# Question 2:

## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $20\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = 15\begin{pmatrix}20\\0\end{pmatrix} + 3\begin{pmatrix}0\\100\end{pmatrix} + 2\begin{pmatrix}25\\200\end{pmatrix}$ | M1 | Method to obtain at least 1 coordinate |
| | B1 | '100' or '25' correct |
| | A1 | Either one RHS term correct or one component of two RHS terms correct |
| $\bar{x} = 17.5$ | A1 | |
| $\bar{y} = 35$ | A1 | |

## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $25\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix} = \begin{pmatrix}350\\700\end{pmatrix} + 5\begin{pmatrix}40\\200\end{pmatrix}$ | M1 | Using (i) or starting again |
| so $\bar{x} = 22$, $\bar{y} = 68$ | E1 | Clearly shown |

## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| We need the edge that the $\bar{x}$ position is nearest | M1 | This may be implied |
| $\bar{x} = 22$; distances are 22 to PQ, 18 to SR, 15 to QR | B1 | One distance correct |
| | B1 | All distances correct |
| so edge QR | A1 | |

## Part (iv)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about RS | M1 | Moments about RS attempted |
| In sense $xOy$ | B1 | Use of weight not mass below. FT mass from here |
| $T\sin 50 \times 200 - T\cos 50 \times 40$ | M1 | Attempt to find moment of $T$ about RS, including attempt at resolution. May try to find perp dist from G to line of action of the force. |
| $-20g \times (40 - 17.5) = 0$ | A1 | $40 - 17.5$ |
| | B1 | All correct allowing sign errors |
| $T = 34.5889\ldots$ so $34.6$ N (3 s.f.) | A1 | cao (except for use of mass) |

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