A cube of uniform density \(\rho\) is placed on a horizontal plane and a second cube, also of uniform density \(\rho\), is placed on top of it. The lower cube has side length \(1\) and the upper cube has side length \(a\), with \(a \leqslant 1\). The centre of mass of the upper cube is vertically above the centre of mass of the lower cube and all the edges of the upper cube are parallel to the corresponding edges of the lower cube. The contacts between the two cubes, and between the lower cube and the plane, are rough, with the same coefficient of friction \(\mu < 1\) in each case. The midpoint of the base of the upper cube is \(X\) and the midpoint of the base of the lower cube is \(Y\).
A horizontal force \(P\) is exerted, perpendicular to one of the vertical faces of the upper cube, at a point halfway between the two vertical edges of this face, and a distance \(h\), with \(h < a\), above the lower edge of this face.
- Show that, if the two cubes remain in equilibrium, the normal reaction of the plane on the lower cube acts at a point which is a distance
\[\frac{P(1+h)}{(1+a^3)\rho g}\]
from \(Y\), and find a similar expression for the distance from \(X\) of the point at which the normal reaction of the lower cube on the upper cube acts.
The force \(P\) is now gradually increased from zero.
- Show that, if neither cube topples, equilibrium will be broken by the slipping of the upper cube on the lower cube, and not by the slipping of the lower cube on the ground.
- Show that, if \(a = 1\), then equilibrium will be broken by the slipping of the upper cube on the lower cube if \(\mu(1+h) < 1\) and by the toppling of the lower and upper cube together if \(\mu(1+h) > 1\).
- Show that, in a situation where \(a < 1\) and \(h\bigl(1 + a^3(1-a)\bigr) > a^4\), and no slipping occurs, equilibrium will be broken by the toppling of the upper cube.
- Show, by considering \(a = \frac{1}{2}\) and choosing suitable values of \(h\) and \(\mu\), that the situation described in (iv) can in fact occur.