| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Maximum/minimum force for equilibrium |
| Difficulty | Standard +0.3 This is a standard A-level mechanics equilibrium problem requiring resolution of forces and application of friction law F ≤ μR. Part (a) is trivial force diagram drawing. Part (b) involves resolving horizontally and vertically, finding the normal reaction R = 50 - 15sin(θ), then applying F ≤ μR to get the required inequality. The steps are routine and follow a standard textbook approach with no novel insight required, making it slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 0 |
\includegraphics{figure_10}
A block $D$ of weight 50 N lies at rest in equilibrium on a fixed rough horizontal surface. A force of magnitude 15 N is applied to $D$ at an angle $\theta$ to the horizontal (see diagram).
\begin{enumerate}[label=(\alph*)]
\item Complete the diagram in the Printed Answer Booklet showing all the forces acting on $D$. [1]
\end{enumerate}
It is given that $D$ remains at rest and the coefficient of friction between $D$ and the surface is 0.2.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that
$$15\cos\theta - 3\sin\theta \leqslant 10.$$ [5]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2021 Q10 [6]}}