Question

A uniform ladder of length 10⋅0 m and mass 16⋅0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60⋅0 kg climbs up the ladder. If he stays on the ladder at a point 8⋅00 m from the lower end, what will be the normal force and the force of friction on the ladder by the ground? What should be the minimum coefficient of friction for the electrician to work safely?

Answer 1

Let N₂ be the normal force on the ladder by the ground.
Let N₁ be the normal force on the ladder by the wall.
Let f_r be the force of friction on the ladder by the ground.
As system is in translation equilibrium, we have:
$$\begin{gathered} N_1=f_r=\mu N_2; \\ {N}_2=16g+60g=745\mathrm{N} \end{gathered}$$
Applying condition of rotational equilibrium at point O,
i.e., about point O, we have:
${\Gamma }_{total}=0$
$N_1\times 10\cos 37°=16g\times 5\sin 37°+60g\times 8\sin 37°$
$$\begin{gathered} \Rightarrow 8N_1=48g+288g \\ \Rightarrow N_1=\frac{336\times 9.8}{8}=412\mathrm{N} \\ \therefore \mu =\frac{N_1}{N_2}=\frac{412}{745}=0.553 \end{gathered}$$