An object with a mass of #12 kg# is on a plane with an incline of # - pi/6 #. If it takes #6 N# to start pushing the object down the plane and #4 N# to keep pushing it, what are the coefficients of static and kinetic friction?

Question

1520 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-14T17:44:36

coefficient of static friction #mu s = 0.366 .sqrt(3)#
coefficient of kinetic friction #mu k = 0.355 .sqrt(3)#

the incline is making an angle of #-pi/6# = -30 degrees from the horizontal.

the forces operating on the body is the weight m.g vertically

down = #12 kg . 10 m/s^2# = 120 N

the normal reaction of the plane acting perpendicular to the plane. say R

breaking the weight in components along the plane and

perpendicular to the plane inclined at 30 degree to horizontal

#mg.cos 30# = #120 . sqrt(3) /2 # = R and

mg. sin30 will act along the plane .

as a force of 6 N is needed to start the motion

so frictional force f s = 6 N + mg sin 30 = 6 N + 120/2 = 66 N

#fs = mu s . R # = 66N so, #mus = {66 / (60. sqrt (3))}#

#mu s# = #0.366 sqrt(3)#

however ,when the body starts moving a force of 4 N is sufficient to maintain the motion , therefore

#f k = mu k .R # = 64N so, #mu k = {64 / (60. sqrt (3))}#

#mu k = 0.355. sqrt(3)#