| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Hemisphere or sphere resting on plane or wall |
| Difficulty | Challenging +1.8 This is a challenging Further Maths 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 setup is standard (disc against wall and floor), the tangential force at angle θ and the requirement to handle limiting equilibrium at both contacts elevates this beyond typical A-level mechanics questions. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F_A + F_B = P\) | M1 A1 | Take moments about \(O\) (\(a\) may be cancelled out) |
| \(R_A + R_B = W - P\sin\theta\) | M1 | Find 2 indep. moment or resolution eqns (M1 needs 2 eqns) |
| \(R_B - R_A = P\cos\theta\) | ||
| \(R_A = 2F_A\) and \(R_B = 2F_B\) | B1 | Relate \(R_A\), \(F_A\), \(R_B\) and \(F_B\) |
| \(2F_A + F_B = W - 4P/5\) | A1 | Substitute in above eqns |
| \(2F_B - F_A = 3P/5\) | A1 | |
| \([F_A = 7P/15 = 7W/34\), \(F_B = 8P/15 = 8W/34]\) | Solve for \(P\) | |
| \(P = 15W/34\) | M1 A1 | A.G. |
| Subtotal: 6 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R_A / R_B = F_A / F_B = 7/8\) | M1 A1 | Find \(R_A/R_B\) by any valid method |
| Subtotal: 2 marks | ||
| Total: 10 marks |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F_A + F_B = P$ | M1 A1 | Take moments about $O$ ($a$ may be cancelled out) |
| $R_A + R_B = W - P\sin\theta$ | M1 | Find 2 indep. moment or resolution eqns (M1 needs 2 eqns) |
| $R_B - R_A = P\cos\theta$ | | |
| $R_A = 2F_A$ and $R_B = 2F_B$ | B1 | Relate $R_A$, $F_A$, $R_B$ and $F_B$ |
| $2F_A + F_B = W - 4P/5$ | A1 | Substitute in above eqns |
| $2F_B - F_A = 3P/5$ | A1 | |
| $[F_A = 7P/15 = 7W/34$, $F_B = 8P/15 = 8W/34]$ | | Solve for $P$ |
| $P = 15W/34$ | M1 A1 | **A.G.** |
| **Subtotal: 6 marks** | | |
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_A / R_B = F_A / F_B = 7/8$ | M1 A1 | Find $R_A/R_B$ by any valid method |
| **Subtotal: 2 marks** | | |
| **Total: 10 marks** | | |
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\includegraphics[max width=\textwidth, alt={}, center]{b486decd-75b8-44bd-889f-2472f1163871-3_567_575_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 [10]}}