| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Range of equilibrium positions |
| Difficulty | Standard +0.3 This is a standard M1 moments question requiring systematic application of equilibrium conditions (sum of moments = 0, sum of forces = 0). Part (a) is routine calculation, part (b) introduces a parameter requiring algebraic manipulation, and part (c) requires solving an inequality. While multi-step, it follows a predictable template with no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| M(\(A\)): \(T \times 4 = 12g \times 2.5\) | M1 A1 | |
| \(T = 7.5g \text{ or } 73.5\text{ N}\) | A1 | |
| R(\(\uparrow\)) \(S + T = 12g\) | M1 | |
| \(\Rightarrow S = 4.5g \text{ or } 44.1\text{ N}\) | A1 (5) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| M(\(A\)): \(V \times 4 = 16g \times y + 12g \times 2.5\) | M1 A1 | |
| \(V = 4gy + 7.5g \text{ or } 39.2y + 73.5\text{ N}\) | A1 (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Notes |
| \(V \leq 98 \Rightarrow 39.2y + 73.5 \leq 98\) | M1 | |
| \(\Rightarrow y \leq 0.625 = \frac{5}{8}\) | DM1 | |
| Hence "load must be no more than \(\frac{5}{8}\) m from \(A\)" | A1 (3) | Total: 11 |
## Question 5:
### Part (a)
| Working | Marks | Notes |
|---------|-------|-------|
| M($A$): $T \times 4 = 12g \times 2.5$ | M1 A1 | |
| $T = 7.5g \text{ or } 73.5\text{ N}$ | A1 | |
| R($\uparrow$) $S + T = 12g$ | M1 | |
| $\Rightarrow S = 4.5g \text{ or } 44.1\text{ N}$ | A1 (5) | |
### Part (b)
| Working | Marks | Notes |
|---------|-------|-------|
| M($A$): $V \times 4 = 16g \times y + 12g \times 2.5$ | M1 A1 | |
| $V = 4gy + 7.5g \text{ or } 39.2y + 73.5\text{ N}$ | A1 (3) | |
### Part (c)
| Working | Marks | Notes |
|---------|-------|-------|
| $V \leq 98 \Rightarrow 39.2y + 73.5 \leq 98$ | M1 | |
| $\Rightarrow y \leq 0.625 = \frac{5}{8}$ | DM1 | |
| Hence "load must be no more than $\frac{5}{8}$ m from $A$" | A1 (3) | **Total: 11** |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7ba14d10-1b57-4930-8d65-f21088c5d513-08_315_817_255_587}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A beam $A B$ has mass 12 kg and length 5 m . It is held in equilibrium in a horizontal position by two vertical ropes attached to the beam. One rope is attached to $A$, the other to the point $C$ on the beam, where $B C = 1 \mathrm {~m}$, as shown in Figure 2. The beam is modelled as a uniform rod, and the ropes as light strings.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the tension in the rope at $C$,
\item the tension in the rope at $A$.
A small load of mass 16 kg is attached to the beam at a point which is $y$ metres from $A$. The load is modelled as a particle. Given that the beam remains in equilibrium in a horizontal position,
\end{enumerate}\item find, in terms of $y$, an expression for the tension in the rope at $C$.
The rope at $C$ will break if its tension exceeds 98 N. The rope at $A$ cannot break.
\item Find the range of possible positions on the beam where the load can be attached without the rope at $C$ breaking.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2008 Q5 [11]}}