372_Physics ProblemsTechnical Physics

372_Physics ProblemsTechnical Physics - 374 P12.60 Static...

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374 Static Equilibrium and Elasticity P12.60 When it is on the verge of slipping, the cylinder is in equilibrium. F x = 0: fn n s 12 1 == µ and s 22 = F y = Pn f F g ++= τ = Pf f =+ As P grows so do f 1 and f 2 Therefore, since s = 1 2 , f n 1 1 2 = and f nn 2 21 24 FIG. P12.60 then Pn n F g ++ = 1 1 4 (1) and P n =+= 11 1 3 4 (2) So F g += 5 4 1 becomes PP F g + F H G I K J = 5 4 4 3 or 8 3 PF g = Therefore, g = 3 8 P12.61 (a) FkL =∆ af , Young’s modulus is Y FL AL F A L L i i Thus, Y kL A i = and k YA L i = (b) W Fdx kx dx YA L xdx YA L L LL i L i =− = = zz z 00 0 2 2 ∆∆ P12.62 (a) Take both balls together. Their weight is 3.33 N and their CG is at their contact point. F x = 0: +−= 31 0 F y = +− = P 2 333 0 . N P 2 = N A = −+ + ° PR PR R R 32 450 .c o s . N ++ ° = PR R 1 5 0 0 cos . Substituting, + ° ° = °= ° P RRR PR P 1 1 1 13 1 12 4 5 0 0 2 167 ..c o s . cos . . cos . cos . .. N N N s o N a f F g P 1 P 2 P 3 3.33 N FIG. P12.62(a) (b) Take the upper ball. The lines of action of its weight, of
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .

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