M2 January 2008 Q5
5.
A ladder \(AB\), of mass \(m\) and length \(4a\), has one end \(A\) resting on rough horizontal ground. The other end \(B\) rests against a smooth vertical wall. A load of mass \(3m\) is fixed on the ladder at the point \(C\), where \(AC = a\). The ladder is modelled as a uniform rod in a vertical plane perpendicular to the wall and the load is modelled as a particle. The ladder rests in limiting equilibrium making an angle of 30\(^\circ\) with the wall, as shown in Figure 2.
Find the coefficient of friction between the ladder and the ground. (10)
| Scheme | Marks |
|---|---|
| M\((A)\) \(N \times 4a\cos 30^\circ = 3mg \times a\sin 30^\circ + mg \times 2a\sin 30^\circ\) | M1 A2(1,0) |
| \(N = \dfrac{5}{4}mg\tan 30^\circ\ \ \left(= \dfrac{5}{4\sqrt{3}}mg = 7.07\ldots m\right)\) | DM1 A1 |
| \(\rightarrow\ F_r = N\) , \(\uparrow\ R = 4mg\) | B1, B1 |
| Using \(F_r = \mu R\) | B1 |
| \(\dfrac{5}{4\sqrt{3}}mg = \mu R\) for their \(R\) | M1 |
| \(\mu = \dfrac{5}{16\sqrt{3}}\) awrt 0.18 | A1 |
| (10) | |
| (10 marks) |
Alternative method:
| M\((B)\): \(mg \times 2a\sin 30 + 3mg \times 3a\sin 30 + F \times 4a\cos 30 = R \times 4a\sin 30\) | |
| \(11mga\sin 30 + F \times 4a\cos 30 = R \times 4a\sin 30\) | M1A3(2,1,0) |
| \(\dfrac{11mg}{2} + F\dfrac{4\sqrt{3}}{2} = 2R\) | DM1A1 |
| \(\uparrow\ R = 4mg\), | B1 |
| Using \(F_r = \mu R\) | B1 |
| \(8\mu\sqrt{3} = \dfrac{5}{2}\), \(\mu = \dfrac{5}{16\sqrt{3}}\) | M1 A1 |