: By Eq. (3), F (Δt ) = (0.04658)(20 − 0) = 0.932 lb ⋅ s The impulse exerted on the rivet the same magnitude but opposite to direction. F Δt = 0.932 lb ⋅ s (b) WA = 9 lb: By Eq. (1), v2 = (0.04658)(20) = 2.857 ft/s 0.32608 By Eq. (2), T2 = (0.04658)(9.316) = 1.331 ft ⋅ lb 0.32608 T1 − T2 = 7.99 ft ⋅ lb Energy absorbed: By Eq. (3), F (Δt ) = (0.04658)(20 − 2.857) F (Δt ) = 0.799 lb ⋅ s 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. 734 PROBLEM 13.149 Bullet B weighs 0.5 oz and blocks A and C both weigh 3 lb. The coefficient of friction between the blocks and the plane is μk = 0.25. Initially the bullet is moving at v0 and blocks A and C are at rest (Figure 1). After the bullet passes through A it becomes embedded in block C and all three objects come to stop in the positions shown (Figure 2). Determine the initial speed of the bullet v0. SOLUTION Masses: 0.5 = 970.5 × 10−6 lb ⋅ s 2 /ft (16)(32.2) Bullet: mB = Blocks A and C: m A = mC = Block C + bullet: 3 = 93.168 × 10−3 lb ⋅ s 2 /ft 32.2 mC + mB = 94.138 × 10−3 lb ⋅ s 2 /ft Normal forces for sliding blocks from N − mg = 0 Block A: N A = m A g = 3.00 lb. Block C + bullet: N C = (mC + mB ) g = 3.03125 lb. Let v0 be the initial speed of the bullet; v1 be the speed of the bullet after it passes through block A; vA be the speed of block A immediately after the bullet passes through it; vC be the speed block C immediately after the bullet becomes embedded in it. Four separate processes and their governing equations are described below. 1. The bullet hits block A and passes through it. Use the principle of conservation of momentum. (v A )0 = 0 mB v0 + mA (v A )0 = mB v1 + m Av A v0 = v1 + 2. mAv A mB (1) The bullet hits block C and becomes embedded in it. Use the principle of conservation of momentum. (vC )0 = 0 mB v1 + mC (vC )0 = (mB + mC )vC v1 = (mB + mC )vC mB (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. 735 PROBLEM 13.149 (Continued) 3. Block A slides on the plane. Use principle of work and energy. T1 + U1→2 = T2 1 m Av A2 − μ k N A d A = 0 or v A = 2 4. 2μk N A d A mA (3) Block C with embedded bullet slides on the plane. Use principle of work and energy. dC = 4 in. = 0.33333 ft T1 + U1→ 2 = T2 1 (mC + mB )vC2 − μ k N C dC = 0 or vC = 2 2 μk NC dC mC + mB (4) Applying the numerical data: (2)(0.25)(3.03125)(0.33333) 94.138 × 10−3 = 2.3166 ft/s From Eq. (4), vC = From Eq. (3), vA = From Eq. (2), v1 = From Eq. (1), v0 = 224.71 + (2)(0.25)(3.00)(0.5) 93.168 × 10−3 = 2.8372 ft/s (94.138 × 10−3 )(2.3166) 970.5 × 10−6 = 224.71 ft/s (93.138 × 10−3 )(2.8372) 970.5 × 10−6 v0 = 497 ft/s 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. 736 PROBLEM 13.150 A 180-lb man and a 120-lb woman stand at opposite ends of a 300-lb boat, ready to dive, each with a 16-ft/s velocity relative to the boat. Determine the velocity of the boat after they have both dived, if (a) the woman dives first, (b) the man dives first. SOLUTION (a) Woman dives first: Conservation of momentum: − 120 300 + 180 (16 − v1 ) + v1 = 0 g g v1 = (120)(16) = 3.20 ft/s 600 Man dives next. Conservation of momentum: 300 + 180 300 180 v1 = − v2 + (16 − v2 ) g g g v2 = (b) 480v1 − (180)(16) = 2.80 ft/s 480 v 2 = 2.80 ft/s Man dives first: Conservation of momentum: 180 300 + 120 (16 − v1′ ) − v1′ = 0 g g v1′ = (180)(16) = 4.80 ft/s 600 Woman dives next. Conservation of momentum: − 300 + 120 300 120 v1′ = v2′ + (16 − v2′ ) g g g −420v1′ + (120)(16) v2′ = = −0.229 ft/s 420 v′2 = 0.229 ft/s 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. 