| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with string support |
| Difficulty | Standard +0.3 This is a standard M2 moments problem requiring resolution of forces and taking moments about a point. Students must handle the geometry of the 30° angle and apply equilibrium conditions, but the setup is straightforward with clear force directions and a single rigid body. The multi-part structure and geometric component add slight complexity beyond basic moments questions, but it follows a well-practiced method. |
| Spec | 3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T\cos 30 \times 1.5\sin 30 = 15g \times 2\) | M1 | Attempt at moments about \(A\), \(g\) can be omitted at this stage |
| \(T = 453\) | A1 | |
| A1 | ||
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X = T_c\sin 30\ (=226)\) | B1ft | Using their value \(T\) or taking moments about \(P\) |
| M1 | Attempt to resolve vertically or taking appropriate moments | |
| \(Y + T_c\cos 30 = 15g\) | A1ft | Using their \(T\); expect \(Y = -245\) or better |
| Either or both of these equations can be replaced with moments about an appropriate point e.g. \(P, Q, B\), c of m of beam | ||
| \(R = \sqrt{(226^2 + 245^2)}\) or \(\tan\theta = 245/226\) | M1 | Any relevant angle |
| \(R = 334\) | A1 | Allow \(333\) |
| \(\theta = 47.3°\) below horizontal (to the left) | A1 | Allow \(47.2\), \(42.7\) to the downward vertical |
| [6] | SC: If \(392\) in (i) leading to \(Y = \pm 245\) only in (ii) max M1A1 |
## Question 3:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\cos 30 \times 1.5\sin 30 = 15g \times 2$ | M1 | Attempt at moments about $A$, $g$ can be omitted at this stage |
| $T = 453$ | A1 | |
| | A1 | |
| **[3]** | | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X = T_c\sin 30\ (=226)$ | B1ft | Using their value $T$ or taking moments about $P$ |
| | M1 | Attempt to resolve vertically or taking appropriate moments |
| $Y + T_c\cos 30 = 15g$ | A1ft | Using their $T$; expect $Y = -245$ or better |
| | | Either or both of these equations can be replaced with moments about an appropriate point e.g. $P, Q, B$, c of m of beam |
| $R = \sqrt{(226^2 + 245^2)}$ or $\tan\theta = 245/226$ | M1 | Any relevant angle |
| $R = 334$ | A1 | Allow $333$ |
| $\theta = 47.3°$ below horizontal (to the left) | A1 | Allow $47.2$, $42.7$ to the downward vertical |
| **[6]** | | SC: If $392$ in (i) leading to $Y = \pm 245$ only in (ii) max M1A1 |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{d1eb99a1-04e5-43bc-87b4-d0f7c962135c-2_599_677_1151_696}
A uniform beam $A B$ of mass 15 kg and length 4 m is freely hinged to a vertical wall at $A$. The beam is held in equilibrium in a horizontal position by a light rod $P Q$ of length $1.5 \mathrm {~m} . P$ is fixed to the wall vertically below $A$ and $P Q$ makes an angle of $30 ^ { \circ }$ with the vertical (see diagram). The force exerted on the beam at $Q$ by the rod is in the direction $P Q$. Find\\
(i) the magnitude of the force exerted on the beam at $Q$,\\
(ii) the magnitude and direction of the force exerted on the beam at $A$.
\hfill \mbox{\textit{OCR M2 2012 Q3 [9]}}