| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | January |
| Marks | 10 |
| 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 taking moments about a point, resolving forces vertically, and applying equilibrium conditions. Part (a) is routine bookwork showing a given result, part (b) involves simple algebra with the 4:1 ratio condition, and part (c) is a straightforward tilting problem. All techniques are standard textbook exercises with no novel insight required, making it slightly easier than average. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(M(D)\): \((R_C \times 2.2) = 55g(2.2-x) + 30g(1.1)\) | M1A1 | Forms equation in \(R_C\) and \(x\) only; moments about \(D\) or two equations combined to eliminate \(R_D\) |
| \(R_C = (686 - 245x)\) (N) | A1* | Correctly reaches given answer with at least one line of intermediate working. NB: M mark not available if \(R_C = 4R_D\) used |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Use of \(R_C = 4R_D\) | M1 | |
| Vert: \(R_C + R_D = 55g + 30g \Rightarrow \frac{5}{4}(686-245x) = 55g + 30g\) | M1A1 | Complete method for equation in \(x\) only; \(R_C\) replaced with expression from (a), \(R_D\) replaced with \(\frac{1}{4}R_C\) |
| \(x = 0.08\) | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(M(C)\): \(Mg(0.4) = 30g(1.1)\) | M1A1 | Use \(S_D = 0\); forms equation in \(M\) only; M0 if \(S_D \neq 0\) |
| \(M = 83\) or \(82.5\) or \(\frac{165}{2}\) oe | A1 | Correct answer |
## Question 5:
**Part (a):**
| Working | Mark | Guidance |
|---------|------|----------|
| $M(D)$: $(R_C \times 2.2) = 55g(2.2-x) + 30g(1.1)$ | M1A1 | Forms equation in $R_C$ and $x$ only; moments about $D$ or two equations combined to eliminate $R_D$ |
| $R_C = (686 - 245x)$ (N) | A1* | Correctly reaches given answer with **at least one line of intermediate working**. NB: M mark not available if $R_C = 4R_D$ used |
**Part (b):**
| Working | Mark | Guidance |
|---------|------|----------|
| Use of $R_C = 4R_D$ | M1 | |
| Vert: $R_C + R_D = 55g + 30g \Rightarrow \frac{5}{4}(686-245x) = 55g + 30g$ | M1A1 | Complete method for equation in $x$ only; $R_C$ replaced with expression from (a), $R_D$ replaced with $\frac{1}{4}R_C$ |
| $x = 0.08$ | A1 | Correct answer |
**Part (c):**
| Working | Mark | Guidance |
|---------|------|----------|
| $M(C)$: $Mg(0.4) = 30g(1.1)$ | M1A1 | Use $S_D = 0$; forms equation in $M$ only; M0 if $S_D \neq 0$ |
| $M = 83$ or $82.5$ or $\frac{165}{2}$ oe | A1 | Correct answer |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e59a66b8-c2ad-41fd-9959-9d21e9455c37-12_412_1529_242_267}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A beam $A B$ has mass 30 kg and length 3 m .\\
The beam rests on supports at $C$ and $D$ where $A C = 0.4 \mathrm {~m}$ and $D B = 0.4 \mathrm {~m}$, as shown in Figure 4.
A person of mass 55 kg stands on the beam between $C$ and $D$.\\
The person is modelled as a particle at the point $P$, where $C P = x$ metres and $0 < x < 2.2$
The beam is modelled as a uniform rod resting in equilibrium in a horizontal position.\\
Using the model,
\begin{enumerate}[label=(\alph*)]
\item show that the magnitude of the reaction at $C$ is $( 686 - 245 x ) \mathrm { N }$.
The magnitude of the reaction at $C$ is four times the magnitude of the reaction at $D$.\\
Using the model,
\item find the value of $x$
The person steps off the beam and places a package of mass $M \mathrm {~kg}$ at $A$.\\
The package is modelled as a particle at the point $A$.\\
The beam is now on the point of tilting about $C$.\\
Using the model,
\item find the value of $M$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2024 Q5 [10]}}