Edexcel M1 2018 January — Question 7 12 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2018
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeNon-uniform beam on supports
DifficultyStandard +0.3 This is a standard M1 moments question requiring equilibrium conditions (sum of forces = 0, sum of moments = 0) applied twice. Part (a) involves straightforward algebra with given reaction ratio to find centre of mass position. Part (b) adds a particle and repositions a support, requiring similar methodology. The techniques are routine for M1 students with no novel insight needed, making it slightly easier than average.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

7. A non-uniform rod \(A B\) has length 6 m and mass 8 kg . The rod rests in equilibrium, in a horizontal position, on two smooth supports at \(C\) and at \(D\), where \(A C = 1 \mathrm {~m}\) and \(D B = 1 \mathrm {~m}\), as shown in Figure 3. The magnitude of the reaction between the rod and the support at \(D\) is twice the magnitude of the reaction between the rod and the support at \(C\). The centre of mass of the rod is at \(G\), where \(A G = x \mathrm {~m}\).
  1. Show that \(x = \frac { 11 } { 3 }\). The support at \(C\) is moved to the point \(F\) on the rod, where \(A F = 2 \mathrm {~m}\). A particle of mass 3 kg is placed on the rod at \(A\). The rod remains horizontal and in equilibrium. The magnitude of the reaction between the rod and the support at \(D\) is \(k\) times the magnitude of the reaction between the rod and the support at \(F\).
  2. Find the value of \(k\).
    7. \begin{figure}[h]
    \captionsetup{labelformat=empty} \caption{Figure 3} \includegraphics[alt={},max width=\textwidth]{04b73f81-3316-4f26-ad98-a7be3a4b738f-20_223_1262_127_338}
    \end{figure}

7.

A non-uniform rod $A B$ has length 6 m and mass 8 kg . The rod rests in equilibrium, in a horizontal position, on two smooth supports at $C$ and at $D$, where $A C = 1 \mathrm {~m}$ and $D B = 1 \mathrm {~m}$, as shown in Figure 3. The magnitude of the reaction between the rod and the support at $D$ is twice the magnitude of the reaction between the rod and the support at $C$. The centre of mass of the rod is at $G$, where $A G = x \mathrm {~m}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $x = \frac { 11 } { 3 }$.

The support at $C$ is moved to the point $F$ on the rod, where $A F = 2 \mathrm {~m}$. A particle of mass 3 kg is placed on the rod at $A$. The rod remains horizontal and in equilibrium. The magnitude of the reaction between the rod and the support at $D$ is $k$ times the magnitude of the reaction between the rod and the support at $F$.
\item Find the value of $k$.\\
7.

\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
  \includegraphics[alt={},max width=\textwidth]{04b73f81-3316-4f26-ad98-a7be3a4b738f-20_223_1262_127_338}
\end{center}
\end{figure}

\begin{center}

\end{center}

\begin{center}

\end{center}
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2018 Q7 [12]}}