Static Equilibrium

[Quipzley]

A meter stick blances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g coins stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

I know it is going to be having equal torques. The torque on one side should be .01 kg*9.8 m/s^2* .335 m = .03283 N*m + the torque of the .335 m of meter stick. The other side is just the torque of the .455 m of stick. I'm not sure how to calculate the torque of the meter stick.

[Quipzley]

The answer is supposed to be 74.4 g.

[arizonian]

Pick a point, any point, and sum moments to zero. Define your variables; mass of stick and coins, ms and mc; forces, a, b, and g. Solve for ms.

[jdstokes]

A meter stick blances horizontally on a knife-edge at the 50.0 cm mark. With two 5.00 g coins stacked over the 12.0 cm mark, the stick is found to balance at the 45.5 cm mark. What is the mass of the meter stick?

I know it is going to be having equal torques. The torque on one side should be .01 kg*9.8 m/s^2* .335 m = .03283 N*m + the torque of the .335 m of meter stick. The other side is just the torque of the .455 m of stick. I'm not sure how to calculate the torque of the meter stick.

The stick's centre of mass experiences a weight force mg, (50.0 - 45.5) cm from the knife-edge, producing a torque of mg(0.045 m). This torque is balanced by the torque of the coins (0.01000 kg)g(0.335 m) = mg(0.045 m). Therefore m = 74.4 g.