Edexcel M1 2008 June — Question 6 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2008
SessionJune
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeNon-uniform beam on supports
DifficultyStandard +0.3 This is a standard M1 moments question with two parts: (a) requires taking moments about a point with uniform rod assumption (routine calculation), and (b) extends to non-uniform rod requiring algebraic manipulation with two equations. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

6. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{9dbbbc01-fb66-460d-a42e-2c37ec8b451a-08_392_678_260_614} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} A plank \(A B\) has mass 12 kg and length 2.4 m . A load of mass 8 kg is attached to the plank at the point \(C\), where \(A C = 0.8 \mathrm {~m}\). The loaded plank is held in equilibrium, with \(A B\) horizontal, by two vertical ropes, one attached at \(A\) and the other attached at \(B\), as shown in Figure 2. The plank is modelled as a uniform rod, the load as a particle and the ropes as light inextensible strings.
  1. Find the tension in the rope attached at \(B\). The plank is now modelled as a non-uniform rod. With the new model, the tension in the rope attached at \(A\) is 10 N greater than the tension in the rope attached at \(B\).
  2. Find the distance of the centre of mass of the plank from \(A\).

AnswerMarks Guidance
(a) \(M(A): 8g \times 0.8 + 12g \times 1.2 = X \times 2.4\) giving \(X \approx 85 \text{ (N)}\)M1 A1 DM1 A1 (4) Accept 84.9, \(\frac{26g}{3}\)
(b) \(R(\uparrow): (X + 10) + X = 8g + 12g\) giving \((X = 93)\)M1 B1 A1
\(M(A): 8g \times 0.8 + 12g \times x = X \times 2.4\) giving \(x = 1.4 \text{ (m)}\)M1 A1 (6) [10] Accept 1.36
**(a)** $M(A): 8g \times 0.8 + 12g \times 1.2 = X \times 2.4$ giving $X \approx 85 \text{ (N)}$ | M1 A1 DM1 A1 | (4) Accept 84.9, $\frac{26g}{3}$

**(b)** $R(\uparrow): (X + 10) + X = 8g + 12g$ giving $(X = 93)$ | M1 B1 A1 | 

$M(A): 8g \times 0.8 + 12g \times x = X \times 2.4$ giving $x = 1.4 \text{ (m)}$ | M1 A1 | (6) [10] Accept 1.36

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6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{9dbbbc01-fb66-460d-a42e-2c37ec8b451a-08_392_678_260_614}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

A plank $A B$ has mass 12 kg and length 2.4 m . A load of mass 8 kg is attached to the plank at the point $C$, where $A C = 0.8 \mathrm {~m}$. The loaded plank is held in equilibrium, with $A B$ horizontal, by two vertical ropes, one attached at $A$ and the other attached at $B$, as shown in Figure 2. The plank is modelled as a uniform rod, the load as a particle and the ropes as light inextensible strings.
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the rope attached at $B$.

The plank is now modelled as a non-uniform rod. With the new model, the tension in the rope attached at $A$ is 10 N greater than the tension in the rope attached at $B$.
\item Find the distance of the centre of mass of the plank from $A$.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2008 Q6 [10]}}