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1990 Paper 2 Q13
D: 1600.0 B: 1484.0
centre of mass non-uniform rod equilibrium string over peg triangle geometry inequality statics suspended body

A thin non-uniform rod \(PQ\) of length \(2a\) has its centre of gravity a distance \(a+d\) from \(P\). It hangs (not vertically) in equilibrium suspended from a small smooth peg \(O\) by means of a light inextensible string of length \(2b\) which passes over the peg and is attached at its ends to \(P\) and \(Q\). Express \(OP\) and \(OQ\) in terms of \(a,b\) and \(d\). By considering the angle \(POQ\), or otherwise, show that \(d < a^{2}/b\).

Solution:

TikZ diagram
Resolving horizontally, it's clear that \(\angle POG = \angle GOQ\), in particular applying the sine rule: \begin{align*} && \sin \angle POG &= \frac{a+d}{2b-x} \sin \angle PGO \\ && \sin \angle GOP &= \frac{a-d}{x} \sin \angle OGQ \\ \Rightarrow && \frac{a+d}{2b-x} &= \frac{a-d}{x} \\ \Rightarrow && x(a+d) &= (2b-x)(a-d) \\ \Rightarrow && 2ax &= 2b(a-d) \\ \Rightarrow && x &= b - \frac{db}{a} \\ \Rightarrow && PO &= b+\frac{db}{a} \\ && OQ &= b - \frac{d}{a} \end{align*} Applying the cosine rule: \begin{align*} && \cos POQ &= \frac{(b + \frac{db}{a})^2 + (b - \frac{db}{a})^2 -4a^2}{2(b^2 - \frac{d^2b^2}{a^2})} \\ &&&= \frac{2b^2 + \frac{2d^2b^2}{a^2}-4a^2}{2(b^2 - \frac{d^2b^2}{a^2})} \\ &&&= \frac{2a^2b^2 + 2d^2b^2-4a^4}{2b^2(a^2 - d^2)} \\ &&&< 1 \\ \Leftrightarrow && 2a^2b^2 + 2d^2b^2-4a^4 &< 2b^2(a^2-d^2) \\ \Leftrightarrow && 2d^2b^2-4a^4 &< -2b^2d^2 \\ \Leftrightarrow && 4d^2b^2&< 4a^4 \\ \Leftrightarrow && d^2&< \frac{a^4}{b^2} \\ \Leftrightarrow && d&< \frac{a^2}{b} \\ \end{align*}

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