M2 January 2013 Q3
3.

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 \(\tan\alpha = \dfrac{4}{3}\), 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 \(AC = 3\) 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 \(\mu\). (9)
| Scheme | Marks | ||||||||||
|---|---|---|---|---|---|---|---|---|---|---|---|
![]() | |||||||||||
| \(F = \mu N\) | B1 | ||||||||||
| R\((\uparrow)\ \ \ 18g + 60g = N\) \(= 78g\) | M1 A1 | ||||||||||
| R\((\rightarrow)\ \ R = F = \mu N\) | |||||||||||
| M1A2 | ||||||||||
| \(45 \times \dfrac{3}{5}g + 180 \times \dfrac{3}{5}g = 4R\) | DM1 | ||||||||||
| \(R = \dfrac{135}{4}g\) | |||||||||||
| \(78g\mu = \dfrac{135}{4}g\) | DM1 | ||||||||||
| \(\mu = \dfrac{135}{4 \times 78} = \dfrac{135}{312} = 0.432\ldots = 0.43\) | A1 | ||||||||||
| (9 marks) |
Notes
B1 Used. Condone an inequality.
M1 Resolve vertically
M1A2 Moments equation. Condone sign errors. Condone sin/cos confusion -1 each error
DM1 Eliminate \(\alpha\). Dependent on the second M1.
DM1 Equation in \(\mu\) only. (Dependent on the first two M marks.)
A1 NB \(g\) cancels. 0.43269..., \(\dfrac{225}{520}\), \(\dfrac{45}{104}\), awrt 0.433 Do not accept an inequality.
NB If use just two moments equations, M1A2 for the better attempt, M1A1 for the other. Remaining marks as above.
(In the moments equation about \(B\) the final \(g\) of \(60g\) is cut off at the edge of the printed mark scheme.)
