| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam suspended by vertical ropes |
| Difficulty | Moderate -0.3 This is a standard M1 moments problem requiring equilibrium conditions (sum of forces = 0, sum of moments = 0) with straightforward algebra. Part (a) involves taking moments about one point with given tension ratio, part (b) adds a particle load. Both parts follow textbook methods with no conceptual challenges, making it slightly easier than average A-level mechanics questions. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(T + 2T\ (= 3T) = W\) | M1A1 | M1 for equation in \(W\) and \(T\) and possibly \(d\); either resolve vertically or moments about any point other than mid-pt; if \(Wg\) used, mark as misread |
| Moments about \(A\): \(2W = 2T \times d\) | M1A1 | Second M1 for second equation in \(W\) and \(T\) and possibly \(d\) |
| Substitute and solve: \(2W = 2\frac{W}{3}d\) | DM1 | Third M1 dependent on first and second M marks, for solving for \(d\) |
| \(d = 3\) | A1 | NB: if single equation used by taking moments about mid-point of rod: \(2T = 2T(d-2)\), scores M2A2 (-1 each error) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(T + 4T = W + kW\ \left(5T = W(1+k)\right)\) | M1A1 ft | ft on their \(d\); if \(Wg\) and \(kWg\) used, mark as misread; if \(T\) and \(4T\) are reversed, mark as per scheme NOT as misread |
| Moments about \(A\): \(2W + 4kW = 3 \times 4T\) | M1A1 ft | Second M1 for equation in \(W\) and same tension \(T_1\) and possibly \(d\) and \(k\) |
| Substitute and solve: \(2W + 4kW = \frac{12}{5}W(1+k)\) | DM1 | Third M1 dependent on first and second M marks |
| \(2 + 4k = \frac{12}{5} + \frac{12}{5}k\) | ||
| \(\frac{8}{5}k = \frac{2}{5}\), \(k = \frac{1}{4}\) | A1 |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $T + 2T\ (= 3T) = W$ | M1A1 | M1 for equation in $W$ and $T$ and possibly $d$; either resolve vertically or moments about any point other than mid-pt; if $Wg$ used, mark as misread |
| Moments about $A$: $2W = 2T \times d$ | M1A1 | Second M1 for second equation in $W$ and $T$ and possibly $d$ |
| Substitute and solve: $2W = 2\frac{W}{3}d$ | DM1 | Third M1 dependent on first and second M marks, for solving for $d$ |
| $d = 3$ | A1 | NB: if single equation used by taking moments about mid-point of rod: $2T = 2T(d-2)$, scores M2A2 (-1 each error) |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $T + 4T = W + kW\ \left(5T = W(1+k)\right)$ | M1A1 ft | ft on their $d$; if $Wg$ and $kWg$ used, mark as misread; if $T$ and $4T$ are reversed, mark as per scheme NOT as misread |
| Moments about $A$: $2W + 4kW = 3 \times 4T$ | M1A1 ft | Second M1 for equation in $W$ and **same tension** $T_1$ and possibly $d$ and $k$ |
| Substitute and solve: $2W + 4kW = \frac{12}{5}W(1+k)$ | DM1 | Third M1 dependent on first and second M marks |
| $2 + 4k = \frac{12}{5} + \frac{12}{5}k$ | | |
| $\frac{8}{5}k = \frac{2}{5}$, $k = \frac{1}{4}$ | A1 | |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{ed659098-c1cf-4ee1-a12a-bf8b6c42db95-07_513_993_276_479}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A beam $A B$ has weight $W$ newtons and length 4 m . The beam is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. One rope is attached to $A$ and the other rope is attached to the point $C$ on the beam, where $A C = d$ metres, as shown in Figure 3. The beam is modelled as a uniform rod and the ropes as light inextensible strings. The tension in the rope attached at $C$ is double the tension in the rope attached at $A$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $d$.
A small load of weight $k W$ newtons is attached to the beam at $B$. The beam remains in equilibrium in a horizontal position. The load is modelled as a particle. The tension in the rope attached at $C$ is now four times the tension in the rope attached at $A$.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2014 Q4 [12]}}