737 PROBLEM 13.151 A 75-g ball is projected from a height of 1.6 m with a horizontal velocity of 2 m/s and bounces from a 400-g smooth plate supported by springs. Knowing that the height of the rebound is 0.6 m, determine (a) the velocity of the plate immediately after the impact, (b) the energy lost due to the impact. SOLUTION Just before impact vy = Just after impact vy = 2 g (1.6) = 5.603 m/s 2 g (0.6) = 3.431 m/s (a) Conservation of momentum: (+ y ) mballv y + 0 = −mballv′y + mplatev′plate (0.075)(5.603) + 0 = −0.075(3.431) + 0.4v′plate v′plate = 1.694 m/s (b) Energy loss Initial energy Final energy (T + V )1 = (T + V ) 2 = 1 (0.075)(2) 2 + 0.075 g (1.6) 2 1 1 (0.075)(2)2 + 0.075 g (0.6) + (0.4)(1.694) 2 2 2 Energy lost = (1.3272 − 1.1653) J = 0.1619 J 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. 738 PROBLEM 13.152 A 2-kg sphere A is connected to a fixed Point O by an inextensible cord of length 1.2 m. The sphere is resting on a frictionless horizontal surface at a distance of 1.5 ft from O when it is given a velocity v 0 in a direction perpendicular to line OA. It moves freely until it reaches position A′, when the cord becomes taut. Determine the maximum allowable velocity v 0 if the impulse of the force exerted on the cord is not to exceed 3 N ⋅ s. SOLUTION For the sphere at A′ immediately before and after the cord becomes taut mv0 + F Δt = mv A′ mv0 sin θ − F Δt = 0 F Δt = 3 N ⋅ s m = 2 kg 2(sin 65.38°)v0 = 3 v0 = 1.650 m/s 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. 739 PROBLEM 13.153 A 1-oz bullet is traveling with a velocity of 1400 ft/s when it impacts and becomes embedded in a 5-lb wooden block. The block can move vertically without friction. Determine (a) the velocity of the bullet and block immediately after the impact, (b) the horizontal and vertical components of the impulse exerted by the block on the bullet. SOLUTION Weight and mass. Bullet: w = 1 oz = Block: W = 5 lb (a) 1 lb 16 m = 0.001941 lb ⋅ s 2 /ft. M = 0.15528 lb ⋅ s 2 /ft. Use the principle of impulse and momentum applied to the bullet and the block together. Σm v1 + ΣImp1→ 2 = mv 2 mv0 cos 30° + 0 = (m + M )v′ Components : mv0 cos 30° (0.001941)(1400)cos 30° = 0.157221 m+M v′ = 14.968 ft/s v′ = v′ = 14.97 ft/s 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. 740 PROBLEM 13.153 (Continued) (b) Use the principle of impulse and momentum applied to the bullet alone. x-components: −mv0 sin 30° + Rx Δt = 0 Rx Δt = mv0 sin 30° = (0.001941)(1400) sin 30° = 1.3587 lb ⋅ s y-components: Rx Δt = 1.359 lb ⋅ s −mv0 cos 30° + Ry Δt = −mv′ Ry Δt = m(v0 cos 30° − v′) = (0.001941)(1400 cos 30° − 14.968) Ry Δt = 2.32 lb ⋅ s 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. 741 PROBLEM 13.154 In order to test the resistance of a chain to impact, the chain is suspended from a 240-lb rigid beam supported by two columns. A rod attached to the last link is then hit by a 60-lb block dropped from a 5-ft height. Determine the initial impulse exerted on the chain and the energy absorbed by the chain, assuming that the block does not rebound from the rod and that the columns supporting the beam are (a) perfectly rigid, (b) equivalent to two perfectly elastic springs. SOLUTION Velocity of the block just before impact: T1 = 0 V1 = Wh = (60 lb)(5 ft) = 300 lb ⋅ ft 1 2 mv V2 = 0 2 T1 + V1 = T2 + V2 T2 = 0 + 300 = 1 60 2 v 2 g (600)(32.2) 60 = 17.94 ft/s v= (a) Rigid columns: − mv + F Δt = 0 60 (17.94) = F Δt g F Δt = 33.43 lb ⋅ s on the block. F Δt = 33.4 lb ⋅ s All of the kinetic energy of the block is absorbed by the chain. T= 1 60 2 (17.94) 2 g = 300 ft ⋅ lb E = 300 ft ⋅ lb 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. 742 PROBLEM 13.154 (Continued) (b) Elastic columns: Momentum of system of block and beam is conserved. mv = ( M + m)v′ m 60 v′ = v= (17.94 ft/s) (m + M ) 300 v′ = 3.59 ft/s Referring to figure in part (a), − mv + F Δt = − mv′ F Δt = m(v − v′) 60 = (17.94 − 3.59) g 1 2 1 1 mv − mv′2 − Mv′2 2 2 2 60 240 = [(17.94)2 − (3.59)2 ] − (3.59)2 2g 2g F Δt = 26.7 lb ⋅ s E= E = 240 ft ⋅ lb 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. 