An object with a mass of #8 kg# is on a plane with an incline of # - pi/12 #. If it takes #12 N# to start pushing the object down the plane and #5 N# to keep pushing it, what are the coefficients of static and kinetic friction?

1 Answer
Jan 23, 2017

Answer:

the static coefficient of friction is #0.4264# (4dp)
the kinetic coefficient of friction is #0.3340# (4dp)

Explanation:

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For our diagram, #m=8kg#, #theta=pi/12#

If we apply Newton's Second Law up perpendicular to the plane we get:

#R-mgcostheta=0#
#:. R=8gcos(pi/12) \ \ N#

Initially it takes #12N# to start the object moving, so #D=12#. If we Apply Newton's Second Law down parallel to the plane we get:

# D+mgsin theta -F = 0 #
# :. F = 12+8gsin (pi/12) \ \ N#

And the friction is related to the Reaction (Normal) Force by

# F = mu R => 12+8gsin (pi/12) = mu (8gcos(pi/12)) #
# :. mu = (12+8gsin (pi/12))/(8gcos(pi/12)) #
# :. mu = 0.426409 ... #

Once the object is moving the driving force is reduced from #12N# to #5N#. Now #D=5#, reapply Newton's Second Law down parallel to the plane and we get:

# D+mgsin theta -F = 0 #
# :. F = 5+8gsin (pi/12) \ \ N#

And the friction is related to the Reaction (Normal) Force by

# F = mu R => 5+8gsin (pi/12) = mu (8gcos(pi/12)) #
# :. mu = (5+8gsin (pi/12))/(8gcos(pi/12)) #
# :. mu = 0.333974 ... #

So the static coefficient of friction is #0.4264# (4dp)
the kinetic coefficient of friction is #0.3340# (4dp)