A uniform cube with edge a rests on a horizontal plane whose friction coefficient equals k. The cube is set in motion with an intial velocity, travels some distance over the plane and comes to a stand still. Explain the disappearance of the angular momentum of the cube relative to the axis lying in the plane at right angles to the cube's motion direction. Find the distance between the resultants of gravitational forces and the reaction forces exerted by the supporting plane.

Explanation

When the cube is given an initial velocity on the table in some direction (as shown) it acquires an angular momentum about an axis on the table perpendicular to the initial velocity and (say) just below the C.G. This angular momentum will dissappear when the cube stops and this can only by due to a torque. Frictional forces cannot do this by themselves because they act in the plain containing the axis. But if the force of normal reaction act eccentrically (as shown), their torque can bring about the vanishing of the angular momentum. We can calculate the distance `Deltax` between the point of application of the normal reaction and the C.G. of the cube as follows. Take the moment about C.G. of all the forces. This must vanish because the cube does not turn or turnble on the table.
Then if the force of friction is `f r`
`f r a/2=NDeltax`
But `N=mg` and `f r=kmg`, so
`Deltax=ka//2`