Question 3 - A-Level Maths

CAIE M1 2008 June — Question 3 5 marks

Exam Board	CAIE
Module	M1 (Mechanics 1)
Year	2008
Session	June
Marks	5
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 and solving simultaneous equations. While it involves trigonometry and algebraic manipulation, it's a routine textbook exercise with a clear method (resolve horizontally and vertically) that M1 students practice extensively. Slightly easier than average due to its straightforward approach and limited conceptual demand.
Spec	3.03e Resolve forces: two dimensions 3.03m Equilibrium: sum of resolved forces = 0 3.03n Equilibrium in 2D: particle under forces

3 \includegraphics[max width=\textwidth, alt={}, center]{ee138c3f-51e1-4a69-9750-9eb49ac87e22-2_520_565_1009_792} Three horizontal forces of magnitudes \(F \mathrm {~N} , 13 \mathrm {~N}\) and 10 N act at a fixed point \(O\) and are in equilibrium. The directions of the forces are as shown in the diagram. Find, in either order, the value of \(\theta\) and the value of \(F\).

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Question 3:

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\([F\cos\theta° = 10,\ F\sin\theta° = 13]\)	M1	For resolving forces in i and j directions or sketching a triangle of forces (with 10, 13 and F shown)
\([\tan\theta° = 13/10,\ \sqrt{269}\sin\theta° = 13]\)	M1	For an equation in \(\theta\) only
\(\theta = 52.4\)	A1
\([F^2 = 10^2 + 13^2,\ F\cos52.4° = 10]\)	M1	For an equation in F only
\(F = 16.4\)	A1 [5]
Alternative (scale drawing):
M1	For scale drawing of correct triangle
M1	For measuring \(\theta\) and finding value in range \([51, 54]\)
\(\theta = 52.4\)	A1
M1	For measuring F and finding value in range \([15.5, 17.5]\)
\(F = 16.4\)	A1 [5]

## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[F\cos\theta° = 10,\ F\sin\theta° = 13]$ | M1 | For resolving forces in **i** and **j** directions or sketching a triangle of forces (with 10, 13 and F shown) |
| $[\tan\theta° = 13/10,\ \sqrt{269}\sin\theta° = 13]$ | M1 | For an equation in $\theta$ only |
| $\theta = 52.4$ | A1 | |
| $[F^2 = 10^2 + 13^2,\ F\cos52.4° = 10]$ | M1 | For an equation in F only |
| $F = 16.4$ | A1 [5] | |
| **Alternative (scale drawing):** | | |
| | M1 | For scale drawing of correct triangle |
| | M1 | For measuring $\theta$ and finding value in range $[51, 54]$ |
| $\theta = 52.4$ | A1 | |
| | M1 | For measuring F and finding value in range $[15.5, 17.5]$ |
| $F = 16.4$ | A1 [5] | |

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3\\
\includegraphics[max width=\textwidth, alt={}, center]{ee138c3f-51e1-4a69-9750-9eb49ac87e22-2_520_565_1009_792}

Three horizontal forces of magnitudes $F \mathrm {~N} , 13 \mathrm {~N}$ and 10 N act at a fixed point $O$ and are in equilibrium. The directions of the forces are as shown in the diagram. Find, in either order, the value of $\theta$ and the value of $F$.

\hfill \mbox{\textit{CAIE M1 2008 Q3 [5]}}

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Q1 4 Q2 4 Q3 5 Q4 7 Q5 8 Q6 9 Q7 13