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MCE 263 Final Exam July 25, 2014 Please do all the work independently and do not write on back of this paper. Problem 1 (50 points): The pitcher throws the baseball horizontally with a speed of 140 ft/s from a height of 5 ft. The batter is 60 ft away. Determine: (a) The initial horizontal and vertical velocity components of the ball. 9 8:28 AM Page 71 (b) Equation of motion in the horizontal direction. (c) Equation of motion in the vertical direction. 6/8/09 8:28 AM Page 71 91962_01_s12-p0001-0176 (d) The time needed for the ball to arrive at the batter. n, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently f this material may be reproduced, any form or by any means, without permission in writing from the publisher. (e)in The height h at which it passes the batter. © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. throws the baseball horizontally with m a height of 5 ft. If the batter is 60 ft h me for the ball to arrive at the The batter *12–88. pitcher throws the baseball horizontally with ich it passes the batter. a speed of 140 ft>s from a height of 5 ft. If the batter is 60 ft 60 = 140t 5 ft t = 0.4286 = 0.429 s + A;B 1 t + ac t2 2 s = vt; 60 = 140t 5 ft h away, determine the time for the ball to arrive at the batter60 ft and the height h at which it passes the batter. 60 ft Ans. Ans. t = 0.4286 = 0.429 s 1 s = s0 + v0 t + ac t2 1 A + c 2B 2 + 0 + (-32.2)(0.4286) = 2.04 ft Ans. 2 1 h = 5 + 0 + (-32.2)(0.4286)2 = 2.04 ft 2 Ans. vA hrown off the top of the building. If it uA in 3 s, determine the initial velocity vA le uA at which it was thrown. Also, find A •12–89. The ball is thrown off the top of the building. If it all’s velocity when it strikes the ground. strikes the ground at B in 3 s, determine the initial velocity v vA uA A and the inclination angle uA at which it was thrown. Also, find the magnitude of the velocity when it strikes the ground. he x–y coordinate system will be set so thatball’s its origin coincides A Coordinate System: The x–y coordinate system will be set so that its origin coincides 75 ft = vA cos u, xA = 0, and B = 60 withxpoint A.ft, and t = 3 s. Thus, (vA)xt A cos u(3) 20 75 ft x-Motion: Here, (vA)x = vA cos u, xA = 0, and xB = 60 ft, and t = 3 s. Thus, + B A: xB = xA + (vA)xt 60 = 0 + vA cos u(3) (1) B 60 ft 2 = vA sin u, ay = -g = -32.2 ft>s , yAu == 020 , and yB = -75 ft, vA cos (1) 2 y-Motion: Here, (vA)y = vA sin u, ay = -g = -32.2 ft>s , yA = 0, and yB = -75 ft, 1 2 ay t and t = 3 s. Thus, 2 1 1 vA sin u(3) + (-32.2) A +Ac32B B yB = yA + (vA)y t + ay t2 2 2 1(2) 3.3 -75 = 0 + vA sin u(3) + (-32.2) A 32 B 2 ) yields v sin u = 23.3 (2) (vA)y t + B 60 ft Problem 2 (50 points): At the instant shown car A is traveling with a velocity of 30 m/s and has an acceleration of 2 m/s2 along the highway. At the same instant B is traveling on the trumpet interchange curve with a speed of 15 m/s, which is decreasing at 0.8 m/s2 . At this particular instant determine: (a) The velocity vectors of A and B in your chosen coordinate system. /09 9:33 AM Page 174 (b) The relative velocity of B with respect to A at this instant, its magnitude and direction. (c) The acceleration vector of A in your chosen coordinate system. (d) The acceleration vector of B in your chosen coordinate system. , Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently this material may be reproduced, or by any means, withoutof permission writing from (e)in any theform relative acceleration B withinrespect tothe A,publisher. its magnitude and direction. nt shown car A is traveling with a 91962_01_s12-p0001-0176 6/8/09 9:33 AM Page 174 has an acceleration of 2 m>s2 along ame instant B is traveling on the rve with a speed of 15 m>s, which is © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently Determine the relative velocity and exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. B with respect to A at this instant. *12–228. At the instant shown car A is traveling with a velocity of 30 m>s and has an acceleration of 2 m>s2 along the highway. At the same instant B is traveling on the trumpet interchange curve with a speed of 15 m>s, which is decreasing at 0.8 m>s2. Determine the relative velocity and relative acceleration of B with respect to A at this instant. A B 60" r ! 