Edexcel M1 2013 January — Question 2 9 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2013
SessionJanuary
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeUniform beam on two supports
DifficultyModerate -0.3 This is a standard M1 moments question requiring taking moments about a point and resolving vertically. Part (a) is routine (take moments about D), part (b) uses equilibrium to find reactions when equal, and part (c) requires setting up a moments equation with unknown x. All techniques are textbook standard 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

A steel girder \(AB\), of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at \(C\) and at \(D\), where \(AC = 2\) m and \(DB = 2\) m. A man of mass 80 kg stands on the girder at the point \(P\), where \(AP = 4\) m, as shown in Figure 1.
The man is modelled as a particle and the girder is modelled as a uniform rod.
  1. Find the magnitude of the reaction on the girder at the support at \(C\). [3]
The support at \(D\) is now moved to the point \(X\) on the girder, where \(XB = x\) metres. The man remains on the girder at \(P\), as shown in Figure 2.
Given that the magnitudes of the reactions at the two supports are now equal and that the girder again rests horizontally in equilibrium, find
  1. the magnitude of the reaction at the support at \(X\), [2]
  2. the value of \(x\). [4]

Part (a)
AnswerMarks Guidance
\(M(D)\), \(8R = (80g \times 6) + (200g \times 4)\)M1 A1
\(R = 160g, 1600, 1570\)A1 (3)
Part (b)
AnswerMarks Guidance
\((\uparrow)\), \(2S = 80g + 200g\)M1
\(S = 140g, 1400, 1370\)A1 (2)
Part (c)
AnswerMarks Guidance
\(M(B)\), \(Sx + (S \times 10) = (80g \times 8) + (200g \times 6)\)M1 A2
\(140x + 1400 = 640 + 1200\)
\(140x = 440\)
\(x = \frac{22}{7}\)A1 (4)
Total for Question 2: 9
### Part (a)
$M(D)$, $8R = (80g \times 6) + (200g \times 4)$ | M1 A1 |
$R = 160g, 1600, 1570$ | A1 | (3)

### Part (b)
$(\uparrow)$, $2S = 80g + 200g$ | M1 |
$S = 140g, 1400, 1370$ | A1 | (2)

### Part (c)
$M(B)$, $Sx + (S \times 10) = (80g \times 8) + (200g \times 6)$ | M1 A2 |
$140x + 1400 = 640 + 1200$ | |
$140x = 440$ | |
$x = \frac{22}{7}$ | A1 | (4)

**Total for Question 2: 9**

---
A steel girder $AB$, of mass 200 kg and length 12 m, rests horizontally in equilibrium on two smooth supports at $C$ and at $D$, where $AC = 2$ m and $DB = 2$ m. A man of mass 80 kg stands on the girder at the point $P$, where $AP = 4$ m, as shown in Figure 1.

\begin{tikzpicture}[>=stealth, thick]
  % Girder
  \draw (0,0) -- (12,0);
  \node[left] at (0,0) {$A$};
  \node[right] at (12,0) {$B$};

  % Support at C (AC = 2 m)
  \draw (2,0) -- (1.6,-0.6) -- (2.4,-0.6) -- cycle;
  \node[above] at (2,0.05) {$C$};

  % Support at D (DB = 2 m)
  \draw (10,0) -- (9.6,-0.6) -- (10.4,-0.6) -- cycle;
  \node[above] at (10,0.05) {$D$};

  % Man at P (AP = 4 m) - stick figure
  \draw (4,0.15) -- (4,0.55);
  \draw (4,0.55) -- (3.8,0.35);
  \draw (4,0.55) -- (4.2,0.35);
  \draw (4,0.15) -- (3.85,-0.0);
  \draw (4,0.15) -- (4.15,-0.0);
  \draw[fill=white] (4,0.7) circle (0.15);
  \node[below] at (4,-0.15) {$P$};

  % Dimensions
  \draw[<->] (0,0.9) -- (2,0.9);
  \node[above] at (1,0.9) {2\,m};
  \draw[<->] (0,-0.9) -- (4,-0.9);
  \node[below] at (2,-0.9) {4\,m};
  \draw[<->] (10,0.9) -- (12,0.9);
  \node[above] at (11,0.9) {2\,m};
  \draw[<->] (0,-1.7) -- (12,-1.7);
  \node[below] at (6,-1.7) {12\,m};

  \node[font=\bfseries] at (6,-2.8) {Figure 1};
\end{tikzpicture}

The man is modelled as a particle and the girder is modelled as a uniform rod.

\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the reaction on the girder at the support at $C$. [3]
\end{enumerate}

The support at $D$ is now moved to the point $X$ on the girder, where $XB = x$ metres. The man remains on the girder at $P$, as shown in Figure 2.

\begin{tikzpicture}[>=stealth, thick]
  % Girder
  \draw (0,0) -- (12,0);
  \node[left] at (0,0) {$A$};
  \node[right] at (12,0) {$B$};

  % Support at C (AC = 2 m)
  \draw (2,0) -- (1.6,-0.6) -- (2.4,-0.6) -- cycle;
  \node[above] at (2,0.05) {$C$};

  % Support at X
  \draw (9.5,0) -- (9.1,-0.6) -- (9.9,-0.6) -- cycle;
  \node[above] at (9.5,0.05) {$X$};

  % Man at P
  \draw (4,0.15) -- (4,0.55);
  \draw (4,0.55) -- (3.8,0.35);
  \draw (4,0.55) -- (4.2,0.35);
  \draw (4,0.15) -- (3.85,-0.0);
  \draw (4,0.15) -- (4.15,-0.0);
  \draw[fill=white] (4,0.7) circle (0.15);
  \node[below] at (4,-0.15) {$P$};

  % Dimensions
  \draw[<->] (0,0.9) -- (2,0.9);
  \node[above] at (1,0.9) {2\,m};
  \draw[<->] (9.5,0.9) -- (12,0.9);
  \node[above] at (10.75,0.9) {$x$\,m};
  \draw[<->] (0,-1.7) -- (12,-1.7);
  \node[below] at (6,-1.7) {12\,m};

  \node[font=\bfseries] at (6,-2.8) {Figure 2};
\end{tikzpicture}

Given that the magnitudes of the reactions at the two supports are now equal and that the girder again rests horizontally in equilibrium, find

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the magnitude of the reaction at the support at $X$, [2]
\item the value of $x$. [4]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2013 Q2 [9]}}