# Question 3:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| a.c. moments about A: $1 \times T - 2 \times 300 = 0$ so $T = 600$ | E1 | |
| Resolving $\rightarrow$: $X = 0$ | B1 | Justified |
| $\uparrow$: $T - Y = 300$ | M1 | |
| so $Y = 300$ | A1 | |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Diagram; the working below sets all internal forces as tensions; candidates need not do this. | B1 | All external forces marked consistent with (i) |
| | B1 | All internal forces with arrows and labels |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let angle DAB be $\theta$; $\cos\theta = \frac{1}{2}$, $\sin\theta = \frac{\sqrt{3}}{2}$ | B1 | Or equivalent seen |
| | M1 | Attempt at equilibrium at pin-joints |
| | M1 | 1 equilibrium correct, allowing sign errors |
| A $\uparrow$: $-300 - T_{AB}\sin\theta = 0$ so $T_{AB} = -200\sqrt{3}$, force is $200\sqrt{3}$ (C) | A1 | |
| A $\rightarrow$: $T_{AD} + T_{AB}\cos\theta = 0$ so $T_{AD} = 100\sqrt{3}$, force is $100\sqrt{3}$ (T) | F1 | |
| C $\uparrow$: $T_{CD}\sin\theta - 300 = 0$ so $T_{CD} = 200\sqrt{3}$, force is $200\sqrt{3}$ (T) | F1 | |
| C $\leftarrow$: $T_{BC} + T_{CD}\cos\theta = 0$ so $T_{BC} = -100\sqrt{3}$, force is $100\sqrt{3}$ (C) | F1 | |
| B $\uparrow$: $T_{AB}\sin\theta + T_{BD} = 0$ so $T_{BD} = 300$, force is $300$ (T) | F1 | |
| | F1 | All T/C consistent with their calculations and diagrams |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| AD, AB, BC, CD: $300$ N, $X$ and $Y$ not changed. Equilibrium equations at A and C are not altered | B1 | |
| | E1 | |
| B $\uparrow$: $T_{AB}\sin\theta + T'_{BD} + 600 = 0$ | M1 | C not needed. |
| so $T'_{BD} = -300$, force is $300$ (C) | A1 | [If 300 N (C) given WWW, award SC1 (NB it must be made clear that this is a compression)] |

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