Question

A uniform ladder having length 10.0 m and mass 24 kg is resting against a vertical wall making and angle of ${53}^0$ with it. The vertical wall is smooth but the ground surface is rough. A painter weighing 75 kg climbs up the ladder. If he stays on the ladder at a point 2 m from the upper 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 between ground and ladder for the painter to work safely? $(g=10m/s^2)$

Answer 1

Given,

Length of ladder is $l=10m$

Position of man on ladder from bottom, $d=8m$

Weight of ladder=$24gN$

Weight of painter$=75gN$

Angle with horizontal is ${53}^o$

Angle with vertical is ${90}^o-{53}^o={37}^o$

Acceleration of gravity is $g$

From diagram

${\text{R}}_2=24\text{g}+75\text{g}=99g\text{N}...........\text{(1)}$

Net moment at bottom of ladder

${\text{R}}_1\times l\cos {37}^{{}^{\circ }}=24\text{g}\times \frac{l}{2}\sin {37}^{{}^{\circ }}+75\text{g}\times (d\times \sin {37}^{{}^{\circ }})$

${\text{R}}_1\times 10\cos {37}^{{}^{\circ }}=24\text{g}\times 5\sin {37}^{{}^{\circ }}+75\text{g}\times 8\times \sin {37}^{{}^{\circ }}$

$\Rightarrow R_1=54.25gN........\left(2\right)$

${\text{R}}_1=\mu {\text{R}}_2$

From equation (1) and (2)

$\mu ={\text{R}}_1/{\text{R}}_2=54.25g/99g=0.55$

Hence, coefficient of friction is $0.55$