| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam suspended by vertical ropes |
| Difficulty | Moderate -0.3 This is a standard M1 moments question requiring taking moments about two points and resolving vertically. Part (a) is routine bookwork, while part (b) requires recognizing limiting cases (tension = 0) but uses the same techniques. Slightly easier than average due to straightforward setup and clear methodology. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T_A + T_C = 85g\) | M1 A1 | M1 for moments or vertical resolution equation, correct no. of terms, dimensionally correct; A1 for correct equation |
| OR \(M(A)\): \(25g \times 2.5 + 60g \times 5 = 4.5 \times T_C\) | M1 A1 | Second M1 for moments equation, correct no. of terms, dimensionally correct |
| OR \(M(C)\): \(T_A \times 4.5 + 60g \times 0.5 = 25g \times 2\) | ||
| OR \(M(B)\): \(T_A \times 5 + T_C \times 0.5 = 25g \times 2.5\) | ||
| OR \(M(G)\): \(T_A \times 2.5 + 60g \times 2.5 = 2 \times T_C\) | ||
| \(T_A = \frac{40g}{9} = 44\) N or \(43.6\) N; \(T_C = \frac{725g}{9} = 790\) N or \(789\) N | A1; A1 (6) | Deduct 1 mark for inexact multiples of \(g\); If both tensions assumed equal, max M1 only |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(M(C)\): \(25g \times 2 = Mg \times 0.5\) | M1 A1 | M1 for moments equation with \(T_A = 0\) |
| \(M = 100\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T_C = 25g + 100g\) | M1 A1 | Second M1 for another moments or vertical resolution equation with \(T_A = 0\) |
| \(T_C = 125g\) \((1200\) or \(1230)\) N | B1 (6) | N.B. No marks in (b) if answers from (a) used or \(M = 60\) |
## Question 5:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T_A + T_C = 85g$ | M1 A1 | M1 for moments or vertical resolution equation, correct no. of terms, dimensionally correct; A1 for correct equation |
| OR $M(A)$: $25g \times 2.5 + 60g \times 5 = 4.5 \times T_C$ | M1 A1 | Second M1 for moments equation, correct no. of terms, dimensionally correct |
| OR $M(C)$: $T_A \times 4.5 + 60g \times 0.5 = 25g \times 2$ | | |
| OR $M(B)$: $T_A \times 5 + T_C \times 0.5 = 25g \times 2.5$ | | |
| OR $M(G)$: $T_A \times 2.5 + 60g \times 2.5 = 2 \times T_C$ | | |
| $T_A = \frac{40g}{9} = 44$ N or $43.6$ N; $T_C = \frac{725g}{9} = 790$ N or $789$ N | A1; A1 (6) | Deduct 1 mark for inexact multiples of $g$; If both tensions assumed equal, max M1 only |
### Part (b)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $M(C)$: $25g \times 2 = Mg \times 0.5$ | M1 A1 | M1 for moments equation with $T_A = 0$ |
| $M = 100$ | A1 | |
### Part (b)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T_C = 25g + 100g$ | M1 A1 | Second M1 for another moments or vertical resolution equation with $T_A = 0$ |
| $T_C = 125g$ $(1200$ or $1230)$ N | B1 (6) | N.B. No marks in (b) if answers from (a) used or $M = 60$ |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{de3245a7-cf6e-423e-8689-9a074bdbc23b-08_582_1230_271_374}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A beam $A B$ has length 5 m and mass 25 kg . The beam is suspended in equilibrium in a horizontal position by two vertical ropes. One rope is attached to the beam at $A$ and the other rope is attached to the point $C$ on the beam where $C B = 0.5 \mathrm {~m}$, as shown in Figure 3. A particle $P$ of mass 60 kg is attached to the beam at $B$ and the beam remains in equilibrium in a horizontal position. The beam is modelled as a uniform rod and the ropes are modelled as light strings.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the tension in the rope attached to the beam at $A$,
\item the tension in the rope attached to the beam at $C$.
Particle $P$ is removed and replaced by a particle $Q$ of mass $M \mathrm {~kg}$ at $B$. Given that the beam remains in equilibrium in a horizontal position,
\end{enumerate}\item find
\begin{enumerate}[label=(\roman*)]
\item the greatest possible value of $M$,
\item the greatest possible tension in the rope attached to the beam at $C$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2015 Q5 [12]}}