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 ${}^{\circ }$ 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

Ladder is in horizontal equilibrium

$R_1=m{R}_2$

Ladder is in vertical equilibrium

$R_2=16g+60g=745\text{N}$

Ladder is in rotational equilibrium and hence taking moments about the point of contact at the ground we get

$R_1\times 10cos{37}^{\circ }=16g\times 5sin{37}^{\circ }$

$+80g\times 8\times sin{37}^{\circ }$

$\Rightarrow 8R_1=48g+288g$

$\Rightarrow R_1=\frac{336g}{8}=412\text{N}$

$\text{Therefore,}\mu =\frac{R_1}{R_2}=\frac{412}{745}=0.553$