| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2010 |
| Session | June |
| Marks | 4 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Pulley systems |
| Type | Two particles over pulley, vertical strings |
| Difficulty | Moderate -0.8 This is a straightforward statics problem requiring basic resolution of forces and understanding of tension in ropes over pulleys. Part (i) involves simple trigonometry to resolve the tension into components, and part (ii) is a standard two-particle equilibrium setup where the tension equals half the weight. Both parts are routine applications of fundamental mechanics principles with no problem-solving insight required. |
| Spec | 3.03b Newton's first law: equilibrium3.03k Connected particles: pulleys and equilibrium3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolving | M1 | Resolving in at least 1 of horiz or vert. Accept \(\sin \leftrightarrow \cos\). No extra terms. |
| \(\leftarrow 250\sin 70 = 234.92...\) so \(235 \text{ N}\) (3 s.f.) | A1 | Either both expressions correct (neglect direction) or one correct in correct direction |
| \(\uparrow 250\cos 70 = 85.5050...\) so \(85.5 \text{ N}\) (3 s.f.) | A1 | cao. Both evaluated and directions correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(250 \div 2 = 125 \text{ N}\) | B1 | Accept \(125g\) only if tension taken to be \(250g\) in (i) |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolving | M1 | Resolving in at least 1 of horiz or vert. Accept $\sin \leftrightarrow \cos$. No extra terms. |
| $\leftarrow 250\sin 70 = 234.92...$ so $235 \text{ N}$ (3 s.f.) | A1 | Either both expressions correct (neglect direction) or one correct in correct direction |
| $\uparrow 250\cos 70 = 85.5050...$ so $85.5 \text{ N}$ (3 s.f.) | A1 | cao. Both evaluated and directions correct |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $250 \div 2 = 125 \text{ N}$ | B1 | Accept $125g$ only if tension taken to be $250g$ in (i) |
---2 Fig. 2 shows a sack of rice of weight 250 N hanging in equilibrium supported by a light rope AB . End A of the rope is attached to the sack. The rope passes over a small smooth fixed pulley.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6cca1e5e-82b0-487d-8048-b9db7745dea6-2_458_479_705_833}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
Initially, end B of the rope is attached to a vertical wall as shown in Fig. 2.\\
(i) Calculate the horizontal and the vertical forces acting on the wall due to the rope.
End B of the rope is now detached from the wall and attached instead to the top of the sack. The sack is in equilibrium with both sections of the rope vertical.\\
(ii) Calculate the tension in the rope.
\hfill \mbox{\textit{OCR MEI M1 2010 Q2 [4]}}