## Question 3(i):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Correct triangle of forces shape with $F$ N, $250$ N, $25$ N, angle $\theta$ | B1 | Shape of triangle; ignore position of $\theta$ if marked in diagram |
| Correct arrows | B1 | 2 marks -1 per error but penalise no arrows only once and penalise no labels only once. Condone $T$ written for $F$. |
| Correct labels | B1 | In the case of a force diagram showing $F$, $25$ and $250$ allow maximum of 2 marks with -1 per error |
| [3] | | |

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## Question 3(ii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan \alpha = \dfrac{25}{250}$ | M1 | M1 for recognising and using $\alpha$ in the triangle |
| $\Rightarrow \alpha = 5.7°$ | A1 | |
| $F = \sqrt{25^2 + 250^2}$ | M1 | Use of Pythagoras |
| $F = 251.2$ | A1 | At least 3 significant figures required |
| Distance $= 30\tan\alpha = 30 \times 0.1 = 3$ m | B1 | CAO |
| [5] | | |

**Alternative:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $F\cos\theta = 250$, $F\sin\theta = 25$, $\tan\theta = \dfrac{25}{250}$ | M1 | |
| $\Rightarrow \theta = 5.7°$ | A1 | |
| $F\cos 5.7° = 250$ | M1 | |
| $F = 251.2$ | A1 | At least 3 significant figures required |
| Distance $= 30\tan\alpha = 30 \times 0.1 = 3$ m | B1 | CAO |

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## Question 3(iii):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Vertical equilibrium: $S\cos\alpha = T\cos\beta + 250$ | M1, A1 | M1 for attempt at resolution in an equation involving both $S$ and $T$; condone sin-cos errors for the M mark only |
| Horizontal equilibrium: $S\sin\alpha = T\sin\beta$ | A1 | |
| [3] | | |

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## Question 3(iv):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $S\sin 8.5° = T\sin 35° \Rightarrow S = 3.8805T$ | M1 | Using one equation to make $S$ or $T$ the subject in terms of the other |
| $(3.8805T)\cos 8.5° = T\cos 35° + 250$ | M1 | Substituting in the other equation |
| $T = 82.8$ | A1 | CAO |
| $S = 321.4$ | A1 | CAO |
| [4] | | |

**Alternative (linear simultaneous equations):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $S\sin 8.5° - T\sin 35° = 0$ and $S\cos 8.5° - T\cos 35° = 250$, valid elimination of $S$ or $T$ | M1 | Valid method that has eliminated terms in either $S$ or $T$ (execution need not be perfect) |
| $S = 321.4$ | A1 | CAO First answer |
| Substituting in either equation | M1 | Substituting to find the second answer |
| $T = 82.8$ | A1 | CAO Second answer |

**Alternative (Triangle of forces / Lami's Theorem):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| Drawing and using a triangle of forces, or quoting and using Lami's Theorem | M1 | |
| $\dfrac{S}{\sin 145°} = \dfrac{T}{\sin 8.5°} = \dfrac{250}{\sin 26.5°}$ | M1 | Correct form of these equations |
| $S = 321.4$ | A1 | CAO |
| $T = 82.8$ | A1 | CAO |

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## Question 3(v):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Abi's weight is $40g = 392$ N | M1 | Consideration of Abi's weight |
| When $\alpha = 60°$, $S\cos 60° > 250 \Rightarrow S > 500$ | M1 | Consideration of vertical forces on the object. Condone no mention of Bob's rope |
| The tension in rope A would be greater than Abi's weight and so she would be lifted off the ground | A1 | The argument must be of high quality and must include consideration of the tension in Bob's rope |
| [3] | | |

**Alternative:**

| Answer | Marks | Guidance |
|--------|-------|----------|
| If Abi is on the ground, the maximum possible tension in rope A is Abi's weight of $392$ N | M1 | Consideration of Abi's weight |
| So the maximum upward force on the object is $392 \times \cos 60° = 192$ N. This is less than the weight of the object, and the tension in Bob's rope is pulling the box down. | M1 | Consideration of vertical forces on the object. Condone no mention of Bob's rope. Or: the box accelerated downwards |
| So Abi would be lifted off the ground | A1 | The argument must be of high quality and must include consideration of the tension in Bob's rope |