Question 5 - A-Level Maths

OCR MEI Further Mechanics Minor 2020 November — Question 5 13 marks

Exam Board	OCR MEI
Module	Further Mechanics Minor (Further Mechanics Minor)
Year	2020
Session	November
Marks	13
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Moments
Type	Rod or block on rough surface in limiting equilibrium (no wall)
Difficulty	Challenging +1.2 This is a multi-part mechanics problem requiring force diagrams, moment equations, and limiting equilibrium conditions. While it involves several steps (moments about A, resolving forces, applying friction law), the techniques are standard for Further Maths mechanics. The algebra is somewhat involved but follows predictable patterns. It's harder than a basic statics problem but doesn't require novel insight—just systematic application of equilibrium principles.
Spec	3.03u Static equilibrium: on rough surfaces 3.04a Calculate moments: about a point 3.04b Equilibrium: zero resultant moment and force

5 A uniform rod AB , of mass \(3 m\) and length \(2 a\), rests with the end A on a rough horizontal surface. A small object of mass \(m\) is attached to the rod at B . The rod is maintained in equilibrium at an angle of \(60 ^ { \circ }\) to the horizontal by a force acting at an angle of \(\theta\) to the vertical at a point C , where the distance \(\mathrm { AC } = \frac { 6 } { 5 } a\). The force acting at C is in the same vertical plane as the rod (see Fig. 5). \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{6418c1b7-092a-4747-bc88-1b57815c6ad9-4_800_648_932_255} \captionsetup{labelformat=empty} \caption{Fig. 5}

\end{figure}

On the copy of Fig. 5 in the Printed Answer Booklet, mark all the forces acting on the rod. [2]
Show that the magnitude of the force acting at C can be expressed as \(\frac { 25 m g } { 6 ( \cos \theta + \sqrt { 3 } \sin \theta ) }\).
Given that the rod is in limiting equilibrium and the coefficient of friction between the rod and the surface is \(\frac { 3 } { 4 }\), determine the value of \(\theta\).

Show mark scheme Show mark scheme source

Question 5:

Answer	Marks
5(a)	Weight of small object acting vertically downwards at

B. Weight of the rod acting vertically downwards at the

midpoint of AB.

Normal contact force acting vertically upwards at A.

Frictional contact force acting horizontally to the right

Answer	Marks
at A.	B2
[2]	1.2
1.2	B1 for any three correct

PPMMTT

Y431/01 Mark Scheme November 2020

Left-hand side correct

oe for right-hand side e.g.

A1 1.1

6 3 3 6√3 A1 3.1b 𝑎𝑎� acos60cos𝜃𝜃

𝑚𝑚𝑔𝑔𝑎𝑎+𝑚𝑚𝑔𝑔𝑎𝑎=𝑎𝑎� 𝑎𝑎cos𝜃𝜃+ 𝑎𝑎sin𝜃𝜃� 5

2 5 10 6

or + sin60 𝑎𝑎sin𝜃𝜃�

5

6 5𝑎𝑎𝑎𝑎sin(30+𝜃𝜃)

∴ 25𝑚𝑚𝑔𝑔=𝑎𝑎�6cos𝜃𝜃+6√3sin𝜃𝜃�

A1 2.2a AG

25𝑚𝑚𝑔𝑔

𝑎𝑎 = [4]

6(cos𝜃𝜃+√3sin𝜃𝜃)

F is the frictional contact

5(c) B1 1.1 Resolving horizontally force at A where T is the

force applied at C

𝐹𝐹 =𝑎𝑎 𝑠𝑠𝑠𝑠𝑎𝑎𝜃𝜃

Resolving vertically – correct number

M1* 3.3

of terms from their force diagram

is the normal contact

A1 1.1

force at A

𝑅𝑅

𝑅𝑅+𝑎𝑎cos𝜃𝜃 =3𝑚𝑚𝑔𝑔+𝑚𝑚𝑔𝑔 Applying with correct T and

M1* 3.4 their F and R

𝐹𝐹 =𝜇𝜇𝑅𝑅

𝐹𝐹= 𝜇𝜇𝑅𝑅

𝑎𝑎sin𝜃𝜃 = 𝜇𝜇(4𝑚𝑚𝑔𝑔−𝑎𝑎cos𝜃𝜃) 

25mgsinθ 3 25mgcosθ

 