743 PROBLEM 13.CQ6 A 5 kg ball A strikes a 1 kg ball B that is initially at rest. Is it possible that after the impact A is not moving and B has a speed of 5v? (a) Yes (b) No Explain your answer. SOLUTION Answer: (b) No. Conservation of momentum is satisfied, but the coefficient of restitution equation is not. The coefficient of restitution must be less than 1. 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. 744 PROBLEM 13.F6 A sphere with a speed v0 rebounds after striking a frictionless inclined plane as shown. Draw impulse-momentum diagrams that could be used to find the velocity of the sphere after the impact. SOLUTION Answer: 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. 745 PROBLEM 13.F7 An 80-Mg railroad engine A coasting at 6.5 km/h strikes a 20-Mg flatcar C carrying a 30-Mg load B which can slide along the floor of the car (μk = 0.25). The flatcar was at rest with its brakes released. Instead of A and C coupling as expected, it is observed that A rebounds with a speed of 2 km/h after the impact. Draw impulsemomentum diagrams that could be used to determine (a) the coefficient of restitution and the speed of the flatcar immediately after impact, and (b) the time it takes the load to slide to a stop relative to the car. SOLUTION Answer: (a) Look at A and C (the friction force between B and C is not impulsive) to find the velocity after impact. (b) Consider just B and C to find their final velocity. Consider just B to find the time. 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. 746 PROBLEM 13.F8 Two frictionless balls strike each other as shown. The coefficient of restitution between the balls is e. Draw the impulse-momentum diagrams that could be used to find the velocities of A and B after the impact. SOLUTION Answer: 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. 747 PROBLEM 13.F9 A 10-kg ball A moving horizontally at 12 m/s strikes a 10-kg block B. The coefficient of restitution of the impact is 0.4 and the coefficient of kinetic friction between the block and the inclined surface is 0.5. Draw impulsemomentum diagrams that could be used to determine the speeds of A and B after the impact. SOLUTION Answer: 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. 748 PROBLEM 13.F10 Block A of mass mA strikes ball B of mass mB with a speed of vA as shown. Draw impulse-momentum diagrams that could be used to determine the speeds of A and B after the impact and the impulse during the impact. SOLUTION Answer: 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. 749 PROBLEM 13.155 The coefficient of restitution between the two collars is known to be 0.70. Determine (a) their velocities after impact, (b) the energy loss during impact. SOLUTION Impulse-momentum principle (collars A and B): Σmv1 + ΣImp1→2 = Σmv 2 Horizontal components Using data, : m A v A + mB vB = mA v′A + mB vB′ (5)(1) + (3)( −1.5) = 5v′A + 3vB′ 5v′A + 3vB′ = 0.5 or (1) Apply coefficient of restitution. vB′ − v′A = e(v A − vB ) vB′ − v′A = 0.70[1 − ( −0.5)] vB′ − v′A = 1.75 (a) (2) Solving Eqs. (1) and (2) simultaneously for the velocities, v′A = −0.59375 m/s v A = 0.594 m/s vB′ = 1.15625 m/s v B = 1.156 m/s 1 1 1 1 m Av 2A + mB vB2 = (5)(1)2 + (3)(−1.5)2 = 5.875 J 2 2 2 2 1 1 1 1 T2 = m A (v′A )2 + mB (vB′ )2 = (5)( −0.59375) 2 + (3)(1.15625) 2 = 2.8867 J 2 2 2 2 Kinetic energies: T1 = (b) T1 − T2 = 2.99 J Energy loss: 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. 