| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform rod on supports or with strings |
| Difficulty | Standard +0.3 This is a straightforward moments equilibrium problem requiring students to set up two equations (vertical forces and moments about a point) and solve for the unknown distance. While it involves a non-uniform rod, the solution method is standard M1 fare with no conceptual surprises—slightly easier than average due to the clear setup and routine algebraic manipulation. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| resolve \(\uparrow\): \(R + T = 10g\); \(R + \frac{3}{2}R = 10g\) | M2 | |
| \(\frac{5}{2}R = 10g\); \(\therefore R = 4g\) so \(T = 6g\) | A1 | |
| moments about pivot: \(10gx - 4(6g) = 0\) | M1 | |
| \(10gx = 24g\), so \(x = 2.4\) and hence c.o.m. is 4.4 m from A | M1 A1 | (6) |
resolve $\uparrow$: $R + T = 10g$; $R + \frac{3}{2}R = 10g$ | M2 |
$\frac{5}{2}R = 10g$; $\therefore R = 4g$ so $T = 6g$ | A1 |
moments about pivot: $10gx - 4(6g) = 0$ | M1 |
$10gx = 24g$, so $x = 2.4$ and hence c.o.m. is 4.4 m from A | M1 A1 | (6)1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c762bd90-5b57-428a-a7a8-291a1a643a14-2_286_933_203_452}
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\caption{Fig. 1}
\end{center}
\end{figure}
Figure 1 shows a non-uniform beam $A B$ of mass 10 kg and length 6 m resting in a horizontal position on a single support 2 m from $A$. The beam is supported at $B$ by a vertical string.
Given that the magnitude of the tension in the string is 1.5 times the magnitude of the reaction at the support, find the distance of the centre of mass of the beam from $A$.\\
(6 marks)\\
\hfill \mbox{\textit{Edexcel M1 Q1 [6]}}