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 (see the drawing). Two forces, both parallel to the tabletop, are applied to the stick in such a way that the net torque is zero. One force has a magnitude of 4.00 N and is applied perpendicular to the stick at the free end. The other force has a magnitude of 6.00 N and acts at an angle ? = 56.6°, measured with respect to the stick. Where along the stick is the 6.00-N force applied? Express this distance with respect to the axis of rotation.

Explanation / Answer

You just need to resolve the force applied into its rectangular components. The force applied perpendicular to the stick is F * sin60 = 6 * 0.866 N = 5.196 N Now the torques are balanced. Torque 1: F = 4 N d = 1 m (Since it is a meter scale) Torque = 4 Nm Torque 2: F = 5.196 N d = ? Torque = 4 Nm Therefore 4 = 5.196 * d d = 0.770 m

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