| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Lamina hinged at point with string support |
| Difficulty | Standard +0.3 This is a standard moments equilibrium problem requiring knowledge that the center of mass of a quadrant is at distance 4r/(3π) from the center, Pythagoras to find AG, and taking moments about the hinge. While it involves multiple steps and the quadrant formula, it follows a routine template for lamina equilibrium problems that M2 students practice extensively. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(OG = 2 \times 0.8\sin(\pi/4)/(3\pi/4)\) \((0.48016\ldots\text{m})\) | B1 | |
| \(AG^2 = (0.8\sin 45)^2 + (0.8\cos 45 - OG)^2\) OR \(AG^2 = 0.8^2 + OG^2 - 2 \times 0.8 \times OG\cos 45\) | M1 | Uses Pythagoras's Theorem OR the cosine formula. |
| \(AG = 0.572(11\ldots)\text{ m}\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan BAG = (0.8\cos 45 - OG)/(0.8\sin 45)\) | M1 | Uses trigonometry to find angle \(BAG\). |
| \(BAG = 8.5965^\circ = 8.6(0)^\circ\) | A1 | |
| \(W \times AG = 12 \times 2 \times 0.8\sin 45 \times \sin BAG\) | M1 | Takes moments about \(A\). |
| \(0.572W = 12 \times 2 \times 0.8\sin 45 \times \sin 8.6\) | A1FT | |
| \(W = 3.55\text{ N}\) | A1 | |
| Total | 5 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $OG = 2 \times 0.8\sin(\pi/4)/(3\pi/4)$ $(0.48016\ldots\text{m})$ | B1 | |
| $AG^2 = (0.8\sin 45)^2 + (0.8\cos 45 - OG)^2$ OR $AG^2 = 0.8^2 + OG^2 - 2 \times 0.8 \times OG\cos 45$ | M1 | Uses Pythagoras's Theorem OR the cosine formula. |
| $AG = 0.572(11\ldots)\text{ m}$ | A1 | |
| **Total** | **3** | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan BAG = (0.8\cos 45 - OG)/(0.8\sin 45)$ | M1 | Uses trigonometry to find angle $BAG$. |
| $BAG = 8.5965^\circ = 8.6(0)^\circ$ | A1 | |
| $W \times AG = 12 \times 2 \times 0.8\sin 45 \times \sin BAG$ | M1 | Takes moments about $A$. |
| $0.572W = 12 \times 2 \times 0.8\sin 45 \times \sin 8.6$ | A1FT | |
| $W = 3.55\text{ N}$ | A1 | |
| **Total** | **5** | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{6b220343-1d64-4dbc-a42d-77967eef9c6d-08_449_890_262_630}\\
$O A B$ is a uniform lamina in the shape of a quadrant of a circle with centre $O$ and radius 0.8 m which has its centre of mass at $G$. The lamina is smoothly hinged at $A$ to a fixed point and is free to rotate in a vertical plane. A horizontal force of magnitude 12 N acting in the plane of the lamina is applied to the lamina at $B$. The lamina is in equilibrium with $A G$ horizontal (see diagram).\\
(i) Calculate the length $A G$.\\
(ii) Find the weight of the lamina.\\
\hfill \mbox{\textit{CAIE M2 2017 Q5 [8]}}