# Question 3:

## Part (a)(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical through C intersects AB at X; $BX = 1$ and $XA = 3$ | B1 | May be implied |
| $\uparrow R - Y - 60 = 0$ | B1 | Must have a correct equation involving $Y$ |
| ac moments about A: $60 \times 4 - R \times 3 = 0$ so $R = 80$ | B1 | AG |
| $Y = R - 60 = 20$ and $X = 0$ | B1 | Both. Can be awarded independent of previous B1 |
| **[4]** | | MR-1 for $AB = 2$ |

## Part (a)(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| All (8 forces, with labelled pairs of arrows for internal forces) present and consistent. $R$ and $Y$ can be used | B1 | |
| In the solutions below all internal forces are set as tensions | | |
| **[1]** | | |

## Part (a)(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For example B $\downarrow$: $60 + T_{BC}\cos 30 = 0$ | M1 | Attempt an equation for the equilibrium in any direction at any pin-joint (all correct (resolved) terms present, allow sign errors, $s \leftrightarrow c$) |
| $T_{BC} = -40\sqrt{3}$ (Force of $40\sqrt{3}$ N (C)) | A1 | Ignore T/C; sign of force must be consistent with their T/C convention |
| A $\downarrow$: $20 + T_{AC}\sin 30 = 0$ | M1 | 2nd equilibrium equation attempted |
| $T_{AC} = -40$ Force of 40 N (C) | A1 | Ignore T/C; sign of force must be consistent with their T/C convention |
| A $\leftarrow$: $T_{AB} + T_{AC}\cos 30 = 0$ | M1 | 3rd equilibrium equation attempted. Ignore T/C |
| $T_{AB} = 20\sqrt{3}$ Force of $20\sqrt{3}$ N (T) | | |
| All three internal forces correct, including T/C | A1 | NOTE: Award first A1 for ANY force correct (need not be first one calculated). Award second A1 for a second force correct, FT if dependent on first one. Award third A1 as cao for everything correct, including T/C |
| **[6]** | | |

# Question 3(b)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| $\cos\theta = \frac{8}{17}$ | B1 | Seen or implied, e.g. in $\cos 61.9°$ |
| $2P + 68 \times 2 \times \cos\theta - 102 \times 4 \times \sin\theta = 0$ | M1 | Moments equation with all terms attempted and no extras. Allow $s \leftrightarrow c$ and sign errors. Moments about other points must include all relevant forces |
| | A1 | Substitution of sin/cos not required |
| $P = 148$ | A1 | cao |
| | **[4]** | |

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# Question 3(b)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| $2Q\sin\theta + 68 \times 2 \times \cos\theta - 102 \times 4 \times \sin\theta = 0$ | M1 | Moments equation with all terms attempted and no extras. Allow $s \leftrightarrow c$ and sign errors. Moments about other points must include all relevant forces |
| $Q = 167.7333\ldots$ so $168$ (3 s.f.) | F1 | FT errors in 2,4,cos,sin, sign from part(i) in 2nd and 3rd terms |
| Horiz force at B is $102 + 167.733\ldots$ | B1 | Adding. FT their $Q$ |
| Magnitude is $\sqrt{269.7333\ldots^2 + 68^2}$ | M1 | FT their horizontal force at B; Must use 68 |
| $= 278.172\ldots$ so $278$ N (3 s.f.) | A1 | cao |
| *Alternative:* $Y = 102\sin\theta - 68\cos\theta + Q\sin\theta$ | | |
| $X = 102\cos\theta - 68\sin\theta + Q\cos\theta$ | | |
| $X = 187.06;\ Y = 206.34$ | B1 | FT their $Q$ |
| Magnitude is $\sqrt{187.06^2 + 206.34^2}$ | M1 | FT their $X$ and $Y$ |
| $= 278.172\ldots$ so $278$ N (3 s.f.) | A1 | cao |
| | **[5]** | |

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