Question

A uniform ladder which is $5m$ long has a mass of $30kg$ with its upper end against a smooth vertical wall and its lower end is on a rough ground. The bottom of the ladder is $3m$ from the wall. Calculate the frictional force between the ladder and the ground. $(g=10m/s^2)$

Answer 1

length of ladder $=5m$

distance of base form vertical wall $=3m$

mass of ladder $=30kg=m$

friction act only on point $A$

let friction force at

$A=f_r$

from figure

$\cos \theta =\frac{3}{5}$

$\theta ={53}^o$

making free body diagram of ladder

Then for vertical balance

$=N^1=mg.....\left(1\right)$

for horizontal balance

$fr=N.....\left(2\right)$

Taking rotational equilibrium about $C$

as torque is only due to perpendicular component

$\Rightarrow AC=BC=\frac{5}{2}$

$AC\times fr\sin \theta +BC\times N\sin \theta =AC\times N^1\cos \theta$

$=\frac{5}{2}\times fr\sin 53+\frac{5}{2}\times fr\sin \theta =\frac{5}{2}\times mg\cos \theta$

as ($N=f_r,N^1=mg$ from $\left(1\right)$ & $\left(2\right)$)

$=2f_r\sin 53=30\times 10\cos 53$

$=fr=\frac{30\times 10}{2}\times \frac{4}{3}=200N$

Hence friction between ladder and ground is $200N$