750 PROBLEM 13.156 Collars A and B, of the same mass m, are moving toward each other with identical speeds as shown. Knowing that the coefficient of restitution between the collars is e, determine the energy lost in the impact as a function of m, e and v. SOLUTION Impulse-momentum principle (collars A and B): Σmv1 + ΣImp1→2 = Σmv 2 Horizontal components : m A v A + mB vB = mA v′A + mB vB′ mv + m(−v) = mv′A + mvB′ Using data, v′A + vB′ = 0 or (1) Apply coefficient of restitution. vB′ − v′A = e(v A − vB ) vB′ − v′A = e [v − (−v)] vB′ − v′A = 2ev Subtracting Eq. (1) from Eq. (2), −2v A = 2ev v A = −ev Adding Eqs. (1) and (2), (2) v A = ev 2vB = 2ev vB = ev v B = ev 1 1 1 1 m A v 2A + mB vB2 = mv 2 + m(−v) 2 = mv 2 2 2 2 2 1 1 1 1 T2 = m A (v′A ) 2 + mB (vB′ )2 = m(ev)2 + m(ev)2 = e 2 mv 2 2 2 2 2 Kinetic energies: T1 = T1 − T2 = (1 − e2 ) mv 2 Energy loss: 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. 751 PROBLEM 13.157 One of the requirements for tennis balls to be used in official competition is that, when dropped onto a rigid surface from a height of 100 in., the height of the first bounce of the ball must be in the range 53 in. ≤ h ≤ 58 in. Determine the range of the coefficient of restitution of the tennis balls satisfying this requirement. SOLUTION Uniform accelerated motion: v = 2 gh v′ = 2 gh′ Coefficient of restitution: e= e= Height of drop Height of bounce Thus, v′ v h′ h h = 100 in. 53 in. ≤ h′ ≤ 58 in. 53 58 ≤e≤ 100 100 0.728 ≤ e ≤ 0.762 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. 752 PROBLEM 13.158 Two disks sliding on a frictionless horizontal plane with opposite velocities of the same magnitude v0 hit each other squarely. Disk A is known to have a weight of 6-lb and is observed to have zero velocity after impact. Determine (a) the weight of disk B, knowing that the coefficient of restitution between the two disks is 0.5, (b) the range of possible values of the weight of disk B if the coefficient of restitution between the two disks is unknown. SOLUTION Total momentum conserved: m A v A + mB vB = m A v′A + mB v′ (m A )v0 + mB (−v0 ) = 0 + mB v′ m v′ = A − 1 v0 mB (1) Relative velocities: vB′ − v′A = e(v A − vB ) v′ = 2ev0 (2) Subtracting Eq. (2) from Eq. (1) and dividing by v0 , mA − 1 − 2e = 0 mB mA = 1 + 2e mB mB = mA 1 + 2e WB = Since weight is proportional to mass, (a) (b) WA (3) 1 + 2e With WA = 6 lb and e = 0.5, WB = 6 1 + (2)(0.5) WB = 6 = 2 lb 1 + (2)(1) WB = 6 = 6 lb 1 + (2)(0) WB = 3.00 lb With WA = 6 lb and e = 1, With WA = 6 lb and e = 0, 2.00 lb ≤ WB ≤ 6.00 lb Range: 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. 753 PROBLEM 13.159 To apply shock loading to an artillery shell, a 20-kg pendulum A is released from a known height and strikes impactor B at a known velocity v0. Impactor B then strikes the 1-kg artillery shell C. Knowing the coefficient of restitution between all objects is e, determine the mass of B to maximize the impulse applied to the artillery shell C. SOLUTION m A = 20 kg, mB = ? First impact: A impacts B. Σmv + ΣImp1→2 = Σmv 2 Impulse-momentum: m A v0 = m Av′A + mB vB′ Components directed left: 20v0 = 20v′A + mB vB′ (1) vB′ − v′A = e(v A − vB ) Coefficient of restitution: vB′ − v′A = ev0 v′A = vB′ − ev0 (2) Substituting Eq. (2) into Eq. (1) yields 20v0 = 20(vB′ − ev0 ) + mB vB′ 20v0 (1 + e) = (+ mB )vB′ vB′ = Second impact: B impacts C. Impulse-momentum: 20v0 (1 + e) 20 + mB (3) mB = ?, mC = 1 kg Σmv 2 + ΣImp 2→3 = Σmv3 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. 