OCR Further Mechanics 2018 December — Question 7 13 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2018
SessionDecember
Marks13
TopicCentre of Mass 1
TypeToppling on inclined plane
DifficultyChallenging +1.2 This is a multi-part Further Maths mechanics question requiring centre of mass calculation, equilibrium analysis on an inclined plane, and friction conditions. While it involves several steps and concepts (centre of mass, moments, elastic strings, friction), each part follows standard procedures without requiring novel insight. The 'show that' format provides targets to work towards, making it more structured than open-ended problems.
Spec3.03v Motion on rough surface: including inclined planes6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

7 Particles \(A , B\) and \(C\) of masses \(2 \mathrm {~kg} , 3 \mathrm {~kg}\) and 5 kg respectively are joined by light rigid rods to form a triangular frame. The frame is placed at rest on a horizontal plane with \(A\) at the point ( 0,0 ), \(B\) at the point ( \(0.6,0\) ) and \(C\) at the point ( \(0.4,0.2\) ), where distances in the coordinate system are measured in metres (see Fig. 1). \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-5_304_666_434_251} \captionsetup{labelformat=empty} \caption{Fig. 1}
\end{figure} \(G\), which is the centre of mass of the frame, is at the point \(( \bar { x } , \bar { y } )\).
  1. - Show that \(\bar { x } = 0.38\).
    A rough plane, \(\Pi\), is inclined at an angle \(\theta\) to the horizontal where \(\sin \theta = \frac { 3 } { 5 }\). The frame is placed on \(\Pi\) with \(A B\) vertical and \(B\) in contact with \(\Pi\). \(C\) is in the same vertical plane as \(A B\) and a line of greatest slope of \(\Pi . C\) is on the down-slope side of \(A B\). The frame is kept in equilibrium by a horizontal light elastic string whose natural length is \(l \mathrm {~m}\) and whose modulus of elasticity is \(g \mathrm {~N}\). The string is attached to \(A\) at one end and to a fixed point on \(\Pi\) at the other end (see Fig. 2). \begin{figure}[h]
    \includegraphics[alt={},max width=\textwidth]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-5_611_842_1649_248} \captionsetup{labelformat=empty} \caption{Fig. 2}
    \end{figure} The coefficient of friction between \(B\) and \(\Pi\) is \(\mu\).
  2. Show that \(l = 0.3\).
  3. Show that \(\mu \geqslant \frac { 14 } { 27 }\). \section*{OCR} Oxford Cambridge and RSA

(a)
AnswerMarks Guidance
Use of \(\bar{x} = \frac{\sum mx}{\sum m}\) or \(\bar{y} = \frac{\sum my}{\sum m}\)M1 [1.2] Values must be substituted
\(\bar{x} = \frac{2 \times 0 + 3 \times 0.6 + 5 \times 0.4}{2 + 3 + 5} = \frac{3.8}{10} = 0.38\) (AG)A1 [3.3] Intermediate step must be seen
\(\bar{y} = \frac{2 \times 0 + 3 \times 0 + 5 \times 0.2}{2 + 3 + 5} = \frac{1}{10} = 0.1\)A1 [3.3]
[3]
(b)
AnswerMarks
G is not on the frame so the weight would have an unbalanced moment about the support pointE1 [2.4]
[1]
(c)
AnswerMarks Guidance
Moments about B: \(0.6T =0g \times 0.1\) M1 [3.4]
\(T = \frac{g(0.8 - l)}{l}\)M1 [3.4] Correct form for Hooke's law
Eliminate T: \(\frac{5g}{3} = \frac{g(0.8 - l)}{l} \Rightarrow 8l = 2.4 \Rightarrow l = 0.3\) (AG)A1 [1.1] Intermediate working must be seen
[3]
(d)
Any two resolving equations oe, eg:
(vertically): \(R\cos\theta + F\sin\theta = 10g\)
(horizontally): \(R\sin\theta = F\cos\theta + T\)
(parallel to plane): \(F + T\cos\theta = 10g\sin\theta\)
AnswerMarks Guidance
(perpendicular to plane): \(R = T\sin\theta + 10g\cos\theta\)M1 [3.4], M1 [3.4] Or taking moments eg about the point where the string is attached to the plane \((1 \times N = (0.8 + 0.1) \times 10g)\) or the point where the lines of action of \(R\) and the weight intersect: \(\frac{0.1}{\sin\theta} \times F = \left(0.6 - \frac{0.1}{\tan\theta}\right) \times T\)
Use \(T = \frac{3}{5}g\), \(\sin\theta = \frac{3}{5}\), \(\cos\theta = \frac{4}{5}\) to derive equations \(4R + 3F = 50g\) and \(9R = 12F + 25g\) oeM1 [1.1a] Forming equations that lead to numerical values for \(F\) and \(R\)
\(R = 9g\) and \(F = \frac{14}{3}g\)A1 [1.1] Both
\(F \leq \mu R \Rightarrow \frac{14}{3}g \leq \mu \times 9g\)M1 [3.4] \(F \leq \mu R\) with values substituted in
\(\therefore \mu \geq \frac{14g}{3 \times 9g} = \frac{14}{27}\) (AG)A1 [1.1] Intermediate working must be seen
[6]
### (a)
Use of $\bar{x} = \frac{\sum mx}{\sum m}$ or $\bar{y} = \frac{\sum my}{\sum m}$ | M1 [1.2] | Values must be substituted

