hw15_sol

# hw15_sol - = 60.85 lb Since B r < B r (max), tipping is...

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- ENGINEERING MECHANICS - STATICS, 2nd. Ed. W. F. RILEY ANDL. D. ~TURGES 9-35 A 100-lb uniform beam 16 ft long lies against a corner as shown in Fig. P9-35. Determine the maximum force p for which the beam will be in equilibrium if the coefficient of friction ~ s p is: (a) 0.60 at both surfaces. (b) 0.75 at the bottom surface and 0.40 at the corner surface. Fig. P9-35 -1 4 ° e = tan - = 26.57 8 d = I 82 + 42 = 8.944 ft «o~~ Sf I .) ,\ B - -- ~- - -~_1\; :~~.L )'t. R.Ry 8~t _u ." )t p From a free-body diagram for the beam when motion is impending: (a) For tipping: (An = Ar = 0) + C LMB = 100(8 - 8 cos 26.57°) - P(16 cos 26.57° - 8) = a + r'LF = B - 100 cos 26.57° y n P = 13.39 lb 13.39 cos 26.57° = 0 B = 101. 41 lb n +? LF = B - 100 sin 26.57° - 13.39 sin 26.57° = a x r B = 50.'72 lb r Bf(max) = ~Bn = 0.60(101.41)
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Unformatted text preview: = 60.85 lb Since B r < B r (max), tipping is impending and P = 13.39 lb max Ans. (b) If ~B= 0.4: Br(max) = ~BBn = 0.40(101.4) = 40.46 lb Since Br > Br(max), tipping is not possible and slipping occurs. For slipping: (A = 0.75A and B = 0.40B ) r n r n + ~ LF h = 0.75A + 0.40B cos 26.57-B sin 26.57 = 0 n n n + t LF = A + 0.40B sin 26.57 + B cos 26.57-100 -P = 0 v n n n + C LM = B (8.944) A n 100(8 cos 26.57)-P(16 cos 26.57) = 0 Solving yields: A = 10.515 lb .n B = 88.08 lb n P = P = 5.05 lb max Ans. as}...
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## This note was uploaded on 09/24/2011 for the course MEEN 221 taught by Professor Mcvay during the Spring '08 term at Texas A&M.

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hw15_sol - = 60.85 lb Since B r < B r (max), tipping is...

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