HWS4 - flww— nee/Van - Fat/06 ENGINEERING MECHANICS —...

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Unformatted text preview: flww— nee/Van - Fat/06 ENGINEERING MECHANICS — STATICS, 2nd. Ed. w. F. RILEY AND L. D. STURGES 3-3 A homogeneous cylinder weighing 500 lb rests against two smooth planes that form a trough as shown in Fig. P3-3. Determine the forces exerted on the cylinder by the planes at contact points A and B. ‘ r" :. v :;:“-‘:‘.’,r¢.‘.‘.’f? 5’? 1.:- SOLUTION ‘From a free-body diagram for the cylinder: + T 28 = F cos 30° - w y B cos 30o — - 577.35 sin 30° = 0 = 288.68 lb a 289 1b A = 289 lb -% 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. in cables A, B, C, D, and E. Determine the tensions SOLUTION From a free-body diagram for the body: +T2FszE—mg:0 { ll §' L0 TE = mg = 300(9.807) = 2942 N g 2940 N Ans. From a free—body diagram for the lower cable joint: + T ZFY 2 TC sin 600 — 2942 = 0 ‘Tb L TC = 3397 N a 3400 N ‘ I o + ‘+ 2F — TC cos 60 — TD = 3397 cos 60° — TD = 0 1698.5 N g 1699 N i From a free-body diagram for E the upper cable joint: + T 29y = TB‘sin 30° + TA sin 40° — 3397 sin 60° = 0 '+ —+ EFX = TB cos 30° —‘TA cos 40° — 3397 cos 60° = 0 1807.53 N 3 1808 N Ans. - Solving yields: T 3560.03 N a 3560 N ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. 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: . ‘ 0 + \u EFX -NB + W s1n 30 —NB + 980.7 sin 30° = 0 NB 490.35 N From a free-body diagram for cylinder C: + /” 2F N — W sin 45° y C - NC ~ 980.7 sin 45° = NC 693.46 N F From a free-body diagram ; for'cylinder A: ' 5 + /” 2§;‘; NL cos 15° + 490.35 sin 15° - 980.7 cos 45° — 693.46 1304.46 N z-1304 N NR — 980.7 sin 45° + 1304.46 sin 15° - 490.35 cos 15° 829.48 N a 029 N )27 ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 3—23* Determine the forces in legs AB, AC, and AD of the tripod shown in Fig. P3-23 when force F = 75 lb. The legs can transmit only axial forces. SOLUTION (0.3982 3 — 0.3186 3 — 0.8602 2)FB <25>2 + («20)2 + (—54)2 ‘ 2 1 + 40 - 54 E j = (0,3437 3 + 0.5579 3 — 0.7531 E)FC (25)2 + (40)2 + (-54) "2° 1 (—0.4180 3 + 0.1003 3 — 0.9029 E)FD 1—25)2 -75 2 lb FB+FC+FD+F‘ = (0.3982FB + 0.3487FC - 0.4180F ) i D + (-0.3186FB + 0.5579FC + 0.1003FD) + (-0.8602FB - 0.7531FC - 0.9029FD 0.3982FB .3487FC - 0.41801?D —0.3186FB .5579FC + 0.1003FD -0.8602FB v .7531FC - 0.9029FD Solving yields: 2 —33.42 lb 3 33.4 lb (C) —11.62 1b = 11.62 lb (C) ~41.53 lb 3 41.5 lb (C) ENGINEERING MECHANICS - STATICS, 2nd. Ed. \ W. F. RILEY AND L. D. STURGES 3-37 The hot—air balloon shown in Fig. P3-37 is tethered with, three mooring cables. If the net lift of the balloon is 900 1b, determine the force exerted on the balloon by each of the cables. wlfiilggfiggflw ~.- 'w9 _ ‘ :1». 1'; ,1. i . ‘9_ 1. SOLUTION 2 2 — 2 (20) + (30) + ('00) Fig.P3-37 = (0.3244 1 + 0.4867 3 — 0.8111 ErrA 1 - 25 ' 50 E ] = (0.2752 1 - 0.4299 3 — 0.8599 E)TB /—-—————-———-———-—————-—-—fi I (16)2 + (—2512 + (—50)2 J ‘2” 1 ' 15 " 50 E '] = (-0.4319 1 - 0.2592 “ — 0.8639 E1TC (—25)2 + (—15)2 + (-50)2 900 E EF=TA+TB+TC+E = (0.32441A + 0.2752TB —_0.4319TC) i + (0.4867TA - 0.42991B — 0.2592TC) + (—0.811111‘S — 0.8599TB - 0.8639TC 0.3244TA + 0.2752TB 0.4867TA - 0.4299TB -0.8111TA — 0.8599‘1‘B 0.4319TC 0.2592TC 0.063%C Solving yields: 418.2 lb 205.2 lb = 444.9 lb ...
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HWS4 - flww— nee/Van - Fat/06 ENGINEERING MECHANICS —...

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