| Exam Board | SPS |
|---|---|
| Module | SPS FM Mechanics (SPS FM Mechanics) |
| Year | 2026 |
| Session | January |
| Marks | 12 |
| Topic | Moments |
| Type | Rod on smooth peg or cylinder |
| Difficulty | Challenging +1.2 This is a multi-part statics problem requiring standard mechanics techniques: geometry (Pythagorean theorem), resolving forces, and taking moments. While it has multiple steps and requires careful setup, the methods are all standard A-level Further Maths mechanics procedures with no novel insights needed. The geometry is straightforward (5-12-13 triangle), and the moment equation about a well-chosen point leads directly to the result. Part (c) requires systematic resolution and use of previous results but follows established patterns. Slightly above average difficulty due to the extended working and coordination of multiple techniques. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
\includegraphics{figure_1}
A smooth solid hemisphere is fixed with its flat surface in contact with rough horizontal ground. The hemisphere has centre $O$ and radius $5a$. A uniform rod $AB$, of length $16a$ and weight $W$, rests in equilibrium on the hemisphere with end $A$ on the ground. The rod rests on the hemisphere at the point $C$, where $AC = 12a$ and angle $CAO = \alpha$, as shown in Figure 1.
Points $A$, $C$, $B$ and $O$ all lie in the same vertical plane.
\begin{enumerate}[label=(\alph*)]
\item Explain why $AO = 13a$ [1]
\end{enumerate}
The normal reaction on the rod at $C$ has magnitude $kW$
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $k = \frac{8}{13}$ [3]
\end{enumerate}
The resultant force acting on the rod at $A$ has magnitude $R$ and acts upwards at $\theta°$ to the horizontal.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find
\begin{enumerate}[label=(\roman*)]
\item an expression for $R$ in terms of $W$
\item the value of $\theta$
\end{enumerate}
[8]
\end{enumerate}
\hfill \mbox{\textit{SPS SPS FM Mechanics 2026 Q8 [12]}}