| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| 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 requiring resolution of forces in two directions and taking moments about one point. The setup is straightforward with given angle, masses, and distances. Students follow a routine procedure: resolve horizontally (R=F), resolve vertically (N=total weight), take moments about A to find R, then use F=μN. While it requires careful bookkeeping of multiple forces and distances, it involves no novel insight—just systematic application of standard mechanics techniques taught in M2. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03v Motion on rough surface: including inclined planes3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(F = \mu N\) | B1 | Used. Condone an inequality |
| \(R(\uparrow)\): \(18g + 60g = N = 78g\) | M1, A1 | Resolve vertically |
| \(R(\rightarrow)\): \(R = F = \mu N\) | ||
| Moments equation: \(2.5 \times 18g\cos\alpha + 3 \times 60g\cos\alpha = 5F\sin\alpha\) | M1A2 | Moments equation. Condone sign errors. Condone sin/cos confusion. \(-1\) each error |
| Eliminate \(\alpha\): \(45 \times \frac{3}{5}g + 180 \times \frac{3}{5}g = 4R\) | DM1 | Eliminate \(\alpha\). Dependent on second M1 |
| \(R = \frac{135}{4}g\) | ||
| \(78g\mu = \frac{135}{4}g\) | DM1 | Equation in \(\mu\) only. Dependent on first two M marks. NB \(g\) cancels |
| \(\mu = \frac{135}{4 \times 78} = \frac{135}{312} = 0.432... = 0.43\) | A1 | \(\frac{225}{520}\), \(\frac{45}{104}\), awrt 0.433. Do not accept an inequality |
## Question 3:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $F = \mu N$ | B1 | Used. Condone an inequality |
| $R(\uparrow)$: $18g + 60g = N = 78g$ | M1, A1 | Resolve vertically |
| $R(\rightarrow)$: $R = F = \mu N$ | | |
| Moments equation: $2.5 \times 18g\cos\alpha + 3 \times 60g\cos\alpha = 5F\sin\alpha$ | M1A2 | Moments equation. Condone sign errors. Condone sin/cos confusion. $-1$ each error |
| Eliminate $\alpha$: $45 \times \frac{3}{5}g + 180 \times \frac{3}{5}g = 4R$ | DM1 | Eliminate $\alpha$. Dependent on second M1 |
| $R = \frac{135}{4}g$ | | |
| $78g\mu = \frac{135}{4}g$ | DM1 | Equation in $\mu$ only. Dependent on first two M marks. NB $g$ cancels |
| $\mu = \frac{135}{4 \times 78} = \frac{135}{312} = 0.432... = 0.43$ | A1 | $\frac{225}{520}$, $\frac{45}{104}$, awrt 0.433. Do not accept an inequality |
---3.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{ad18c22c-2fc5-4844-99b8-492f758bb24e-05_876_757_125_589}
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\caption{Figure 1}
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\end{figure}
A ladder, of length 5 m and mass 18 kg , has one end $A$ resting on rough horizontal ground and its other end $B$ resting against a smooth vertical wall. The ladder lies in a vertical plane perpendicular to the wall and makes an angle $\alpha$ with the horizontal ground, where $\tan \alpha = \frac { 4 } { 3 }$, as shown in Figure 1. The coefficient of friction between the ladder and the ground is $\mu$. A woman of mass 60 kg stands on the ladder at the point $C$, where $A C = 3 \mathrm {~m}$. The ladder is on the point of slipping. The ladder is modelled as a uniform rod and the woman as a particle.
Find the value of $\mu$.\\
\hfill \mbox{\textit{Edexcel M2 2013 Q3 [9]}}