# Question 5:

## Part 5a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $B$ | M1 | Dimensionally correct. Condone sin/cos confusion and errors in angles. OR: Correct moments equation and resolution where required |
| $T \times 2.5\sin\alpha = 70 \times 1.25\sin 2\alpha$ | A1 | Unsimplified equation in $\alpha$ with at most one error |
| Or $T \times 2.5\sin\alpha = 70 \times 2\sin\alpha$ | A1 | Correct unsimplified equation in $\alpha$ |
| Or use similar triangles $T \times \frac{3}{2} = 70 \times \frac{6}{5}$ | | |
| $T = 70 \times \frac{4}{5} = 56\ \text{(N)}$ | A1* | Obtain given answer from correct exact working and no errors seen |
| **Total: (4)** | | |

## Part 5b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve horizontally | M1 | First equation |
| $H = T\sin\alpha\ (= 33.6\ \text{(N)})$ | A1ft | Correct unsimplified equation |
| Resolve vertically | M1 | Second equation |
| $V + T\cos\alpha = 70\ \ (V = 25.2\ \text{(N)})$ | A1ft | Correct unsimplified equation |
| $V = \mu H$ | M1 | Use of $F = \mu R$ with their $V$, $H$ |
| $\mu = \frac{3}{4}$ | A1 | Correct only (no substitution for $g$ required) |
| **Total: (6)** | | |

## Part 5b (Alternative):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve parallel to the rod | M1 | |
| $H\sin 2\alpha + 70\cos 2\alpha = 56\cos\alpha + V\cos 2\alpha$ | A1ft | $(24H - 7V = 630)$ |
| Resolve perpendicular to the rod | M1 | |
| $70\sin 2\alpha = 56\sin\alpha + V\sin 2\alpha + H\cos 2\alpha$ | A1ft | $(24V + 7H = 840)$ |
| $V = \mu H$ | M1 | Use of $F = \mu R$ with their $V$, $H$ |
| $\mu = \frac{3}{4}$ | A1 | Correct only (no substitution for $g$ required) |
| **Total: (6) [10]** | | |

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