| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| 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 -0.8 This is a straightforward two-dimensional equilibrium problem requiring resolution of forces in perpendicular directions and basic trigonometry. Students need only apply F cos θ = 60 and F sin θ = 40, then use Pythagoras and inverse tan—standard M1 bookwork with no problem-solving insight required. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(F\cos\theta = 60\) and \(F\sin\theta = 40\) | M1 | Resolving horizontally and vertically |
| \(F = \sqrt{60^2 + 40^2}\) | ||
| \(F = \sqrt{5200} = 72.1 \text{ N}\) | A1 | Accept \(20\sqrt{13}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \frac{40}{60}\) | M1 | Correct trig ratio |
| A1 | Correct equation | |
| \(\theta = 33.7°\) | A1 | cao |
# Question 2:
## Part (a)
| $F\cos\theta = 60$ and $F\sin\theta = 40$ | M1 | Resolving horizontally and vertically |
|---|---|---|
| $F = \sqrt{60^2 + 40^2}$ | | |
| $F = \sqrt{5200} = 72.1 \text{ N}$ | A1 | Accept $20\sqrt{13}$ |
## Part (b)
| $\tan\theta = \frac{40}{60}$ | M1 | Correct trig ratio |
|---|---|---|
| | A1 | Correct equation |
| $\theta = 33.7°$ | A1 | cao |
---2 Three forces are in equilibrium in a vertical plane, as shown in the diagram. There is a vertical force of magnitude 40 N and a horizontal force of magnitude 60 N . The third force has magnitude $F$ newtons and acts at an angle $\theta$ above the horizontal.\\
\includegraphics[max width=\textwidth, alt={}, center]{788534a5-abbb-4d6a-87b2-c54e859a128a-04_490_894_456_571}
\begin{enumerate}[label=(\alph*)]
\item $\quad$ Find $F$.
\item $\quad$ Find $\theta$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2014 Q2 [5]}}