754 PROBLEM 13.159 (Continued) Components directed left: mB vB′ = mB vB′′ + mC vC′′ mB vB′ = mB vB′′ + vC′′ (4) vC′′ − vB′′ = e(vB′ − vC′ ) Coefficient of restitution: vC′′ − vB′′ = evB′ vB′′ − vC′′ = evC′ (5) Substituting Eq. (4) into Eq. (5) yields mB vB′ = mB (vC′′ − evB′ ) + mC vC′′ mB vB′ (1 + e) = (1 + mB )vC′′ vC′′ = mB vB′ (1 + e) 1 + mB vC′′ = 20mB v0 (1 + e) 2 (20 + mB )(1 + mB ) (6) Substituting Eq. (3) for vB′ in Eq. (6) yields The impulse applied to the shell C is mC vC′′ = (1)(20) mB v0 (1 + e) 2 (20 + mB )(1 + mB ) To maximize this impulse choose mB such that Z= mB (20 + mB )(1 + mB ) is maximum. Set dZ /dmB equal to zero. (20 + mB )(1 + mB ) − mB [(20 + mB ) + (1 + mB )] dZ = =0 dmB (20 + mB ) 2 (1 + mB )2 20 + 21mB + mB2 − mB (21 + 2mB ) = 0 20 − mB2 = 0 mB = 4.47 kg 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. 755 PROBLEM 13.160 Two identical cars A and B are at rest on a loading dock with brakes released. Car C, of a slightly different style but of the same weight, has been pushed by dockworkers and hits car B with a velocity of 1.5 m/s. Knowing that the coefficient of restitution is 0.8 between B and C and 0.5 between A and B, determine the velocity of each car after all collisions have taken place. SOLUTION m A = mB = mC = m Collision between B and C: The total momentum is conserved: mvB′ + mvC′ = mvB + mvC vB′ + vC′ = 0 + 1.5 (1) Relative velocities: (vB − vC )(eBC ) = (vC′ − vB′ ) (−1.5)(0.8) = (vC′ − vB′ ) −1.2 = vC′ − vB′ (2) Solving (1) and (2) simultaneously, vB′ = 1.35 m/s vC′ = 0.15 m/s v′C = 0.150 m/s Since vB′ > vC′ , car B collides with car A. Collision between A and B: mv′A + mvB′′ = mv A + mvB′ v′A + vB′′ = 0 + 1.35 (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. 756 PROBLEM 13.160 (Continued) Relative velocities: (v A − vB′ )eAB = (vB′′ − v′A ) (0 − 1.35)(0.5) = vB′′ − v′A v′A − vB′′ = 0.675 (4) Solving (3) and (4) simultaneously, 2v′A = 1.35 + 0.675 v′A = 1.013 m/s v′′B = 0.338 m/s Since vC′ < vB′′ < v′A , there are no further collisions. 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. 757 PROBLEM 13.161 Three steel spheres of equal weight are suspended from the ceiling by cords of equal length which are spaced at a distance slightly greater than the diameter of the spheres. After being pulled back and released, sphere A hits sphere B, which then hits sphere C. Denoting by e the coefficient of restitution between the spheres and by v 0 the velocity of A just before it hits B, determine (a) the velocities of A and B immediately after the first collision, (b) the velocities of B and C immediately after the second collision. (c) Assuming now that n spheres are suspended from the ceiling and that the first sphere is pulled back and released as described above, determine the velocity of the last sphere after it is hit for the first time. (d ) Use the result of Part c to obtain the velocity of the last sphere when n = 5 and e = 0.9. SOLUTION (a) First collision (between A and B): The total momentum is conserved: mv A + mvB = mv′A + mvB′ v0 = v′A + vB′ (1) Relative velocities: (v A − vB )e = (vB′ − v′A ) v0 e = vB′ − v′A (2) Solving Equations (1) and (2) simultaneously, (b) v′A = v0 (1 − e) 2 vB′ = v0 (1 + e) 2 Second collision (between B and C): The total momentum is conserved. mvB′ + mvC = mvB′′ + mvC′ Using the result from (a) for vB′ v0 (1 + e) + 0 = vB′′ + vC′ 2 (3) Relative velocities: (vB′ − 0)e = vC′ − vB′′ 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. 758 PROBLEM 13.161 (Continued) Substituting again for vB′ from (a) v0 (1 + e) (e) = vC′ − vB′′ 2 (4) Solving equations (3) and (4) simultaneously, vC′ = (c) 1 v0 (1 + e) (e) + v0 (1 + e) 2 2 2 vC′ = v0 (1 + e) 2 4 vB′′ = v0 (1 − e2 ) 4 For n spheres n balls (n − 1)th collision, we note from the answer to part (b) with n = 3 vn′ = v3′ = vC′ = v3′ = or v0 (1 + e)2 4 v0 (1 + e)(3−1) 2(3−1) Thus, for n balls vn′ = (d ) v0 (1 + e)( n −1) 2( n −1) For n = 5, e = 0.90, from the answer to part (c) with n = 5 vB′ = = v0 (1 + 0.9)(5−1) 2(5−1) v0 (1.9)4 (2) 4 vB′ = 0.815 v0 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. 