Edexcel M1 2011 January — Question 3 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2011
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeUniform beam on two supports
DifficultyModerate -0.8 This is a straightforward M1 moments question requiring standard application of equilibrium conditions (sum of moments = 0, sum of forces = 0) with clearly defined geometry. Part (a) involves taking moments about one point to find reactions, while part (b) requires setting two reactions equal and solving for a distance. No novel insight or complex problem-solving is needed—just routine application of textbook methods.
Spec3.04b Equilibrium: zero resultant moment and force

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{4878b6c2-0c62-4398-8a8f-913139bc8a14-04_245_860_260_543} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A uniform beam \(A B\) has mass 20 kg and length 6 m . The beam rests in equilibrium in a horizontal position on two smooth supports. One support is at \(C\), where \(A C = 1 \mathrm {~m}\), and the other is at the end \(B\), as shown in Figure 1. The beam is modelled as a rod.
  1. Find the magnitudes of the reactions on the beam at \(B\) and at \(C\). A boy of mass 30 kg stands on the beam at the point \(D\). The beam remains in equilibrium. The magnitudes of the reactions on the beam at \(B\) and at \(C\) are now equal. The boy is modelled as a particle.
  2. Find the distance \(A D\).

AnswerMarks Guidance
PartAnswer/Working Marks
(a)Taking moments about B: \(5 \times R_C = 20g \times 3\) → \(R_C = 12g\) or \(60g/5\) or \(118\) or \(120\) M1 A1 A1
Resolving vertically: \(R_C + R_B = 20g\) → \(R_B = 8g\) or \(78.4\) or \(78\)M1 A1 (5 marks)
(b)Resolving vertically: \(50g = R + R\) B1
Taking moments about B: \(5 \times 25g = 3 \times 20g + (6-x) \times 30g\)M1 A1 A1
\(30x = 115\) → \(x = 3.8\) or better or \(23/6\) oeA1 (5 marks)
[10 marks total]
| **Part** | **Answer/Working** | **Marks** | **Guidance** |
|----------|-------------------|----------|-------------|
| (a) | Taking moments about B: $5 \times R_C = 20g \times 3$ → $R_C = 12g$ or $60g/5$ or $118$ or $120$ | M1 A1 A1 | |
| | Resolving vertically: $R_C + R_B = 20g$ → $R_B = 8g$ or $78.4$ or $78$ | M1 A1 | (5 marks) |
| (b) | Resolving vertically: $50g = R + R$ | B1 | |
| | Taking moments about B: $5 \times 25g = 3 \times 20g + (6-x) \times 30g$ | M1 A1 A1 | |
| | $30x = 115$ → $x = 3.8$ or better or $23/6$ oe | A1 | (5 marks) |
| | | | **[10 marks total]** |
3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{4878b6c2-0c62-4398-8a8f-913139bc8a14-04_245_860_260_543}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A uniform beam $A B$ has mass 20 kg and length 6 m . The beam rests in equilibrium in a horizontal position on two smooth supports. One support is at $C$, where $A C = 1 \mathrm {~m}$, and the other is at the end $B$, as shown in Figure 1. The beam is modelled as a rod.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitudes of the reactions on the beam at $B$ and at $C$.

A boy of mass 30 kg stands on the beam at the point $D$. The beam remains in equilibrium. The magnitudes of the reactions on the beam at $B$ and at $C$ are now equal. The boy is modelled as a particle.
\item Find the distance $A D$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2011 Q3 [10]}}