Challenging +1.2 This is a standard two-rod statics problem requiring moments about two points, resolution of forces, and friction analysis. While it involves multiple steps and careful bookkeeping of forces at the hinge, the techniques are routine for Further Maths mechanics: take moments about A for each rod separately, resolve forces, then apply friction conditions. The geometry is straightforward (right angle at A), and the method is well-practiced in FM courses.
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\includegraphics[max width=\textwidth, alt={}, center]{3daca234-9b7f-41d4-bbaa-d35615a120fc-2_419_1102_1859_520}
The diagram shows two uniform rods \(B A\) and \(A C\), smoothly hinged at \(A\). The rod \(B A\) has length \(8 a\) and weight \(W\); the rod \(A C\) has length \(6 a\) and weight \(2 W\). The rods are in equilibrium in a vertical plane with \(B\) and \(C\) resting on a rough horizontal floor and angle \(C A B\) equal to \(90 ^ { \circ }\). Show that the normal contact force at \(B\) is \(\frac { 26 } { 25 } W\).
The coefficient of friction between each rod and the floor is \(\mu\). Find the least possible value of \(\mu\).
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\includegraphics[max width=\textwidth, alt={}, center]{3daca234-9b7f-41d4-bbaa-d35615a120fc-2_419_1102_1859_520}
The diagram shows two uniform rods $B A$ and $A C$, smoothly hinged at $A$. The rod $B A$ has length $8 a$ and weight $W$; the rod $A C$ has length $6 a$ and weight $2 W$. The rods are in equilibrium in a vertical plane with $B$ and $C$ resting on a rough horizontal floor and angle $C A B$ equal to $90 ^ { \circ }$. Show that the normal contact force at $B$ is $\frac { 26 } { 25 } W$.
The coefficient of friction between each rod and the floor is $\mu$. Find the least possible value of $\mu$.
\hfill \mbox{\textit{CAIE FP2 2011 Q3 [12]}}