29. The bar is in equilibrium, so the forces and the torques acting on it each sum to zero. LetTlbe the tension force of the left–hand cord, Trbe the tension force of the right–hand cord, and mbe the mass of the bar. The equations for equilibrium are: vertical force componentshorizontal force componentstorquescoscos0sinsin0cos0.lrrTTmgTTmgx T Lθφθφ+−=−+=−=The origin was chosen to be at the left end of the bar for purposes of calculating the torque. The unknown quantities are Tl,Tr, and x. We want to eliminate Tland Tr, then solve for x. The second equation yields Tl= Trsin /sin and when this is substituted
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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.