One end of a meter stick is pinned to a table, so the stick can rotate freely in
ID: 1352950 • Letter: O
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 56.5o 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
let L be the length of the stick
L = 1m
torque due to the first force F1=2N is = F1*L = 2 Nm
torque due to the second force = F2*l*sin29.9 = 6*l*sin29.9 = 2.99*l Nm
given net torque = 0
2 + 2.99*l = 0
l = -2/2.99 = -0.67 m
the distance of the second force from the pinned end = l = 0.67 m <---answer
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