Edexcel M1 2014 June — Question 6 11 marks

Exam BoardEdexcel
ModuleM1 (Mechanics 1)
Year2014
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeNon-uniform beam on supports
DifficultyStandard +0.3 This is a standard M1 moments question requiring equilibrium conditions and taking moments about a point. Part (a) involves basic moment calculations with two unknowns (tensions in 2:1 ratio). Parts (b) and (c) extend this with an added load, requiring algebraic manipulation and inequality solving for taut rope conditions. All techniques are routine for M1 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]{b896c631-00a0-46c5-bce9-16d65f6e3095-11_600_969_127_491} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure} A non-uniform beam \(A D\) has weight \(W\) newtons and length 4 m . It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. The ropes are attached to two points \(B\) and \(C\) on the beam, where \(A B = 1 \mathrm {~m}\) and \(C D = 1 \mathrm {~m}\), as shown in Figure 3. The tension in the rope attached to \(C\) is double the tension in the rope attached to \(B\). The beam is modelled as a rod and the ropes are modelled as light inextensible strings.
  1. Find the distance of the centre of mass of the beam from \(A\). A small load of weight \(k W\) newtons is attached to the beam at \(D\). The beam remains in equilibrium in a horizontal position. The load is modelled as a particle. Find
  2. an expression for the tension in the rope attached to \(B\), giving your answer in terms of \(k\) and \(W\),
  3. the set of possible values of \(k\) for which both ropes remain taut.

Question 6:
Part (a):
AnswerMarks Guidance
Working/AnswerMarks Guidance
Resolving vertically: \(T + 2T\ (=3T) = W\)M1A1
Moments about \(B\): \(2 \times 2T = (d-1)W\)M1A1
Substitute and solve for \(d\): \(2 \times 2T = (d-1)3T\)DM1 Dependent on first and second M marks
\(d = \frac{7}{3}\) (m)A1 \(2.3\) (m) or better
N.B. If \(Wg\) used, mark as misread. Above 4 marks can be scored if \(d\) measured from different point.
Alternative: Single equation taking moments about centre of mass: \(2T(3-d) = T(d-1)\) scores M2A2 (-1 each error); then DM1 A1 for \(d = \frac{7}{3}\).
Part (b):
AnswerMarks Guidance
Working/AnswerMarks Guidance
Moments about \(C\): \((T_B \times 2) + (kW \times 1) = W \times \frac{2}{3}\)M1A1 Equation in \(T_B\) and \(W\) only
\(T_B = W\dfrac{(2-3k)}{6}\) or equivalentA1 N.B. M0 if tensions from part (a) are used
Part (c):
AnswerMarks Guidance
Working/AnswerMarks Guidance
Solving \(T_B \geq 0\) or \(T_B > 0\) for \(k\)M1
\(0 < k \leq \frac{2}{3}\) or \(0 < k < \frac{2}{3}\) onlyA1 N.B. If \(T=0 \Rightarrow k=\frac{2}{3}\) then answer is M0; if \(T_C \geq 0\) or \(T_C > 0\) also solved, can still score M1 and possibly A1
## Question 6:

### Part (a):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Resolving vertically: $T + 2T\ (=3T) = W$ | M1A1 | |
| Moments about $B$: $2 \times 2T = (d-1)W$ | M1A1 | |
| Substitute and solve for $d$: $2 \times 2T = (d-1)3T$ | DM1 | Dependent on first and second M marks |
| $d = \frac{7}{3}$ (m) | A1 | $2.3$ (m) or better |

**N.B.** If $Wg$ used, mark as misread. Above 4 marks can be scored if $d$ measured from different point.

**Alternative:** Single equation taking moments about centre of mass: $2T(3-d) = T(d-1)$ scores M2A2 (-1 each error); then DM1 A1 for $d = \frac{7}{3}$.

---

### Part (b):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Moments about $C$: $(T_B \times 2) + (kW \times 1) = W \times \frac{2}{3}$ | M1A1 | Equation in $T_B$ and $W$ only |
| $T_B = W\dfrac{(2-3k)}{6}$ or equivalent | A1 | **N.B.** M0 if tensions from part (a) are used |

---

### Part (c):

| Working/Answer | Marks | Guidance |
|---|---|---|
| Solving $T_B \geq 0$ or $T_B > 0$ for $k$ | M1 | |
| $0 < k \leq \frac{2}{3}$ or $0 < k < \frac{2}{3}$ only | A1 | N.B. If $T=0 \Rightarrow k=\frac{2}{3}$ then answer is M0; if $T_C \geq 0$ or $T_C > 0$ also solved, can still score M1 and possibly A1 |

---
6.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{b896c631-00a0-46c5-bce9-16d65f6e3095-11_600_969_127_491}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

A non-uniform beam $A D$ has weight $W$ newtons and length 4 m . It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. The ropes are attached to two points $B$ and $C$ on the beam, where $A B = 1 \mathrm {~m}$ and $C D = 1 \mathrm {~m}$, as shown in Figure 3. The tension in the rope attached to $C$ is double the tension in the rope attached to $B$. The beam is modelled as a rod and the ropes are modelled as light inextensible strings.
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the beam from $A$.

A small load of weight $k W$ newtons is attached to the beam at $D$. The beam remains in equilibrium in a horizontal position. The load is modelled as a particle.

Find
\item an expression for the tension in the rope attached to $B$, giving your answer in terms of $k$ and $W$,
\item the set of possible values of $k$ for which both ropes remain taut.
\end{enumerate}

\hfill \mbox{\textit{Edexcel M1 2014 Q6 [11]}}