| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Particle suspended by strings |
| Difficulty | Standard +0.3 This is a multi-part connected particles question with standard equilibrium resolution. Parts (i)-(ii) involve basic resolution of forces with given trig values, (iii) requires conceptual understanding of equilibrium, (iv)-(v) extend to more complex configurations but still use routine mechanics techniques. The question is slightly easier than average due to given trig values, clear diagrams, and scaffolded structure, though it requires careful bookkeeping across multiple parts. |
| Spec | 3.03a Force: vector nature and diagrams3.03e Resolve forces: two dimensions3.03k Connected particles: pulleys and equilibrium3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_{AB}\sin\alpha = 147\) | M1 | Attempt at resolving. Accept \(\sin \leftrightarrow \cos\). Must have \(T\) resolved and equated to 147. |
| so \(T_{AB} = \dfrac{147}{0.6}\) | B1 | Use of 0.6. Accept correct substitution for angle in wrong expression. |
| \(= 245\) so \(245\) N | A1 | Only accept answers agreeing to 3 s.f. [Lami: M1 pair of ratios attempted; B1 correct sub; A1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_{BC} = 245\cos\alpha\) | M1 | Attempt to resolve 245 and equate to \(T\), or equivalent. Accept \(\sin \leftrightarrow \cos\) |
| \(= 245 \times 0.8 = 196\) | E1 | Substitution of 0.8 clearly shown. [SC1: \(245 \times 0.8 = 196\)] [Lami: M1 pair of ratios attempted; E1] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Geometry of A, B and C and weight of B the same | E1 | Mention of two of: same weight; same direction AB; same direction BC |
| and these determine the tension | E1 | Specific mention of same geometry & weight or recognition of same force diagram |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Diagram: \(196\) N horizontal, \(90\) N vertical, \(T\) at angle | B1 B1 | No extra forces. Correct orientation and arrows. \(T\), 196 and 90 labelled. Accept 'tension' written out. |
| either Realise 196 N and 90 N are horizontal and vertical forces; resultant has magnitude and line of action of tension | M1 | Allow for only \(\beta\) or \(T\) attempted |
| \(\tan\beta = 90/196\) | B1 | Use of arctan (196/90) or arctan (90/196) or equivalent |
| \(\beta = 24.6638...\) so \(24.7°\) (3 s.f.) | A1 | |
| \(T = \sqrt{196^2 + 90^2}\) | M1 | Use of Pythagoras |
| \(T = 215.675...\) so \(216\) N (3 s.f.) | E1 | |
| or \(\uparrow\; T\sin\beta - 90 = 0\) | B1 | Allow if \(T = 216\) assumed |
| \(\rightarrow\; T\cos\beta - 196 = 0\) | B1 | Allow if \(T = 216\) assumed |
| Solving \(\tan\beta = \dfrac{90}{196} = 0.45918...\) | M1 | Eliminating \(T\), or... |
| \(\beta = 24.6638...\) so \(24.7°\) (3 s.f.) | A1 | [If \(T = 216\) assumed, B1 for \(\beta\); B1 for check in 2nd equation; E0] |
| \(T = 215.675...\) so \(216\) N (3 s.f.) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Tension on block is \(215.675...\) N (pulley is smooth and string is light) | B1 | May be implied. Reasons not required. |
| \(M \times 9.8 \times \sin 40 = 215.675... + 20\) | M1 | Equating their tension on the block unresolved \(\pm\, 20\) to weight component. If equation in any other direction, normal reaction must be present. |
| \(M = 37.4128...\) so \(37.4\) (3 s.f.) | A1 A1 | Correct. Accept answers rounding to 37 and 38. |
## Question 6:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{AB}\sin\alpha = 147$ | M1 | Attempt at resolving. Accept $\sin \leftrightarrow \cos$. Must have $T$ resolved and equated to 147. |
| so $T_{AB} = \dfrac{147}{0.6}$ | B1 | Use of 0.6. Accept correct substitution for angle in wrong expression. |
| $= 245$ so $245$ N | A1 | Only accept answers agreeing to 3 s.f. [Lami: M1 pair of ratios attempted; B1 correct sub; A1] |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_{BC} = 245\cos\alpha$ | M1 | Attempt to resolve 245 and equate to $T$, or equivalent. Accept $\sin \leftrightarrow \cos$ |
