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# 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?

Relevance

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Mass of man = Mm

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

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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