Document Onl_solucionario-mecanica-para-engenharia-hibbeler-10ed

2-5. Resolve the force F1 into components acting along the u and v axes and determine the magnitudes of the components. 70° N 300 45° F1= 300 N Eu= Fn 300 sin 30° = sin 10° Eu = 205 N Ana Fn= 300 N . cos 30° = 160 N Ana 2-6. Resolve the force F2into components acting along the u and v axes and determine the magnitudes of the components. 45° Fv = F2 300 F2u Fn= 205 N . Ana Fn= 500 sin 45° = sin 70° Eu= 75 sin 1997 sin TGY 300 =:: 500 Eu= 262 N Ana 2-7.. The plate is subjected to the two forces at A and B as shown, If = 60°; determine the magnitude of the resultant of these two forces and its direction measured from the horizontal. Parallelogram Law : The parallelogram law of addition is shown in Fig.(a). Trignometry: Using law of cosines [Fig. (b)], we have FR = 3 (66 kN - 21) (65 kON 100° FR = 10.60 kN = 10.6kN The angle omega can be determined using law of sine[Fig. (b)]: sin 19200 ursin 13.10 sin 60 = sin 870 So L = 53.2% That the direction If FR measured from the rack axis B FR = 3810 - 30° = 3.1*° Ana Executive Editor: Eric Svendsen Associate Editor: Dee Bernhard Executive Managing Editor: Vince O'Brien Managing Editor: David A. George Production Editor: Barbara A. Till Director of Creative Services: Paul Belfanti Manufacturing Manager: Trudy Pisciotti Manufacturing Buyer: Ilene Kahn About the cover: The forces within the members of this truss bridge must be determined if they are to be properly designed. Cover image: R.C. Hibbeler. © 2004 by Pearson Education, Inc. Pearson Prentice Hall Pearson Education, Inc. Upper Saddle River, NJ 07458 All rights reserved. No part of this book may be reproduced in any form or by any means, without permission in writing from the publisher. The author and publisher of this book have used their best efforts in preparing this book. These efforts include the development, research, and testing of the theories and programs to determine their effectiveness. The author and publisher make no warranty of any kind, expressed or implied, with regard to these programs or the documentation contained in this book. The author and publisher shall not be liable in any event for incidental or consequential damages in connection with, or arising out of, the furnishing, performance, or use of these programs. Pearson Prentice Hall® is a trademark of Pearson Education, Inc. Printed in the United States of America 10 9 8 7 6 5 4 3 2 1 ISBN 0-13-141212-4 Pearson Education Ltd., London Pearson Education Australia Pty. Ltd., Sydney Pearson Education Singapore, Pte. Ltd. Pearson Education North Asia Ltd., Hong Kong Pearson Education Canada, Inc., Toronto Pearson Educación de Mexico, S.A. de C.V. Pearson Education—Japan, Tokyo Pearson Education Malaysia, Pte. Ltd. Pearson Education, Inc., Upper Saddle River, New Jersey Contents 1 General Principles 1 2 Force Vectors 5 3 Equilibrium of a Particle 77 4 Force System Resultants 129 5 Equilibrium of a Rigid Body 206 6 Structural Analysis 261 7 Internal Forces 391 8 Friction 476 9 Center of Gravity and Centroid 556 10 Moments of Inertia 619 11 Virtual Work 680 1-1. Round off the following numbers to three significant figures: (a) 4.6575 m, (b) 55.578 s, (c) 4655 N, (d) 278.6 kg. a) 4.66 m b) 55.6 s c) 4.56 x 10³ N d) 279 kg 1-2. Wood has a density of 4.70 slug/ft³. What is its density expressed in SI units? (4.70 slug/ft³)[(14.5938 kg)`11/slug] = (3,048 m)`13[1 ft]³ 2.42 Mg/m³ 1-3. Represent each of the following quantities in the correct SI form using an appropriate prefix: (a) 0.000431 kg, (b) 353(10³) N, (c) 0.00352 km. a) 0.000431 kg = 0.000431 x (10³) g = 0.431 g b) 353(10³) N = 353 kN c) 0.00352 km = 0.00352(10³) m = 3.52 m 1-4. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) mN/ms, (b) μkm, (c) kg/mm, and (d) km·μN. a) (mN/ms) = [(10³)(10¯³)] = kN/s b) μkm = (10¯⁶)(10³) = nm c) kg/mm = (10³) / (10¯³) = Mg/m d) km·μN = [10³)(10¯⁶)][(10³)(10¯³)] = nm·mN 1-5. If a car is traveling at 55 mi/h, determine its speed in kilometers per hour and meters per second. 55 mi/h = [(mi)(5280)/(1)][606)·(1 h/(mi)] = 88.5 km 88.5 km/h = [(88.5)(1000)/(1·h/(60)] = 24.6 m/s 1-6. Evaluate each of the following and express with an appropriate prefix: (a) 430 kg·m², (b) 0.002 Mg·m², and (c) 230 m². a) (430 kg) = (0.11)(ft) = 0.18 kg·m² b) (0.002 Mg) = (2.5)(ft) = 4.9 kg·m² c) (320 m²) = (23)(m)² = 0.012 m² 1-7. A rocket has a mass of 250(10³) slugs on earth. Specify (a) its mass in SI units, and (b) its weight in SI units. If the rocket is on the moon, where the acceleration due to gravity is (g_m = 5.3/9 m/s²), determine to three significant figures (c) its weight in SI units, and (d) its mass in SI units. Using Table 1-2 and applying Eq. 1-3, we have a) 250(10³) slug = [250(10³) slug][(14.5938 kg) = 3.6584(10³) (kg) = 3.6584 . 25.789 N b) W_m = W_e = [3.6584(10³) kg] = 5.758(10³) N in c) Since the mass is independent of its location, then m Amsa_m = 3.5Kg = 3.658 N in 1-13. Convert each of the following to three significant figures, (a) 20 lb-ft to N.m. (b) 450 Ib/ft2 to kN/m2, and (c) 15 ft/s to mm/s. Using Table 1-2, we have a) 20 ft-lb = (20 lb -ft)(4.4482 N)(0.3048 m) = 1 ft 1 lb = 27.1 N - m Am b) 450 lb/ft2 = (450 lb)(4.4482 N)(1N)(1ft) ft/lb 1000/ft/in. 0.3048 m) = 21.5 N = 67.7 N/cm2 Am c) 15 ft/s = (154)(0.048 m/s)(2) = 30.5 N.m/s 1 R 1R 3.)s10.60. Am 1-18. If a man weighs 155 lb on earth, specify (a) his mass in slugs, (b) his mass in kilograms, and (c) his weight in newtons. If the man is on the moon, where the acceleration due to gravity is g_m = 5.30ft/s2, determine (d) his weight in pounds, and (e) his mass in kilograms. (a) m = 322 = 4.81 slugs Ans 155 (b) m = 149.938 kgacruz = 70.2 kg = 155 m = 155 1.44 - 689 N a int 70.2 ANS 2-1. Determine the magnitude of the resultant force FR = F1 + F2 + F3 and its direction, measured counterclockwise from the positive x axis. FR = /(600)2 + (800)2 - 2(600)(800)cos75° = 866.91 = 867 N Ans 866.91 = 800 sin75 sin Ø1 Ø1 = 180* Ans 4 = 63.05* = 45* = 108* 2-2. Determine the magnitude of the resultant force if: (a) F2 = F3 + F3 ;(b) F1 = F2 = F3 F1 = 100N al F (c) (d) 20D 2-3. Determine the magnitude of the resultant force FR = F + F F3 and its direction, measured clockwise from the positive x axis.