Problems

A wedge of mass \(km\) has the shape (in cross-section) of a right-angled triangle. It stands on a smooth horizontal surface with one face vertical. The inclined face makes an angle \(\theta\) with the horizontal surface. A particle \(P\), of mass \(m\), is placed on the inclined face and released from rest. The horizontal face of the wedge is smooth, but the inclined face is rough and the coefficient of friction between \(P\) and this face is \(\mu\).

1. When \(P\) is released, it slides down the inclined plane at an acceleration \(a\) relative to the wedge. Show that the acceleration of the wedge is \[ \frac {a \cos\theta}{k+1}\,. \] To a stationary observer, \(P\) appears to descend along a straight line inclined at an angle~\(45^\circ\) to the horizontal. Show that \[ \tan\theta = \frac k {k+1}\,. \] In the case \(k=3\), find an expression for \(a\) in terms of \(g\) and \(\mu\).
2. What happens when \(P\) is released if \(\tan\theta \le \mu\)?

A wedge of mass \(M\) rests on a smooth horizontal surface. The face of the wedge is a smooth plane inclined at an angle \(\alpha\) to the horizontal. A particle of mass \(m\) slides down the face of the wedge, starting from rest. At a later time \(t\), the speed \(V\) of the wedge, the speed \(v\) of the particle and the angle \(\beta\) of the velocity of the particle below the horizontal are as shown in the diagram.

1. \(V\sin\alpha=v\sin(\beta-\alpha)\);
2. \(\tan\beta=(1+m/M)\tan\alpha\);
3. \(2gy=v^2(M+m\cos^2\beta)/M\).