OCR MEI Further Mechanics Minor 2023 June — Question 5 12 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2023
SessionJune
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeParticle suspended by strings
DifficultyStandard +0.8 This is a two-part mechanics problem requiring equilibrium analysis with forces at angles, followed by a moments problem with a beam involving friction. Part (a) requires resolving forces in two directions for a standard pulley system. Parts (b) and (c) require taking moments about a point, resolving forces including friction, and working with a beam at an angle—more sophisticated than basic mechanics but still follows standard A-level Further Maths procedures without requiring novel insight.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

5 Fig. 5.1 shows a particle P, of mass 5 kg , and a particle Q, of mass 11 kg , which are attached to the ends of a light, inextensible string. The string is taut and passes over a small smooth pulley fixed to the ceiling. \begin{figure}[h]
\captionsetup{labelformat=empty} \caption{Fig. 5.1} \includegraphics[alt={},max width=\textwidth]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-5_367_707_495_251}
\end{figure} When a force of magnitude \(H \mathrm {~N}\), acting at an angle \(\theta\) to the upward vertical, is applied to Q the particles hang in equilibrium, with the part of the string connecting the pulley to Q making an angle of \(40 ^ { \circ }\) with the upward vertical. It is given that the force acts in the same vertical plane in which the string lies.
  1. Determine the values of \(H\) and \(\theta\). Particle Q is now removed. The string is instead attached to one end of a uniform beam B of length 3 m and mass 7 kg . The other end of B is in contact with a rough horizontal floor. The situation is shown in Fig. 5.2. \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Fig. 5.2} \includegraphics[alt={},max width=\textwidth]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-5_504_978_1557_251}
    \end{figure} With B in equilibrium, at an angle \(\phi\) to the horizontal, the part of the string connecting the pulley to B makes an angle of \(30 ^ { \circ }\) with the upward vertical. It is given that the string and B lie in the same vertical plane.
  2. Determine the smallest possible value for the coefficient of friction between B and the floor.
  3. Determine the value of \(\phi\).

