OCR MEI Further Mechanics Major 2022 June — Question 1 5 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Major (Further Mechanics Major)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeTriangle of forces method
DifficultyModerate -0.8 This is a straightforward equilibrium problem requiring either a triangle of forces diagram or resolving forces in two perpendicular directions. The methods are standard A-level techniques with minimal problem-solving required—students simply apply sine/cosine rules or component resolution to find two unknowns. While it's a Further Maths question, the mechanics content is routine and below average difficulty for A-level.
Spec3.03m Equilibrium: sum of resolved forces = 0

\includegraphics{figure_1} Three forces of magnitudes 4 N, 7 N and \(P\) N act at a point in the directions shown in the diagram. The forces are in equilibrium.
  1. Draw a closed figure to represent the three forces. [1]
  2. Hence, or otherwise, find the following.
    1. The value of \(\theta\). [2]
    2. The value of \(P\). [2]

Question 1:
AnswerMarks Guidance
1(a) e.g.
[1]1.1 All three lengths and both angles must be correctly
placed (arrows not required)
AnswerMarks Guidance
(b)(i) sinθ sin40
=
4 7
AnswerMarks
θ=21.5 M1
A1
AnswerMarks
[2]1.1
1.1Correct application of sine rule (oe) for their closed
figure or resolving horizontally 7sinθ=4sin40
21.54964123…
AnswerMarks Guidance
(b)(ii) P2 =42 +72 −2(4)(7)cos(180−40−θ)
P=9.57M1
A1
AnswerMarks
[2]1.1
1.1Correct application of cosine rule (oe) with their θ(or
just θ) or correctly resolving vertically
P=4cos40+7cosθoe for example,
72 =42 +P2 −2(4)(P)cos40
sin(140−θ)
sin40
=
P 7
9.57487554… - accept 9.58 (from using 21.5 for θ)
Question 1:
1 | (a) | e.g. | B1
[1] | 1.1 | All three lengths and both angles must be correctly
placed (arrows not required)
(b) | (i) | sinθ sin40
=
4 7
θ=21.5  | M1
A1
[2] | 1.1
1.1 | Correct application of sine rule (oe) for their closed
figure or resolving horizontally 7sinθ=4sin40
21.54964123…
(b) | (ii) | P2 =42 +72 −2(4)(7)cos(180−40−θ)
P=9.57 | M1
A1
[2] | 1.1
1.1 | Correct application of cosine rule (oe) with their θ(or
just θ) or correctly resolving vertically
P=4cos40+7cosθoe for example,
72 =42 +P2 −2(4)(P)cos40
sin(140−θ)
sin40
=
P 7
9.57487554… - accept 9.58 (from using 21.5 for θ)
\includegraphics{figure_1}

Three forces of magnitudes 4 N, 7 N and $P$ N act at a point in the directions shown in the diagram.

The forces are in equilibrium.

\begin{enumerate}[label=(\alph*)]
\item Draw a closed figure to represent the three forces. [1]

\item Hence, or otherwise, find the following.
\begin{enumerate}[label=(\roman*)]
\item The value of $\theta$. [2]
\item The value of $P$. [2]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics Major 2022 Q1 [5]}}