Question 6 - A-Level Maths

OCR MEI Further Mechanics A AS 2018 June — Question 6 11 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2018
Session	June
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Moments
Type	Rod on inclined plane
Difficulty	Standard +0.3 This is a standard equilibrium problem requiring moments and resolution of forces. While it involves multiple steps (moments about B, horizontal resolution, vertical resolution), each step follows routine mechanics procedures. The geometry with the 30° inclined plane adds mild complexity, but the question guides students through each stage explicitly, making it slightly easier than average for A-level mechanics.
Spec	3.03m Equilibrium: sum of resolved forces = 0 3.03n Equilibrium in 2D: particle under forces 3.04a Calculate moments: about a point 3.04b Equilibrium: zero resultant moment and force

6 A uniform rod AB has length \(2 a\) and weight \(W\). The rod is in equilibrium in a horizontal position. The end A rests on a smooth plane which is inclined at an angle of \(30 ^ { \circ }\) to the horizontal. The force exerted on AB by the plane is \(R\). The end B is attached to a light inextensible string inclined at an angle of \(\theta\) to AB as shown in Fig. 6. The rod and string are in the same vertical plane, which also contains the line of greatest slope of the plane on which A lies. The tension in the string is \(T\). \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-5_474_862_479_616} \captionsetup{labelformat=empty} \caption{Fig. 6}

\end{figure}

Add the forces \(R\) and \(T\) to the copy of Fig. 6 in the Printed Answer Booklet.
By taking moments about B , find an expression for \(R\) in terms of \(W\).
By resolving horizontally, show that \(6 T \cos \theta = W \sqrt { 3 }\).
By finding a second equation connecting \(T\) and \(\theta\), determine
- the value of \(\theta\),
- an expression for \(T\) in terms of \(W\).

Show mark scheme Show mark scheme source

Question 6:

Answer	Marks	Guidance
6	(i)	R at A perp’r to plane; T along string from B
[1]	1.2	By eye; arrows needed
(ii)	𝑊×𝑎 = 𝑅×2𝑎sin60°

𝑊

𝑅 = √3

Answer	Marks
3	M1

A1

Answer	Marks	Guidance
[2]	3.3
1.1	Allow sin / cos confusion, sign error	a can be cancelled; ft R from (i)

if R through A (not horiz)

𝑊

Accept

√3

Answer	Marks
(iii)	𝑇cos𝜃 = 𝑅cos60°

𝑊

𝑇cos𝜃 = √3 cos60°

3

Answer	Marks
6𝑇cos𝜃 = 𝑊√3	M1

M1ft

A1

Answer	Marks
[3]	3.4

1.1

Answer	Marks
2.1	Must have sin/cos in each term;

allow sin / cos confusion

Using their R from part (ii); allow

sin / cos confusion

AG

Answer	Marks
(iv)	𝑇sin𝜃+𝑅sin60° = 𝑊

𝑊√3 𝑊 √3

sin𝜃+ √3× = 𝑊

6cos𝜃 3 2

6 √3×√3

tan𝜃 = ×(1− )

√3 3×2

𝑊

OR 𝑇sin𝜃+ √3sin60° = 𝑊

3 𝑊 √3

𝑇sin𝜃 = ; Solve with 𝑇cos𝜃 = 𝑊 from

2 6

(iii)

𝜃 = 60°

𝑊

𝑇 = √3

Answer	Marks
3	M1*

A1

*M1

(A1)

(M1)

A1

Answer	Marks
[5]	3.1b

3.4

1.1

Answer	Marks
1.1	Resolve vertically for AB; must

have T, R and W terms, comps of T

and R; allow sign error

Elim T and subst for R

Attempt to re-arrange to find tan θ

Answer	Marks
Subst for R Must be correct	OR Moments about A

𝑊𝑎 = 𝑇×2𝑎sin𝜃 M1

𝑇×2𝑎sin𝜃 𝑊𝑎

= A1

6𝑇cos𝜃 𝑊√3

tan𝜃 = √3 M1

PPMMTT

OCR (Oxford Cambridge and RSA Examinations)

