and the force exerted there is the tension T. Computing torques about the hinge, we find ()()221112250.0 kg9.8 m/s1.00 m(50.0 kg)(9.8m/s )(3.00 m)=sin3.00 m 0.800408 N.mgxmgxTxθ++==(b) Equilibrium of horizontal forces requires the horizontal hinge force beFx= Tcos = 245 N. (c) The direction of the horizontal force is rightward. (d) Equilibrium of vertical forces requires the vertical hinge force be Fy= mg– Tsin = 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.