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
Jan 16, 2017

Answer:

The static coefficient of friction is #0.9112# (4dp)
The kinetic coefficient of friction is #0.8284# (4dp)

Explanation:

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For our diagram, #m=4kg#, #theta=pi/8#

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 #18N# to start the object moving, so #D=18#. If we Apply Newton's Second Law down parallel to the plane we get:

# 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 #18N# to #15N#. Now #D=15#, reapply Newton's Second Law down parallel to the plane and we get:

# 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 #0.9112# (4dp)
the kinetic coefficient of friction is #0.8284# (4dp)