Question

A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 58 ◦ with the floor. The ladder has a mass 26.4 kg. A 68.64 kg man climbs 82% of the way to the top of the ladder when it slips and falls to the floor. What is the coefficient of static friction between the ladder and the floor?

Answer 1

the ladder are the friction at the floor (let's call it "f") and the normal force at the wall (Fw). For horizontal equilibrium, these must be equal -- find Fw and found f.

So sum the moments about the base of the ladder:
ΣM = 0 = Fw*L*sin58º - (26.4kg*(L/2) + 68.64kg*0.82L)*cos58º*9.8m/s²
Notice that it doesn't matter WHAT the length of the ladder is -- it cancels.
Solve this for Fw.

We know Fw, so we know f.

f=425.5 N

But f = µ*Fn
where Fn is the normal force at the floor --
Fn = (26.4+68.64)kg * 9.8m/s² = 931.392 N

so
µ = f / Fn =0.457