| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Standard +0.3 This is a standard M1 moments question with straightforward application of equilibrium principles. Part (a) is trivial (reaction = 0 when tilting), parts (b-d) involve taking moments about a pivot with clear setups, and part (e) tests understanding of modeling assumptions. The two-scenario approach adds mild complexity but follows a predictable pattern typical of M1 exam questions. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| 0 | B1 | (1 mark) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(D): \quad 2w = 1500 \times 5\) | M1 A1 | |
| \(\Rightarrow w = 3750 \text{ N}\) | A1 | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(M(D): \quad 1500 \times 5 = w'(4-x)\) | M1 A1 | |
| \(M(C): \quad 1000 \times 5 = w' x\) | M1 A1 | |
| Solve \(\Rightarrow w' = 3125 \text{ N}\) | M1 A1 | (6 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 1.6 \text{ m}\) | M1 A1 | (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB\) remains a straight line (o.e.) | B1 | (1 mark) Total: 13 |
## Question 5:
### Part (a):
0 | B1 | (1 mark)
### Part (b):
$M(D): \quad 2w = 1500 \times 5$ | M1 A1 |
$\Rightarrow w = 3750 \text{ N}$ | A1 | (3 marks)
[If moments about another point: M1 for complete method to get $w$, A1 for moments equation correct]
### Part (c):
$M(D): \quad 1500 \times 5 = w'(4-x)$ | M1 A1 |
$M(C): \quad 1000 \times 5 = w' x$ | M1 A1 |
Solve $\Rightarrow w' = 3125 \text{ N}$ | M1 A1 | (6 marks)
### Part (d):
$x = 1.6 \text{ m}$ | M1 A1 | (2 marks)
### Part (e):
$AB$ remains a straight line (o.e.) | B1 | (1 mark) **Total: 13**
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5.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\includegraphics[alt={},max width=\textwidth]{218383c1-0875-46f2-9416-8e827065a7a6-5_328_993_491_483}
\end{center}
\end{figure}
A large $\log A B$ is 6 m long. It rests in a horizontal position on two smooth supports $C$ and $D$, where $A C = 1 \mathrm {~m}$ and $B D = 1 \mathrm {~m}$, as shown in Figure 4. David needs an estimate of the weight of the log, but the log is too heavy to lift off both supports. When David applies a force of magnitude 1500 N vertically upwards to the $\log$ at $A$, the $\log$ is about to tilt about $D$.
\begin{enumerate}[label=(\alph*)]
\item State the value of the reaction on the $\log$ at $C$ for this case.
David initially models the log as uniform rod. Using this model,
\item estimate the weight of the log
The shape of the log convinces David that his initial modelling assumption is too simple. He removes the force at $A$ and applies a force acting vertically upwards at $B$. He finds that the log is about to tilt about $C$ when this force has magnitude 1000 N. David now models the log as a non-uniform rod, with the distance of the centre of mass of the $\log$ from $C$ as $x$ metres. Using this model, find
\item a new estimate for the weight of the log,
\item the value of $x$.
\item State how you have used the modeling assumption that the log is a rod.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q5 [13]}}