| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | October |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam suspended by vertical ropes |
| Difficulty | Moderate -0.3 This is a standard M1 equilibrium problem requiring resolution of forces and taking moments about a point. The setup is straightforward with a uniform beam, vertical forces only, and a simple relationship between tensions (one is double the other). It requires two equations (vertical equilibrium and moments) to solve for two unknowns, which is typical textbook material, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(A): (2T \times x) + T(x+2) = (24 \times 3)\) | M1 A1 | Forms moments equation in \(x\) and \(T\) only with correct no. of terms. Allow consistent extra \(g\)'s. M0 if no \(x\). A1 for correct unsimplified equation |
| \(3T = 24\) (vertical resolution) | M1 | Resolves vertically to give equation in \(T\) only, or second moments equation in \(x\) and \(T\). Must be dimensionally correct with correct no. of terms |
| \(T = 8\) (N) | A1 | Correct value for tension at \(D\) |
| \(x = \frac{7}{3}\) (accept 2.3 or better) | A1 | Correct value for \(x\) |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A): (2T \times x) + T(x+2) = (24 \times 3)$ | M1 A1 | Forms moments equation in $x$ and $T$ only with correct no. of terms. Allow consistent extra $g$'s. M0 if no $x$. A1 for correct unsimplified equation |
| $3T = 24$ (vertical resolution) | M1 | Resolves vertically to give equation in $T$ only, or second moments equation in $x$ and $T$. Must be dimensionally correct with correct no. of terms |
| $T = 8$ (N) | A1 | Correct value for tension at $D$ |
| $x = \frac{7}{3}$ (accept 2.3 or better) | A1 | Correct value for $x$ |
Alternative moment equations: $M(C): 24(3-x) = T \times 2$; $M(G): 2T(3-x) = T(x-1)$; $M(D): (2T \times 2) = 24 \times (x-1)$; $M(B): 2T(6-x)+T(4-x)=(24\times3)$; $M(C): \frac{24x}{6}\times\frac{x}{2}+2T=\frac{24(6-x)}{6}\times\frac{(6-x)}{2}$
**N.B.** If $T$ and $2T$ wrong way round or use $24g$, can score max M1A0M1A0A0.
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{017cc2b0-9ec3-45ff-94c0-9d989badfd5d-02_529_1362_246_349}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a beam $A B$ with weight 24 N and length 6 m .\\
The beam is suspended by two light vertical ropes. The ropes are attached to the points $C$ and $D$ on the beam where $A C = x$ metres and $C D = 2 \mathrm {~m}$.
The tension in the rope attached to the beam at $C$ is double the tension in the rope attached to the beam at $D$.
The beam is modelled as a uniform rod, resting horizontally in equilibrium.\\
Find\\
(i) the tension in the rope attached to the beam at $D$.\\
(ii) the value of $x$.
\hfill \mbox{\textit{Edexcel M1 2023 Q1 [5]}}