hw17a_sol

# hw17a_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 9-1* Determine the horizontal force F required to start moving the 250-lb block shown in Fig. P9-1 up the inclined surface. The coefficient of friction between the inclined surface and the block is u = 0.30. S SOLUTION .a A free—body diagram for the block is shown at the right. For impending motion: Ff = HF“ = 0.30Fn + T 2F = F cos 30° - F sin 300 — H y n f = F cos 30° - 0.3017n sin 30° — 250 = o n = 349.15 lb = P - Fn sin 300 - Fr cos 300 P - 349.15 sin 30° - 0.30(349.15) cos 30° = 0 265.29 lb 3 265 lb 8J6 “INEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Two blocks with masses mA = 20 kg and m8 = 80 kg are connected with a flexible cable that passes over a frictionless pulley as shown in Fig. P9-4. The coefficient of friction between the blocks is 0.25. If motion of the blocks is impending. determine the coefficient of friction between block B and the inclined surface and the tension in the cable between the two blocks. SOLUTION A free-body diagram for block A is shown at the right. ' mAg = 20(9.807) = 196.14 N For impending motion of block A: + /” BE? A — 196.14 cos 35° = o - = 160.67 N n + \y 2Fx —T + 0.25(160.67) + 196.14 sin 350 = 0 = 152.67 N i ‘3 52.47 N 784.54 N ,r' A free-body diagram for block 35° )LO.(.'7 N B is shown at the right. ‘4c>.)7 )1 W3 = mag = 80(9.807) = 784.56 N For impending motion of block B: + /” ZFY = B - 160.67 - 784.56 cos 35° = 0 303.34 N n + \y ZFX -Br — 40.17 - 152.67 + 784.56 sin 350 = 0 257.17 N B . _ r _ 257.17 _ ‘ B ' 303.34 ' 0'320 Ans' n ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY AND L. D. STURGES Two blocks with masses mA = 50 kg and m8 = 100 kg are connected with a flexible cable that passes over a frictionless pulley as shown in Fig. P9-6. Determine the force F required to start moving the blocks if the coefficient of friction between the blocks and the surface is 0.10. SOLUTION ‘h free-body diagram for block A is shown at the right. WA = mAg + 50(9.807) = 490.4 N For impending motion: + “\ Epy F - 490.4 cos 30° = o n + /” 2F, _ T - Fr — w sin 30° T - 0.10(424.7) - 490.4 sin 30° = o A free—body diagram for block B is shown at the right. W8 = mBg + 100(9.807) = 980.7 N For impending motion: T + EFY = 9 sin 25° + Fn — 980.7 = o —a + 2Fx P cos 25° - Ff - T 9 cos 25° - 0.10Fn - 287.7 = 0 Solving yields: P = 406.7 N E 407 N ENGINEERING MECHANICS - STATICS, 2nd. Ed. A homogeneous, triangular block of weight W has height h and base width b as shown in Fig. P9-7. Develop an expression for the coefficient of friction between the block and the surface for which impending motion by slipping and tipping occur simultaneously if (a) Force F acts in the direction shown on the figure. (b) The direction of force F is reversed. SOLUTION (a; From the free-body diagram for the triangular block: +T2F = Y +-—)2F : x For slipping: = Af(max) = HA“ 5 b For tipping: + C 2MB = Pth - Wtﬂ - For simultaneous slipping and tipping: From which: (b) From the free-body diagram for the triangular block: + T 2F y + -e 2F X For slipping: For tipping: For simultaneous slipping and tipping: From which: 82?. wt F. RILEY AND L. D. STURGES ENGINEERING MECHANICS - STATICS. 2nd. Ed. W. F. RILEY AND L. D. STURGES The masses of blocks A and B of Fig. P9-24 are mA 100 kg and mB = 50 kg. If the coefficient of friction is 0.15 for block A and 0.25 for block B, determine the maximum angle 6 for equilibrium of the blocks. SOLUTION From a free-body diagram for block B when motion is impending: = 50(9.807) 490.35 = Bn — WB cos 6 B - 490.35 cos 6 = 0 n 490.35 cos 6 0.25Bn = 0.25(490.35 cos 6) = 122.59 cos 6 + /' 2F. -T - wB sin 6 + 131‘ = -T - 490.35 sin 6 + 122.59 cos 6 = o T = 122.59 cos 6 - 490.35 sin 6 From a free—body diagram for block wﬁ=980.7N A when motion is impending: = 100(9.807) = 980. Ar(max) = uAAn = A - WA cos 9 n A - 980.7 cos 6 n 980.7 cos 6 = 0.15A = 0.15(980.7 cos 9) = 147.11 cos 6 n = T + Ar - WA sin 6 = 122.59 cos 6 — 490.35 sin 6 + 147.11 cos 6 - 980.7 sin 6 = 269.70 cos 6 - 1471.05 sin 9 = 0 -1 260.70 n _______ 0 0 1471.05 - 10.389 E 10.39 ...
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