# Rotation of a block when not in equilibrium

Tags:
1. Apr 2, 2016

### Soren4

1. The problem statement, all variables and given/known data
Consider a block placed on a surface, in two different configuration, a and b. Explain the condition for which the mass is in equilibrium and describe qualitatively the rotation it follows when it falls.

2. Relevant equations
Center of mass theorem $\sum F = M a_{cm}$

3. The attempt at a solution
Of course the condition for the equilibrium is that the vertical line passing through the CM is inside the basis of the mass (situation a). In both the situation the weight $P$ equals the normal reaction force in magnitude $R$, but when the condition of equilibrium is not satisfied, weight $P$ and the normal reaction force $R$ becomes a force couple and they exert a torque, which makes the block rotate. The problem is: about which point does it rotate? On my book it says that the rotation is about point $B$ but I don't see how can this be possible, since, as said before, $P=R$, so $\sum F=0=M a_{cm}$. The center of mass does not accelerate, while it should do if the block would rotate about $B$ (centripetal acceleration would be needed in that case). How can that be?

2. Apr 2, 2016

### TSny

If this were to remain true in case B as the block rotates, would the CM ever change its vertical height?

If the block is to rotate about B, what other force besides R and P must act on the block?

3. Apr 2, 2016

### PeroK

Unless you lift the block off the ground, it can only rotate about point A or point B. That's a physical constraint. Imagine trying to move a wardrobe: it can only rotate about those two points and/or slide along the ground.

If you push from the right, it will rotate about A; and, if you push from the left it will rotate about point B.

4. Apr 2, 2016

### haruspex

Moreover, even before it starts to rotate they cannot be equal.

5. Apr 2, 2016

### TSny

Right, the instant it is released the normal force decreases.