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2004 Paper 1 Q11
D: 1500.0 B: 1500.0
moments ladder friction equilibrium hinged two ladders limiting friction rigid body

Two uniform ladders \(AB\) and \(BC\) of equal length are hinged smoothly at \(B\). The weight of \(AB\) is \(W\) and the weight of \(BC\) is \(4W \). The ladders stand on rough horizontal ground with \(\angle ABC=60^\circ\,\). The coefficient of friction between each ladder and the ground is \(\mu\). A decorator of weight \(7W\) begins to climb the ladder \(AB\) slowly. When she has climbed up \(\frac13\) of the ladder, one of the ladders slips. Which ladder slips, and what is the value of \(\mu\)?

Solution:

TikZ diagram
\begin{align*} \text{N2}(\rightarrow): && F_A - F_C &= 0\\ && F_A &= F_C \\ \text{N2}(\uparrow): && R_A + R_C - 7W - W - 4W &= 0\\ && R_A + R_C &= 12W \\ \overset{\curvearrowright}{A}: && \frac{1}{6}7W + \frac{1}{4}W + \frac{3}{4}4W - R_C &= 0 \\ \Rightarrow && R_C &= \frac{53}{12}W\\ \Rightarrow && R_A = 12W - \frac{53}{12}W &= \frac{91}{12}W \\ \overset{\curvearrowleft}{B}(AB): && \frac{1}{2}W + \frac{2}{3}7W - R_A+\sqrt{3}F_A &= 0 \\ \Rightarrow && F_A = \frac{1}{\sqrt{3}} \l \frac{91}{12}-\frac12-\frac{14}3\r W &= \frac{29}{12\sqrt{3}}W \end{align*} We know that the system is about to slip, so equality holds in one of \(F_A \leq \mu R_A\) or \(F_C \leq \mu R_C\). Since \(F_A = F_C\), we know it must occur for whichever of \(\mu R_A\) and \(\mu R_C\) is smaller. Since \(R_C\) is much smaller, this must be the ladder about to slip \(BC\) and \[ \mu = \frac{F_C}{R_C} = \frac{\frac{29}{12\sqrt{3}}W}{\frac{53}{12}W} = \boxed{\frac{29}{53\sqrt{3}}}\]

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