Find minimum force to tip block

[vu10758]

A force F applied at point A is just large enough to tip over a block of mass M. The block will rotate about its bottom right corner. Find the magnitude of F.

The picture for this is problem 13 at this link http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460084048

The know that torque is force x radius, and the radius here is L.
The correct answer is MgL/2H, so I suspect that gravity somehow plays a role in this. However, I don't know what to do. How is this horizontal force F be related to gravity? Where does the 2H come from?

[ponjavic]

A force F applied at point A is just large enough to tip over a block of mass M. The block will rotate about its bottom right corner. Find the magnitude of F.

The picture for this is problem 13 at this link http://viewmorepics.myspace.com/index.cfm?fuseaction=viewImage&friendID=128765607&imageID=1460084048

The know that torque is force x radius, and the radius here is L.
The correct answer is MgL/2H, so I suspect that gravity somehow plays a role in this. However, I don't know what to do. How is this horizontal force F be related to gravity? Where does the 2H come from?
Could you draw where the gravity acts on the block? And thus make a FBD

[vu10758]

I would have mg pointing down from the center of the block and N pointing up. I would also have friction pointing to the left. The distance between the center and the rotational axis is H. However, why do we multiply this mg by L when L is the horizontal distance. Does this have anything to do with friction, but I don't see mu in the answer. I know that normal force is equal to mg, but I don't know if that will help. When looking at torque, shouldn't I have F*L = mg*H and then have F = mg*H/L. However, this is not right.

[ponjavic]

Because what you are interested is the perpendicular distance. The vector L is perpendicular to the force mg, thus the torque mg applies is mg*L/2

What is the torque that the force F creates? force * perpendicular distance

[vu10758]

Oh I see now. Thanks very much.