Edexcel M1 2002 January — Question 5 10 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2002
SessionJanuary
Marks10
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeBeam suspended by vertical ropes
DifficultyModerate -0.3 This is a standard M1 moments problem requiring taking moments about a point and resolving vertically. The setup is straightforward with clearly defined forces and positions, though it involves some algebraic manipulation across multiple steps. Slightly easier than average due to being a textbook application of equilibrium principles with no conceptual surprises.
Spec3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force

\includegraphics{figure_1} A heavy uniform steel girder \(AB\) has length 10 m. A load of weight 150 N is attached to the girder at \(A\) and a load of weight 250 N is attached to the girder at \(B\). The loaded girder hangs in equilibrium in a horizontal position, held by two vertical steel cables attached to the girder at the points \(C\) and \(D\), where \(AC = 1\) m and \(DB = 3\) m, as shown in Fig. 1. The girder is modelled as a uniform rod, the loads as particles and the cables as light inextensible strings. The tension in the cable at \(D\) is three times the tension in the cable at \(C\).
  1. Draw a diagram showing all the forces acting on the girder. [2]
Find
  1. the tension in the cable at \(C\), [5]
  2. the weight of the girder. [2]
  3. Explain how you have used the fact that the girder is uniform. [1]

Part (a)
\(1\uparrow T\) 4 O 2 \(\uparrow 3T\)
AnswerMarks Guidance
\(150 \downarrow\) \(\omega\) \(\downarrow w\) \(\downarrow 250\) all rest \(\omega\)B1 G1 (2)
Part (b)
\(M(o): 150.5 + 3T.2 = T.4 + 250.5\)
Solve: \(T = 250\text{ N}\)
AnswerMarks Guidance
[Allow M1 A2, 1,0 for moments eqn at any pt. Then m1 A1 for complete soln \(\Rightarrow T =\) ]M1 A2,1,0 M1 A1 (5)
Part (c)
\(R(\uparrow)\) \(4T = 450 + W\) \(\Rightarrow W = 600\text{ N}\)
AnswerMarks
(M1 needs complete soln \(\Rightarrow W\text{R} =\) )M1 A1 (2)
Part (d)
AnswerMarks
By having weight act at centre/mid-pt.B1 (1) (18)
## Part (a)
$1\uparrow T$ 4 O 2 $\uparrow 3T$ 

$150 \downarrow$ $\omega$ $\downarrow w$ $\downarrow 250$ all rest $\omega$ | B1 | G1 (2)

## Part (b)
$M(o): 150.5 + 3T.2 = T.4 + 250.5$

Solve: $T = 250\text{ N}$

[Allow M1 A2, 1,0 for moments eqn at any pt. Then m1 A1 for complete soln $\Rightarrow T =$ ] | M1 A2,1,0 | M1 A1 (5)

## Part (c)
$R(\uparrow)$ $4T = 450 + W$ $\Rightarrow W = 600\text{ N}$

(M1 needs complete soln $\Rightarrow W\text{R} =$ ) | M1 A1 (2)

## Part (d)
By having weight act at centre/mid-pt. | B1 (1) (18)

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\includegraphics{figure_1}

A heavy uniform steel girder $AB$ has length 10 m. A load of weight 150 N is attached to the girder at $A$ and a load of weight 250 N is attached to the girder at $B$. The loaded girder hangs in equilibrium in a horizontal position, held by two vertical steel cables attached to the girder at the points $C$ and $D$, where $AC = 1$ m and $DB = 3$ m, as shown in Fig. 1. The girder is modelled as a uniform rod, the loads as particles and the cables as light inextensible strings. The tension in the cable at $D$ is three times the tension in the cable at $C$.

\begin{enumerate}[label=(\alph*)]
\item Draw a diagram showing all the forces acting on the girder. [2]
\end{enumerate}

Find
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item the tension in the cable at $C$, [5]
\item the weight of the girder. [2]
\item Explain how you have used the fact that the girder is uniform. [1]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2002 Q5 [10]}}