| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with rough contact at free end |
| Difficulty | Standard +0.3 This is a standard M2 moments problem with clear geometry and straightforward application of equilibrium conditions. While it requires resolving forces, taking moments, and using limiting equilibrium (F=μR), all steps follow routine procedures with no conceptual surprises. The given sin θ = 3/5 simplifies calculations, making this slightly easier than average for M2. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(18(10) - T(20\sin\theta) + 3(6) = 0\) | M1 | Moments about \(P\) |
| A1 | Need a value for \(\sin\theta\) or \(\theta\) | |
| \(T = 16.5\) N | A1 | Exact |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(X = T\cos\theta\) | B1ft | ft candidates value of \(T\). Resolve horizontally (\(X = 13.2\) N) or moments; Need a value for \(\cos\theta\) or \(\theta\) |
| \(Y + T\sin\theta - 18 - 3 = 0\) | M1 | Resolve vertically or moments |
| A1ft | ft candidates value of \(T\). \(Y = 11.1\) N; Need a value for \(\sin\theta\) or \(\theta\) | |
| \(R = \sqrt{13.2^2 + 11.1^2} = 17.2\) N | A1 | \(R = 17.2467\ldots\) |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mu = cv(\vert Y\vert)/cv(\vert X\vert) = 11.1/13.2\) | M1 | Use of \(Fr = \mu R\) |
| \(\mu = 0.841\) | A1 | \(\mu = 0.8409\ldots\); allow \(\frac{37}{44}\) |
| [2] |
## Question 4:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $18(10) - T(20\sin\theta) + 3(6) = 0$ | M1 | Moments about $P$ |
| | A1 | Need a value for $\sin\theta$ or $\theta$ |
| $T = 16.5$ N | A1 | Exact |
| **[3]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $X = T\cos\theta$ | B1ft | ft candidates value of $T$. Resolve horizontally ($X = 13.2$ N) or moments; Need a value for $\cos\theta$ or $\theta$ |
| $Y + T\sin\theta - 18 - 3 = 0$ | M1 | Resolve vertically or moments |
| | A1ft | ft candidates value of $T$. $Y = 11.1$ N; Need a value for $\sin\theta$ or $\theta$ |
| $R = \sqrt{13.2^2 + 11.1^2} = 17.2$ N | A1 | $R = 17.2467\ldots$ |
| **[4]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mu = cv(\vert Y\vert)/cv(\vert X\vert) = 11.1/13.2$ | M1 | Use of $Fr = \mu R$ |
| $\mu = 0.841$ | A1 | $\mu = 0.8409\ldots$; allow $\frac{37}{44}$ |
| **[2]** | | |
---4
A uniform rod $P Q$ has weight 18 N and length 20 cm . The end $P$ rests against a rough vertical wall. A particle of weight 3 N is attached to the rod at a point 6 cm from $P$. The rod is held in a horizontal position, perpendicular to the wall, by a light inextensible string attached to the rod at $Q$ and to a point $R$ on the wall vertically above $P$, as shown in the diagram. The string is inclined at an angle $\theta$ to the horizontal, where $\sin \theta = \frac { 3 } { 5 }$. The system is in limiting equilibrium.\\
(i) Find the tension in the string.\\
(ii) Find the magnitude of the force exerted by the wall on the rod.\\
(iii) Find the coefficient of friction between the wall and the rod.
\hfill \mbox{\textit{OCR M2 2014 Q4 [9]}}