CAIE M1 2016 June — Question 4 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2016
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeCoplanar forces in equilibrium
DifficultyStandard +0.3 This is a standard coplanar forces equilibrium problem requiring resolution in two perpendicular directions and use of given trigonometric information. While it involves multiple forces and angles, the method is routine for M1: resolve horizontally and vertically, use the given tan ratio to find sin/cos values, then solve simultaneous equations. Slightly above average due to the trigonometric manipulation required, but still a textbook-style question with a well-established solution method.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces
4 \includegraphics[max width=\textwidth, alt={}, center]{099c81e0-a95a-4f98-801c-32d905ef7c7d-2_446_752_1521_699} Coplanar forces of magnitudes \(50 \mathrm {~N} , 48 \mathrm {~N} , 14 \mathrm {~N}\) and \(P \mathrm {~N}\) act at a point in the directions shown in the diagram. The system is in equilibrium. Given that \(\tan \alpha = \frac { 7 } { 24 }\), find the values of \(P\) and \(\theta\).
Question 4:
AnswerMarks Guidance
AnswerMark Guidance
\(P\cos\theta = 48\cos\alpha - 14\sin\alpha\) and/or \(P\sin\theta = 50 - 48\sin\alpha - 14\cos\alpha\)M1 For resolving forces horizontally and/or vertically
\(P\cos\theta = 48(24/25) - 14(7/25) = 42.16\)A1 Allow \(\alpha = 16.3\) used throughout
\(P\sin\theta = 50 - 48(7/25) - 14(24/25) = 23.12\)A1
M1For attempting to find \(P\) or \(\theta\)
\(P = \sqrt{42.16^2 + 23.12^2} = 48.1\)A1 Allow \(P = 34\sqrt{2}\)
\(\tan\theta = \frac{23.12}{42.16}\), \(\theta = 28.7\)B1 [6] 
## Question 4:
| Answer | Mark | Guidance |
|--------|------|----------|
| $P\cos\theta = 48\cos\alpha - 14\sin\alpha$ and/or $P\sin\theta = 50 - 48\sin\alpha - 14\cos\alpha$ | M1 | For resolving forces horizontally and/or vertically |
| $P\cos\theta = 48(24/25) - 14(7/25) = 42.16$ | A1 | Allow $\alpha = 16.3$ used throughout |
| $P\sin\theta = 50 - 48(7/25) - 14(24/25) = 23.12$ | A1 | |
| | M1 | For attempting to find $P$ or $\theta$ |
| $P = \sqrt{42.16^2 + 23.12^2} = 48.1$ | A1 | Allow $P = 34\sqrt{2}$ |
| $\tan\theta = \frac{23.12}{42.16}$, $\theta = 28.7$ | B1 | [6] |

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\includegraphics[max width=\textwidth, alt={}, center]{099c81e0-a95a-4f98-801c-32d905ef7c7d-2_446_752_1521_699}

Coplanar forces of magnitudes $50 \mathrm {~N} , 48 \mathrm {~N} , 14 \mathrm {~N}$ and $P \mathrm {~N}$ act at a point in the directions shown in the diagram. The system is in equilibrium. Given that $\tan \alpha = \frac { 7 } { 24 }$, find the values of $P$ and $\theta$.

\hfill \mbox{\textit{CAIE M1 2016 Q4 [6]}}
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