Question 5:
AnswerMarks Guidance
5(a) H c o s 5 g c o s 4 0 1 1 g  +  =
H s i n 5 g s i n 4 0  = M1
A13.3
1.1Attempt to resolve both
horizontally and verticallyCorrect
number of
terms
tan= 5gsin40 =24.14487 
AnswerMarks Guidance
11g−5gcos40A1 1.1
and H  7 7 .0A1 1.1
Or triangle of forces method:
AnswerMarks
Force triangle withM1
• sides H, 5g and 11g,
• 40° between 5g and 11g,
•  between H and 11g.
AnswerMarks
H 2 = ( 5 g ) 2 + ( 1 1 g ) 2 − 2 ( 5 g ) ( 1 1 g ) c o s 4 0 A1
 H  7 7 .0A1
(11g)2+T2−(5g)2
= =24.14487 
AnswerMarks Guidance
2T11gA1 A1
[4]
AnswerMarks
(b)Friction = 5 g s i n 3 0 
Normal contact = 7 g − 5 g c o s 3 0 M1
A13.3
1.1Resolving vertically and
horizontally for the beam
5 sinc 3o 0s 0 .9 3 6 3 7 4  = g−  =
AnswerMarks Guidance
m in 7 5 3 0 g g A1 1.1
[3]
AnswerMarks
(c)Taking moments about the point of contact between beam
and floor:
3
5gcos30° ∙3𝑐𝑜𝑠𝜙 = 5g𝑠𝑖𝑛 30°∙3𝑠𝑖𝑛𝜙+7g∙ 𝑐𝑜𝑠 𝜙
2
or 5g𝑐𝑜𝑠(𝜙+30 °)∙3 = 7g∙3𝑐𝑜𝑠 𝜙
AnswerMarks
2M1*
A23.1b
1.1
AnswerMarks
1.1Taking moments: dimensionally
consistent; correct number of terms
for their method.
All correct. A1 for one error.
AnswerMarks Guidance
5 3 5 t a n 7   = +M1dep* 1.1
t a n 5 35 7 1 8 .3 6 8 7 8    = −  = A1 2.2a
[5]
Question 5:
5 | (a) | H c o s 5 g c o s 4 0 1 1 g  +  =
H s i n 5 g s i n 4 0  =  | M1
A1 | 3.3
1.1 | Attempt to resolve both
horizontally and vertically | Correct
number of
terms
tan= 5gsin40 =24.14487 
11g−5gcos40 | A1 | 1.1
and H  7 7 .0 | A1 | 1.1 | H = 7 7 .0 0 0 2 6 0 so accept 77.
Or triangle of forces method:
Force triangle with | M1
• sides H, 5g and 11g,
• 40° between 5g and 11g,
•  between H and 11g.
H 2 = ( 5 g ) 2 + ( 1 1 g ) 2 − 2 ( 5 g ) ( 1 1 g ) c o s 4 0  | A1
 H  7 7 .0 | A1
(11g)2+T2−(5g)2
= =24.14487 
2T11g | A1 | A1
[4]
(b) | Friction = 5 g s i n 3 0 
Normal contact = 7 g − 5 g c o s 3 0  | M1
A1 | 3.3
1.1 | Resolving vertically and
horizontally for the beam
5 sinc 3o 0s 0 .9 3 6 3 7 4  = g−  =
m in 7 5 3 0 g g  | A1 | 1.1
[3]
(c) | Taking moments about the point of contact between beam
and floor:
3
5gcos30° ∙3𝑐𝑜𝑠𝜙 = 5g𝑠𝑖𝑛 30°∙3𝑠𝑖𝑛𝜙+7g∙ 𝑐𝑜𝑠 𝜙
2
or 5g𝑐𝑜𝑠(𝜙+30 °)∙3 = 7g∙3𝑐𝑜𝑠 𝜙
2 | M1*
A2 | 3.1b
1.1
1.1 | Taking moments: dimensionally
consistent; correct number of terms
for their method.
All correct. A1 for one error.
5 3 5 t a n 7   = + | M1dep* | 1.1 | Obtaining tan (oe)
t a n 5 35 7 1 8 .3 6 8 7 8    = −  =  | A1 | 2.2a
[5]
5 Fig. 5.1 shows a particle P, of mass 5 kg , and a particle Q, of mass 11 kg , which are attached to the ends of a light, inextensible string. The string is taut and passes over a small smooth pulley fixed to the ceiling.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 5.1}
  \includegraphics[alt={},max width=\textwidth]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-5_367_707_495_251}
\end{center}
\end{figure}

When a force of magnitude $H \mathrm {~N}$, acting at an angle $\theta$ to the upward vertical, is applied to Q the particles hang in equilibrium, with the part of the string connecting the pulley to Q making an angle of $40 ^ { \circ }$ with the upward vertical. It is given that the force acts in the same vertical plane in which the string lies.
\begin{enumerate}[label=(\alph*)]
\item Determine the values of $H$ and $\theta$.

Particle Q is now removed. The string is instead attached to one end of a uniform beam B of length 3 m and mass 7 kg . The other end of B is in contact with a rough horizontal floor. The situation is shown in Fig. 5.2.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Fig. 5.2}
  \includegraphics[alt={},max width=\textwidth]{cad8805d-59f6-4ed2-81f4-9e8c749461f5-5_504_978_1557_251}
\end{center}
\end{figure}

With B in equilibrium, at an angle $\phi$ to the horizontal, the part of the string connecting the pulley to B makes an angle of $30 ^ { \circ }$ with the upward vertical.

It is given that the string and B lie in the same vertical plane.
\item Determine the smallest possible value for the coefficient of friction between B and the floor.
\item Determine the value of $\phi$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2023 Q5 [12]}}