759 PROBLEM 13.162 At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s and car C has a velocity vB = 1.5 m/s to the left, but car B is initially at rest. The coefficient of restitution between each car is 0.8. Determine the final velocity of each car, after all impacts, assuming (a) cars A and C hit car B at the same time, (b) car A hits car B before car C does. SOLUTION Assume that each car with its rider may be treated as a particle. The masses are: m A = 200 + 40 = 240 kg, mB = 200 + 60 = 260 kg, mC = 200 + 35 = 235 kg. Assume velocities are positive to the right. The initial velocities are: v A = 2 m/s vB = 0 vC = −1.5 m/s Let v′A , vB′ , and vC′ be the final velocities. (a) Cars A and C hit B at the same time. Conservation of momentum for all three cars. m Av A + mB vB + mC vC = m Av′A + mB vB′ + mC vC′ (240)(2) + 0 + (235)(−1.5) = 240v′A + 260vB′ + 235vC′ (1) Coefficient of restition for cars A and B. vB′ − v′A = e(v A − vB ) = (0.8)(2 − 0) = 1.6 (2) Coefficient of restitution for cars B and C. vC′ − vB′ = e(vB − vC ) = (0.8)[0 − ( −1.5)] = 1.2 (3) Solving Eqs. (1), (2), and (3) simultaneously, v′A = −1.288 m/s vB′ = 0.312 m/s vC′ = 1.512 m/s vA′ = 1.288 m/s vB′ = 0.312 m/s vC′ = 1.512 m/s 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. 760 PROBLEM 13.162 (Continued) (b) Car A hits car B before C does. First impact. Car A hits car B. Let v′A and vB′ be the velocities after this impact. Conservation of momentum for cars A and B. m A v A + mB vB = m A v′A + mB vB′ (240)(2) + 0 = 240v′A + 260vB′ (4) Coefficient of restitution for cars A and B. vB′ − v′A = e(v A − vB ) = (0.8)(2 − 0) = 1.6 (5) Solving Eqs. (4) and (5) simultaneously, v′A = 0.128 m/s, vB′ = 1.728 m/s v′A = 0.128 m/s v′B = 1.728 m/s Second impact. Cars B and C hit. Let vB′′ and vC′′ be the velocities after this impact. Conservation of momentum for cars B and C. mB vB′ + mC vC = mB vB′′ + mC vC′′ (260)(1.728) + (235)(−1.5) = 260vB′′ + 235vC′′ (6) Coefficient of restitution for cars B and C. vC′′ − vB′′ = e(vB′ − vC) = (0.8)[1.728 − (−1.5)] = 2.5824 (7) Solving Eqs. (6) and (7) simultaneously, vB′′ = −1.03047 m/s vC′′ = 1.55193 m/s v′′B = 1.03047 m/s v′′C = 1.55193 m/s Third impact. Cars A and B hit again. Let v′′′A and vB′′′ be the velocities after this impact. Conservation of momentum for cars A and B. mA v′A + mB vB′′ = mA v′′′A + mB vB′′′ (240)(0.128) + (260)(−1.03047) = 240vA′′′ + 260vB′′′ (8) Coefficient of restitution for cars A and B. vB′′′ − v′′′A = e(v′A − vB′′ ) = (0.8)[0.128 − (−1.03047)] = 0.926776 (9) Solving Eqs. (8) and (9) simultaneously, v′′′A = −0.95633 m/s vB′′′ = −0.02955 m/s v′′′A = 0.95633 m/s v′′′B = 0.02955 m/s 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. 761 PROBLEM 13.162 (Continued) There are no more impacts. The final velocities are: v′′′A = 0.956 m/s v′′′B = 0.0296 m/s v′′C = 1.552 m/s We may check our results by considering conservation of momentum of all three cars over all three impacts. m A v A + mB vB + mC vC = (240)(2) + 0 + (235)(−1.5) = 127.5 kg ⋅ m/s m A v′′′A + mB vB′′′ + mC vC′′ = (240)( −0.95633) + (260)(−0.02955) + (235)(1.55193) = 127.50 kg ⋅ m/s. 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. 