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 57.0o 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

given data

F1=2N

F2=6N

angle= 57deg

from rotational dynamics

You just need to resolve the force applied into its rectangular components. The force applied perpendicular to the stick is F * sin57 = 6* 0.8386 N = 5.0320N

Now the torques are balanced.

Torque 1:
F1 = 2 N
d = 1 m (Since it is a meter scale)
Torque = 2 Nm

Torque 2:
F2 = 5.0320 N
d = ?
Torque = 2Nm

Therefore
2= 5.0320 * d
d = 0.3974 m

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