An object with a mass of #8 kg# is on a plane with an incline of # - pi/6 #. If it takes #9 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?

1 Answer
Jan 13, 2017

Answer:

The static coefficient of friction is #0.7099# (4dp)
The kinetic coefficient of friction is #0.6068# (4dp)

Explanation:

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For our diagram, #m=8kg#, #theta=pi/6#

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

#R-mgcostheta=0#
#:. R=8gcos(pi/6) \ \ N#

Initially it takes #9N# to start the object moving, so #D=9#. If we Apply Newton's Second Law down parallel to the plane we get:

# D+mgsin theta -F = 0 #
# :. F = 9+8gsin (pi/6) \ \ N#

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

# F = mu R => 9+8gsin (pi/6) = mu (8gcos(pi/6)) #
# :. mu = (9+8gsin (pi/6))/(8gcos(pi/6)) #
# :. mu = 0.7099051 ... #

Once the object is moving the driving force is reduced from #9N# to #2N#. Now #D=2#, reapply Newton's Second Law down parallel to the plane and we get:

# D+mgsin theta -F = 0 #
# :. F = 2+8gsin (pi/6) \ \ N#

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

# F = mu R => 2+8gsin (pi/6) = mu (8gcos(pi/6)) #
# :. mu = (2+8gsin (pi/6))/(8gcos(pi/6)) #
# :. mu = 0.6068069 ... #

So the static coefficient of friction is #0.7099# (4dp)
the kinetic coefficient of friction is #0.6068# (4dp)