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

2-5. Resolve the force F1 into components acting along the n and v axes and determine the magnitudes of the components. 70° 74 0 7y Fуд 450 Єо V Fn- 300 F-oo N Fn V 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 México, S.A. de C.V. Pearson Education—Japan, Tokyo Ltd. 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) 4.63575 m, (b) 55.578 s, (c) 4555 N, (d) 278.6 kg. 1-2. Wood has a density of 4.70 slug/ft³. What is its density expressed in SI units? 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) 0.00532 km. 1-4. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) mNm/s, (b) μkm, (c) kN·mg, and (d) km·μN. 1-5. If a car is traveling at 55 mi/h, determine its speed in kilometers per hour and meters per second. 1-6. Evaluate each of the following and express with an appropriate prefix: (a) 430 kg/m³, (b) 0.002 mg·m³, and (c) 230 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 = 5.30 m/s², determine to three significant figures (c) its weight in SI units, and (d) its mass in SI units. 1-13. Convert each of the following to three significant figures: (a) 20 lb-f to N, (b) 450 lb/ft^2 to kN/m^2, and (c) 15 ft/s to mm/s. Using Table 1-2, we have a) 20 lb-f = (20 lb-f) (4.4482 N/lb-f) = 89.0048 N 1 N = 27.0 N Ans b) 450 lb/ft^2 = (450 lb/ft^2) (4.4482 N/lb-f)(1 ft^2) 1 ft^2 1000.00 0.3048^2 m^2 = 2.15 kN/m^2 Ans c) 15 ft/s = (15 ft)(0.3048 m)(1 m)(1 mm) 1 ft 1 1 R 3600 S = 4.572 mm/s Ans 1-14. If an object has a mass of 40 slugs, determine its mass in kilograms. 40 slugs (14.5939 kg/slug) = 584 kg Ans 1-15. Water has a density of 1.94 slug/ft^3. What is the density expressed in SI units? Express the answer to three significant figures. Using Table 1-2, we have p_w = (1.94 slugs/ft^3)(14.5939 kg/slug)(1 ft^3) 1 1 kg 0.3048^3 m^3 = 999.8 kg/m^3 = 1.00 Mg/m^3 Ans *1-16. Two particles have a mass of 8 kg and 12 kg, respectively. If they are 800 mm apart, determine the force of gravity acting between them. Compare this result with the weight of each particle. F = Gm_1m_2 r Where G = 6.673(10^-11) m^3/kg*s^2 F = 6.673(10^-11)(8)(12) (0.80)^3 N = 100.0 nN W_1 = (8)(9.81) = 78.5 N W_2 = (12)(9.81) = 118 N Ans 1-17. Determine the mass of an object that has a weight of (a) 20 kN, (b) 150 kN, (c) 60 MN. Express the answer to three significant figures. Applying Eq. 1–3, we have m = W/g 9.81 m/s^2 a) m = 20(10^3) kg·m/s^2 = 2.04 t Ans b) m = 150(10^3) kg·m/s^2 = 15.3 Mg Ans (9.81 m/s^2) c) m = 60(10^6) kg·m/s^2 = 612 Mg (9.81 m/s^2) Ans 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 = 5.30 ft/s^2, determine (d) his weight in pounds, and (e) his mass in kilograms. (a) m = 155 32.2 = 4.81 slug Ans (b) m = 155 32.2 = = = 70.2 kg Ans (c) W = 155 (4.448 N) = 689 N Ans (d) W_m = 155 (5.30 ft/s^2) = 25.5 lb Ans (e) m = = = 70.2 kg Ans m = [×15 ] or m = 70.2 kg 1-19 Using the base units of the SI system, show that Eq. 1.2 is dimensionally homogeneous equation which gives F in newtons Determine to three significant figures the gravitational force acting between two spheres that are touching each other. The mass of each sphere is 200 kg and the radius is 300 mm. Using Eq. 1–2, F = Gm_1m_2 r N F = Gm_1m_2 r N = (3)(3) = 7.41 An = 66.7 = 7.41 (10^-5) N = 7.41 µN Ans *1-20. Evaluate each of the following to three significant figures and express each answer in SI units using an appropriate prefix: (a) (0.631 Mnton)(8.60 kg)^2, (b) (35 mm)^2(48 kg)^2. (a) 0.631 Mnton/8.60 kg\u0252\ 1} = 8332 m (50 kg)^2 * = 8.33 km^2 Ans (b) (35 mm)^2(48 kg)^2 = 153 m2*kg^2 Ans 2-1. Determine the magnitude of the resultant force F_R = F_1 + F_2 and its direction, measured counterclockwise from the positive x axis. 863.9 534 sin 60 F_s = \(600 - (600) - (200)(600)(600)cos 75 = 866.91 = 867 N Ans theta sin 75 cos 60ϕ = 63.05° + 45° = 108° Ans ϕ = 2-2. Determine the magnitude of the resultant force if: (a) F_R = F_1 + F_2; (b) F_R = F_1 - F_2. F_1 = 100 N F_2 = 180 N 80N F_2 - Trigonometry : Using law of cosines (Fig. 6b and 6d), we have F_1 b) a) 111 N b-c) c) c) F_1 ≠ F_2(d) 110 N + F_2)(1 - cos 75°) = 150 N Ans a = 100 N 2-3. Determine the magnitude of the resultant force F_R = F_1 + F_2 and its direction, measured counterclockwise from the positive x axis. F_R = (250 lb) + (375) - 250(250)(375)cos75 9=-9 393.2 387.8 383.9 393.2 = 393 lb Ans sin θ = 75 θ = 7.39° 30° F_s+306° = 45° +7.5 360° -45° + 37.89° = 337° F_R = 375 lb 2-4. Determine the magnitude of the resultant force F_R = F_x + F_y and its direction, measured clockwise from the positive Y axis . F_s = \(300\) = 500 (200) cos 7 - 500 (500)(500) cos 75 = 6051 = 605 N Ans F_s\( 6051 50) 5250 N = 8.5 θ= 55.45° + 30° = 85.4° Ans