CAIE M1 2019 June — Question 1 3 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2019
SessionJune
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyModerate -0.5 This is a straightforward equilibrium problem requiring resolution of forces in two perpendicular directions using given trigonometric ratios. While it involves multiple forces and angles, it's a standard textbook exercise with a clear method (resolve horizontally and vertically, show both sum to zero) and no problem-solving insight required beyond routine application of equilibrium conditions.
Spec3.03n Equilibrium in 2D: particle under forces
1 \includegraphics[max width=\textwidth, alt={}, center]{555678d3-f37d-4822-a005-de8c6094dc50-03_563_503_262_820} Given that \(\tan \alpha = \frac { 12 } { 5 }\) and \(\tan \theta = \frac { 4 } { 3 }\), show that the coplanar forces shown in the diagram are in equilibrium.
Question 1:
AnswerMarks Guidance
\((X =) 78 \times \frac{5}{13} - 50 \times \frac{3}{5} = 78\cos 67.4 - 50\cos 53.1\)M1 Attempt to resolve forces either horizontally (2 terms) or vertically (3 terms)
\((Y =) 78 \times \frac{12}{13} + 50 \times \frac{4}{5} - 112 = 78\sin 67.4 + 50\sin 53.1 - 112\)
\([X = 30 - 30 = 0, \; Y = 72 + 40 - 112 = 0]\)A1 Correct expressions horizontally and vertically
\(X = 0\) and \(Y = 0\)A1 From convincing exact calculations
*Alternative method:*
AnswerMarks Guidance
\(\frac{112}{\sin 59.5} = \frac{50}{\sin 157.4} = \frac{78}{\sin 143.1}\)M1 Attempt to use Lami, one pair of terms
A1All terms correct
\(\frac{112}{56/65} = \frac{50}{5/13} = \frac{78}{3/5} = 130\)A1 Exact values seen and used and shown to be \(= 130\); \(\cos[180-(\theta+\alpha)] = \frac{33}{65}\) and \(\sin[180-(\theta+\alpha)] = \frac{56}{65}\)
[3] 
**Question 1:**

$(X =) 78 \times \frac{5}{13} - 50 \times \frac{3}{5} = 78\cos 67.4 - 50\cos 53.1$ | **M1** | Attempt to resolve forces either horizontally (2 terms) or vertically (3 terms)

$(Y =) 78 \times \frac{12}{13} + 50 \times \frac{4}{5} - 112 = 78\sin 67.4 + 50\sin 53.1 - 112$ | |

$[X = 30 - 30 = 0, \; Y = 72 + 40 - 112 = 0]$ | **A1** | Correct expressions horizontally and vertically

$X = 0$ and $Y = 0$ | **A1** | From convincing exact calculations

*Alternative method:*

$\frac{112}{\sin 59.5} = \frac{50}{\sin 157.4} = \frac{78}{\sin 143.1}$ | **M1** | Attempt to use Lami, one pair of terms

| **A1** | All terms correct

$\frac{112}{56/65} = \frac{50}{5/13} = \frac{78}{3/5} = 130$ | **A1** | Exact values seen and used and shown to be $= 130$; $\cos[180-(\theta+\alpha)] = \frac{33}{65}$ and $\sin[180-(\theta+\alpha)] = \frac{56}{65}$

| **[3]** |

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1\\
\includegraphics[max width=\textwidth, alt={}, center]{555678d3-f37d-4822-a005-de8c6094dc50-03_563_503_262_820}

Given that $\tan \alpha = \frac { 12 } { 5 }$ and $\tan \theta = \frac { 4 } { 3 }$, show that the coplanar forces shown in the diagram are in equilibrium.\\

\hfill \mbox{\textit{CAIE M1 2019 Q1 [3]}}
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