Question 2 - A-Level Maths

OCR MEI Further Mechanics A AS 2022 June — Question 2 7 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2022
Session	June
Marks	7
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Forces, equilibrium and resultants
Type	Triangle of forces method
Difficulty	Standard +0.3 This is a straightforward application of the triangle of forces method for equilibrium. Part (a) requires drawing a force triangle, part (b) uses the cosine rule to find an angle given three sides, and part (c) uses the sine rule to find possible force magnitudes. While it requires understanding of equilibrium and triangle geometry, these are standard techniques with no novel problem-solving required, making it slightly easier than average.
Spec	3.03a Force: vector nature and diagrams 3.03m Equilibrium: sum of resolved forces = 0 3.03n Equilibrium in 2D: particle under forces

2 Three forces, of magnitudes \(33 \mathrm {~N} , 45 \mathrm {~N}\) and \(P \mathrm {~N}\), act at a point in the directions shown in the diagram. The system is in equilibrium. \includegraphics[max width=\textwidth, alt={}, center]{d1ec7861-dc8b-450b-8e05-c70479ab0dc2-3_501_703_342_239}

Draw a triangle of forces for the system shown above. Your diagram should include the magnitudes of the forces ( \(33 \mathrm {~N} , 45 \mathrm {~N}\) and \(P \mathrm {~N}\) ) and angle \(\theta\).
If \(P = 38\), find, in degrees, the value of \(\theta\).
If \(\theta = 40 ^ { \circ }\), determine the possible values for \(P\).

Show mark scheme Show mark scheme source

Question 2:

Answer	Marks	Guidance
2	(a)	45 N
33 N P N	B1
B1	1.1
1.1	Closed triangle with all lengths

present, and sides parallel to the

relevant forces. Arrowheads and

units not needed. Allow magnitudes

written at arrowheads

θ correctly indicated on diagram

(not necessarily as internal angle of

the triangle)

[2]

Answer	Marks
(b)	452+382−332

cos𝜃 =

Answer	Marks	Guidance
2⋅45⋅38	M1	1.1a

e.g 332=452+382˗2(45)(38)cosθ

Answer	Marks	Guidance
4 5 .9 0 0 5   	A1	1.1

[2]

Answer	Marks
(c)	Let the angle between the 33 and P sides be A.

sin40°

sin𝐴 = ×45

Answer	Marks	Guidance
33	M1	1.1

𝐴 = 61.2263…° or 𝐴 = 118.7736…°

Answer	Marks	Guidance
So the third angle is either 2 1 .2 2 6 3  or 7 8 .7 7 3 6 	M1	1.1

complement of each other in 180°

and two values for third angle:

1 4 0  − t h e i r p o s s i b i li t i e s f o r A

So either P = 3 34  s i n ( 2 1 .2 2 6 3  ) = 1 8 .5 8 7 4

sin 0 

or P = 3 34  s i n ( 7 8 .7 7 3 6  ) = 5 0 .3 5 6 5

Answer	Marks	Guidance
sin 0 	A1	1.1

Alternative to (c):

Answer	Marks	Guidance
3 3 2 = P 2 + 4 5 2 − 2  P  4 5 c o s 4 0 	M1	M1
 P 2 − 9 0 c o s 4 0   P + 9 3 6 = 0	 P 2 − 9 0 c o s 4 0   P + 9 3 6 = 0	M1

form, e.g. P2 ˗ 68.94P = ˗936

Answer	Marks	Guidance
P = 1 8 .5 8 7 4 or 5 0 .3 5 6 5	A1	Both

[3]

Question 2:
2 | (a) | 45 N
33 N P N | B1
B1 | 1.1
1.1 | Closed triangle with all lengths
present, and sides parallel to the
relevant forces. Arrowheads and
units not needed. Allow magnitudes
written at arrowheads
θ correctly indicated on diagram
(not necessarily as internal angle of
the triangle)
[2]
(b) | 452+382−332
cos𝜃 =
2⋅45⋅38 | M1 | 1.1a | Correct application of cos rule
e.g 332=452+382˗2(45)(38)cosθ
4 5 .9 0 0 5    | A1 | 1.1 | Condone 0.80111… radians
[2]
(c) | Let the angle between the 33 and P sides be A.
sin40°
sin𝐴 = ×45
33 | M1 | 1.1 | Correct application of sine rule
𝐴 = 61.2263…° or 𝐴 = 118.7736…°
So the third angle is either 2 1 .2 2 6 3  or 7 8 .7 7 3 6  | M1 | 1.1 | Two possibilities for A, the
complement of each other in 180°
and two values for third angle:
1 4 0  − t h e i r p o s s i b i li t i e s f o r A
So either P = 3 34  s i n ( 2 1 .2 2 6 3  ) = 1 8 .5 8 7 4
sin 0 
or P = 3 34  s i n ( 7 8 .7 7 3 6  ) = 5 0 .3 5 6 5
sin 0  | A1 | 1.1 | Both
Alternative to (c):
3 3 2 = P 2 + 4 5 2 − 2  P  4 5 c o s 4 0  | M1 | M1 | Correct application of cos rule | Correct application of cos rule
 P 2 − 9 0 c o s 4 0   P + 9 3 6 = 0 |  P 2 − 9 0 c o s 4 0   P + 9 3 6 = 0 | M1 | Rearranging to standard quadratic
form, e.g. P2 ˗ 68.94P = ˗936
P = 1 8 .5 8 7 4 or 5 0 .3 5 6 5 | A1 | Both
[3]

Show LaTeX source

2 Three forces, of magnitudes $33 \mathrm {~N} , 45 \mathrm {~N}$ and $P \mathrm {~N}$, act at a point in the directions shown in the diagram. The system is in equilibrium.\\
\includegraphics[max width=\textwidth, alt={}, center]{d1ec7861-dc8b-450b-8e05-c70479ab0dc2-3_501_703_342_239}
\begin{enumerate}[label=(\alph*)]
\item Draw a triangle of forces for the system shown above. Your diagram should include the magnitudes of the forces ( $33 \mathrm {~N} , 45 \mathrm {~N}$ and $P \mathrm {~N}$ ) and angle $\theta$.
\item If $P = 38$, find, in degrees, the value of $\theta$.
\item If $\theta = 40 ^ { \circ }$, determine the possible values for $P$.
\end{enumerate}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2022 Q2 [7]}}

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