= 4mg− A1 1.1 Correct equation in cosine and sine

6 (

cosθ+ 3sinθ

) 4

6 (

cosθ+ 3sinθ

)

 

3( ) 3 Dependent on all M marks

25sinθ= 24 3sinθ−cosθ ⇒tanθ= 4 M1dep* 3.1a Re-arranges and uses tan = sin / cos

4 18 3−25 and the B mark

9

Question 5:
--- 5(a) ---
5(a) | Weight of small object acting vertically downwards at
B.
Weight of the rod acting vertically downwards at the
midpoint of AB.
Normal contact force acting vertically upwards at A.
Frictional contact force acting horizontally to the right
at A. | B2
[2] | 1.2
1.2 | B1 for any three correct
PPMMTT
Y431/01 Mark Scheme November 2020
Left-hand side correct
oe for right-hand side e.g.
A1 1.1
6
3 3 6√3 A1 3.1b 𝑎𝑎� acos60cos𝜃𝜃
𝑚𝑚𝑔𝑔𝑎𝑎+𝑚𝑚𝑔𝑔𝑎𝑎=𝑎𝑎� 𝑎𝑎cos𝜃𝜃+ 𝑎𝑎sin𝜃𝜃� 5
2 5 10 6
or + sin60 𝑎𝑎sin𝜃𝜃�
5
6
5𝑎𝑎𝑎𝑎sin(30+𝜃𝜃)
∴ 25𝑚𝑚𝑔𝑔=𝑎𝑎�6cos𝜃𝜃+6√3sin𝜃𝜃�
A1 2.2a AG
25𝑚𝑚𝑔𝑔
𝑎𝑎 = [4]
6(cos𝜃𝜃+√3sin𝜃𝜃)
F is the frictional contact
5(c) B1 1.1 Resolving horizontally force at A where T is the
force applied at C
𝐹𝐹 =𝑎𝑎 𝑠𝑠𝑠𝑠𝑎𝑎𝜃𝜃
Resolving vertically – correct number
M1* 3.3
of terms from their force diagram
is the normal contact
A1 1.1
force at A
𝑅𝑅
𝑅𝑅+𝑎𝑎cos𝜃𝜃 =3𝑚𝑚𝑔𝑔+𝑚𝑚𝑔𝑔 Applying with correct T and
M1* 3.4 their F and R
𝐹𝐹 =𝜇𝜇𝑅𝑅
𝐹𝐹= 𝜇𝜇𝑅𝑅
𝑎𝑎sin𝜃𝜃 = 𝜇𝜇(4𝑚𝑚𝑔𝑔−𝑎𝑎cos𝜃𝜃) 
25mgsinθ 3 25mgcosθ
 
= 4mg− A1 1.1 Correct equation in cosine and sine
6
(
cosθ+ 3sinθ
) 4
6
(
cosθ+ 3sinθ
)
 
3( ) 3 Dependent on all M marks
25sinθ= 24 3sinθ−cosθ ⇒tanθ= 4 M1dep* 3.1a Re-arranges and uses tan = sin / cos
4 18 3−25 and the B mark
9

Show LaTeX source

5 A uniform rod AB , of mass $3 m$ and length $2 a$, rests with the end A on a rough horizontal surface. A small object of mass $m$ is attached to the rod at B . The rod is maintained in equilibrium at an angle of $60 ^ { \circ }$ to the horizontal by a force acting at an angle of $\theta$ to the vertical at a point C , where the distance $\mathrm { AC } = \frac { 6 } { 5 } a$. The force acting at C is in the same vertical plane as the rod (see Fig. 5).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{6418c1b7-092a-4747-bc88-1b57815c6ad9-4_800_648_932_255}
\captionsetup{labelformat=empty}
\caption{Fig. 5}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item On the copy of Fig. 5 in the Printed Answer Booklet, mark all the forces acting on the rod. [2]
\item Show that the magnitude of the force acting at C can be expressed as $\frac { 25 m g } { 6 ( \cos \theta + \sqrt { 3 } \sin \theta ) }$.
\item Given that the rod is in limiting equilibrium and the coefficient of friction between the rod and the surface is $\frac { 3 } { 4 }$, determine the value of $\theta$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2020 Q5 [13]}}

This paper (6 questions)

View full paper

Q1 6 Q2 7 Q3 9 Q4 8 Q5 13 Q6 17