Problem 10.103 Rod AB is attached to a block 160 N at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of q corresponding to equilibrium. 160 N 200 mm A B q C 800 N 200 mm D 100 mm

Solving Problems on Your Own Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of q corresponding to equilibrium. Problem 10.103 160 N 800 N 200 mm 100 mm A B C D q 1. Apply the principle of virtual work. If a system of connected rigid bodies is in equilibrium, the total virtual work of the external forces applied to the system is zero for any virtual displacement of the system. 1a. Define a virtual displacement. By using a single variable, define a virtual displacement of all the forces that do work.

Solving Problems on Your Own Rod AB is attached to a block at A that can slide freely in the vertical slot shown. Neglecting the effect of friction and the weights of the rods, determine the value of q corresponding to equilibrium. Problem 10.103 160 N 800 N 200 mm 100 mm A B C D q 1b. Express the total virtual work done by the forces and couples during the virtual displacement. dU = F . d r and dU = M dq where dU is the virtual work, F is a force undergoing a virtual displacement d r, and M is a couple undergoing a virtual rotation dq. 1c. Set the virtual work to zero and solve for the variable.

yA = 400 sin q dyA = 400 cos q dq xD = 100 cos q dxD = _ 100 sin q dq Problem 10.103 Solution 160 N 800 N 200 mm 100 mm A B C D q Define a virtual displacement. 160 N 800 N A B C D q 200 sin q yA xD yA = 400 sin q dyA = 400 cos q dq xD = 100 cos q dxD = _ 100 sin q dq

_ ( 160 )( 400 cos q d q ) _ (800)(_ 100 sinq dq) = 0 A B C D q 200 sin q yA xD Problem 10.103 Solution Express the total virtual work done by the forces. Set the virtual work to zero and solve for the variable. Virtual work: dU = _ ( 160 N ) d yA _ ( 800 N ) dxD = 0 _ ( 160 )( 400 cos q d q ) _ (800)(_ 100 sinq dq) = 0 = 0.8 ; tan q = 0.8 q = 38.7o sin q cos q