| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform rod on supports or with strings |
| Difficulty | Standard +0.3 This is a straightforward non-uniform rod equilibrium problem requiring standard moment and force resolution techniques. Students must take moments about a point (typically A or B) and resolve forces vertically—both routine procedures for M2. The non-uniform aspect (centre of mass at 1.5m rather than 1.25m) adds minimal complexity, and the angle calculation is direct once T is found. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For taking moments about A | |
| \(3g \times 1.5 = 2.5(T\cos15°)\) | A1 | |
| \(T = 18.6\) | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T_A \sin\alpha = T\sin15°\) | M1 | For resolving forces horizontally and vertically or taking moments about B |
| \(T_A \cos\alpha + T\cos15° = 3g\) or \(T_A \cos\alpha \times 2.5 = 3g \times 1\) | ||
| \(\tan\alpha = T\sin15° \div (3g - T\cos15°)\) | M1 | For eliminating \(T_A\) |
| \(\alpha = 21.8\) or \(21.9\) | A1 (3) |
## Question 5:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For taking moments about A |
| $3g \times 1.5 = 2.5(T\cos15°)$ | A1 | |
| $T = 18.6$ | A1 (3) | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A \sin\alpha = T\sin15°$ | M1 | For resolving forces horizontally and vertically or taking moments about B |
| $T_A \cos\alpha + T\cos15° = 3g$ or $T_A \cos\alpha \times 2.5 = 3g \times 1$ | | |
| $\tan\alpha = T\sin15° \div (3g - T\cos15°)$ | M1 | For eliminating $T_A$ |
| $\alpha = 21.8$ or $21.9$ | A1 (3) | |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{0cb05368-9ddf-4564-8428-725c77193a1e-3_383_1031_543_557}
A non-uniform rod $A B$ of length 2.5 m and mass 3 kg has its centre of mass at the point $G$ of the rod, where $A G = 1.5 \mathrm {~m}$. The rod hangs horizontally, in equilibrium, from strings attached at $A$ and $B$. The strings at $A$ and $B$ make angles with the vertical of $\alpha ^ { \circ }$ and $15 ^ { \circ }$ respectively. The tension in the string at $B$ is $T \mathrm {~N}$ (see diagram). Find\\
(i) the value of $T$,\\
(ii) the value of $\alpha$.
\hfill \mbox{\textit{CAIE M2 2006 Q5 [6]}}