$\bar{x} = \frac{2 \times 0 + 3 \times 0.6 + 5 \times 0.4}{2 + 3 + 5} = \frac{3.8}{10} = 0.38$ (AG) | A1 [3.3] | Intermediate step must be seen

$\bar{y} = \frac{2 \times 0 + 3 \times 0 + 5 \times 0.2}{2 + 3 + 5} = \frac{1}{10} = 0.1$ | A1 [3.3]

| [3]

### (b)
G is not on the frame so the weight would have an unbalanced moment about the support point | E1 [2.4]

| [1]

### (c)
Moments about B: $0.6T = |0g \times 0.1$ | M1 [3.4] | Each term must be of the form $F \times d$

$T = \frac{g(0.8 - l)}{l}$ | M1 [3.4] | Correct form for Hooke's law

Eliminate T: $\frac{5g}{3} = \frac{g(0.8 - l)}{l} \Rightarrow 8l = 2.4 \Rightarrow l = 0.3$ (AG) | A1 [1.1] | Intermediate working must be seen

| [3]

### (d)
Any two resolving equations oe, eg:
(vertically): $R\cos\theta + F\sin\theta = 10g$
(horizontally): $R\sin\theta = F\cos\theta + T$
(parallel to plane): $F + T\cos\theta = 10g\sin\theta$
(perpendicular to plane): $R = T\sin\theta + 10g\cos\theta$ | M1 [3.4], M1 [3.4] | Or taking moments eg about the point where the string is attached to the plane $(1 \times N = (0.8 + 0.1) \times 10g)$ or the point where the lines of action of $R$ and the weight intersect: $\frac{0.1}{\sin\theta} \times F = \left(0.6 - \frac{0.1}{\tan\theta}\right) \times T$

Use $T = \frac{3}{5}g$, $\sin\theta = \frac{3}{5}$, $\cos\theta = \frac{4}{5}$ to derive equations $4R + 3F = 50g$ and $9R = 12F + 25g$ oe | M1 [1.1a] | Forming equations that lead to numerical values for $F$ and $R$

$R = 9g$ and $F = \frac{14}{3}g$ | A1 [1.1] | Both

$F \leq \mu R \Rightarrow \frac{14}{3}g \leq \mu \times 9g$ | M1 [3.4] | $F \leq \mu R$ with values substituted in

$\therefore \mu \geq \frac{14g}{3 \times 9g} = \frac{14}{27}$ (AG) | A1 [1.1] | Intermediate working must be seen

| [6]
7 Particles $A , B$ and $C$ of masses $2 \mathrm {~kg} , 3 \mathrm {~kg}$ and 5 kg respectively are joined by light rigid rods to form a triangular frame. The frame is placed at rest on a horizontal plane with $A$ at the point ( 0,0 ), $B$ at the point ( $0.6,0$ ) and $C$ at the point ( $0.4,0.2$ ), where distances in the coordinate system are measured in metres (see Fig. 1).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-5_304_666_434_251}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

$G$, which is the centre of mass of the frame, is at the point $( \bar { x } , \bar { y } )$.
\begin{enumerate}[label=(\alph*)]
\item - Show that $\bar { x } = 0.38$.

\begin{itemize}
  \item Find $\bar { y }$.
\item Explain why it would be impossible for the frame to be in equilibrium in a horizontal plane supported at only one point.
\end{itemize}

A rough plane, $\Pi$, is inclined at an angle $\theta$ to the horizontal where $\sin \theta = \frac { 3 } { 5 }$. The frame is placed on $\Pi$ with $A B$ vertical and $B$ in contact with $\Pi$. $C$ is in the same vertical plane as $A B$ and a line of greatest slope of $\Pi . C$ is on the down-slope side of $A B$.

The frame is kept in equilibrium by a horizontal light elastic string whose natural length is $l \mathrm {~m}$ and whose modulus of elasticity is $g \mathrm {~N}$. The string is attached to $A$ at one end and to a fixed point on $\Pi$ at the other end (see Fig. 2).

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{493f11f4-e25c-4eeb-a0ab-20ec6d7a7a7d-5_611_842_1649_248}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}

The coefficient of friction between $B$ and $\Pi$ is $\mu$.
\item Show that $l = 0.3$.
\item Show that $\mu \geqslant \frac { 14 } { 27 }$.

\section*{OCR}
Oxford Cambridge and RSA
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2018 Q7 [13]}}