| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Standard +0.3 This is a standard M1 moments question requiring taking moments about two points and using the 2:1 reaction ratio condition. Part (b) adds a particle but follows the same method with the 4:1 ratio. The setup is straightforward with clear numerical values, requiring only systematic application of equilibrium conditions (ΣF=0, ΣM=0) without geometric complexity or novel insight. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
\includegraphics{figure_5}
A uniform rod $AB$ has length 2 m and mass 50 kg. The rod is in equilibrium in a horizontal position, resting on two smooth supports at $C$ and $D$, where $AC = 0.2$ metres and $DB = x$ metres, as shown in Figure 5. Given that the magnitude of the reaction on the rod at $D$ is twice the magnitude of the reaction on the rod at $C$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $x$. [6]
\end{enumerate}
The support at $D$ is now moved to the point $E$ on the rod, where $EB = 0.4$ metres. A particle of mass $m$ kg is placed on the rod at $B$, and the rod remains in equilibrium in a horizontal position. Given that the magnitude of the reaction on the rod at $E$ is four times the magnitude of the reaction on the rod at $C$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item find the value of $m$. [7]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q8 [13]}}