Q:

If I have an equalateral triangle ABC, where C is on the center of mass of an object, and I apply a constant force to B, perpendicular to BC, and the same level of force to A, but perpendicular to AC (where both forces are facing the 'outside' of the triangle, then how does the mass behave? I would expect that the mass would accelerate away from the center of AB, as when you add the forces together, they give an upward result. On the other hand, the forces are perpendicular to the line to center of gravity, they would constitute torque. The two torques would cancel themselves out, meaning the system would not move. Which is correct, and why?

- John (age 30)

Canada

- John (age 30)

Canada

A:

As you say, the two torques about the center of mass would cancel out. That means that the angular momentum won't change. The two forces, however, don't cancel, as you also note. So the momentum does change. The momentum can change without contributing to the angular momentum about the original center of mass because the acceleration is along a fixed direction. So the center of mass travels in a straight line going through the original location. There's no tangential velocity.

Mike W.

*(published on 01/06/2015)*