## Question 9:

### Part (a):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Moments about $A$ (or any other complete method) | M1 | 3.3 |
| $T \cdot 2a\sin\alpha = Mga + 3Mgx$ | A1 | 1.1b |
| $T = \frac{Mg(a+3x)}{2a \times \frac{3}{5}} = \frac{5Mg(3x+a)}{6a}$ * GIVEN ANSWER | A1* | 2.1 |

### Part (b):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{5Mg(3x+a)}{6a}\cos\alpha = 2Mg$ **OR** $2Mg \cdot 2a\tan\alpha = Mga + 3Mgx$ | M1 | 3.1b |
| $x = \frac{2a}{3}$ | A1 | 2.2a |

### Part (c):

| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve vertically **OR** Moments about $B$ | M1 | 3.1b |
| $Y = 3Mg + Mg - \frac{5Mg(3\cdot\frac{2a}{3}+a)}{6a}\sin\alpha$ or $2aY = Mga + 3Mg(2a - \frac{2a}{3})$ **Or:** $Y = 3Mg + Mg - \left(\frac{2Mg}{\cos\alpha}\right)\sin\alpha$ | A1ft | 1.1b |
| $Y = \frac{5Mg}{2}$ | A1 | 1.1b |
| $\tan\beta = \frac{Y}{X}$ or $\frac{R\sin\beta}{R\cos\beta} = \frac{\frac{5Mg}{2}}{2Mg}$ | M1 | 3.4 |
| $= \frac{5}{4}$ | A1 | 2.2a |

**Notes:**
- N.B. May use $R\sin\beta$ for $Y$ and/or $R\cos\beta$ for $X$ throughout

### Part (d):

| Working/Answer | Mark | Guidance |
|---|---|---|
| $\frac{5Mg(3x+a)}{6a} \leq 5Mg$ and solve for $x$ | M1 | 2.4 |
| $x \leq \frac{5a}{3}$ | A1 | 2.4 |
| For rope not to break, block can't be more than $\frac{5a}{3}$ from $A$ | B1 A1 | 2.4 |

**Notes:**
- N.B. If the correct inequality is not found, their comment must mention 'distance from $A$'
- Or just: $x \leq \frac{5a}{3}$, if no incorrect statement seen

# Question (from first page - appears to be Q9 based on context):

## Part (a):
| $\sin\alpha = \frac{3}{5}$ used to obtain correct equation in $a$, $M$, $x$ and $T$ only | M1 | Using M(A) with usual rules or any other complete method |
|---|---|---|
| Correct equation | A1 | |
| Correct printed answer, correctly obtained, need to see $\sin\alpha = \frac{3}{5}$ used | A1* | Given answer |

## Part (b):
| Accept $0.67a$ or better | A1 | Using appropriate strategy e.g. resolve horizontally OR moments about $C$; must use given expression for $T$ |

## Part (c):
| $Y$ (or $R\sin\beta$) $= \frac{5Mg}{2}$ or $2.5Mg$ or $2.50Mg$ | A1 | Correct equation with $x$ substituted in $T$ expression or using $T = \frac{2Mg}{\cos\alpha}$ |
| Equation in $\tan\beta$ only using $\tan\beta = \frac{Y}{X}$ or $\tan\beta = \frac{X}{Y}$ | M1 | Independent but must have found a $Y$ |
| Accept $\frac{-5}{4}$ if it follows from working | A1 | |

## Part (d):
| Allow $x = \frac{5a}{3}$ or $x < \frac{5a}{3}$; accept $1.7a$ or better | A1 | Allow $T = 5Mg$ or $T < 5Mg$, solve for $x$ showing all steps; M0 for $T > 5Mg$ |
| Maximum value of $x$ is $\frac{5a}{3}$ | B1 | Treat as A1; any appropriate equivalent fully correct statement |

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