| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2017 |
| Session | Specimen |
| Marks | 7 |
| Topic | Motion on a slope |
| Type | Limiting equilibrium both directions |
| Difficulty | Standard +0.3 This is a standard two-part friction problem on an inclined plane requiring resolution of forces and limiting equilibrium conditions. Part (a) involves straightforward application of F=ma and friction laws to derive a given result. Part (b) requires setting up a second equation and solving simultaneously—routine algebraic manipulation with no novel insight required. Slightly easier than average due to the structured guidance and standard mechanics techniques. |
| Spec | 3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces3.03v Motion on rough surface: including inclined planes |
A body of mass 20 kg is on a rough plane inclined at angle $\alpha$ to the horizontal. The body is held at rest on the plane by the action of a force of magnitude $P$ N. The force is acting up the plane in a direction parallel to a line of greatest slope of the plane. The coefficient of friction between the body and the plane is $\mu$.
\begin{enumerate}[label=(\alph*)]
\item When $P = 100$, the body is on the point of sliding down the plane.
Show that $g \sin \alpha = g\mu \cos \alpha + 5$. [4]
\item When $P$ is increased to 150, the body is on the point of sliding up the plane.
Use this, and your answer to part (a), to find an expression for $\alpha$ in terms of $g$. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR H240/03 2017 Q10 [7]}}