| Exam Board | Edexcel |
|---|---|
| Module | Paper 3 (Paper 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against smooth wall in limiting equilibrium |
| Difficulty | Challenging +1.2 This is a standard mechanics problem requiring resolution of forces and taking moments about a point, with the added complexity of limiting equilibrium in two directions (friction both ways). While it involves multiple steps and careful bookkeeping of forces, the techniques are routine for A-level mechanics students. The 'show that' part (a) guides students through the setup, and part (b) follows standard inequality methods for friction problems. Part (c) is conceptual recall. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks |
|---|---|
| Take moments about \(A\) (or any other complete method to produce an equation in \(S\), \(W\) and \(\alpha\) only) | M1 |
| \(Wa\cos\alpha + 7W\cdot 2a\cos\alpha = S\cdot 2a\sin\alpha\) | A1, A1 |
| Use of \(\tan\alpha = \dfrac{5}{2}\) to obtain \(S\) | M1 |
| \(S = 3W\) * | A1* |
| Answer | Marks |
|---|---|
| \(R = 8W\) | B1 |
| \(F = \dfrac{1}{4}R\ (= 2W)\) | M1 |
| \(P_{\text{MAX}} = 3W + F\) or \(P_{\text{MIN}} = 3W - F\) | M1 |
| \(P_{\text{MAX}} = 5W\) or \(P_{\text{MIN}} = W\) | A1 |
| \(W \leq P \leq 5W\) | A1 |
| Answer | Marks |
|---|---|
| \(M(A)\) shows that the reaction on the ladder at \(B\) is unchanged | M1 |
| Also \(R\) increases (resolving vertically) | M1 |
| Which increases max \(F\) available | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Equation in \(S\), \(W\) and \(\alpha\) only | M1 | Producing an equation in \(S\), \(W\) and \(\alpha\) only |
| Equation correct, or with one error/omission | A1 | |
| Fully correct equation | A1 | |
| Use of \(\tan\alpha = \frac{5}{2}\) to obtain \(S\) in terms of \(W\) only | M1 | |
| \(S = 3W\) | A1* | Given answer correctly obtained |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = 8W\) | B1 | |
| Use of \(F = \frac{1}{4}R\) | M1 | |
| \(P = (3W + \text{their } F)\) or \(P = (3W - \text{their } F)\) | M1 | |
| Correct max or min value for correct range of \(P\) | A1 | |
| Correct range for \(P\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Show reaction at \(B\) unchanged by taking moments about \(A\) | M1 | |
| Show \(R\) increases by resolving vertically | M1 | |
| Conclude limiting friction at \(A\) increases | M1 |
## Question 9:
### Part (a):
| Take moments about $A$ (or any other complete method to produce an equation in $S$, $W$ and $\alpha$ only) | M1 | |
| $Wa\cos\alpha + 7W\cdot 2a\cos\alpha = S\cdot 2a\sin\alpha$ | A1, A1 | |
| Use of $\tan\alpha = \dfrac{5}{2}$ to obtain $S$ | M1 | |
| $S = 3W$ * | A1* | |
### Part (b):
| $R = 8W$ | B1 | |
| $F = \dfrac{1}{4}R\ (= 2W)$ | M1 | |
| $P_{\text{MAX}} = 3W + F$ or $P_{\text{MIN}} = 3W - F$ | M1 | |
| $P_{\text{MAX}} = 5W$ or $P_{\text{MIN}} = W$ | A1 | |
| $W \leq P \leq 5W$ | A1 | |
### Part (c):
| $M(A)$ shows that the reaction on the ladder at $B$ is unchanged | M1 | |
| Also $R$ increases (resolving vertically) | M1 | |
| Which increases max $F$ available | M1 | |
## Question 9 (continued):
### Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Equation in $S$, $W$ and $\alpha$ only | M1 | Producing an equation in $S$, $W$ and $\alpha$ only |
| Equation correct, or with one error/omission | A1 | |
| Fully correct equation | A1 | |
| Use of $\tan\alpha = \frac{5}{2}$ to obtain $S$ in terms of $W$ only | M1 | |
| $S = 3W$ | A1* | Given answer correctly obtained |
### Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = 8W$ | B1 | |
| Use of $F = \frac{1}{4}R$ | M1 | |
| $P = (3W + \text{their } F)$ or $P = (3W - \text{their } F)$ | M1 | |
| Correct max or min value for correct range of $P$ | A1 | |
| Correct range for $P$ | A1 | |
### Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Show reaction at $B$ unchanged by taking moments about $A$ | M1 | |
| Show $R$ increases by resolving vertically | M1 | |
| Conclude limiting friction at $A$ increases | M1 | |
---9.
Figure 1
A uniform ladder $A B$, of length $2 a$ and weight $W$, has its end $A$ on rough horizontal ground. The coefficient of friction between the ladder and the ground is $\frac { 1 } { 4 }$.\\
The end $B$ of the ladder is resting against a smooth vertical wall, as shown in Figure 1.\\
A builder of weight $7 W$ stands at the top of the ladder.\\
To stop the ladder from slipping, the builder's assistant applies a horizontal force of magnitude $P$ to the ladder at $A$, towards the wall.\\
The force acts in a direction which is perpendicular to the wall.\\
The ladder rests in equilibrium in a vertical plane perpendicular to the wall and makes an angle $\alpha$ with the horizontal ground, where $\tan \alpha = \frac { 5 } { 2 }$.\\
The builder is modelled as a particle and the ladder is modelled as a uniform rod.
\begin{enumerate}[label=(\alph*)]
\item Show that the reaction of the wall on the ladder at $B$ has magnitude $3 W$.
\item Find, in terms of $W$, the range of possible values of $P$ for which the ladder remains in equilibrium.
Often in practice, the builder's assistant will simply stand on the bottom of the ladder.
\item Explain briefly how this helps to stop the ladder from slipping.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 3 Q9 [13]}}