CAIE M1 2021 June — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2021
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeCoplanar forces in equilibrium
DifficultyModerate -0.3 This is a standard two-dimensional equilibrium problem requiring resolution of forces in perpendicular directions and solving simultaneous equations. While it involves multiple forces at various angles and requires careful trigonometry (using given sine values to find cosines), it follows a routine procedure taught in M1 with no novel insight needed. The calculation is somewhat involved but methodical, making it slightly easier than average.
Spec3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0

3 \includegraphics[max width=\textwidth, alt={}, center]{ba29ddb2-3558-4be1-a8a8-134e27a70149-04_456_767_260_689} Four coplanar forces act at a point. The magnitudes of the forces are \(20 \mathrm {~N} , 30 \mathrm {~N} , 40 \mathrm {~N}\) and \(F \mathrm {~N}\). The directions of the forces are as shown in the diagram, where \(\sin \alpha ^ { \circ } = 0.28\) and \(\sin \beta ^ { \circ } = 0.6\). Given that the forces are in equilibrium, find \(F\) and \(\theta\).

Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(F\sin\theta + 20\sin 60 - 30\sin\alpha - 40\sin\beta = 0\)M1 For resolving in either direction
Vertical: \(F\sin\theta + 20\sin 60 - 30 \times 0.28 - 40 \times 0.6 = 0\) \([F\sin\theta = 15.07949...]\)A1
Horizontal: \(F\cos\theta + 40 \times 0.8 - 30 \times 0.96 - 20\cos 60 = 0\) \([F\cos\theta = 6.8]\)A1
\(\theta = \tan^{-1}\dfrac{15.0794...}{6.8}\)M1 For method for finding \(\theta\)
\(F = \sqrt{15.07949...^2 + 6.8^2}\)M1 For method for finding \(F\)
\(\theta = 65.7,\ F = 16.5\)A1
6
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $F\sin\theta + 20\sin 60 - 30\sin\alpha - 40\sin\beta = 0$ | M1 | For resolving in either direction |
| Vertical: $F\sin\theta + 20\sin 60 - 30 \times 0.28 - 40 \times 0.6 = 0$ $[F\sin\theta = 15.07949...]$ | A1 | |
| Horizontal: $F\cos\theta + 40 \times 0.8 - 30 \times 0.96 - 20\cos 60 = 0$ $[F\cos\theta = 6.8]$ | A1 | |
| $\theta = \tan^{-1}\dfrac{15.0794...}{6.8}$ | M1 | For method for finding $\theta$ |
| $F = \sqrt{15.07949...^2 + 6.8^2}$ | M1 | For method for finding $F$ |
| $\theta = 65.7,\ F = 16.5$ | A1 | |
| | **6** | |

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\includegraphics[max width=\textwidth, alt={}, center]{ba29ddb2-3558-4be1-a8a8-134e27a70149-04_456_767_260_689}

Four coplanar forces act at a point. The magnitudes of the forces are $20 \mathrm {~N} , 30 \mathrm {~N} , 40 \mathrm {~N}$ and $F \mathrm {~N}$. The directions of the forces are as shown in the diagram, where $\sin \alpha ^ { \circ } = 0.28$ and $\sin \beta ^ { \circ } = 0.6$.

Given that the forces are in equilibrium, find $F$ and $\theta$.\\

\hfill \mbox{\textit{CAIE M1 2021 Q3 [6]}}