| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2020 |
| Session | January |
| Marks | 7 |
| 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 requiring two equilibrium equations (vertical forces and moments about a point) to find two unknowns. The setup is straightforward with clearly defined positions, and the solution involves routine algebraic manipulation of simultaneous equations. Slightly easier than average due to the simple geometry and standard technique. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(T + (T+20) = W\) | M1 A1 | M1 first equation (vertical resolution or moments) with usual rules. A1 correct equation (\(T\) may be replaced by \(T-20\)) |
| \(M(A)\): \(4.5(T+20) = 2.625W\) | M1 A1 | M1 second equation (vertical resolution or moments) with usual rules. A1 correct equation |
| \(M(G)\): \(2.625T = 1.875(T+20)\) | ||
| \(M(C)\): \(4.5T = 1.875W\) | ||
| \(M(B)\): \(6T + 1.5(T+20) = 3.375W\) | ||
| Solve for \(W\) | DM1 | Dependent on previous two M marks |
| \(W = 120\) | A1 (6) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| The beam remains straight, or rigid, or in a straight line or 1-dimensional or it doesn't bend | B1 (1) | Any appropriate comment. N.B. Penalise incorrect extras |
## Question 2:
### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $T + (T+20) = W$ | M1 A1 | M1 first equation (vertical resolution or moments) with usual rules. A1 correct equation ($T$ may be replaced by $T-20$) |
| $M(A)$: $4.5(T+20) = 2.625W$ | M1 A1 | M1 second equation (vertical resolution or moments) with usual rules. A1 correct equation |
| $M(G)$: $2.625T = 1.875(T+20)$ | | |
| $M(C)$: $4.5T = 1.875W$ | | |
| $M(B)$: $6T + 1.5(T+20) = 3.375W$ | | |
| Solve for $W$ | DM1 | Dependent on previous two M marks |
| $W = 120$ | A1 (6) | cao |
**N.B.** A marks and DM1 can **only** be scored if candidate is using $T$ and $T+20$ or $T$ and $T-20$ in **both** equations.
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| The beam remains straight, or rigid, or in a straight line or 1-dimensional or it doesn't bend | B1 (1) | Any appropriate comment. **N.B.** Penalise incorrect extras |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{916543cb-14f7-486c-ba3c-eda9be134045-04_473_1254_221_346}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A non-uniform beam $A B$ has length 6 m and weight $W$ newtons. The beam is supported in equilibrium in a horizontal position by two vertical ropes, one attached to the beam at $A$ and the other attached to the beam at $C$, where $C B = 1.5 \mathrm {~m}$, as shown in Figure 1 .
The centre of mass of the beam is 2.625 m from $A$.
The ropes are modelled as light strings. The beam is modelled as a non-uniform rod.
Given that the tension in the rope attached at $C$ is 20 N greater than the tension in the rope attached at $A$,
\begin{enumerate}[label=(\alph*)]
\item find the value of $W$.
\item State how you have used the fact that the beam is modelled as a rod.\\
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2020 Q2 [7]}}