An object with a mass of #4 kg# is on a plane with an incline of # - pi/8 #. If it takes #18 N# to start pushing the object down the plane and #15 N# to keep pushing it, what are the coefficients of static and kinetic friction?
1 Answer
The static coefficient of friction is
The kinetic coefficient of friction is
Explanation:
For our diagram,
If we apply Newton's Second Law up perpendicular to the plane we get:
#R-mgcostheta=0#
#:. R=4gcos(pi/8) \ \ N#
Initially it takes
# D+mgsin theta -F = 0 #
# :. F = 18+4gsin (pi/8) \ \ N#
And the friction is related to the Reaction (Normal) Force by
# F = mu R => 18+4gsin (pi/8) = mu (4gcos(pi/8)) #
# :. mu = (18+4gsin (pi/8))/(4gcos(pi/8)) #
# :. mu = 0.911230 ... #
Once the object is moving the driving force is reduced from
# D+mgsin theta -F = 0 #
# :. F = 15+4gsin (pi/8) \ \ N#
And the friction is related to the Reaction (Normal) Force by
# F = mu R => 15+4gsin (pi/8) = mu (4gcos(pi/8)) #
# :. mu = (15+4gsin (pi/8))/(4gcos(pi/8)) #
# :. mu = 0.828394 ... #
So the static coefficient of friction is
the kinetic coefficient of friction is