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1989 Paper 2 Q12
D: 1600.0 B: 1500.0
centre of mass lamina rectangular equilibrium smooth pegs trigonometric identity rigid body geometric proof inequality

A uniform rectangular lamina of sides \(2a\) and \(2b\) rests in a vertical plane. It is supported in equilibrium by two smooth pegs fixed in the same horizontal plane, a distance \(d\) apart, so that one corner of the lamina is below the level of the pegs. Show that if the distance between this (lowest) corner and the peg upon which the side of length \(2a\) rests is less than \(a\), then the distance between this corner and the other peg is less than \(b\). Show also that \[ b\cos\theta-a\sin\theta=d\cos2\theta, \] where \(\theta\) is the acute angle which the sides of length \(2b\) make with the horizontal.

Solution:

TikZ diagram
We must have \(G\) between the two pegs (vertically), otherwise we will induce a moment. Considering moments about the peg, if the second peg is outside the centre then we must induce a moment and therefore we cannot be in equilibrium. \begin{align*} \text{N2}(\nearrow):&& 0 &= R_1-mg\sin\theta \\ \text{N2}(\nwarrow):&&0&= R_2-mg \cos \theta \\ \Rightarrow && R_1 &= mg \sin\theta \\ && R_2 &= mg \cos\theta \\ \\ \overset{\curvearrowleft}{G}: && 0 &= R_1(a-d\sin\theta) -R_2(b-d \cos\theta) \\ \Rightarrow && 0&= a \sin\theta -d \sin^2\theta - b\cos \theta+d \cos^2 \theta \\ \Rightarrow && b \cos \theta - a \sin \theta &= d \cos 2 \theta \end{align*}

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