Standard +0.3 This is a straightforward moments problem requiring students to identify the center of mass of a uniform triangular lamina (at 2/3 height from base), take moments about point A, and solve a single equation. The geometry is clearly specified and the calculation is direct with no conceptual surprises, making it slightly easier than average for A-level mechanics.
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\includegraphics[max width=\textwidth, alt={}, center]{9c82b387-8e5e-48b9-973d-5337b4e56a66-2_536_905_520_621}
A uniform lamina \(A B C\) in the shape of an isosceles triangle has weight 24 N . The perpendicular distance from \(A\) to \(B C\) is 12 cm . The lamina rests in a vertical plane in equilibrium, with the vertex \(A\) in contact with a horizontal surface. Angle \(B A C = 100 ^ { \circ }\) and \(A B\) makes an angle of \(10 ^ { \circ }\) with the horizontal. Equilibrium is maintained by a force of magnitude \(F \mathrm {~N}\) acting along \(B C\) (see diagram). Show that \(F = 8\).
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\includegraphics[max width=\textwidth, alt={}, center]{9c82b387-8e5e-48b9-973d-5337b4e56a66-2_536_905_520_621}
A uniform lamina $A B C$ in the shape of an isosceles triangle has weight 24 N . The perpendicular distance from $A$ to $B C$ is 12 cm . The lamina rests in a vertical plane in equilibrium, with the vertex $A$ in contact with a horizontal surface. Angle $B A C = 100 ^ { \circ }$ and $A B$ makes an angle of $10 ^ { \circ }$ with the horizontal. Equilibrium is maintained by a force of magnitude $F \mathrm {~N}$ acting along $B C$ (see diagram). Show that $F = 8$.
\hfill \mbox{\textit{CAIE M2 2014 Q2 [3]}}