# Question 4:

## Part (a)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Write $d = 0.8$ | | |
| $(2.5+1.2+1.3+2.4)\times d \times \begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}$ | M1 | Method for c.m (length is 7.4 m, mass is 5.92 kg) |
| $= 2.5d\begin{pmatrix}1.2\\-0.35\end{pmatrix}+1.2d\begin{pmatrix}2.4\\-0.1\end{pmatrix}+1.3d\begin{pmatrix}1.8\\0.25\end{pmatrix}+2.4d\begin{pmatrix}1.2\\0\end{pmatrix}$ | B1 | One rod mass and cpts correct or if done by separate $x$ and $y$ equations 2 rod components and masses correct. (Allow length used instead of mass) |
| OR: $(2+0.96+1.04+1.92)\times\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}$ | | |
| $= 2\begin{pmatrix}1.2\\-0.35\end{pmatrix}+0.96\begin{pmatrix}2.4\\-0.1\end{pmatrix}+1.04\begin{pmatrix}1.8\\0.25\end{pmatrix}+1.92\begin{pmatrix}1.2\\0\end{pmatrix}$ | B1 | Another rod dealt with correctly or if done by separate $x$ and $y$ equations, the other equation attempted with 2 rod components and masses correct. (Allow length used instead of mass) |
| $7.4\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}=\begin{pmatrix}3+2.88+2.34+2.88\\-0.875-0.12+0.325+0\end{pmatrix}=\begin{pmatrix}11.1\\-0.67\end{pmatrix}$ | | |
| OR: $5.92\begin{pmatrix}\bar{x}\\\bar{y}\end{pmatrix}=\begin{pmatrix}2.4+2.304+2.304+1.872\\-0.7-0.096+0.26+0\end{pmatrix}=\begin{pmatrix}8.88\\-0.536\end{pmatrix}$ | | |
| $\bar{x}=1.5$ | E1 | Clearly shown, with at least one intermediate step |
| $\bar{y}=-0.090540... = -0.0905$ (3 s.f.) | A1 | Condone $-0.09$ |
| | **[5]** | |

## Part (a)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| EITHER: New c.m. has $\bar{x}=1.2$ | M1 | Identifying and using a suitable condition |
| $(5.92+m)\times 1.2 = 5.92\times 1.5 + m\times 0$ | M1 | Complete method |
| $m=1.48$ | A1 | cao |
| | **[3]** | |
| OR: Moment about any point is zero | M1 | Identifying a suitable condition |
| e.g. about S: $1.2mg = 0.3\times 5.92g$ | M1 | Allow $g$ omitted. Correct number of terms must be included |
| $m=1.48$ | A1 | cao |
| | **[3]** | |

## Part (b)(i):

| Answer | Mark | Guidance |
|--------|------|----------|
| Consider the equilibrium at R | | |
| Resolving horizontally gives $T_{QR}=0$ | E1 | |
| Then resolving vertically gives $T_{OR}=0$ | E1 | |
| | **[2]** | |

## Part (b)(ii):

| Answer | Mark | Guidance |
|--------|------|----------|
| c.w. moments about O | | |
| $120\times 1 + 60\times 2 = 3T$ | M1 | May also be argued by first considering internal forces |
| so $T=80$ | A1 | |
| Resolve to give $X=80$ and $Y=180$ | A1 | FT $X=T$. Only $Y=180$ scores 0 |
| | **[3]** | |

## Part (b)(iii):

| Answer | Mark | Guidance |
|--------|------|----------|
| [diagram] | B1 | All correct. Accept $T$, $X$ and $Y$ labelled but not substituted. Accept mixes of T and C. Require pairs of arrows with label on OQ, OP and PQ. |
| | **[1]** | |

## Part (b)(iv):

| Answer | Mark | Guidance |
|--------|------|----------|
| Take angle OPQ as $\alpha$ | | Forces internal to the rods have been taken to be tensions |
| At P $\downarrow$ $60 + T_{OP}\sin\alpha = 0$ | M1 | Equilibrium at ANY pin-joint (not R) |
| | A1 | Correct equation(s) that leads directly to finding $T_{OP}$ or $T_{QP}$ |
| $\sin\alpha = \dfrac{3}{\sqrt{13}}$: $\alpha = 56.3°$ | | |
| $T_{OP} = -\dfrac{60}{\sin\alpha} = -20\sqrt{13}$ so $20\sqrt{13}$ N (C) | A1 | o.e. Accept 72.1 N |
| At P $\leftarrow$ $T_{QP} + T_{OP}\cos\alpha = 0$ | M1 | A second equilibrium equation leading to a second internal force |
| so $T_{QP} = 40$ so 40 N (T) | A1 | cao T/C correct for both rods |
| | **[5]** | |