OCR MEI Paper 1 2021 November — Question 5 5 marks

Exam BoardOCR MEI
ModulePaper 1 (Paper 1)
Year2021
SessionNovember
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeResultant force on lamina
DifficultyStandard +0.3 This is a standard mechanics equilibrium problem requiring resolution of forces and taking moments about a point. Part (a) involves straightforward application of ΣF=0 and ΣM=0 to find two unknowns. Part (b) requires understanding that moving the force creates a resultant couple. While it requires careful bookkeeping of multiple forces, the techniques are routine for A-level mechanics students.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

5 ABCD is a rectangular lamina in which AB is 30 cm and AD is 10 cm .
Three forces of 20 N and one force of 30 N act along the sides of the lamina as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-05_558_981_1263_233} An additional force \(F \mathrm {~N}\) is also applied at right angles to AB to a point on the edge \(\mathrm { AB } x \mathrm {~cm}\) from A .
  1. Given that the lamina is in equilibrium, calculate the values of \(F\) and \(x\). The point of application of the force \(F \mathrm {~N}\) is now moved to B , but the magnitude and direction of the force remain the same.
  2. Explain the effect of the new system of forces on the lamina.

Question 5(a):
AnswerMarks Guidance
AnswerMarks Guidance
\(F = 50\)B1 cao
Take moments about A (or D): \(Fx = 20 \times 30 + 20 \times 10\); OR Take moments about B (or C): \(F(30-x) + 20 \times 10 = 30 \times 30\)M1 Forming an equation for moments about any point – allow their value of \(F\). Allow one incorrect distance or a missing term. Do not allow for moment = force
\(x = \dfrac{800}{50} = 16\)A1 [3] Allow 0.16 m if unit given
Question 5(b):
AnswerMarks Guidance
AnswerMarks Guidance
Forces have a resultant momentE1 Allow 'resultant moment' or 'not in equilibrium' soi. Do not allow for 'a resultant force' on its own
which have a clockwise turning effect on the laminaE1 [2] Clockwise must be indicated
## Question 5(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $F = 50$ | B1 | cao |
| Take moments about A (or D): $Fx = 20 \times 30 + 20 \times 10$; OR Take moments about B (or C): $F(30-x) + 20 \times 10 = 30 \times 30$ | M1 | Forming an equation for moments about any point – allow their value of $F$. Allow one incorrect distance or a missing term. Do not allow for moment = force |
| $x = \dfrac{800}{50} = 16$ | A1 [3] | Allow 0.16 m if unit given |

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## Question 5(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| Forces have a resultant moment | E1 | Allow 'resultant moment' or 'not in equilibrium' soi. Do not allow for 'a resultant force' on its own |
| which have a clockwise turning effect on the lamina | E1 [2] | Clockwise must be indicated |

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5 ABCD is a rectangular lamina in which AB is 30 cm and AD is 10 cm .\\
Three forces of 20 N and one force of 30 N act along the sides of the lamina as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fac72cb-85cb-48d9-8817-899ef3f80a0f-05_558_981_1263_233}

An additional force $F \mathrm {~N}$ is also applied at right angles to AB to a point on the edge $\mathrm { AB } x \mathrm {~cm}$ from A .
\begin{enumerate}[label=(\alph*)]
\item Given that the lamina is in equilibrium, calculate the values of $F$ and $x$.

The point of application of the force $F \mathrm {~N}$ is now moved to B , but the magnitude and direction of the force remain the same.
\item Explain the effect of the new system of forces on the lamina.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q5 [5]}}