Question 1 - A-Level Maths

OCR MEI Further Mechanics A AS 2018 June — Question 1 6 marks

Exam Board	OCR MEI
Module	Further Mechanics A AS (Further Mechanics A AS)
Year	2018
Session	June
Marks	6
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Forces, equilibrium and resultants
Type	Equilibrium of particle under coplanar forces
Difficulty	Moderate -0.8 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions and solving two simultaneous equations. While it involves trigonometry and multiple forces, it's a standard textbook exercise with a clear method (resolve horizontally and vertically, set sums to zero) that any student who has learned the topic should be able to execute routinely.
Spec	3.03m Equilibrium: sum of resolved forces = 0 3.03n Equilibrium in 2D: particle under forces

1 Forces of magnitude \(4 \mathrm {~N} , 3 \mathrm {~N} , 5 \mathrm {~N}\) and \(R \mathrm {~N}\) act on a particle in the directions shown in Fig. 1. \begin{figure}[h]

\includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-2_697_780_443_639} \captionsetup{labelformat=empty} \caption{Fig. 1}

\end{figure} The particle is in equilibrium. Find each of the following.

The value of \(R\).
The value of \(\theta\).

Show mark scheme Show mark scheme source

Question 1:

Answer	Marks
1	3+4cos60°+5cos150°+𝑅cos𝜃(= 0)

𝑅cos𝜃 = 2.5√3−5 (= −0.670)

4sin60°+5sin150°−𝑅sin𝜃 (= 0)

𝑅sin𝜃 = 2.5+2√3 (= 5.964)

R = 6.0(0)

Answer	Marks
θ = 96.4°	M1

A1

M1

A1

Answer	Marks
[6]	2.5

1.1

Answer	Marks
1.1	Oe; resolve horizontally; allow sign

errors and sin / cos confusion for

both M marks

− 0.66987

Oe; resolve vertically

6.00160…Dep M2A2

Answer	Marks
96.40846…Dep M2A2	X or 𝑅cos(180°−𝜃) for

𝑅cos𝜃 oe

Accept Y for 𝑅sin𝜃 oe

Possibly BC

Question 1:
1 | 3+4cos60°+5cos150°+𝑅cos𝜃(= 0)
𝑅cos𝜃 = 2.5√3−5 (= −0.670)
4sin60°+5sin150°−𝑅sin𝜃 (= 0)
𝑅sin𝜃 = 2.5+2√3 (= 5.964)
R = 6.0(0)
θ = 96.4° | M1
A1
M1
A1
A1
A1
[6] | 2.5
1.1
1.1
1.1
1.1
1.1 | Oe; resolve horizontally; allow sign
errors and sin / cos confusion for
both M marks
− 0.66987
Oe; resolve vertically
6.00160…Dep M2A2
96.40846…Dep M2A2 | X or 𝑅cos(180°−𝜃) for
𝑅cos𝜃 oe
Accept Y for 𝑅sin𝜃 oe
Possibly BC
Possibly BC

Show LaTeX source

1 Forces of magnitude $4 \mathrm {~N} , 3 \mathrm {~N} , 5 \mathrm {~N}$ and $R \mathrm {~N}$ act on a particle in the directions shown in Fig. 1.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{fa99d9e6-e174-42dd-ac92-7b7d112c08be-2_697_780_443_639}
\captionsetup{labelformat=empty}
\caption{Fig. 1}
\end{center}
\end{figure}

The particle is in equilibrium.

Find each of the following.

\begin{itemize}
  \item The value of $R$.
  \item The value of $\theta$.
\end{itemize}

\hfill \mbox{\textit{OCR MEI Further Mechanics A AS 2018 Q1 [6]}}

This paper (6 questions)

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Q1 6 Q2 12 Q3 9 Q4 9 Q5 13 Q6 11