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Question

A uniform ladder which is 5 m long has a mass of 30 kg with its upper end against a smooth vertical wall and its lower end is on a rough ground. The bottom of the ladder is 3 m from the wall. Calculate the frictional force between the ladder and the ground. (g=10 m/s2)

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Solution

length of ladder =5 m

distance of base form vertical wall =3 m

mass of ladder =30 kg=m

friction act only on point A

let friction force at

A=fr

from figure

cosθ=35

θ=53o

making free body diagram of ladder

Then for vertical balance

=N1=mg.....(1)

for horizontal balance

fr=N.....(2)

Taking rotational equilibrium about C

as torque is only due to perpendicular component

⇒AC=BC=52

AC×fr sinθ+BC×N sinθ=AC×N1 cosθ

=52×frsin53+52×frsinθ=52×mg cosθ

as (N=fr,N1=mg from (1) & (2))

=2frsin53=30×10cos53

=fr=30×102×43=200 N

Hence friction between ladder and ground is 200 N

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