| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Triangle of forces method |
| Difficulty | Moderate -0.3 This is a standard M1 statics problem using triangle of forces method with clear geometric setup (90° rack angle simplifies the force triangle). Part (i) is routine diagram drawing, part (ii) requires basic trigonometry on the force triangle, and part (iii) involves sketching and comparing trigonometric functions. The problem is slightly easier than average because the 90° angle makes the geometry straightforward and the method is explicitly specified in the question type. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Closed triangle with cycling arrows | B1 | Accept any consistent orientation |
| All forces labelled | B1 | |
| Correct angles; \(90°\) may be implied; \(\alpha\) may be shown between \(S\) and the horizontal | B1 | SC1 for force diagram with no extra forces and all labels and directions correct |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(R=W\cos\alpha\) | B1 | Allow FT for sin-cos interchange following wrong angle in triangle being marked \(\alpha\) in part (i) for both marks |
| \(S=W\sin\alpha\) | B1 | SC1 if both \(S\) and \(R\) are given negative signs |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Sketch graph of \(R\) against \(\alpha\) | B1 | Condone no explicit vertical scale. Do not accept straight lines |
| Correct sketch graph of \(S\) against \(\alpha\) | B1 | Must be consistent with graph of \(R\) |
| \(45°<\alpha\ (\leq90°)\) | B1 | Condone \(45°\leq\alpha\) |
| [3] |
# Question 3:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Closed triangle with cycling arrows | B1 | Accept any consistent orientation |
| All forces labelled | B1 | |
| Correct angles; $90°$ may be implied; $\alpha$ may be shown between $S$ and the horizontal | B1 | SC1 for force diagram with no extra forces and all labels and directions correct |
| **[3]** | | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $R=W\cos\alpha$ | B1 | Allow FT for sin-cos interchange following wrong angle in triangle being marked $\alpha$ in part (i) for both marks |
| $S=W\sin\alpha$ | B1 | SC1 if both $S$ and $R$ are given negative signs |
| **[2]** | | |
## Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Sketch graph of $R$ against $\alpha$ | B1 | Condone no explicit vertical scale. Do not accept straight lines |
| Correct sketch graph of $S$ against $\alpha$ | B1 | Must be consistent with graph of $R$ |
| $45°<\alpha\ (\leq90°)$ | B1 | Condone $45°\leq\alpha$ |
| **[3]** | | |
---3 Fig. 3 shows a smooth ball resting in a rack. The angle in the middle of the rack is $90 ^ { \circ }$. The rack has one edge at angle $\alpha$ to the horizontal.
The weight of the ball is $W \mathrm {~N}$. The reaction forces of the rack on the ball at the points of contact are $R \mathrm {~N}$ and $S \mathrm {~N}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{63a2dc41-5e8b-4275-8653-ece5067c4306-3_314_460_484_813}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}
(i) Draw a fully labelled triangle of forces to show the forces acting on the ball. Your diagram must indicate which angle is $\alpha$.\\
(ii) Find the values of $R$ and $S$ in terms of $W$ and $\alpha$.\\
(iii) On the same axes draw sketches of $R$ against $\alpha$ and $S$ against $\alpha$ for $0 ^ { \circ } \leqslant \alpha \leqslant 90 ^ { \circ }$.
For what values of $\alpha$ is $R < S$ ?
\hfill \mbox{\textit{OCR MEI M1 2014 Q3 [8]}}