| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Ladder against wall |
| Difficulty | Standard +0.3 This is a standard M2 ladder equilibrium problem with routine application of resolving forces, taking moments, and friction at limiting equilibrium. Part (a) requires resolving vertically (straightforward), part (b) involves taking moments when μ=0.4 (standard technique), and part (c) is a modelling statement. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{M}(A): 3 \times 30g \times \frac{1}{2} + 70g \times 2 \times \frac{1}{2} = N \times \frac{6\sqrt{3}}{2}\) | M1 | All terms required. Must be dimensionally correct. Condone sin/cos confusion and sign errors. Allow with sin/cos \(60°\) |
| \((45g + 70g = 3\sqrt{3}N)\) | A1 | Correct unsimplified |
| \(\uparrow:\ R = 100g\) | B1 | B0 if they have \(F_B \neq 0\) |
| \(\leftrightarrow:\ F = N = 217\ \text{(N)}\ \left(\frac{115g}{3\sqrt{3}}\right)\) | B1 | Solve for \(F\) (\(216.891...\) seen or implied) |
| \(\sqrt{(100g)^2 + 217^2}\) | DM1 | Use of Pythagoras with their \(R\), \(F\). Dependent on the preceding M mark |
| \(= 1000\ \text{(N)}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(F = 0.4 \times 100g\ (= 392)\) | M1 | Use of \(F = \mu R\) with their value for \(R\) |
| \(\text{M}(A):\ F \times 3\sqrt{3} = 70g \times \frac{x}{2} + 30g \times \frac{3}{2}\) | M1 | \((F \neq 217)\) Allow for moments about \(B\) to find distance from the top |
| \(40g \times 3\sqrt{3} = 35gx + 45g\) | A1 | Equation in \(x\) (distance from ground) only |
| \((AD =)\ x = 4.65\ \text{(m)}\) | A1 | \(4.7\) or better \((4.65274....)\) |
| Answer | Marks | Guidance |
|---|---|---|
| e.g. The ladder does not bend / The ladder meets the wall/floor at a point / The weight acts at a single point | B1 | With no incorrect statement(s) seen |
## Question 6(a):
$\text{M}(A): 3 \times 30g \times \frac{1}{2} + 70g \times 2 \times \frac{1}{2} = N \times \frac{6\sqrt{3}}{2}$ | M1 | All terms required. Must be dimensionally correct. Condone sin/cos confusion and sign errors. Allow with sin/cos $60°$
$(45g + 70g = 3\sqrt{3}N)$ | A1 | Correct unsimplified
$\uparrow:\ R = 100g$ | B1 | B0 if they have $F_B \neq 0$
$\leftrightarrow:\ F = N = 217\ \text{(N)}\ \left(\frac{115g}{3\sqrt{3}}\right)$ | B1 | Solve for $F$ ($216.891...$ seen or implied)
$\sqrt{(100g)^2 + 217^2}$ | DM1 | Use of Pythagoras with their $R$, $F$. Dependent on the preceding M mark
$= 1000\ \text{(N)}$ | A1 |
**[6]**
## Question 6(b):
$F = 0.4 \times 100g\ (= 392)$ | M1 | Use of $F = \mu R$ with their value for $R$
$\text{M}(A):\ F \times 3\sqrt{3} = 70g \times \frac{x}{2} + 30g \times \frac{3}{2}$ | M1 | $(F \neq 217)$ Allow for moments about $B$ to find distance from the top
$40g \times 3\sqrt{3} = 35gx + 45g$ | A1 | Equation in $x$ (distance from ground) only
$(AD =)\ x = 4.65\ \text{(m)}$ | A1 | $4.7$ or better $(4.65274....)$
**[4]**
## Question 6(c):
e.g. The ladder does not bend / The ladder meets the wall/floor at a point / The weight acts at a single point | B1 | With no incorrect statement(s) seen
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3eb71ecb-fa88-4cca-a2b6-bcf11f1d689b-16_639_561_246_689}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A ladder $A B$ has length 6 m and mass 30 kg . The ladder rests in equilibrium at $60 ^ { \circ }$ to the horizontal with the end $A$ on rough horizontal ground and the end $B$ against a smooth vertical wall, as shown in Figure 3.
A man of mass 70 kg stands on the ladder at the point $C$, where $A C = 2 \mathrm {~m}$, and the ladder remains in equilibrium. The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall. The man is modelled as a particle.
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the force exerted on the ladder by the ground.
The man climbs further up the ladder. When he is at the point $D$ on the ladder, the ladder is about to slip.
Given that the coefficient of friction between the ladder and the ground is 0.4
\item find the distance $A D$.
\item State how you have used the modelling assumption that the ladder is a rod.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2021 Q6 [11]}}