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M2 Ladder question (please help!!)
A uniform ladder rests in limiting equilibrium with its top against a smooth vertical wall and its base on a rough horizontal floor. The coefficient of friction between the floor and the ladder is µ. Given that the ladder makes an angle theta with the floor show that:
2µ tan theta = 1
2µ tan theta = 1
futureaussiecto
A uniform ladder rests in limiting equilibrium with its top against a smooth vertical wall and its base on a rough horizontal floor. The coefficient of friction between the floor and the ladder is µ. Given that the ladder makes an angle theta with the floor show that:
2µ tan theta = 1
2µ tan theta = 1
So there are 4 forces involved, the reaction on the wall, the reaction on the floor, the weight from the centre of the ladder and friction. Since there's equilibrium:
Reaction of wall = Friction
Reaction of floor = Weight
Friction = µ*Reaction of floor
Friction = µ*Weight
Resolve around the point on the floor (assume ladder is length 2)
2*Reaction of wall*sinθ = 1*Weight*cosθ
2*Friction*sinθ = Weight*cosθ
2*µ*Weight*sinθ = Weight*cosθ
Divide both sides by cosθ
2*u*tanθ = 1
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