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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: f n n s 1 2 1 = = µ and f n s 2 2 = µ F y = 0: P n f F g + + = 1 2 τ = 0: P f f = + 1 2 As P grows so do f 1 and f 2 Therefore, since µ s = 1 2 , f n 1 1 2 = and f n n 2 2 1 2 4 = = FIG. P12.60 then P n n F g + + = 1 1 4 (1) and P n n n = + = 1 1 1 2 4 3 4 (2) So P n F g + = 5 4 1 becomes P P F g + F H G I K J = 5 4 4 3 or 8 3 P F g = Therefore, P F g = 3 8 P12.61 (a) F k L = a f , Young’s modulus is Y FL A L F A L L i i = = a f Thus, Y kL A i = and k YA L i = (b) W Fdx kx dx YA L xdx YA L L L L i L i = − = − = = z z z 0 0 0 2 2 a f a f P12.62 (a) Take both balls together. Their weight is 3.33 N and their CG is at their contact point. F x = 0: + = P P 3 1 0 F y = 0: + = P 2 3 33 0 . N P 2 3 33 = . N τ A = 0: + + ° P R P R R R 3 2 3 33 45 0 . cos . N a f + + ° = P R R 1 2 45 0 0 cos . a f Substituting, + + ° + + ° = °= ° = = P R R R P R P P P 1 1 1 1 3 3 33 3 33 1 45 0 1 2 45 0 0 3 33 45 0 2 45 0 1 67 1 67 . . cos . cos . . cos . cos . . . N N N
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