| $= 245 \times 0.8 = 196$ | E1 | Substitution of 0.8 clearly shown. [SC1: $245 \times 0.8 = 196$] [Lami: M1 pair of ratios attempted; E1] |
**(iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Geometry of A, B and C and weight of B the same | E1 | Mention of two of: same weight; same direction AB; same direction BC |
| and these determine the tension | E1 | Specific mention of same geometry & weight or recognition of same force diagram |
**(iv)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram: $196$ N horizontal, $90$ N vertical, $T$ at angle | B1 B1 | No extra forces. Correct orientation and arrows. $T$, 196 and 90 labelled. Accept 'tension' written out. |
| **either** Realise 196 N and 90 N are horizontal and vertical forces; resultant has magnitude and line of action of tension | M1 | Allow for only $\beta$ or $T$ attempted |
| $\tan\beta = 90/196$ | B1 | Use of arctan (196/90) or arctan (90/196) or equivalent |
| $\beta = 24.6638...$ so $24.7°$ (3 s.f.) | A1 | |
| $T = \sqrt{196^2 + 90^2}$ | M1 | Use of Pythagoras |
| $T = 215.675...$ so $216$ N (3 s.f.) | E1 | |
| **or** $\uparrow\; T\sin\beta - 90 = 0$ | B1 | Allow if $T = 216$ assumed |
| $\rightarrow\; T\cos\beta - 196 = 0$ | B1 | Allow if $T = 216$ assumed |
| Solving $\tan\beta = \dfrac{90}{196} = 0.45918...$ | M1 | Eliminating $T$, or... |
| $\beta = 24.6638...$ so $24.7°$ (3 s.f.) | A1 | [If $T = 216$ assumed, B1 for $\beta$; B1 for check in 2nd equation; E0] |
| $T = 215.675...$ so $216$ N (3 s.f.) | E1 | |
**(v)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Tension on block is $215.675...$ N (pulley is smooth and string is light) | B1 | May be implied. Reasons not required. |
| $M \times 9.8 \times \sin 40 = 215.675... + 20$ | M1 | Equating their tension on the block unresolved $\pm\, 20$ to weight component. If equation in any other direction, normal reaction must be present. |
| $M = 37.4128...$ so $37.4$ (3 s.f.) | A1 A1 | Correct. Accept answers rounding to 37 and 38. |
---6 A box of weight 147 N is held by light strings AB and BC . As shown in Fig. 7.1, AB is inclined at $\alpha$ to the horizontal and is fixed at $\mathrm { A } ; \mathrm { BC }$ is held at C . The box is in equilibrium with BC horizontal and $\alpha$ such that $\sin \alpha = 0.6$ and $\cos \alpha = 0.8$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-4_380_542_377_791}
\captionsetup{labelformat=empty}
\caption{Fig. 7.1}
\end{center}
\end{figure}
(i) Calculate the tension in string AB .\\
(ii) Show that the tension in string BC is 196 N .
As shown in Fig. 7.2, a box of weight 90 N is now attached at C and another light string CD is held at D so that the system is in equilibrium with BC still horizontal. CD is inclined at $\beta$ to the horizontal.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-4_378_695_1282_687}
\captionsetup{labelformat=empty}
\caption{Fig. 7.2}
\end{center}
\end{figure}
(iii) Explain why the tension in the string BC is still 196 N .\\
(iv) Draw a diagram showing the forces acting on the box at C .
Find the angle $\beta$ and show that the tension in CD is 216 N , correct to three significant figures.
The string section CD is now taken over a smooth pulley and attached to a block of mass $M \mathrm {~kg}$ on a rough slope inclined at $40 ^ { \circ }$ to the horizontal. As shown in Fig. 7.3, the part of the string attached to the box is still at $\beta$ to the horizontal and the part attached to the block is parallel to the slope. The system is in equilibrium with a frictional force of 20 N acting on the block up the slope.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bf477f61-9f8f-418a-86d8-392bc30323b1-5_436_1049_524_536}
\captionsetup{labelformat=empty}
\caption{Fig. 7.3}
\end{center}
\end{figure}
(v) Calculate the value of $M$.
\hfill \mbox{\textit{OCR MEI M1 Q6 [14]}}