| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Smooth ring on string |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium problem with three forces on a particle. Part (i) requires understanding of tension in a string over a smooth ring (basic concept). Parts (ii) and (iii) involve routine resolution of forces and solving simultaneous equations using Pythagoras. The right-angle geometry simplifies the problem significantly. Slightly easier than average due to the straightforward setup and standard techniques, though the three-part structure and need to explain tension makes it not trivial. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| The ring is smooth so no friction acts on the string; the string is light and inextensible; therefore tension is the same throughout | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Horizontal: \(T\cos\theta - T\cos90° + 5 = 0\)... resolving: since \(\angle ARB = 90°\), part AR makes \((90°-\theta)\) with horizontal | M1 | |
| \(T\cos(90°-\theta) - T\cos\theta = 5\) i.e. \(T\sin\theta - T\cos\theta = 5\) | A1 | |
| Vertical: \(T\sin(90°-\theta) + T\sin\theta = 7\) i.e. \(T\cos\theta + T\sin\theta = 7\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Adding equations: \(2T\sin\theta = 12\), \(T\sin\theta = 6\) | M1 | |
| Subtracting: \(2T\cos\theta = 2\), \(T\cos\theta = 1\) | M1 | |
| \(\tan\theta = 6\), \(\theta = 80.5°\) | A1 | |
| \(T = \sqrt{6^2 + 1^2} = \sqrt{37} = 6.08\) N | M1 A1 |
# Question 6:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| The ring is smooth so no friction acts on the string; the string is light and inextensible; therefore tension is the same throughout | B1 | |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Horizontal: $T\cos\theta - T\cos90° + 5 = 0$... resolving: since $\angle ARB = 90°$, part AR makes $(90°-\theta)$ with horizontal | M1 | |
| $T\cos(90°-\theta) - T\cos\theta = 5$ i.e. $T\sin\theta - T\cos\theta = 5$ | A1 | |
| Vertical: $T\sin(90°-\theta) + T\sin\theta = 7$ i.e. $T\cos\theta + T\sin\theta = 7$ | M1 A1 | |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Adding equations: $2T\sin\theta = 12$, $T\sin\theta = 6$ | M1 | |
| Subtracting: $2T\cos\theta = 2$, $T\cos\theta = 1$ | M1 | |
| $\tan\theta = 6$, $\theta = 80.5°$ | A1 | |
| $T = \sqrt{6^2 + 1^2} = \sqrt{37} = 6.08$ N | M1 A1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{ce4c43e6-da4f-4c02-ab0f-01a21717949c-3_348_1109_1345_516}
A small smooth ring $R$ of weight 7 N is threaded on a light inextensible string. The ends of the string are attached to fixed points $A$ and $B$ at the same horizontal level. A horizontal force of magnitude 5 N is applied to $R$. The string is taut. In the equilibrium position the angle $A R B$ is a right angle, and the portion of the string attached to $B$ makes an angle $\theta$ with the horizontal (see diagram).\\
(i) Explain why the tension $T \mathrm {~N}$ is the same in each part of the string.\\
(ii) By resolving horizontally and vertically for the forces acting on $R$, form two simultaneous equations in $T \cos \theta$ and $T \sin \theta$.\\
(iii) Hence find $T$ and $\theta$.
\hfill \mbox{\textit{OCR M1 2011 Q6 [11]}}