Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition Russell C. Hibbeler

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Solution Manual for Engineering Mechanics Dynamics in SI Units, 14th Edition Russell C. Hibbeler 13–1. The crate has a mass of 80 kg and is being towed by a chain which is always directed at 20° from the horizontal as shown. If the magnitude of P is increased until the crate begins to slide, determine the crate’s initial acceleration if the coefficient of static friction is p 20 and the ms = 0.5 coefficient of kinetic friction is . mk = 0.3 SOLUTION Equations of Equilibrium: If the crate is on the verge of slipping, . Ff = ms N = 0.5N From FBD(a), (1) + c ©Fy = 0; N + P sin 20° - 80(9.81) = 0 (2) :+ ©Fx = 0; P cos 20° - 0.5N = 0 Solving Eqs.(1) and (2) yields P = 353.29 N N = 663.97 N Equations of Motion: The friction force developed between the crate and its contacting surface is since the crate is moving. From FBD(b), Ff = mkN = 0.3N + c ©Fy = may ; N - 80(9.81) + 353.29 sin 20° = 80(0) N = 663.97 N :+ ©Fx = max ; 353.29 cos 20° - 0.3(663.97) = 80a Ans. a = 1.66 m >s 2 The motor lifts the 50-kg crate with an acceleration of 6 m > s y 2 . Determine the components of force reaction and the couple moment at the fixed support A. 4 m B A x 30 Solution 2 6 m/s Equation of Motion. Referring to the FBD of the crate shown in Fig. a, + c ΣFy = may; T - 50(9.81) = 50(6) T = 790.5 N Equations of Equilibrium. Since the pulley is smooth, the tension is constant throughout entire cable. Referring to the FBD of the pulley shown in Fig. b, S + ΣFx = 0; 790.5 cos 30° - Bx = 0 Bx = 684.59 N + c ΣFy = 0; By - 790.5 - 790.5 sin 30° = 0 By = 1185.75 N Consider the FBD of the cantilever beam shown in Fig. c, + ΣFx = 0; 684.59 - Ax = 0 Ax = 684.59 N = 685 N S Ans. + c ΣFy = 0; Ay - 1185.75 = 0 Ay = 1185.75 N = 1.19 kN Ans. # = 4.74 kN m # Ans. a+ ΣMA = 0; MA - 1185.75(4) = 0 MA = 4743 N m

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