Question 3 - A-Level Maths

CAIE M1 2018 June — Question 3 4 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2018
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Forces, equilibrium and resultants
Type	Equilibrium of particle under coplanar forces
Difficulty	Moderate -0.5 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions to find two unknowns. While it involves trigonometry and simultaneous equations, it's a routine textbook exercise with a well-established method, making it slightly easier than average for A-level mechanics.
Spec	3.03m Equilibrium: sum of resolved forces = 0

3 \includegraphics[max width=\textwidth, alt={}, center]{16640429-198d-4ea9-a2f6-6e2ef6ac1b4a-05_535_616_260_762} The three coplanar forces shown in the diagram have magnitudes \(3 \mathrm {~N} , 2 \mathrm {~N}\) and \(P \mathrm {~N}\). Given that the three forces are in equilibrium, find the values of \(\theta\) and \(P\).

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Question 3:

Answer	Marks	Guidance
Answer	Marks	Guidance
\([3\cos 60 = 2\cos\theta]\)	M1	Attempt to resolve forces horizontally (2 terms)
\(\theta = 41.4\)	A1
\([P = 3\sin 60 + 2\sin\theta]\)	M1	Attempt to resolve forces vertically (3 terms)
\(P = 3.92\)	A1
First alternative: \(\frac{P}{\sin(120-\theta)} = \frac{2}{\sin 150} = \frac{3}{\sin(90+\theta)}\)	M1	Attempt two terms of Lami's equation which can be used to find \(\theta\)
\(\theta = 41.4\)	A1
M1	Attempt an equation which can be used to find \(P\)
\(P = 3.92\)	A1
Second alternative: \(\frac{P}{\sin(60+\theta)} = \frac{2}{\sin 30} = \frac{3}{\sin(90-\theta)}\)	M1	Attempt two terms from the triangle of forces which can be used to find \(\theta\)
\(\theta = 41.4\)	A1
M1	Attempt an equation which can be used to find \(P\)
\(P = 3.92\)	A1

## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $[3\cos 60 = 2\cos\theta]$ | M1 | Attempt to resolve forces horizontally (2 terms) |
| $\theta = 41.4$ | A1 | |
| $[P = 3\sin 60 + 2\sin\theta]$ | M1 | Attempt to resolve forces vertically (3 terms) |
| $P = 3.92$ | A1 | |
| **First alternative:** $\frac{P}{\sin(120-\theta)} = \frac{2}{\sin 150} = \frac{3}{\sin(90+\theta)}$ | M1 | Attempt two terms of Lami's equation which can be used to find $\theta$ |
| $\theta = 41.4$ | A1 | |
| | M1 | Attempt an equation which can be used to find $P$ |
| $P = 3.92$ | A1 | |
| **Second alternative:** $\frac{P}{\sin(60+\theta)} = \frac{2}{\sin 30} = \frac{3}{\sin(90-\theta)}$ | M1 | Attempt two terms from the triangle of forces which can be used to find $\theta$ |
| $\theta = 41.4$ | A1 | |
| | M1 | Attempt an equation which can be used to find $P$ |
| $P = 3.92$ | A1 | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{16640429-198d-4ea9-a2f6-6e2ef6ac1b4a-05_535_616_260_762}

The three coplanar forces shown in the diagram have magnitudes $3 \mathrm {~N} , 2 \mathrm {~N}$ and $P \mathrm {~N}$. Given that the three forces are in equilibrium, find the values of $\theta$ and $P$.\\

\hfill \mbox{\textit{CAIE M1 2018 Q3 [4]}}

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