Resolução Mecânica para Engenheiros - Estática - Hibbeler

2-5. Resolve the force F2 into components acting along the n and v axes and determine the magnitudes of the components. Fn = 300 sin 110° Fn = 205 N Ans - 70° - 30° = 300N 45° F = 500 N Fn = 300 Fn sin 110° Fn = 160 N Ans 2-6. Resolve the force F2 into components acting along the n and v axes and determine the magnitudes of the components. Fn = 500 Fn = 375 N Ans Fn = 500 cos 110° Fn = 428 N Ans 2-7. The plate is subjected to the two forces at A and B as shown. If θ = 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. (c). Trigonometry : Using law of cosines [Fig. (b) ], we have Fv = 8 F = 10.68 N 8 FN = 10.68 N = 10.5 kN = = 3 1.6° FS, and Fa. 8 kN The angle 𝜙 can be determined using law of sines [Fig. (b)]. sin θ sin 100° 8 = 10.68 sin 20 = F = 5316 That, the direction 𝜙 of F measured from the x axis is = 3.1 = 30° -. 316° Ans 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 1-13. Convert each of the following to three significant figures. (a) 20 lb-ft to N-m, (b) 450 lb/ft2 to kN/m2, and (c) 15 ft/s to mm/s Using Table 1-2, we have a) 20 lb-ft = (20) lb (4.4482 N/1 lb)(0.3048 m/1 ft) = 27.1 N-m Ans b) 450 lb/ft2 = (450) lb (4.4482 N/1 lb) dm = 1000,001 0.3048 104112 = 2.15 N-m mm1000609 77 m2 Ans c) 15 ft/s = (15 *)(0.048 m11 = 4.8 scm) Ans F = Gnr ma (m,/or 722 F = (6.67 x lettuce)(12 /() 10(0. /Wx12.94]} = 1.18 N () W G X m: (0.8) = 10m (09)(12 m G 9.84 )1 m fr = 2.8.4)13 )