Force Required to Tip a Block of wood?

[melancholy2]

Hi everyone,

Please refer to the attached diagram.

I was playing around with pulling tablecloths out from the bottom of plates and objects resting on it, and was getting good results. I also happened to have a cuboid wooden paperweight which was on the table cloth and I noticed that that wooden block would tip and stand up if I didn't pull hard enough (but enough to make it tip).

This got me thinking how I would calculate the force required to be exerted on the block in order to achieve this effect. I haven't been able to figure out anything yet, so I would like some inputs. I reformed the effect in the form of a question:

A block of wood with mass M, height h and length l is placed on a smooth frictionless surface. A force F is applied at the bottom left edge of the wooden block, causing it to pivot at that edge, counter-clockwise upwards till it is standing up. Find the minimum force required.

If one were to consider moments, where would one use as a pivot? If it were at the bottom edge, then F wouldn't generate a torque, and it wouldn't tip. Anywhere else on the cube and the pivot will also rise (will that cause any problems?).

 

[Doc Al]

Realize that as soon as you apply the force F, the block will accelerate. The easiest way to analyze the motion is from the view of the non-inertial frame co-moving with the block. In that frame you need to add a pseudoforce that acts at the block's center of mass. In effect you'll have two forces exerting a torque about the lower left hand corner (a fine pivot point)--figure out when one is great enough to overcome the other.

 

[melancholy2]

Realize that as soon as you apply the force F, the block will accelerate. The easiest way to analyze the motion is from the view of the non-inertial frame co-moving with the block. In that frame you need to add a pseudoforce that acts at the block's center of mass. In effect you'll have two forces exerting a torque about the lower left hand corner (a fine pivot point)--figure out when one is great enough to overcome the other.

Thanks for the help but I still can't figure out how much that imaginary force would be. Will taking the view of the non-inertial frame comoving with the block, would that also affect how I find force F (which I have to minus the accelerating force?). I have no clue where to place the pivot given that it will rise when the block starts tipping.

 

[Doc Al]

Viewed from a non-inertial frame that accelerates to the right, there will be an imaginary force to the left equal to ma. Since F is the accelerating force, the imaginary force will equal F. As I said before, use the lower left edge as your pivot.