1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
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 ∘ 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 ?

Open in App
Solution

## Ladder is in horizontal equilibriumR1=mR2Ladder is in vertical equilibrium R2=16g+60g=745NLadder is in rotational equilibrium and hence taking moments about the point of contact at the ground we get R1×10 cos 37∘=16g× 5 sin 37∘ +80g×8× sin 37∘ ⇒8R1=48g+288g ⇒R1=336g8=412N Therefore,μ=R1R2=412745=0.553

Suggest Corrections
9
Join BYJU'S Learning Program
Related Videos
Friction: A Quantitative Picture
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program