AQA Paper 2 2024 June — Question 17 4 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2024
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeUniform beam on two supports
DifficultyStandard +0.3 This is a standard mechanics equilibrium problem requiring moment balance about one point and force balance. The setup is straightforward with clearly defined geometry, and the algebraic manipulation to reach the required form is routine. Slightly easier than average due to being a 'show that' question with a clear target and standard technique.
Spec3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces

A uniform rod is resting on two fixed supports at points \(A\) and \(B\). \(A\) lies at a distance \(x\) metres from one end of the rod. \(B\) lies at a distance \((x + 0.1)\) metres from the other end of the rod. The rod has length \(2L\) metres and mass \(m\) kilograms. The rod lies horizontally in equilibrium as shown in the diagram below. \includegraphics{figure_17} The reaction force of the support on the rod at \(B\) is twice the reaction force of the support on the rod at \(A\). Show that $$L - x = k$$ where \(k\) is a constant to be found. [4 marks]

Question 17:
AnswerMarks
17Forms a dimensionally correct
moment in L and x
May see :
(L−x)R
(L−(x+0.1))2R
(L−x)mg
(2L−(2x+0.1))2R
AnswerMarks Guidance
OE3.3 B1
(L−x)R= (L−(x+0.1))2R
L−x=2(L−(x+0.1))
L−x=2L−2x−0.2
L−x=0.2
Forms dimensionally correct
moments equation with at least
one term correct
May also see:
(L−x)mg =(2L−(2x+0.1))2R
Or
(L−(x+0.1))mg =(2L−(2x+0.1))R
AnswerMarks Guidance
Condone missing brackets1.1a M1
Obtains a fully correct equation in
AnswerMarks Guidance
L and x only1.1b A1
Completes reasoned argument to
obtain L – x = 0.2
Must show expansion of all
AnswerMarks Guidance
brackets before the final answer2.1 R1
Question 17 Total4
QMarking instructions AO
Question 17:
17 | Forms a dimensionally correct
moment in L and x
May see :
(L−x)R
(L−(x+0.1))2R
(L−x)mg
(2L−(2x+0.1))2R
OE | 3.3 | B1 | The reaction at A is R
(L−x)R= (L−(x+0.1))2R
L−x=2(L−(x+0.1))
L−x=2L−2x−0.2
L−x=0.2
Forms dimensionally correct
moments equation with at least
one term correct
May also see:
(L−x)mg =(2L−(2x+0.1))2R
Or
(L−(x+0.1))mg =(2L−(2x+0.1))R
Condone missing brackets | 1.1a | M1
Obtains a fully correct equation in
L and x only | 1.1b | A1
Completes reasoned argument to
obtain L – x = 0.2
Must show expansion of all
brackets before the final answer | 2.1 | R1
Question 17 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
A uniform rod is resting on two fixed supports at points $A$ and $B$.

$A$ lies at a distance $x$ metres from one end of the rod.

$B$ lies at a distance $(x + 0.1)$ metres from the other end of the rod.

The rod has length $2L$ metres and mass $m$ kilograms.

The rod lies horizontally in equilibrium as shown in the diagram below.

\includegraphics{figure_17}

The reaction force of the support on the rod at $B$ is twice the reaction force of the support on the rod at $A$.

Show that
$$L - x = k$$
where $k$ is a constant to be found.
[4 marks]

\hfill \mbox{\textit{AQA Paper 2 2024 Q17 [4]}}