## Question 4:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $A$: $T \times \frac{5a}{2}\sin(\pi - 2\theta) - W \times 2a\sin\theta = 0$ | M1 A1 | Take moments for rod about one chosen point; $[\sin\theta = \frac{2}{\sqrt{5}}$, $\cos\theta = \frac{1}{\sqrt{5}}$, $\sin(\pi-2\theta) = \sin 2\theta = \frac{4}{5}]$ |
| $C$: $F_A \times \frac{5a}{2}\sin\theta - R_A \times \frac{5a}{2}\cos\theta - W \times \frac{1}{2}a\sin\theta = 0$ | | |
| $B$: $F_A \times 4a\sin\theta - R_A \times 4a\cos\theta - W \times 2a\sin\theta + T \times \frac{3a}{2}\sin(\pi-2\theta) = 0$ | | |
| $G$: $F_A \times 2a\sin\theta - R_A \times 2a\cos\theta - T \times \frac{1}{2}a\sin(\pi-2\theta) = 0$ ($G$ is midpoint of $AB$) | | |
| $D$: $R_A \times 5a\cos\theta - W \times 2a\sin\theta = 0$ | | |
| $R_A = T\sin\theta$ | B1 | Find two more independent equations |
| $F_A = W - T\cos\theta$ | B1 | e.g. resolution of forces on rod (a second moment equation may be used) |
| $F_A = \mu R_A$ | B1 | Relate $F_A$ and $R_A$ (may be implied) |
| $T = \frac{2W\sin\theta}{\frac{5}{2}\sin 2\theta} = \frac{2W}{5\cos\theta}$ | M1 | Find $T$ by any method (e.g. from moments about $A$) |
| $= \frac{2W}{\sqrt{5}}$ or $\frac{2\sqrt{5}}{5}W$ or $0.894W$ | A1 | |
| $F_A = \frac{3}{5}W$ and $R_A = \frac{4}{5}W$; $\mu = \frac{3}{4}$ or $0.75$ | M1 A1 | Find or imply $F_A$ and $R_A$ by any method (e.g. from resolutions) and hence $\mu$ |
| | A1 | |
| | **10** | |

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