| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2002 |
| Session | November |
| Marks | 6 |
| 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 equilibrium problem requiring resolution of forces in two directions with given angles. The smooth ring means tension is uniform throughout the string, simplifying the problem. While it requires careful trigonometry and simultaneous equations, it follows a routine textbook approach with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For resolving forces on \(R\) vertically (3 terms needed) | M1 | |
| \(T\sin 50° = T\sin 20° + 0.8g\) | A1 | |
| Tension is \(18.9\text{ N}\) (\(18.5\) from \(g = 9.81\) or \(g = 9.8\)) | A1 | Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| For resolving forces on \(R\) horizontally | M1 | |
| \(X = T\cos 50° + T\cos 20°\) | A1 | |
| \(X = 29.9\) (\(8\tan 75°\)) (\(29.3\) from \(g = 9.81\) or \(g = 9.8\)) | A1ft | Total: 3 |
| Answer | Marks |
|---|---|
| \(X/\sin 105 = 8/\sin 165\) | M1 |
| \(X = 29.9\) | A1 |
| \(2T\cos 35/\sin 90 = 8/\sin 165\) | M1 |
| \(T = 18.9\) | A1 |
| Answer | Marks |
|---|---|
| \(\sqrt{8^2 + X^2}\sin 15 = 8\) | A1 |
| \(X = 29.9\) | A1 |
| \(T/\sin 145 = \sqrt{8^2 + 29.9^2}/\sin 70\) | A1 |
| \(T = 18.9\) | A1 |
## Question 3:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For resolving forces on $R$ vertically (3 terms needed) | M1 | |
| $T\sin 50° = T\sin 20° + 0.8g$ | A1 | |
| Tension is $18.9\text{ N}$ ($18.5$ from $g = 9.81$ or $g = 9.8$) | A1 | **Total: 3** |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| For resolving forces on $R$ horizontally | M1 | |
| $X = T\cos 50° + T\cos 20°$ | A1 | |
| $X = 29.9$ ($8\tan 75°$) ($29.3$ from $g = 9.81$ or $g = 9.8$) | A1ft | **Total: 3** |
**Notes:**
- $F_y = T\sin 50 - T\sin 20 - 0.8g$ scores M0 in (i); $F_x = X - T\cos 50 - T\cos 20$ scores M0 in (ii)
- sin/cos mix can score M1 A0 A0 M1 A0 A0 at best
- None of the four A marks can be scored unless $T_1 = T_2$ is stated or implied
*By scale drawing:* Correct quadrilateral drawn to scale scores M1; $18.4 \leq T \leq 19.4$ scores A2; $29.4 \leq X \leq 30.4$ scores A2; $T = 18.9$ and $X = 29.9$ scores A1
*Using Lami's theorem (reducing to 3 forces of magnitudes $2T\cos 35$, $X$ and $8$):*
| $X/\sin 105 = 8/\sin 165$ | M1 | |
|---|---|---|
| $X = 29.9$ | A1 | |
| $2T\cos 35/\sin 90 = 8/\sin 165$ | M1 | |
| $T = 18.9$ | A1 | |
*Alternatively reducing to forces $T$, $T$ and $\sqrt{8^2 + X^2}$:*
| $\sqrt{8^2 + X^2}\sin 15 = 8$ | A1 | |
|---|---|---|
| $X = 29.9$ | A1 | |
| $T/\sin 145 = \sqrt{8^2 + 29.9^2}/\sin 70$ | A1 | |
| $T = 18.9$ | A1 | |
---3\\
\includegraphics[max width=\textwidth, alt={}, center]{fcd2b219-d9b4-4972-b8fe-25cf543b9054-2_438_621_1676_762}
A light inextensible string has its ends attached to two fixed points $A$ and $B$, with $A$ vertically above $B$. A smooth ring $R$, of mass 0.8 kg , is threaded on the string and is pulled by a horizontal force of magnitude $X$ newtons. The sections $A R$ and $B R$ of the string make angles of $50 ^ { \circ }$ and $20 ^ { \circ }$ respectively with the horizontal, as shown in the diagram. The ring rests in equilibrium with the string taut. Find\\
(i) the tension in the string,\\
(ii) the value of $X$.
\hfill \mbox{\textit{CAIE M1 2002 Q3 [6]}}