762 PROBLEM 13.163 At an amusement park there are 200-kg bumper cars A, B, and C that have riders with masses of 40 kg, 60 kg, and 35 kg respectively. Car A is moving to the right with a velocity vA = 2 m/s when it hits stationary car B. The coefficient of restitution between each car is 0.8. Determine the velocity of car C so that after car B collides with car C the velocity of car B is zero. SOLUTION Assume that each car with its rider may be treated as a particle. The masses are: m A = 200 + 40 = 240 kg mB = 200 + 60 = 260 kg mC = 200 + 35 = 235 kg Assume velocities are positive to the right. The initial velocities are: v A = 2 m/s, vB = 0, vC = ? First impact. Car A hits car B. Let v′A and vB′ be the velocities after this impact. Conservation of momentum for cars A and B. m A v A + mB vB = m A v′A + mB vB′ (240)(2) + 0 = 240v′A + 260vB′ (1) Coefficient of restitution for cars A and B. vB′ − v′A = e(v A − vB ) = (0.8)(2 − 0) = 1.6 (2) Solving Eqs. (1) and (2) simultaneously, v′A = 0.128 m/s vB′ = 1.728 m/s v′A = 0.128 m/s v′B = 1.728 m/s Second impact. Cars B and C hit. Let vB′′ and vC′′ be the velocities after this impact. vB′′ = 0. Coefficient of restitution for cars B and C. vC′′ − vB′′ = e(vB′ − vC ) = (0.8)(1.728 − vC ) vC′′ = 1.3824 − 0.8vC 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. 763 PROBLEM 13.163 (Continued) Conservation of momentum for cars B and C. mB vB′ + mC vC = mB vB′′ + mC vC′′ (260)(1.728) + 235vC = (260)(0) + (235)(1.3824 − 0.8vC ) (235)(1.8)vC = (235)(1.3824) − (260)(1.728) vC = −0.294 m/s vC = 0.294 m/s Note: There will be another impact between cars A and 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. 764 PROBLEM 13.164 Two identical billiard balls can move freely on a horizontal table. Ball A has a velocity v0 as shown and hits ball B, which is at rest, at a Point C defined by θ = 45°. Knowing that the coefficient of restitution between the two balls is e = 0.8 and assuming no friction, determine the velocity of each ball after impact. SOLUTION Ball A: t-dir mv0 sin θ = mv′At v′At = v0 sin θ Ball B: t-dir 0 = mB v′Bt v′Bt = 0 Balls A + B: n-dir mv0 cosθ + 0 = m v′An + m v′Bn (1) Coefficient of restitution vBn ′ − v′An = e (v An − vBn ) v′Bn − v′An = e (v0 cosθ − 0) (2) Solve (1) and (2) 1 − e 1 + e v′An = v0 cosθ ; v′Bn = v0 cosθ 2 2 With numbers e = 0.8; θ = 45° v′At = v0 sin 45° = 0.707 v0 1 − 0.8 cos 45° = 0.0707 v0 v′An = v0 2 v′Bt = 0 1 + 0.8 v′Bn = v0 cos 45° = 0.6364 v0 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. 765 PROBLEM 13.164 (Continued) (A) 1 v" A = [(0.707 v0 ) 2 + (0.0707v0 )2 ] 2 = 0.711v0 = 0.711v0 0.0707 β = tan −1 = 5.7106° 0.707 So θ = 45 − 5.7106 = 39.3° (B) v′A = 0.711v0 39.3° v′B = 0.636 v0 45° 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. 766 PROBLEM 13.165 The coefficient of restitution is 0.9 between the two 2.37-in. diameter billiard balls A and B. Ball A is moving in the direction shown with a velocity of 3 ft/s when it strikes ball B, which is at rest. Knowing that after impact B is moving in the x direction, determine (a) the angle θ , (b) the velocity of B after impact. SOLUTION (a) Since vB′ is in the x-direction and (assuming no friction), the common tangent between A and B at impact must be parallel to the y-axis, tan θ = 10 6−D 10 6 − 2.37 = 70.04° θ = tan −1 θ = 70.0° (b) Conservation of momentum in x(n) direction: mv A cos θ + m(vB ) n = m(v′A ) n + mvB′ (3)(cos 70.04) + 0 = (v′A )n + vB′ 1.0241 = (v′A )n + (vB′ ) (1) Relative velocities in the n direction: e = 0.9 (v A cos θ − (vB ) n )e = vB′ − (v′A ) n (1.0241 − 0)(0.9) = vB′ − (v′A ) n (1) + (2) 2vB′ = 1.0241(1.9) (2) v′B = 0.972 ft/s 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. 767 PROBLEM 13.166 A 600-g ball A is moving with a velocity of magnitude 6 m/s when it is hit as shown by a 1-kg ball B which has a velocity of magnitude 4 m/s. Knowing that the coefficient of restitution is 0.8 and assuming no friction, determine the velocity of each ball after impact. SOLUTION Before After v A = 6 m/s (v A ) n = (6)(cos 40°) = 4.596 m/s (v A )t = −6(sin 40°) = −3.857 m/s vB = (vB ) n = −4 m/s (vB )t = 0 t-direction: Total momentum conserved: m A(v A )t + mB (vB )t = mA (vB′ )t + mB (vB′ )t (0.6 kg)( −3.857 m/s) + 0 = (0.6 kg)(v′A )t + (1 kg)(vB′ )t −2.314 m/s = 0.6 (v′A )t + (vB′ )t (1) Ball A alone: Momentum conserved: m A (v A )t = mA (v′A )t −3.857 = (v′A )t (v′A )t = −3.857 m/s (2) Replacing (v′A )t in (2) in Eq. (1) −2.314 = (0.6)(−3.857) + (vB′ )t −2.314 = −2.314 + (vB′ )t (vB′ )t = 0 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. 