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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 mdistance of base form vertical wall =3 mmass of ladder =30 kg=mfriction act only on point Alet friction force atA=frfrom figurecosθ=35 θ=53omaking 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=52AC×fr sinθ+BC×N sinθ=AC×N1 cosθ=52×frsin53+52×frsinθ=52×mg cosθas (N=fr,N1=mg from (1) & (2))=2frsin53=30×10cos53=fr=30×102×43=200 NHence friction between ladder and ground is 200 N

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