| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Non-uniform beam on supports |
| Difficulty | Standard +0.3 This is a standard M1 moments problem with two equilibrium conditions to solve simultaneously for two unknowns (W and x). While it requires setting up moments equations about two different pivot points, the method is routine and well-practiced in M1 courses, 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 |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R_D = 0\) for scenario 1 OR \(R_C = 0\) for scenario 2 | B1 | Seen on diagram or implied by working |
| Scenario 1: \(M(C): (150 \times 1.26) = Wx\) | M1 | Complete method for equation in \(W\) and consistent unknown distance only; all equations dimensionally correct; condone sign errors |
| Correct unsimplified equation in \(W\) and \(x\) | A1 | |
| Scenario 2: \(M(D): (225 \times 0.36) = W \times (0.9 - x)\) | M1 | Complete method for equation in \(W\) and consistent unknown distance only; all equations dimensionally correct; condone sign errors |
| Correct unsimplified equation in \(W\) and \(x\) | A1 | |
| Solve simultaneously e.g. \(0.81W = 243\) | dM1 | Dependent on both M marks; solve for \(W\) or unknown |
| \(W = 300\) | A1 | |
| \(x = 0.63\) | A1 | Must be distance from \(C\) to centre of mass |
# Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R_D = 0$ for scenario 1 OR $R_C = 0$ for scenario 2 | B1 | Seen on diagram or implied by working |
| **Scenario 1:** $M(C): (150 \times 1.26) = Wx$ | M1 | Complete method for equation in $W$ and consistent unknown distance only; all equations dimensionally correct; condone sign errors |
| Correct unsimplified equation in $W$ and $x$ | A1 | |
| **Scenario 2:** $M(D): (225 \times 0.36) = W \times (0.9 - x)$ | M1 | Complete method for equation in $W$ and consistent unknown distance only; all equations dimensionally correct; condone sign errors |
| Correct unsimplified equation in $W$ and $x$ | A1 | |
| Solve simultaneously e.g. $0.81W = 243$ | dM1 | Dependent on both M marks; solve for $W$ or unknown |
| $W = 300$ | A1 | |
| $x = 0.63$ | A1 | Must be distance from $C$ to centre of mass |
**Note:** Including $g$ in moments equation loses the A mark for that equation and both final A marks.
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{84c0eead-0a87-4d87-b33d-794a94bb466c-10_419_1445_283_312}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A branch $A B$, of length 1.5 m , rests horizontally in equilibrium on two supports.\\
The two supports are at the points $C$ and $D$, where $A C = 0.24 \mathrm {~m}$ and $D B = 0.36 \mathrm {~m}$, as shown in Figure 1.
When a force of 150 N is applied vertically upwards at $B$, the branch is on the point of tilting about $C$.
When a force of 225 N is applied vertically downwards at $B$, the branch is on the point of tilting about $D$.
The branch is modelled as a non-uniform rod $A B$ of weight $W$ newtons.\\
The distance from the point $C$ to the centre of mass of the rod is $x$ metres.\\
Use the model to find\\
(i) the value of $W$\\
(ii) the value of $x$
\hfill \mbox{\textit{Edexcel M1 2023 Q4 [8]}}