| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Standard +0.3 This is a standard equilibrium problem requiring resolution of forces in two perpendicular directions and basic trigonometry. Part (i) involves solving simultaneous equations from resolving forces, part (ii) uses Pythagoras, and part (iii) applies vector addition. While multi-part with several steps, it follows a routine textbook approach with no novel insight required, making it slightly easier than average. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\sin 2\theta = 6\sin\theta\) | B1 | Resolving horizontally |
| \(8\sin\theta\cos\theta = 6\sin\theta\) | M1 | Use of double angle formulae |
| \(\cos\theta = \frac{3}{4} \Rightarrow \theta = 41.4°\) | A1, A1 [4] | AG |
| Answer | Marks | Guidance |
|---|---|---|
| \(P = 4\cos 2\theta + 6\cos\theta\) | M1 | Resolving vertically – allow sin/cos errors |
| \(P = 5\) | A1 [2] | Accept 5.002 or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(3\sin\theta\) and \(P - 3\cos\theta\) | B1 | Resolving horizontally and vertically |
| \(\sqrt{(3\sin\theta)^2 + (P - 3\cos\theta)^2}\) | M1 | Pythagoras on two forces – both must include 41.4 |
| Magnitude is \(3.39\text{N}\) | A1 [3] | 3.4 or better |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\alpha = \frac{P - 3\cos\theta}{3\sin\theta}\) | M1 | Where \(\alpha\) is the angle below the horizontal |
| \(54.2°\) below the horizontal | A1 [2] | 54.2 or better – must indicate 'below horizontal' |
## Question 10:
### Part (i):
$4\sin 2\theta = 6\sin\theta$ | B1 | Resolving horizontally |
$8\sin\theta\cos\theta = 6\sin\theta$ | M1 | Use of double angle formulae |
$\cos\theta = \frac{3}{4} \Rightarrow \theta = 41.4°$ | A1, A1 [4] | AG |
### Part (ii):
$P = 4\cos 2\theta + 6\cos\theta$ | M1 | Resolving vertically – allow sin/cos errors |
$P = 5$ | A1 [2] | Accept 5.002 or better |
### Part (iii)(a):
$3\sin\theta$ and $P - 3\cos\theta$ | B1 | Resolving horizontally and vertically |
$\sqrt{(3\sin\theta)^2 + (P - 3\cos\theta)^2}$ | M1 | Pythagoras on two forces – both must include 41.4 |
Magnitude is $3.39\text{N}$ | A1 [3] | 3.4 or better |
### Part (iii)(b):
$\tan\alpha = \frac{P - 3\cos\theta}{3\sin\theta}$ | M1 | Where $\alpha$ is the angle below the horizontal |
$54.2°$ below the horizontal | A1 [2] | 54.2 or better – must indicate 'below horizontal' |
---10 Three forces, of magnitudes $4 \mathrm {~N} , 6 \mathrm {~N}$ and $P \mathrm {~N}$, act at a point in the directions shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{d5ab20c8-afd5-473e-8238-96762bd3786d-7_604_601_306_724}
The forces are in equilibrium.\\
(i) Show that $\theta = 41.4 ^ { \circ }$, correct to 3 significant figures.\\
(ii) Hence find the value of $P$.
The force of magnitude 4 N is now removed and the force of magnitude 6 N is replaced by a force of magnitude 3 N acting in the same direction.\\
(iii) Find
\begin{enumerate}[label=(\alph*)]
\item the magnitude of the resultant of the two remaining forces,
\item the direction of the resultant of the two remaining forces.
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2018 Q10 [11]}}