CAIE FP2 2013 November — Question 4

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2013
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeCoplanar forces in equilibrium
DifficultyChallenging +1.8 This is a challenging mechanics problem requiring resolution of forces in two directions, taking moments about a strategic point, applying limiting friction conditions at two surfaces simultaneously, and solving a system of equations involving trigonometry. While the individual techniques are A-level standard, the combination of multiple equilibrium conditions, the geometric setup with tangential force, and the algebraic manipulation needed to reach the specific numerical results elevates this significantly above average difficulty. It requires careful systematic approach and is more demanding than typical statics problems, though not requiring novel mathematical insight beyond standard Further Maths mechanics methods.
Spec3.03m Equilibrium: sum of resolved forces = 03.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force

4 \includegraphics[max width=\textwidth, alt={}, center]{c1aae41e-530c-4db4-8959-8afe223c4dbc-3_563_572_258_785} A uniform circular disc, with centre \(O\) and weight \(W\), rests in equilibrium on a horizontal floor and against a vertical wall. The plane of the disc is vertical and perpendicular to the wall. The disc is in contact with the floor at \(A\) and with the wall at \(B\). A force of magnitude \(P\) acts tangentially on the disc at the point \(C\) on the edge of the disc, where the radius \(O C\) makes an angle \(\theta\) with the upward vertical, and \(\tan \theta = \frac { 4 } { 3 }\) (see diagram). The coefficient of friction between the disc and the floor and between the disc and the wall is \(\frac { 1 } { 2 }\). Show that the sum of the magnitudes of the frictional forces at \(A\) and \(B\) is equal to \(P\). Given that the equilibrium is limiting at both \(A\) and \(B\),
  1. show that \(P = \frac { 15 } { 34 } \mathrm {~W}\),
  2. find the ratio of the magnitude of the normal reaction at \(A\) to the magnitude of the normal reaction at \(B\).

4\\
\includegraphics[max width=\textwidth, alt={}, center]{c1aae41e-530c-4db4-8959-8afe223c4dbc-3_563_572_258_785}

A uniform circular disc, with centre $O$ and weight $W$, rests in equilibrium on a horizontal floor and against a vertical wall. The plane of the disc is vertical and perpendicular to the wall. The disc is in contact with the floor at $A$ and with the wall at $B$. A force of magnitude $P$ acts tangentially on the disc at the point $C$ on the edge of the disc, where the radius $O C$ makes an angle $\theta$ with the upward vertical, and $\tan \theta = \frac { 4 } { 3 }$ (see diagram). The coefficient of friction between the disc and the floor and between the disc and the wall is $\frac { 1 } { 2 }$. Show that the sum of the magnitudes of the frictional forces at $A$ and $B$ is equal to $P$.

Given that the equilibrium is limiting at both $A$ and $B$,\\
(i) show that $P = \frac { 15 } { 34 } \mathrm {~W}$,\\
(ii) find the ratio of the magnitude of the normal reaction at $A$ to the magnitude of the normal reaction at $B$.

\hfill \mbox{\textit{CAIE FP2 2013 Q4}}