250 m A j = 30i + (vB>A)xi + (vB>A)y j A)x y m>s ; B vB = vA + vB>A 60" 15 cos 60°i + 15 sin 60°j = 30i + (vB>A)xi + (vB>A)y j 15 cos 60° = 30 + (vB>A)x 15 sin 60° = 0 + (vB>A)y (vB>A)x = -22.5 = 22.5 m>s ; (vB>A)y = 12.99 m>s c .99)2 = 26.0 m>s 0° b 26.0 m>s vB>A = 2(22.5)2 + (12.99)2 =Ans. Ans. u = tan - 1 a Ans. (aB)n = 9 m>s2 12.99 b = 30° b 22.5 Ans. v2 152 = = 0.9 m>s2 r 250 aB = aA + aB>A -0.8 cos 60°i - 0.8 sin 60°j + 0.9 sin 60°i - 0.9 cos 60°j = 2i + (aB>A)x i + (aB>A)y j -0.8 cos 60° + 0.9 sin 60° = 2 + (aB>A)x -0.8 sin 60° - 0.9 cos 60° = (a ) B>A y 60°j + 0.9 sin 60°i - 0.9 cos 60°j = 2i + (aB>A)x i + (aB>A )y j (aB>A)x = -1.6206 ft>s = 1.6206 m>s2 ; 2 0° = 2 + (aB>A)x 0° = (aB>A)y = 1.6206 m>s2 ; (aB>A)y = -1.1428 ft>s2 = 1.1428 m>s2 T aB>A = 2(1.6206)2 + (1.1428)2 = 1.98 m>s2 1.1428 f = tan - 1 ¢ ≤ = 35.2° d 1.6206 = 1.1428 m>s2 T 1.1428)2 = 1.98 m>s2 Ans. 35.2° d Ans. Ans. Ans. r ! 250 m Problem 3 (50 points): The man pushes on the 60-lb crate with a force F . The force is always directed down at 30◦ from the horizontal as shown, and its magnitude is increased until the crate begins to slide. The coefficient of static friction is µs = 0.6 and the coefficient of kinetic friction is µk = 0.3. Determine: AM (a) Free-body and mass-acceleration diagrams with the appropriate xy axis. Page 188 (b) The equation of motion in x direction. (c) The equation of motion in y direction. The This minimal force neededunder to produce motion and corresponding Saddle River, NJ. All rights(d) reserved. material is protected all copyright laws as theythe currently may be reproduced, in any form or by any means, without permission in writing from the publisher. normal force. (e) The crate’s initial acceleration right after it begins to slide. 91962_02_s13_p0177-0284 -lb crate with a force F. own at 30° from the de is increased until the ne the crate’s initial atic friction is ms = 0.6 n is mk = 0.3. 6/8/09 10:00 AM Page 188 © 2010 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently exist. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. F 30⬚ *13–16. The man pushes on the 60-lb crate with a force F. The force is always directed down at 30° from the horizontal as shown, and its magnitude is increased until the crate begins to slide. Determine the crate’s initial acceleration if the coefficient of static friction is ms = 0.6 and the coefficient of kinetic friction is mk = 0.3. F 30⬚ Force to produce motion: + ©F = 0; : x Fcos 30° - 0.6N = 0 + c ©Fy = 0; N - 60 - F sin 30° = 0 N = 91.80 lb = 0 F = 63.60 lb Since N = 91.80 lb, + ©F = ma ; : x x 0° = 0 63.60 cos 30° - 0.3(91.80) = a 60 ba 32.2 a = 14.8 ft>s2 Ans. F = 63.60 lb - 0.3(91.80) = a 60 ba 32.2 Ans. 188 Problem 4 (50 points): The ends of bar AB are confined to move along the paths shown. At a given instant, A has a velocity of 8 ft/s and an acceleration of 3 ft/s2 . At this instant, determine: (a) The kinematic diagram of bar AB for the velocity analysis. PM Page 595 (b) The angular velocity of AB and the velocity of B. (c) The kinematic diagram of bar AB for the acceleration analysis. (d) The expressions for the acceleration vectors for A and B. pper Saddle River, NJ. All rights reserved. This material is protected under all copyright laws as they currently rial may be reproduced, in any (e) form The or by any means, without permissionof in AB writing fromthe the publisher. angular acceleration and acceleration are confined to move along ant, A has a velocity of 8 ft>s s2. Determine the angular on of AB at this instant. 