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

Instructor’s Solutions Manual ENGINEERING MECHANICS STATICS TENTH EDITION R. C. Hibbeler PEARSON Prentice Hall Pearson Education, Inc. Upper Saddle River, New Jersey 07458 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. PEARSON Prentice Hall © 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 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 iii 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. \text{Using Table 1 - 2, we have} a) 20 \times \frac{lb\, f}{18} = (20)(\frac{4.4482}{N}) (\frac{9.0048}{1ft}) = 23.71 \, lb \quad \text{Ans} \pm 2 \frac{n}{9.81} = 1.96 \, kN b) 450 \times \frac{(ft)^4}{10^3} = (3600)(10^3)(1 \times (9.007ft^2 = lb)) (\frac{1}{3.048^2}) = 200 \frac{7.87N}{m^2} \quad \text{Ans} c) 15 \times {ft/s} = \frac{(15) (3.048 \, m)}{0} = 1.27 \, m/s \quad \text{Ans} 1-14. If an object has a mass of 40 slugs, determine its mass in kilograms. 40 \times \frac{(14.5939 \frac{kg}{slug})}{m} = 584 \, kg \quad \text{Ans} 1-15. Water has a density of 1.94 \text{ slug/ft}^3. What is the density expressed in SI units? Express the answer to three significant figures. \text{Using Table 1 - 2, we have} \rho _w = \frac{1.94 \times (14.593 kg)}{1 \, m} = 999.8 \cdot \text{ kg/m}^3 \approx 1.00 \cdot \text{ m}^3 \approx 1.00 \cdot \text{ kg/liter} \quad \text{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 = \frac{G m_1 m_2}{r^2} \text{Where } G = 6.67 \times {10}^{-11} \frac{1}{m^2\cdot s^2} F = \frac{6.67 \times (10^{-11})}{{0.89}^2} = 10.00 \pm 0.89 \quad \text{N} W_1 = 8(9.81) = 78.5 \pm 0.89 \quad \text{N} W_k = (12)(9.81) = 118 \pm 0.89 \quad \text{N} 1-17. Determine the mass of an object that has a weight of (a) 20 mN, (b) 150 kN, (c) 60 MN. Express the answer to three significant figures. \text{Applying Eq. 1 - 3, we have} a) \frac{m}{9} = \frac{200(10^3)}{kg \cdot m^2} = 2.04 \pm 45^4 \text{ \quad N} \quad \text{Ans} b) \frac{m}{1.5} = \frac{150(0^{60}) \text{ kg/m}^2}{m^2} = 153 \quad \text{kg} \quad \text{ \quad N} \quad \text{Ans} c) 60 MN \times = 6152 \frac{1}{9.81 \cdot \text{kg}} = 61.2 \pm 0.89 \quad \text{ \quad N} \quad \text{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 = \frac{155}{32.2} = 4.81 \, slug \quad \text{Ans} (b) m = \frac{155}{32.2} (14.5939 \frac{kg}{slug}) = 70.2 \, kg \quad \text{Ans} (c) W = 155 (4.4482) = 689 \, N \quad \text{Ans} (d) W = \frac{155}{32.2} (5.30) = 25.5 \, lb \quad \text{Ans} (e) m = \frac{155}{32.2} (14.5939 \frac{kg}{slug}) = 70.2 \, kg \quad \text{Ans} \text{Alt.} \quad m = \frac{25.5}{5.30} (14.5939 \frac{kg}{slug}) = 70.2 \, kg \quad \text{Ans} 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. \text{Using Eq. 1 - 2.} F = G \frac{m_1 m_2}{r^2} N = \left( \frac{kg \cdot m}{s^2} \right) \quad (Q.E.D.) F = G \frac{m_1 m_2}{r^2} F = 66.7 \left( 10^{-12} \right) \frac{kg^2}{m^2} = 200 kg = 7.41 \left( 10^{-6} \right) \approx 7.41 \mu N \quad \text{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 \quad MNm)/(8.60 \quad kg/m^2) (b) (35 \quad mm^3)/(48 \quad kg^3). (a) 0.631 \quad MNm/(8.60 \quad kg/m^2) = \frac{0.631(10^6) \, m}{8.60 \, kg/m^2} = 8.51(10^4) \approx 8.31 \quad km^2 \pm \text{Ans} (b) 0.3 \quad \text{mm}/48 \quad \text{kg}^3 = \left[ 25.1 \left(10^{-3}\right) \right] = 135 \quad m/kg^2 \quad \text{Ans} 2-1. Determine the magnitude of the resultant force F_R = F_1 + F_2 + F_3 and its direction, measured counterclockwise from the positive x axis. F_R = \sqrt{(-600)^2 + (600)^2 - 2(600)(600)\cos 75^\circ} = 866.91 = 867 \, N \quad \text{Ans} \phi = 63.05\degree \sin \theta / \sin60 = \theta = 63.05\degree + 45\degree = 108\degree \quad \text{Ans} 2-2. Determine the magnitude of the resultant force if: (a) F_1 = F_2 = F_3; (b) F_1 = F_2 = F_3. F_1 = 100 \, N \quad \theta \cos60^\circ = 80 \, N \, \theta \cos45^\circ = 45 \, N \ \text{Parallelogram Law: The parallelogram law of addition is shown in Fig. (a) and (c).} \text{Trigonometry: Using law of cosines [Fig. (b) and (d)], we have} b) \quad = \sqrt{100^2 + 80^2 - 2(100)(60)\cos 75^\circ} \quad \text{Ans} = 111 \, N \quad \text{Ans} = 30^\circ = 143 \degree \pm 135^\circ \pm 45^\circ \pm \degree \pm \text{Ans} = 143 \degree \pm \text{Ans} 2-3. Determine the magnitude of the resultant force F_R = \sqrt{R_x^2 + R_y^2} and its direction, measured counterclockwise from the positive x axis. F_3 = 250 \, lb Q = 30^\circ Q = \left(250 + 375\right) = \left(250 \times 375 \right)\cos 75^\circ = 393.2 = 393 \, lb \quad \text{Ans} = 353 \deg \pm \text{Ans} p = 75 \pm 89 \deg \, \theta \pm b = 353 \degree \pm 0.289 \pm \text{Ans} *2-4. Determine the magnitude of the resultant force F_R = F_1 + F_2 + F_3 and its direction, measured clockwise from the positive y axis. F_2 = (F_1) \left( 300 \right) = - 250(500) (500)\cos39 = 605 \, N \quad \text{Ans} = 30^\circ, 30^\circ F_2 = 300 \, N F_y = 500 \, N = 85 \deg \pm \text{Ans} 2-5. Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. Fu = 300 sin 150º Fu = 205 N Ans cos 150º Fv = 300 cos 150º Fn = 160 N Ans F2 = 300 N Fv 2-6. Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. Fu = 500 sin 70º Fu = 750 N Ans cos 70º Fv = 500 cos 70º Fv = 482 N Ans 2-7. The plate is subjected to the two forces at A and B as shown. If θ = 60º, 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). Trigonometry : Using the law of cosines [Fig. (b)], we have Fr = √(8^2 +6^2 -2(8)(6)cos 100º = 10.84 kN = 10.8 kN The angle θ can be determined using law of sines [Fig. (b)]: sin θ 8 = sin 100º 10.84 θ = 0.830 = 51.7º Thus, the direction θ of Fr measured from the x axis is θ = 39.1º - 30º = 3.1º