- #1

Ghost Repeater

- 32

- 5

## Homework Statement

A block of uniform density experiences a force F to the right, applied 5/3 m from the bottom of the block. The block is 2 m high and 1 m wide. Take the pivot point to be the point at the bottom right of the block. Find the value of the force that is just able to tip the box.

## Homework Equations

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torque = Frsin(theta)

## The Attempt at a Solution

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The force F is the only horizontal force. There are two vertical forces, the normal force and the weight of the block. I take these both to act at the block's CM, which is its geometric center. I calculate torques and get an answer of F is approximately 29.6 N.

However, the solution in the book gets 30 N, but by a very different method that doesn't seem right to me. First, it doesn't account for the normal force at all. How can this be? Don't we take the normal force to act at the CM? If that's so, then if the pivot is the bottom right corner, doesn't the normal force have a nonzero lever arm, and therefore doesn't it exert a torque?

Second, the book solution takes the lever arm for the gravitational torque to be simply .5 m, as if the gravity force were acting midway along the bottom of the block. This doesn't seem right either. Shouldn't the lever arm for the gravity force in this case be the magnitude of the vector from the pivot to the CM, scaled by the angle between that vector and the gravity vector?

My main question here is whether the normal force exerts a torque and if not, why not.

Thanks.