ch12-p028 - 28. (a) The sign is attached in two places: at...

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and the force exerted there is the tension T . Computing torques about the hinge, we find () ( ) 22 11 12 2 50.0 kg 9.8 m/s 1.00 m (50.0 kg)(9.8m/s )(3.00 m) = sin 3.00 m 0.800 408 N. mgx mgx T x θ + + = = (b) Equilibrium of horizontal forces requires the horizontal hinge force be F x = T cos = 245 N. (c) The direction of the horizontal force is rightward. (d) Equilibrium of vertical forces requires the vertical hinge force be F y = mg T sin = 163 N. (e) The direction of the vertical force is upward.
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This note was uploaded on 05/17/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.

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