The static coefficient of friction is
The kinetic coefficient of friction is
For our diagram,
If we apply Newton's Second Law up perpendicular to the plane we get:
#R-mgcostheta=0#
#:. R=12gcos((3pi)/8) \ \ N#
Initially it takes
# D+mgsin theta -F = 0 #
# :. F = 4+12gsin ((3pi)/8) \ \ N#
And the friction is related to the Reaction (Normal) Force by
# F = mu R => 4+12gsin ((3pi)/8) = mu (12gcos((3pi)/8)) #
# :. mu = (4+12gsin ((3pi)/8))/(12gcos((3pi)/8)) #
# :. mu = 2.5030953 ... #
Once the object is moving the driving force is reduced from
# D+mgsin theta -F = 0 #
# :. F = 2+12gsin ((3pi)/8) \ \ N#
And the friction is related to the Reaction (Normal) Force by
# F = mu R => 2+12gsin ((3pi)/8) = mu (12gcos((3pi)/8) #
# :. mu = (2+12gsin ((3pi)/8))/(12gcos((3pi)/8)) #
# :. mu = 2.4586544 ... #
So the static coefficient of friction is
the kinetic coefficient of friction is
An object with a mass of #12 kg# is on a plane with an incline of # -(3 pi)/8 #. If it takes # 4 N# to start pushing the object down the plane and #2 N# to keep pushing it, what are the coefficients of static and kinetic friction?
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