369_Physics ProblemsTechnical Physics

# 369_Physics ProblemsTechnical Physics - Chapter 12 P12.54...

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Chapter 12 371 P12.54 (a), (b) Use the first diagram and sum the torques about the lower front corner of the cabinet. τ =⇒ + = 0 1 00 400 0 300 0 F .. m N m af a f yielding F == 400 0 300 100 120 m m N a f . . Ff x −+ = 01 2 0 0 N, o r f = 120 N Fn y += 04 0 0 0 s o n = 400 N Thus, µ s f n = 120 0300 N 400 N (c) Apply F at the upper rear corner and directed so θφ ° 90 0 . to obtain the largest possible lever arm. θ = F H G I K J tan . . 1 59 0 m 0.600 m Thus, φ ° 90 0 59 0 31 0 . . Sum the torques about the lower front corner of the cabinet: ++ = F 1 00 0 600 400 0 300 0 22 . m N m a f a f so ′= = F 120 103 Nm 1.17 m N. Therefore, the minimum force required to tip the cabinet is 400 N n f F 0.300 m 1.00 m n f 1.00 m 0.600 m 400 N F’ FIG. P12.54 103 N applied at 31.0 above the horizontal at the upper left corner ° . P12.55 (a) We can use FF xy ∑∑ 0 and = 0 with pivot point at the contact on the floor. Then FT n xs =− = 0, FnM gm g y − = 0, and τθ =+ F H G I K J −= Mg L mg L TL cos cos sin 2 0 Solving the above equations gives
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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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