CAIE M1 2006 June — Question 3 6 marks

Exam BoardCAIE
ModuleM1 (Mechanics 1)
Year2006
SessionJune
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicForces, equilibrium and resultants
TypeEquilibrium of particle under coplanar forces
DifficultyModerate -0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions. Students must resolve horizontally and vertically to form two simultaneous equations, then solve for F and G. While it requires careful angle work and simultaneous equations, it's a routine textbook exercise with no novel insight needed—slightly easier than average due to its straightforward method.
Spec3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces

3 \includegraphics[max width=\textwidth, alt={}, center]{b5873699-d207-4cad-9518-1321dc429c15-2_508_1011_1096_568} A particle \(P\) is in equilibrium on a smooth horizontal table under the action of horizontal forces of magnitudes \(F\) N, \(F\) N, \(G\) N and 12 N acting in the directions shown. Find the values of \(F\) and \(G\). [6]

AnswerMarks Guidance
\(F\sin 50° = F\sin 20° + 12\)M1 For resolving forces in the 'j' direction
\(F = 28.3\)A1
A1
\(G = F\cos 50° + F\cos 20°\)M1 For resolving forces in the 'i' direction
\(G = 44.8\)A1 ft 6
$F\sin 50° = F\sin 20° + 12$ | M1 | For resolving forces in the 'j' direction |
$F = 28.3$ | A1 | |
| A1 | |
$G = F\cos 50° + F\cos 20°$ | M1 | For resolving forces in the 'i' direction |
$G = 44.8$ | A1 ft | 6 | Ft value of $1.5825F$ |

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\includegraphics[max width=\textwidth, alt={}, center]{b5873699-d207-4cad-9518-1321dc429c15-2_508_1011_1096_568}

A particle $P$ is in equilibrium on a smooth horizontal table under the action of horizontal forces of magnitudes $F$ N, $F$ N, $G$ N and 12 N acting in the directions shown. Find the values of $F$ and $G$. [6]

\hfill \mbox{\textit{CAIE M1 2006 Q3 [6]}}