One end of a meter stick is pinned to a table, so the stick can rotate freely in

Question

One end of a meter stick is pinned to a table, so the stick can rotate freely in a plane parallel to the tabletop. Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. The first force has a magnitude of 2.00 N and is applied perpendicular to the length of the stick at the free end. The second force has a magnitude of 6.00 N and acts at a 60.1o angle with respect to the length of the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the end of the stick that is pinned.

Explanation / Answer

Sum of torques = 0

F1 * x1 = F2 * x
2 * 1 = 6 * sin(60.1) * x

x = (2*1)/(6*0.866) = 0.384 m   

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