The Triangle Building

Shaftesbury Road

Cambridge

CB2 8EA

OCR Customer Contact Centre

Education and Learning

Telephone: 01223 553998

Facsimile: 01223 552627

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Registered in England

Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA

Registered Company Number: 3484466

OCR is an exempt Charity

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Telephone: 01223 552552

Facsimile: 01223 552553

© OCR 2018

Question 6:
6 | (i) | R at A perp’r to plane; T along string from B | B1
[1] | 1.2 | By eye; arrows needed
(ii) | 𝑊×𝑎 = 𝑅×2𝑎sin60°
𝑊
𝑅 = √3
3 | M1
A1
[2] | 3.3
1.1 | Allow sin / cos confusion, sign error | a can be cancelled; ft R from (i)
if R through A (not horiz)
𝑊
Accept
√3
(iii) | 𝑇cos𝜃 = 𝑅cos60°
𝑊
𝑇cos𝜃 = √3 cos60°
3
6𝑇cos𝜃 = 𝑊√3 | M1
M1ft
A1
[3] | 3.4
1.1
2.1 | Must have sin/cos in each term;
allow sin / cos confusion
Using their R from part (ii); allow
sin / cos confusion
AG
(iv) | 𝑇sin𝜃+𝑅sin60° = 𝑊
𝑊√3 𝑊 √3
sin𝜃+ √3× = 𝑊
6cos𝜃 3 2
6 √3×√3
tan𝜃 = ×(1− )
√3 3×2
𝑊
OR 𝑇sin𝜃+ √3sin60° = 𝑊
3
𝑊 √3
𝑇sin𝜃 = ; Solve with 𝑇cos𝜃 = 𝑊 from
2 6
(iii)
𝜃 = 60°
𝑊
𝑇 = √3
3 | M1*
A1
*M1
(A1)
(M1)
A1
A1
[5] | 3.1b
3.4
1.1
1.1
1.1 | Resolve vertically for AB; must
have T, R and W terms, comps of T
and R; allow sign error
Elim T and subst for R
Attempt to re-arrange to find tan θ
Subst for R Must be correct | OR Moments about A
𝑊𝑎 = 𝑇×2𝑎sin𝜃 M1
𝑇×2𝑎sin𝜃 𝑊𝑎
= A1
6𝑇cos𝜃 𝑊√3
tan𝜃 = √3 M1
PPMMTT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance
programme your call may be recorded or monitored
Oxford Cambridge and RSA Examinations
is a Company Limited by Guarantee
Registered in England
Registered Office; The Triangle Building, Shaftesbury Road, Cambridge, CB2 8EA
Registered Company Number: 3484466
OCR is an exempt Charity
OCR (Oxford Cambridge and RSA Examinations)
Head office
Telephone: 01223 552552
Facsimile: 01223 552553
© OCR 2018

Show LaTeX source

6 A uniform rod AB has length $2 a$ and weight $W$. The rod is in equilibrium in a horizontal position. The end A rests on a smooth plane which is inclined at an angle of $30 ^ { \circ }$ to the horizontal. The force exerted on AB by the plane is $R$. The end B is attached to a light inextensible string inclined at an angle of $\theta$ to AB as shown in Fig. 6. The rod and string are in the same vertical plane, which also contains the line of greatest slope of the plane on which A lies. The tension in the string is $T$.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-5_474_862_479_616}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}

(i) Add the forces $R$ and $T$ to the copy of Fig. 6 in the Printed Answer Booklet.\\
(ii) By taking moments about B , find an expression for $R$ in terms of $W$.\\
(iii) By resolving horizontally, show that $6 T \cos \theta = W \sqrt { 3 }$.\\
(iv) By finding a second equation connecting $T$ and $\theta$, determine

\begin{itemize}
  \item the value of $\theta$,
  \item an expression for $T$ in terms of $W$.
\end{itemize}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2018 Q6 [11]}}

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Q1 6 Q2 12 Q3 9 Q4 9 Q5 13 Q6 11