Posted by **Nicole** on Tuesday, March 20, 2007 at 10:44pm.

Suppose you have a meter stick with two masses attached. The mass m1 is 0.60 kg and it is located at x1 = 30 cm from the left (zero) end of the meter stick. The pivot point is located at x = 45 cm. The mass of the meter stick (mms = 0.40 kg) is located at its geometric center, xms = 50 cm. The mass m2 is 0.25 kg and it is located at x2 = 80 cm from the left end of the meter stick. Calculate the net torque (in N⋅m with the proper sign) due to these three weights. Use g = 9.8 m/s2.

Mass m1 exerts a force m1*g at a distance of 15 cm (0.15 m) left of pivot point. That creates a counterclockwise torque of 0.6*9.8*0.15 N-m. The mass of the meter stick exerts a gravitational force ms*g through a piint 5 cm to the right of the pivot, creating a clockwise torque. Mass m2 exerts a force m2*g that is applied 35 cm to the right of the pivot.

All you have to do is compute the three torques and add them. Treat the counterclockwise torque as negative.

If you need more assistance with this one, please show your work.