| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Equilibrium of particle under coplanar forces |
| Difficulty | Standard +0.3 This is a standard M1 equilibrium/Newton's second law problem requiring resolution of forces in two perpendicular directions. Part (a) is a 'show that' requiring straightforward application of F=ma in horizontal and vertical directions, then algebraic manipulation. Part (b) requires substituting Q=0 into the derived equation. The vector resolution is routine and the algebra is mechanical, making this slightly easier than average for M1. |
| Spec | 3.03c Newton's second law: F=ma one dimension3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 0 |
8 Three forces act in a vertical plane on an object of mass 250 kg , as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{7ac7dfd0-4c3e-4eb7-920f-ce5b24ad1281-5_481_1139_408_447}
The two forces $P$ newtons and $Q$ newtons each act at $80 ^ { \circ }$ to the horizontal. The object accelerates horizontally at $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ under the action of these forces.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$P = 125 \left( \frac { a } { \cos 80 ^ { \circ } } + \frac { g } { \sin 80 ^ { \circ } } \right)$$
\item Find the value of $a$ for which $Q$ is zero.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q8 [8]}}