# Physics Balanced Torque Problem?

I've been absent from my physics class for two weeks, and I'm without a textbook. Can someone please help with this? If a yardstick is balanced at the center, (18 inches), and a .1 pound mass is positioned at the 4 inch mark, and a .3 pound mass is positioned at the 6 inch mark, where should a .3 pound mass be positioned for the yardstick to be balanced and perfectly straight across?

Relevance

When torque is generated by a force applied perpendicular to the beam, things are simpler than when an angle is involved. Torque is generated by a force F being applied at a point on the beam that is distance d from the pivot.

The general torque formula is

T = F*d*sin(theta)

where theta is the angle (0 to 90 degrees) between the force and the beam. In this case theta is 90, that's why I said this case is easier than the general case.

The key to achieving balance is for clockwise (CW) torque to equal CCW torque. A CCW torque can be given a negative sign. Assuming the inches increase from left to right, in this case we can write

Tcw = Tccw

.3*d = -.1*14 inch-pounds - .3*12 inch-pounds

d = -1.4 - 3.6 inch-pound/.3 pounds = -5/.3 inches

d = -16.7 inches

Since the pivot is 18 inches, the 3rd weight will need to be at the 34.7 inch mark.