4 ft 30⬚ B 30⬚ 4 ft Ans. A C a (4) S + C (2)2 (4 ) S c 60° g60° 30° = 0 + a(4) sin 60° + 16 cos 60° 30° = -3 + a(4) cos 60° - 16 sin 60° Ans. >A + ( (8)2 (8)2 ) sin 30°i + ( ) cos 30°j = -3j 4 4 30°j) - (2)2(-4 sin 30°i + 4 cos 30°j) = -3.464a + 8 564 = -3 + 2a - 13.8564 Ans. vA ⫽ 8 ft/s aA ⫽ 3 ft/s2 of B. the paths shown. At a given instant, A has a velocity of 8 ft>s and an acceleration of 3 ft>s2. Determine the angular velocity and angular acceleration of AB at this instant. 4 ft 30⬚ B 30⬚ 8 v = = 2 rad>s 4 4 ft b Ans. vB = 4(2) = 8 ft>s (aB)n = A vA ⫽ 8 ft/s aA ⫽ 3 ft/s2 (8)2 = 16 ft>s2 4 aB = aA + aB>A C 16 S + C (aB) t S = C 3 S + C a (4) S + C (2)2 (4 ) S T c 30° c 60° g60° g30° + ) (: 16 sin 30° + (aB)t cos 30° = 0 + a(4) sin 60° + 16 cos 60° (+ c ) 16 cos 30° - (aB)t sin 30° = -3 + a(4) cos 60° - 16 sin 60° a = 7.68 rad>s2 b Ans. 2 (aB)t = 30.7 ft>s Also, aB = aA + aAB * rB>A - v2rB>A (aB)t cos 30°i - (aB)t sin 30°j + ( (8)2 (8)2 ) sin 30°i + ( ) cos 30°j = -3j 4 4 -(ak) * (-4 sin 30°i + 4 cos 30°j) - (2)2(-4 sin 30°i + 4 cos 30°j) + ) (: A+cB (aB)t cos 30° + 8 = -3.464a + 8 -(aB)t sin 30° + 13.8564 = -3 + 2a - 13.8564 a = 7.68 rad>s2 b Ans. (aB)t = 30.7 ft>s2 595 Problem 5 (50 points): The uniform rod AB of mass m is released from rest when β = 70◦ . Assuming that the friction force between end A and the surface is large enough to prevent sliding, determine immediately after release: (a) The free-body and mass-acceleration diagrams. (b) The linear (force) equations of motion with respect to chosen coordinates. (c) The angular (moment) equation of motion. (d) The angular acceleration of the rod and the acceleration of its center of mass. (e) The normal reaction at A and the friction force at A. PROBLEM 16.157 The uniform rod AB of weight W is re Assuming that the friction force betwee enough to prevent sliding, determine im angular acceleration of the rod, (b) th friction force at A. PROBLEM 16.157 The uniform rod AB of weight W is released from rest when β = 70°. Assuming that the friction force between end A and the surface is large enough to prevent sliding, determine immediately after release (a) the angular acceleration of the rod, (b) the normal reaction at A, (c) the friction force at A. SOLUTION SOLUTION We note that rod rotates about A. ω = 0 We note that rod rotates about A. ω = 0 1 mL2 12 L a= α 2 I = 1 I = mL2 12 L a= α 2 ΣM A = Σ( M A )eff : L L mg cos β = I α + (ma ) 2 2 L L mg cos β = I α + (ma ) 2 2 ΣM = Σ ( M ) : 1 1 A2 L L mgL cos β = mL α + m αA eff 2 12 2 2 1 2 = mL α 3 1 1 L L mgL cos β = mL2α + m α (1) 2 12 2 2 1 FA = ma sin β = mL2α 3 L L 3 g cos β F = m α sin β = m sin β 3 g cos β α= 2 L ΣFx = Σ( Fx )eff : A FA = ΣFy = Σ( Fy )eff : 2 2 2 L 3 mg sin β cos β 4 L NA − mg = −ma cos β = − m α cos β x 2 eff x ΣF = Σ ( F ) : L 3 g cos β 2 2 L cos β 3 2 NA = mg 1 − cos β 4 NA − mg = −m α= 3 g cos β (2) 2 L FA = ma sin β L L 3 g cos β FA = m(3) α sin β = m 2 22 L 3 mg sin β cos β 4 PROPRIETARY MATERIAL. © 2013 The McGraw-Hill Companies, Inc. All rights reserved. No part of this Manual may be displayed, FA = reproduced or distributed in any form or by any means, without the prior written permission of the publisher, or used beyond the limited distribution to teachers and educators permitted by McGraw-Hill for their individual course preparation. If you are a student using this Manual, you are using it without permission. ΣFy 6= Σ( Fy )eff : 1672 L NA − mg = −ma cos β = − m α cos β 2 NA − mg = −m L 3 g cos β 2 2 L cos β