hw4_sol

# hw4_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 3—12 A body with a mass of 300 kg is supported by the flexible cable system shown in Fig. P3-12. Determine the tensions in cables A, B, C, D, and E. SOLUTION From a free—body diagram for the body: T : - — z + . EFY TE mg 0 TE = mg = 300(9.807} = 2942 N 3 2940 N From a free-body diagram for the lower cable joint: + T 29? = TC sin 60° - 2942 = 0 TC = 3397 N 3 3400 N O — TC cos 60 n TD 2 3397 cos 60° — T0 = 0 1698.5 N 3 1699 N From a free—body diagram for the upper cable joint: + T 2Fy = TB_sin 30° + TA sin 40° — 3397 sin 60° = + so EFX = TB cos 30° — TA cos 40° — 3397 cos 60° = 0 Solving yields: 180?.53 N 3560.03 N ENGINEERING MECHANICS — STATICS, 2nd. Ed. . RILEY AND L. D. STURGES 3-14* Three smooth homogeneous cylinders A, B, and C are stacked in a V-shaped trough as shown in Fig. P3—14. Each cylinder has a diameter of 500 mm and a mass of 100 kg. Determine the forces exerted on cylinder A by the inclined surfaces. SOLUTION # W = mg = 100(9.807) = 980.7 N From a free-body diagram for cylinder B: + \y ZFX = -NB + w sin 30° -NB + 980.7 sin 30° NB 490.35 N From a free-body diagram for cylinder C: + /” ZF N - W sin 450 y C Nc - 980.7 sin 45° NC 693.46 N From a free-body diagram for cylinder A: + /” ZFX NL cos 15° + 490.35 sin 15° - 980.7 cos 45° — 693.46 1304.46 N i 1304 N = NR — 980.7 sin 45° + 1304.46 sin 15° — 490.35 cos 15° 829.48 N ﬁ 829 N ’27 W. F. RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS, 2nd. Ed. 3—25 The circular disk shown in Fig. P3—25 weighs 750 lb. tensions in cables A, B, and C. Determine the SOLUTION {0.5963 3 - 0.2981 3 + 0.7454 KITA r-Jl u H _.:§_£_:_§_l_:_§_3___] = (~0.4082 i — 0.4082 3 + 0.8165 EITB H}! n .—l _.:i_1_i_§mJ_:_i_E._] = (—0.624? E + 0.4685 3 + 0.6247 ElTC 'Th -% TA + TB + Tc + W = {0.5963TA - 0.4082TB - 0.6247TC} 1 + {-0.2981TA - 0.4082TB + 0.4685TC1 J + {0.7454TA + 0.8165TB + 0.6247TC - 750) E = G 0.5963TA - 0.4082TB ~ 0.624TTC II D -0.2981TA — 0.4082TB + 0.4685TC 0.7454T + 0.8165T + 0.6247T A B C Solving yields: 525.61 lb % 526 lb (T) Ans. 109.7 lb {T} 109.71 lb 430.02 lb 430 1b {T} ENGINEERING MECHANICS - STATICS, 2nd. Ed. The hot-air balloon shown in Fig. P3-37 is tethered with three mooring cables. If the net lift of the balloon is 900 lb, determine the force exerted on the balloon by each of the cables. SOLUTION ‘ (0.2752 T — 0.4299 (-0.4319 1 900 E ZF=TA+TB+TC+E (0.3244TA + 0.2752TB — 0.4319TC) T + (0.4867TA - 0.4299TB - 0.2592TC) + (-0.8111TA 0.8599TB 0.3244'1‘A 0.4867TA -0.8111TA 0.2752TB 0.4299TB 0.8599TB Solving yields: W. F. RILEY AND L. D. STURGES ng3a7 A J — 0.8599 E)TB A — 0.2592 3 - 0.8639 E)TC 0.8639TC 0.4319TC 0.2592TC 0.8639TC 418.2 lb 205.2 lb 444.9 lb ...
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hw4_sol - ENGINEERING MECHANICS STATICS 2nd Ed W F RILEY...

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