Anonymous
Anonymous asked in Science & MathematicsPhysics · 1 decade ago

Coefficient of static friction?

A 6m long, uniform ladder leans against a frictionless wall and makes an angle of 69.3 degrees with the floor. The ladder has a mass Ml=31.7 kg. An Mm=82.42 kg man climbs 82 percent of the way to the top of the ladder when it slips and falls to the floor.

A. What is the coefficient of static friction between the ladder and the floor?

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  • 1 decade ago
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    Mass of man = Mm

    Mass of ladder =Ml

    As wall is frictionless, reaction of wall is perpendicular to the wall

    Normal reaction of floor= R = [ Mm + Ml ] g

    R= [ Mm + Ml ] g=1118.376 N

    As the man climbs 82 percent of the way to the top of the ladder ,

    distance of man from top of ladder = 0.18L

    force of friction = f

    angle which ladder makes with the floor = O =69.3 degree

    Taking torque about the upper point where ladder touches the wall,

    Torque due to wt of man= t1=Mm*g*0.18LcosO anticlockwise

    Torque due to wt of ladder= t2=Ml*g*0.5LcosO anticlockwise

    Torque due to friction =t3 =fLsinO.......anticlockwise

    Torque due to normal reaction =t4 =[ Mm+Ml ] gLcos O clockwise

    When ladder is about to slide resultant torque is zero which means,

    Sum of anticlockwise torques = sum of clockwise torque

    Mm*g*0.18LcosO + Ml*g*0.5LcosO +fLsinO= [ Mm+Ml ] gLcos O

    Dividing both sides by LcosO and simplifying,

    f tanO = [ Mm+Ml ] g- { Mm*g*0.18 + Ml*g*0.5 }

    f tanO = [0.82 Mm+0.5Ml ] g

    f = [0.82 Mm+0.5Ml ] g / tanO

    f = [0.82*82.42+0.5*31.7 ] *9.8 / tan69.3

    f = [ 67.404 + 16.85 ]*9.8 / 2.6464 =312.0019 N

    coefficient of static friction =f/R=312.0019 / 1118.376=0.279

    coefficient of static friction between ladder and floor is 0.279

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