| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 18 |
| 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 equilibrium question with multiple straightforward parts: drawing a force triangle, resolving forces in two directions, and substituting values. Part (v) requires comparing tension to maximum available force (weight), which is a common textbook-style extension. All techniques are routine for M1 students with no novel problem-solving required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Triangle with three forces: tension in rope A (\(F\) N along rope), horizontal 25 N (rope B), weight 250 N downward | B1 B1 B1 | Fully labelled with correct directions forming closed triangle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal: \(F\sin\theta = 25\) | M1 | |
| Vertical: \(F\cos\theta = 250\) | M1 | |
| \(\tan\theta = \frac{25}{250} = 0.1 \Rightarrow \theta = 5.71°\) | A1 | |
| \(F = \frac{250}{\cos\theta} = \sqrt{250^2+25^2} = 251\) N | A1 | |
| Distance \(= 30\tan\theta = 30 \times \frac{1}{10} = 3\) m ✓ | A1 | Shown |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Vertical: \(S\cos\alpha = 250 + T\sin\beta\)... wait — Vertical: \(S\cos\alpha = 250\); Horizontal: \(S\sin\alpha = T\cos\beta\)... | Checking diagram: rope A at angle \(\alpha\) to vertical (tension \(S\)), rope B horizontal (tension \(T\)) | |
| Vertical: \(S\cos\alpha = 250\) | B1 | |
| Horizontal: \(S\sin\alpha = T\) | B1 B1 | Accept equivalent correct equations |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\alpha = 8.5°\), \(\beta = 35°\): \(S = \frac{250}{\cos 8.5°} = 252.8 \approx 253\) N | M1 A1 | |
| \(T = S\sin\alpha = 252.8 \times \sin 8.5° = 37.3\) N | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| When \(\alpha = 60°\): \(S = \frac{250}{\cos 60°} = 500\) N | M1 | |
| Abi must pull down with force 500 N, but her weight is \(40 \times 10 = 400\) N, which is less than 500 N | M1 | |
| She cannot exert a downward force greater than her own weight, so it is not possible | A1 |
# Question 7:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Triangle with three forces: tension in rope A ($F$ N along rope), horizontal 25 N (rope B), weight 250 N downward | B1 B1 B1 | Fully labelled with correct directions forming closed triangle |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal: $F\sin\theta = 25$ | M1 | |
| Vertical: $F\cos\theta = 250$ | M1 | |
| $\tan\theta = \frac{25}{250} = 0.1 \Rightarrow \theta = 5.71°$ | A1 | |
| $F = \frac{250}{\cos\theta} = \sqrt{250^2+25^2} = 251$ N | A1 | |
| Distance $= 30\tan\theta = 30 \times \frac{1}{10} = 3$ m ✓ | A1 | Shown |
## Part (iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical: $S\cos\alpha = 250 + T\sin\beta$... wait — Vertical: $S\cos\alpha = 250$; Horizontal: $S\sin\alpha = T\cos\beta$... | | Checking diagram: rope A at angle $\alpha$ to vertical (tension $S$), rope B horizontal (tension $T$) |
| Vertical: $S\cos\alpha = 250$ | B1 | |
| Horizontal: $S\sin\alpha = T$ | B1 B1 | Accept equivalent correct equations |
## Part (iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\alpha = 8.5°$, $\beta = 35°$: $S = \frac{250}{\cos 8.5°} = 252.8 \approx 253$ N | M1 A1 | |
| $T = S\sin\alpha = 252.8 \times \sin 8.5° = 37.3$ N | M1 A1 | |
## Part (v):
| Answer | Marks | Guidance |
|--------|-------|----------|
| When $\alpha = 60°$: $S = \frac{250}{\cos 60°} = 500$ N | M1 | |
| Abi must pull down with force 500 N, but her weight is $40 \times 10 = 400$ N, which is less than 500 N | M1 | |
| She cannot exert a downward force greater than her own weight, so it is not possible | A1 | |
---(i) Represent the forces acting on the object as a fully labelled triangle of forces.\\
(ii) Find $F$ and $\theta$.
Show that the distance between the object and the vertical section of rope A is 3 m .
Abi then pulls harder and the object moves upwards. Bob adjusts the tension in rope B so that the object moves along a vertical line.
Fig. 7.2 shows the situation when the object is part of the way up. The tension in rope A is $S \mathrm {~N}$ and the tension in rope B is $T \mathrm {~N}$. The ropes make angles $\alpha$ and $\beta$ with the vertical as shown in the diagram. Abi and Bob are taking a rest and holding the object stationary and in equilibrium.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{83e69140-4abf-4713-85da-922ce7530e47-5_383_360_534_854}
\captionsetup{labelformat=empty}
\caption{Fig. 7.2}
\end{center}
\end{figure}
(iii) Give the equations, involving $S , T , \alpha$ and $\beta$, for equilibrium in the vertical and horizontal directions.\\
(iv) Find the values of $S$ and $T$ when $\alpha = 8.5 ^ { \circ }$ and $\beta = 35 ^ { \circ }$.\\
(v) Abi's mass is 40 kg .
Explain why it is not possible for her to raise the object to a position in which $\alpha = 60 ^ { \circ }$.
\hfill \mbox{\textit{OCR MEI M1 2013 Q7 [18]}}