HW__11_S06_Solutions - Physics 112 - HW #11 Solutions 11-10...

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- 1 - Physics 112 - HW #11 Solutions Spring 2006 11-10 [Leaning Ladder] (a) Maximum frictional force is f s,max = μ s N g Apply N’s 2 nd law in vertical direction: Σ F y = N g – W L - W p = 0 so N g = W L + W p = 160 N + 740 N = 900 N f s,max = μ s N g = (0.40)(900 N) = 360 N (b) and (c): Consider man on ladder, s meters along it. To find friction at bottom, take torques about top of ladder Σ τ top = (W L )(D/2) + W p (D – x ) - N g D + f s H = 0 so f s H = - (W L )(D/2) - W p D L - s L + N g D (f s )(4.0 m) = - (160 N)(1.5 m) - (740 N)(3.0 m) 5.0 m - s 5.0 m + (900N)(3.0 m) f s = 60 N + (111 N/m) s (1) For part (b), let s = 1.0 m in Eq (1) : Then f s = 60 N + (111 N/m)(1.0 m) = 171 N For part (c), put f s = f s,max = 36 0 N in Eq (1) : 360 N = 60 N + (111 N/m) s Solving for s gives s = 2.70 m w p w L N w N g f s L D H x s D = 3.0 m, L = 5.0 m, so H = 4.0 m
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- 2 - 11-11 [Diving Board] Free-body diagram showing forces acting on Person + Board: (a) Torques about left end of board: Σ τ = F e (0) + F s (1.00 m) - w b (1.50 m) - w p (3.00 m) = 0 F s = w b (1.50 m) + w p (3.00 m) 1.00 m = (280 N)(1.50 m) + (500 N)(3.00 m) 1.00 m = 1920 N (b) Σ F y = F s - F e - w b - w p = 0 F e = F s - w b - w p = 1920 N - 280 N - 500 N = 1140 N 11.73 [Farmyard Gate] (a) Take torques about point B (the lower hinge) Σ τ B = T sin 30 ° (4.00 m) + T cos 30 ° (2.00 m) - W (2.00m) = 0 T =
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This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

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HW__11_S06_Solutions - Physics 112 - HW #11 Solutions 11-10...

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