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The diagram shows the cross-section through the centre of mass of a uniform solid prism. The cross-section is a right-angled triangle ABC , with AB perpendicular to AC , which lies in a vertical plane. The length of AB is 3 cm , and the length of AC is 12 cm .
The prism is resting in equilibrium on a horizontal surface and against a vertical wall. The side AC of the prism makes an angle \(\theta\) with the horizontal.
A horizontal force of magnitude \(P \mathrm {~N}\) is now applied to the prism at B . This force acts towards the wall in the vertical plane which passes through the centre of mass G of the prism and is perpendicular to the wall.
The weight of the prism is 15 N and the coefficients of friction between the prism and the surface, and between the prism and the wall, are each \(\frac { 1 } { 2 }\).
- Show that the least value of \(P\) needed to move the prism is given by
$$P = \frac { 40 \cos \theta + 95 \sin \theta } { 16 \sin \theta - 13 \cos \theta } .$$
- Determine the range in which the value of \(\theta\) must lie.