768 PROBLEM 13.166 (Continued) n-direction: Relative velocities: [(v A ) n − (vB ) n ]e = (vB′ )n − (v′A )n [(4.596) − (−4)](0.8) = (vB′ )n − (v′A )n 6.877 = (vB′ )n − (v′A )n (3) Total momentum conserved: m A (v A ) n + mB (vB ) n = m A (v′A ) n + mB (vB′ ) n (0.6 kg)(4.596 m/s) + (1 kg)( − 4 m/s) = (1 kg)(vB′ ) n + (0.6 kg)(v′A ) n −1.2424 = (vB′ ) n + 0.6(v′A ) n (4) Solving Eqs. (4) and (3) simultaneously, (v′A )n = 5.075 m/s (vB′ )n = 1.802 m/s Velocity of A: tan β = |(v A )t | |(v A ) n | 3.857 5.075 β = 37.2° = β + 40° = 77.2° v′A = (3.857) 2 + (5.075) 2 = 6.37 m/s v′A = 6.37 m/s v′B = 1.802 m/s Velocity of B: 77.2° 40° 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. 769 PROBLEM 13.167 Two identical hockey pucks are moving on a hockey rink at the same speed of 3 m/s and in perpendicular directions when they strike each other as shown. Assuming a coefficient of restitution e = 0.9, determine the magnitude and direction of the velocity of each puck after impact. SOLUTION Use principle of impulse-momentum: Σmv1 + ΣImp1→2 = Σmv 2 t-direction for puck A: − mv A sin 20° + 0 = m(v′A )t (v′A )t = v A sin 20° = 3sin 20° = 1.0261 m/s t-direction for puck B: − mvB cos 20° + 0 = m(vB′ )t (vB′ )t = vB cos 20° = −3cos 20° = −2.8191 m/s n-direction for both pucks: mv A cos 20° − mvB sin 20° = m(v′A ) n + m(vB′ ) n (v′A )n + (vB′ ) n = v A cos 20° − vB sin 20° = 3cos 20° − 3sin 20° (1) e = 0.9 Coefficient of restitution: (vB′ )n − (v′A ) n = e[(v A ) n − (vB ) n ] = 0.9[3cos 20° − (−3)sin 20°] (2) Solving Eqs. (1) and (2) simultaneously, (v′A )n = −0.8338 m/s (vB′ )n = 2.6268 m/s 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. 770 PROBLEM 13.167 (Continued) Summary: ( v′A )n = 0.8338 m/s 20° ( v′A )t = 1.0261 m/s 70° ( v′B )n = 2.6268 m/s 20° ( v′B )t = 2.8191 m/s 70° v A = (0.8338) 2 + (1.0261) 2 = 1.322 m/s tan α = 1.0261 0.8338 α = 50.9° α + 20° = 70.9° v′A = 1.322 m/s 70.9° vB′ = (2.6268) 2 + (2.8191) 2 = 3.85 m/s tan β = 2.8191 2.6268 β = 47.0° β − 20° = 27.0° v′B = 3.85 m/s 27.0° 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. 771 PROBLEM 13.168 Two identical pool balls of 57.15-mm diameter, may move freely on a pool table. Ball B is at rest and ball A has an initial velocity v = v0 i. (a) Knowing that b = 50 mm and e = 0.7, determine the velocity of each ball after impact. (b) Show that if e = 1, the final velocities of the balls from a right angle for all values of b. SOLUTION Geometry at instant of impact: b 50 = d 57.15 θ = 61.032° sin θ = Directions n and t are shown in the figure. Principle of impulse and momentum: Ball B: Ball A: Ball A, t-direction: mv0 sin θ + 0 = m(v A )t Ball B, t-direction: 0 + 0 = m(v B )t Balls A and B, n-direction: Coefficient of restitution: (a) (v A )t = v0 sin θ (1) (v B )t = 0 (2) mv0 cos θ + 0 + m(v A ) n + m(vB ) n (v A )n + (vB ) n = v0 cos θ (3) (vB ) n − (v A ) n = e[v0 cos θ ] (4) e = 0.7. From Eqs. (1) and (2), (v A )t = 0.87489v0 (1)′ ( vB ) t = 0 (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. 772 PROBLEM 13.168 (Continued) From Eqs. (3) and (4), (v A ) n + (vB )n = 0.48432v0 (3)′ (vB ) n − (v A ) n = (0.7)(0.48432v0 ) (4)′ Solving Eqs. (5) and (6) simultaneously, (v A )n = 0.072648v0 (vB ) n = 0.41167v0 v A = (v A )n2 + (v A )t2 = (0.072648v0 ) 2 + (0.87489v02 = 0.87790v0 tan β = (v A ) n 0.0