hw15_sol - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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Unformatted text preview: ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7—84* Determine all forces acting on member ABC of the linkage shown in Fig. P7-84. SOLUTION From a free-body diagram for member ABC: .z + C EMA = By(200) — 500(400) o B = 1000 N = 1000 N T Ay + 1000 - 500 = 0 -500 N = 500 N ¢ + _; M ’13 < II A + B = 0 H + —e 2F X X X From a free-body diagram for member BD: + C 2MB = Bx(300) + 1000(400) - 250(200) = 0 B = —1166.7 N 3 1167 N 9— (on ABC) 1166.7 N 3 1167 N —4 /(Ax)2 + (Ay)2 = /(1166.712 + (-500.0)2 = 1269.3 N A = e _ t “-1 f1 _ ta -1 —500.0 _ _23 200 A ‘ a Ax ' " 1166.7 ' ' K a 1269 N % 23.2° Ans. 2 2 2 2 B = (Bx)> (By) = /(-1166.71 + (1000) = 1536.6 N e _ t -1 E1 _ t -1 1000.0 _ 139 400 a ‘ a" B ' 3“ —1166.7 ' ° B a 1537 N z 40.6° (£213 ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. 7-87* A pin-connected system fififififlflflfiéfifit of levers and bars is used as a toggle for a press as shown in Fig. force F exerted on the Determine the can at A when a force fl = 100 lb is applied to the lever at G. SOLUTION From a free-body diagram for the lever: + C EMF = 100(30) - FDE(8) = o F : DE 375 lb From a free-body diagram for the pin at D: cos 67° - F cos 73° — 375 = 0 CD , O Sln 67 + FCD + —+ 2F = —F 80 -F . 0 __ BD Sln 78 — O — -639.5 lb -601.8 lb From a free-body diagram for the piston at B: + T EFy = A- - 639.5 sin 67° = 0 Y Ay = 588.7 lb = 588.7 lb T Force on the can: F E 589 lb ¢ D. STURGES ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-95* Forces of 25 lb are applied to the handles of the pipe pliers shown in Fig. P7-95. Determine the force exerted on the pipe at D and the force exerted on handle F. . P7-95 DAB by the pin at A. % SOLUTION / From a free-body diagram for member DAB: + C EMA = D(1.25) — 25(9) = o D 180.0 lb Ans. D sin 38° + Ax = o _ -D sin 38° -180.0 sin 38° -110.82 lb = 110.82 lb +— Ay - D cos 38° — 25 = o D cos 38° + 25 180.0 cos 38° + 25 166.84 lb = 166.84 lb T /(Ax)2 + (Ay)2 = /(—110.82)2 + (155.84)2 = 200.29 1b a 200 lb -1 166.84 0 m - 133-59 tan ‘ = 200 lb 8 56.4° Ans. ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES 7-103* The fold-down chair of Fig. P7-103 weighs 25 lb and has its center of gravity at G. Determine all forces acting on member ABC. SOLUTION From a free-body diagram for the chair: + C EMA = 25(3) — E(Zl) = o , E = 3 571 lb = 3.571 lb +— = A - 3 571 = o X 3.571 lb = 3.571 lb -4 = 25 lb = 25 lb T | ‘X _ 2 2 - (Ax) + (Av) I /(3.571)2 + (25)? = 25.25 lb —1 25 ex ta“ 3. 71 K = 25.3 lb 8 81.9° = 81.87° Member BD is a two—force member; therefore, the line of action of force 3 is known as shown on the free-body diagram for member ABC: + C inc = B cos 39.81° (9) — 3.571(21) = o B 10.846 lb B = 10.85 lb 8 e 39.8° c + 3.571 - 10.846 cos 39.81° = 0 . c = 4.761 lb X X = Cy + 25 - 10.846 sin 39.810 = 0 CV = -18.056 lb /(cx)2 + (Cy)? = /(4.761)2 + (-18.056)2 = 18.673 lb -1 —18.056 .___~ 0 = o 4.761 - 75.23 C 18.67 lb 2 75.2 tan ...
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hw15_sol - ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F....

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