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

Instructor’s Solutions Manual ENGINEERING MECHANICS STATICS TENTH EDITION R. C. Hibbeler 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. © 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 warran- ty of any kind, expressed or implied, with regard to these programs or the documen- tation 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 iii 1-1. Round off the following numbers to three significant figures: (a) 4.6375 m, (b) 55.578 s, (c) 4555 N, (d) 278.6 kg (a) 4.64 m (b) 55.6 s (c) 4.56 kN (d) 279 Mg Ans 1-2. Wood has a density of 4.70 slug/ft^3. What is its density expressed in SI units? (4.70 slug/ft^3) * [(1 lb)/(4.998 sl)] * [(1 kg)/(2.20462 lb)] = 2.24 Mg/m^3. [0.3048 m/ft] * [1 slug] Ans 1-3. Represent each of the following quantities in the correct SI form using an appropriate prefix (a) 0.000431 kg, (b) 353 [10^3] N, (c) 0.00532 km (a) 0.000431 kg = 0.000431 (10^3) kg = 0.431 g Ans b) 353 (10^3) N = 353 kN Ans (c) 0.00532 km = 0.00532 (10^3) m = 5.32 m 1-4. Represent each of the following combinations of units in the correct SI form using an appropriate prefix: (a) mNm, (b) kJ/km, (c) kN/g.s.mg, and (d) km * μN. (a) mNm = (10^-3 N) (10^-3 m) = km Ans (b) kJ/km = (10^3) J (10^3) m = km Ans (c) kN/g.s.mg = [(10^3 N)/(10^-3 g)] [(1N)/(10^-3 g)] (d) km * μN = (10^3 m) [(10^-6) N] = mN.m = mm Ans 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 * [(5280 ft)/(1 mi)] [(0.3048 m)/(1 ft)] [(1 km)/(1000 m)] = 88.5 km/h Ans 88.5 km/h = (88.5 km)/(1 h) * [(1000 m)/(1 km)] * [(1 h)/(3600 s)] = 24.6 m/s Ans 1-6. Evaluate each of the following and express with an appropriate prefix: (a) 430 kg^2, (b) 0.002 mg^2, and (c) 230 m^3. (a) (430 kg)^2 = (183.700)^2 = 0.186 kg^2 Ans (b) (0.002 mg)^2 = [2(10^-3)] ^ 2 = 4 μm^2 (c) (230 m)^2 = [203(10^-3) ^ 2] = 0.622 um^2 Ans 1-7. A rocket has a mass of 250(10^3) 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/s^2, determine to three significant figures (c) its weight in SI units, and (d) its a) slugs p/slags = [250(10^4) slog](1 kg/0.3 slrog] = 3.56448 (10^4) kg = 3.564 Kilo kg Ans b) W_re = W_in_s = [3.5648(10^2) kg] [(9.81 m/s^2)] = 359.791 (10^3) kg.m/s^2 Ans c) W_m = W_g = [(3.79(10^3)] [(5.30 m/s^2)] = 5.98 = [3.564(10^2) m = 5.98 N Ans Since the mass is independent of its location, then m_m = m_e = 3.56 (10^2)) kg = 3.56 G Ans 1-13. Convert each of the following to three significant figures, (a) 20 lb-ft to N m, (b) 450 lb/ft^2 to kN/m^2, and (e) 15 ft/s to mm/s. Using Table 1-2, we have a) 20 lb-ft = (20 lb)(4.4482 N)(0.3048 m) = 1 ft lb = 1 ft = 1 lb = 1 = 27.1 N m Ans b) 450 lb/ft^2 = (450 lb)(4.4482 N) (1 N) 1 ft = 1 1 N 1 ft = 1 ft = 1 ft^2 (1 m) (1000/0.3048)^2 m2 = 7.07 kN/m2 Ans c) 15 ft/s = (15 ft)(0.3048 m) 1 ft 1 hr = 1 1 ft 3600 s = 1.27 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 slugs/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 slug/ft^3)(14.5939 kg) 1 ft^3 1 slug 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 = G (m1*m2) r^2 Where G = 6.673(10^-11) m^3/kg^2 S F = G(6.673)(10^-11)(8)(12) (0.8)^2 = 100.0 nN W_1 = (8)(9.81) = 78.5 N Ans W_2 = 12(9.81) = 118 N Ans 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. Applying Eq. 1–3, we have w = mg a) m = 20(10^6) N = 2.04 g Ans 9.81 m/s^2 b) m = 150(10^3) kN = 15.3 Mg Ans 9.81 m/s^2 c) m = 60(10^6) N = 6.12 Gg Ans 9.81 m/s^2 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 = 4.81 slug Ans 32.2 (b) m = 155(14.5939 kg) = 70.2 kg Ans 32.2 (c) W = 155(4.4482) = 689 N Ans (d) W = 155(5.3) = 25.5 l b 32.2 (e) m = 155(14.5939 kg) = 70.2 kg Ans 32.2 Altns. m = 35 (14.5939 kg) = 70.2 kg 5.30 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 = G m1*m2 r^2 F = Gm/r2 (m2) (Q. E. D. ) F = Gm*r 2 F = 66.67(10^-11) (200)^2 0.3 = 7.44(10^-9) N = 7.44 nN 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 Msn)/(8.60 kg), (b) (35 mm)^2(48 kg). (a) 0.631 Msn/8.60 kg^2 kg^2/m = 8.93 m Ans = 8.93(10^4) = 8.93 km^2/s (b) (35 mm)^2(48 kg) = [35(10^-3) m]^2[48 kg] = 15.3 m^2 kg Ans 2-1. Determine the magnitude of the resultant force FR = F1 + F2 + F3 and its direction, measured counterclockwise from the positive x axis. Fs = √((600)^2 + (600)^2 - 2(600)(600)cos75°) = 866.91 = 867 N Ans 866.91 = 320 866.91 sin 60 θ = 63.05° + 45° = 108° Ans 2-2. Determine the magnitude of the resultant force if: (a) FR = F1 + F2; (b) FR = F1 – F2. FR = 100 N 60^ F2 = 80 N Parallelogram Law : The parallelogram law of addition is shown in Fig. (a) and (c). Trigonometry : Using law of cosines [Fig. (b) and (d)], we have b) FR = √((100)^2 + 80^2 - 2(100)(80) cos 75°) = 111 N Ans = √((100)^2 + (80)^2cos 105° = 143 N Ans 2-3. Determine the magnitude of the resultant force FR = F1 + F2 + F3 and its direction, measured counterclockwise from the positive x axis. FR=250 lb φ =20° FR=375 lb Fs = (√((250)^2 + (375)^2 - 2(250)(375)cos78°)) = 393 lb Ans 391.2 = 250 291.75 θ = 737.89° φ = 360° – 45° + 37.89° = 353° Ans 2-4. Determine the magnitude of the resultant force FR = F1 + F2 + F3 and its direction, measured clockwise from the positive y axis. Fs= (√((500)^2 + (500)^2 – 2(500)(500)cos75° = 605.1 = 601 N Ans 605.1 = 590 605.1 sin 60 θ = 454.7 = 30° + 55.4° = 355.4° = 30° = 85.4° Ans 2-5. Resolve the force F1 into components acting along the n and v axes and determine the magnitudes of the components. Fn = F1v = 300 N 205 N Ans Fn = 300 sin 110° Angles shown. 160 N Ans 70°m F1v 300 N 45° F1 Tm F1 = 500 N 2-6. Resolve the force F2 into components acting along the u and v axes and determine the magnitudes of the components. F,v – 500 N F2u = 500 sin 70° 136 sin = 705 Fv = 375 N Ans F2u Cos Fv 500 u sin 70° 530 N F2 = 482 N Ans 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). Trigonometry: Using law of cosines [Fig. (b)], we have = (3 (18) = R 30° Answer 6 kN (3) = 6.718 +(6) (18) 100 10.8 kN = 10.8 kN The angle 0 can be determined using law of sinse [Fig. (b)]. A Ans B V sin 100° sin = sin 6 100 - 0.3517 B 60° A sin 0 = 0.3517 = sin that, the direction of R, measured from the x axis is Du <3 A (5) 60 a 60on 0 = 30° = 30° = 3.1° R Ans 8 kN = 6 kN (a) 60 (b) N B 60 A