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14 July, 16:48

A 6 m long, uniform ladder leans against a frictionless wall and makes an angle of 74.3 ◦ with the floor. The ladder has a mass 25.8 kg. A 67.08 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 be - tween the ladder and the floor?

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  1. 14 July, 17:18
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    Answer: µ=0.205

    Explanation:

    The horizontal forces acting on the ladder are the friction (f) at the floor and the normal force (Fw) at the wall. For horizontal equilibrium,

    f=Fw

    The sum of the moments about the base of the ladder Is 0

    ΣM = 0 = Fw*L*sin74.3º - (25.8kg * (L/2) + 67.08kg*0.82L) * cos74.3º*9.8m/s²

    Note that it doesn't matter WHAT the length of the ladder is - - it cancels.

    Solve this for Fw.

    0 = 0.9637FwL - (67.91L) 2.652

    Fw=180.1/0.9637

    Fw=186.87N

    f=186.81N

    Since Fw=f

    We know Fw, so we know f.

    But f = µ*Fn

    where Fn is the normal force at the floor - -

    Fn = (25.8 + 67.08) kg * 9.8m/s² =

    910.22N

    so

    µ = f / Fn

    186.81/910.22

    µ = 0.205
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