Question

a uniform ladder of length 3.25m and weighing 250N is placed against a smooth vertical wall with its lower end 1.25m from the wall. if coefficient of friction between the ladder and the floor is 0.3 , then , find the frictional force acting on the ladder at the point of contact between ladder and floor. ans: 52.1N ????

Answer 1

Given l = 3.25 m, W = 250 N, μ=0.3
let Ff be the frictional force acting on the ladder at the point of contact between the ladder and floor.

Rf = 250 N

BC=√(3.25)^2−1.25)^2=3 mtaking moments about B and equating the samewe get
Ff×3=(Rf×1.25)−(250×0.625)=156.2 NFf=156.2/3=52.1 N

Answer 2

Answer:

Pls see the attachment also.

Explanation:

Given l = 3.25 m, W = 250 N, μ=0.3

let Ff be the frictional force acting on the ladder at the point of contact between the ladder and floor.

Rf = 250 N

BC=(3.25)2−1.252−−−−−−−−−−−−−√=3 mtaking moments about B and equating the samewe getFf×3=(Rf×1.25)−(250×0.625)=156.2 NFf=156.2/3=52.1 N

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