| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform rod or wire suspended in equilibrium |
| Difficulty | Standard +0.3 This is a standard M2 moments problem using the standard result for the center of mass of a semicircular arc (2r/π from center). Part (i) requires resolving forces vertically when θ=0, and part (ii) requires taking moments about A when θ=30. Both are straightforward applications of equilibrium conditions with no novel insight required, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(x_G = (2 \times 2)/\pi\) | B1 | \(x_G = 1.2732\ldots\) May be seen in (ii), mark only once |
| \(P(\text{or } X) \times 4 = 0.3g \times x_G\) | *M1, A1ft | Take moments about \(A\) or \(B\); \(P = 0.9358\ldots\) ft their \(x_G\) |
| \(Y = 0.3g\) | B1 | |
| Use \(R^2 = X^2 + Y^2\) to find \(R\) | dep*M1 | |
| \(R = 3.09\) N | A1 | |
| [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(P \times 4 =\) | M1, A1 | Attempt at moments, force \(\times\) distance \(= 0.3g \times\) distance |
| \(0.3g \times (2\sin 30 + x_G\sin 60)\) | A1 | \(0.3g \times 2.1026\ldots\) |
| \(P = 1.55\) | A1 | \(1.545453\ldots\) |
| [4] |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $x_G = (2 \times 2)/\pi$ | B1 | $x_G = 1.2732\ldots$ May be seen in (ii), mark only once |
| $P(\text{or } X) \times 4 = 0.3g \times x_G$ | *M1, A1ft | Take moments about $A$ or $B$; $P = 0.9358\ldots$ ft their $x_G$ |
| $Y = 0.3g$ | B1 | |
| Use $R^2 = X^2 + Y^2$ to find $R$ | dep*M1 | |
| $R = 3.09$ N | A1 | |
| **[6]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $P \times 4 =$ | M1, A1 | Attempt at moments, force $\times$ distance $= 0.3g \times$ distance |
| $0.3g \times (2\sin 30 + x_G\sin 60)$ | A1 | $0.3g \times 2.1026\ldots$ |
| $P = 1.55$ | A1 | $1.545453\ldots$ |
| **[4]** | | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{74eaa61a-1507-4cef-8f97-5c1860bdc36a-2_542_638_1208_717}
A uniform semicircular arc $A C B$ is freely pivoted at $A$. The arc has mass 0.3 kg and is held in equilibrium by a force of magnitude $P$ N applied at $B$. The line of action of this force lies in the same plane as the arc, and is perpendicular to $A B$. The diameter $A B$ has length 4 cm and makes an angle of $\theta ^ { \circ }$ with the downward vertical (see diagram).\\
(i) Given that $\theta = 0$, find the magnitude of the force acting on the arc at $A$.\\
(ii) Given instead that $\theta = 30$, find the value of $P$.
\hfill \mbox{\textit{OCR M2 2013 Q3 [10]}}