Edexcel M1 2016 October — Question 3 7 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2016
SessionOctober
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeUniform beam on two supports
DifficultyModerate -0.3 This is a standard M1 moments problem requiring equilibrium conditions (sum of moments = 0, sum of vertical forces = 0) with straightforward algebra. The setup is clear, involving a uniform beam with two supports and one additional mass. While it requires careful bookkeeping of distances and forces, it follows a routine textbook approach with no conceptual surprises, making it slightly easier than average.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{6978be48-561b-49a0-a297-c8886ca66c19-06_267_1092_254_428} \captionsetup{labelformat=empty} \caption{Figure 1}
\end{figure} A plank \(A B\) has length 8 m and mass 12 kg . The plank rests on two supports. One support is at \(C\), where \(A C = 3 \mathrm {~m}\) and the other support is at \(D\), where \(A D = x\) metres. A block of mass 3 kg is placed on the plank at \(B\), as shown in Figure 1. The plank rests in equilibrium in a horizontal position. The magnitude of the force exerted on the plank by the support at \(D\) is twice the magnitude of the force exerted on the plank by the support at \(C\). The plank is modelled as a uniform rod and the block is modelled as a particle. Find the value of \(x\).

Question 3:
AnswerMarks Guidance
WorkingMarks Guidance
\((\uparrow)\quad R + 2R = 12g + 3g\)M1 A2 Vertical resolution; correct terms; \(-1\) each error
\(M(A),\quad 2Rx + 3R = 12g(4) + 3g(8)\)M1 A2 Moments equation; all terms dimensionally correct
\(x = 5.7\)A1 (7)
*Additional moments equations noted:*
- \(M(B),\ R \times 5 + S(8-x) = 12g \times 4\)
- \(M(C),\ S(x-3) = 12g \times 1 + 3g \times 5\)
- \(M(D),\ R(x-3) + 3g(8-x) = 12g(x-4)\)
## Question 3:
| Working | Marks | Guidance |
|---------|-------|---------|
| $(\uparrow)\quad R + 2R = 12g + 3g$ | M1 A2 | Vertical resolution; correct terms; $-1$ each error |
| $M(A),\quad 2Rx + 3R = 12g(4) + 3g(8)$ | M1 A2 | Moments equation; all terms dimensionally correct |
| $x = 5.7$ | A1 (7) | |

*Additional moments equations noted:*
- $M(B),\ R \times 5 + S(8-x) = 12g \times 4$
- $M(C),\ S(x-3) = 12g \times 1 + 3g \times 5$
- $M(D),\ R(x-3) + 3g(8-x) = 12g(x-4)$

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3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{6978be48-561b-49a0-a297-c8886ca66c19-06_267_1092_254_428}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}

A plank $A B$ has length 8 m and mass 12 kg . The plank rests on two supports. One support is at $C$, where $A C = 3 \mathrm {~m}$ and the other support is at $D$, where $A D = x$ metres. A block of mass 3 kg is placed on the plank at $B$, as shown in Figure 1. The plank rests in equilibrium in a horizontal position. The magnitude of the force exerted on the plank by the support at $D$ is twice the magnitude of the force exerted on the plank by the support at $C$. The plank is modelled as a uniform rod and the block is modelled as a particle.

Find the value of $x$.\\

\begin{center}

\end{center}

\hfill \mbox{\textit{Edexcel M1 2016 Q3 [7]}}