Question 1 - A-Level Maths

AQA M2 — Question 1

Exam Board	AQA
Module	M2 (Mechanics 2)
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Moments
Type	Beam suspended by vertical ropes
Difficulty	Moderate -0.3 This is a straightforward two-rope equilibrium problem requiring resolution of vertical forces and taking moments about one point. The uniform beam setup is standard, and the calculation involves basic arithmetic with clearly given values. Slightly easier than average due to its routine nature, though it requires understanding of moments and equilibrium conditions.
Spec	3.04a Calculate moments: about a point 3.04b Equilibrium: zero resultant moment and force

1 A uniform beam, \(A B\), has mass 20 kg and length 7 metres. A rope is attached to the beam at \(A\). A second rope is attached to the beam at the point \(C\), which is 2 metres from \(B\). Both of the ropes are vertical. The beam is in equilibrium in a horizontal position, as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{88aec6ab-af83-4d5e-84b6-5fd84c16a6c9-003_298_906_756_552} Find the tensions in the two ropes.

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Question 1:

Part (a)(i)

Answer	Marks	Guidance
\(\mathbf{a} = \frac{d\mathbf{v}}{dt} = -8\sin 2t\,\mathbf{i} + 3\cos t\,\mathbf{j}\)	M1	Differentiate velocity
\(\mathbf{F} = 4\mathbf{a} = (-32\sin 2t\,\mathbf{i} + 12\cos t\,\mathbf{j})\) N	A1A1	A1 each component

Part (a)(ii)

Answer	Marks	Guidance
At \(t = \pi\): \(\mathbf{F} = (-32\sin 2\pi\,\mathbf{i} + 12\cos\pi\,\mathbf{j}) = (0\,\mathbf{i} - 12\,\mathbf{j})\)	M1	Substitute \(t=\pi\)
\(	\mathbf{F}	= 12\) N

Part (b)

Answer	Marks	Guidance
\(\mathbf{r} = \int \mathbf{v}\,dt = 2\sin 2t\,\mathbf{i} - 3\cos t\,\mathbf{j} + \mathbf{c}\)	M1	Integrate velocity
At \(t=0\): \(2\mathbf{i} - 14\mathbf{j} = 0\,\mathbf{i} - 3\mathbf{j} + \mathbf{c}\), so \(\mathbf{c} = 2\mathbf{i} - 11\mathbf{j}\)	M1	Apply initial condition
\(\mathbf{r} = (2\sin 2t + 2)\,\mathbf{i} + (-3\cos t - 11)\,\mathbf{j}\)	A1A1A1	A1 each component correct, A1 constant

# Question 1:

## Part (a)(i)
| $\mathbf{a} = \frac{d\mathbf{v}}{dt} = -8\sin 2t\,\mathbf{i} + 3\cos t\,\mathbf{j}$ | M1 | Differentiate velocity |
|---|---|---|
| $\mathbf{F} = 4\mathbf{a} = (-32\sin 2t\,\mathbf{i} + 12\cos t\,\mathbf{j})$ N | A1A1 | A1 each component |

## Part (a)(ii)
| At $t = \pi$: $\mathbf{F} = (-32\sin 2\pi\,\mathbf{i} + 12\cos\pi\,\mathbf{j}) = (0\,\mathbf{i} - 12\,\mathbf{j})$ | M1 | Substitute $t=\pi$ |
|---|---|---|
| $|\mathbf{F}| = 12$ N | A1 | |

## Part (b)
| $\mathbf{r} = \int \mathbf{v}\,dt = 2\sin 2t\,\mathbf{i} - 3\cos t\,\mathbf{j} + \mathbf{c}$ | M1 | Integrate velocity |
|---|---|---|
| At $t=0$: $2\mathbf{i} - 14\mathbf{j} = 0\,\mathbf{i} - 3\mathbf{j} + \mathbf{c}$, so $\mathbf{c} = 2\mathbf{i} - 11\mathbf{j}$ | M1 | Apply initial condition |
| $\mathbf{r} = (2\sin 2t + 2)\,\mathbf{i} + (-3\cos t - 11)\,\mathbf{j}$ | A1A1A1 | A1 each component correct, A1 constant |

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Show LaTeX source

1 A uniform beam, $A B$, has mass 20 kg and length 7 metres. A rope is attached to the beam at $A$. A second rope is attached to the beam at the point $C$, which is 2 metres from $B$. Both of the ropes are vertical. The beam is in equilibrium in a horizontal position, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{88aec6ab-af83-4d5e-84b6-5fd84c16a6c9-003_298_906_756_552}

Find the tensions in the two ropes.

\hfill \mbox{\textit{AQA M2  Q1}}

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