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2016 Paper 2 Q10
D: 1600.0 B: 1516.0
centre of mass friction equilibrium isosceles triangle wire peg inequality rigid body

A thin uniform wire is bent into the shape of an isosceles triangle \(ABC\), where \(AB\) and \(AC\) are of equal length and the angle at \(A\) is \(2\theta\). The triangle \(ABC\) hangs on a small rough horizontal peg with the side \(BC\) resting on the peg. The coefficient of friction between the wire and the peg is \(\mu\). The plane containing \(ABC\) is vertical. Show that the triangle can rest in equilibrium with the peg in contact with any point on \(BC\) provided \[ \mu \ge 2\tan\theta(1+\sin\theta) \,. \]

Solution:

TikZ diagram
Clearly the centre of mass will lie on the perpendicular from \(A\). We can also consider each side's wire as equivalent to a point mass at the centre of the side with mass proportional to the length of the side. Recalling that \(b = c\) (the triangle is isoceles we must have (for the \(y\)-coordinate \begin{align*} && a \cdot 0 + b \cdot \frac12 b \cos \theta + c \cdot \frac12 c \cos \theta &= (a+b+c) \overline{y} \\ \Rightarrow && b^2 \cos \theta &= (2b + 2b\sin \theta) \overline{y} \\ \Rightarrow && \overline{y} &= \frac{b \cos \theta}{2(1+\sin \theta)} \end{align*}
TikZ diagram
\begin{align*} \text{N2}(\nearrow): && R - mg \cos \phi &= 0 \\ \text{N2}(\nwarrow): && F -mg \sin \phi &= 0 \\ \Rightarrow && F &\leq \mu R \\ \Rightarrow && \sin \phi &\leq \mu \cos \phi \\ \Rightarrow && \tan \phi &\leq \mu \end{align*} When the peg is at \(C\) \begin{align*} \tan \phi &= \frac{CM}{MG} \\ &= \frac{b\sin \theta}{\frac{b \cos \theta}{2(1+\sin \theta)}} \\ &= 2 \tan \theta(1+\sin \theta) \end{align*} Therefore \(2 \tan \theta(1+\sin \theta) \leq \mu\) as required.

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