An object with a mass of #8 kg# is on a plane with an incline of # - pi/8 #. If it takes #4 N# to start pushing the object down the plane and #1 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 #0.4694# (4dp)
The kinetic coefficient of friction is #0.4280# (4dp)

Explanation:

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

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

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

Initially it takes #4N# to start the object moving, so #D=4#. If we Apply Newton's Second Law down parallel to the plane we get:

# D+mgsin theta -F = 0 #
# :. F = 4+8gsin (pi/8) \ \ N#

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

# F = mu R => 4+8gsin (pi/8) = mu (8gcos(pi/8)) #
# :. mu = (4+8gsin (pi/8))/(8gcos(pi/8)) #
# :. mu = 0.469437 ... #

Once the object is moving the driving force is reduced from #4N# to #1N#. Now #D=1#, reapply Newton's Second Law down parallel to the plane and we get:

# D+mgsin theta -F = 0 #
# :. F = 1+8gsin (pi/8) \ \ N#

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

# F = mu R => 1+8gsin (pi/8) = mu (8gcos(pi/8)) #
# :. mu = (1+8gsin (pi/8))/(8gcos(pi/8)) #
# :. mu = 0.428019 ... #

So the static coefficient of friction is #0.4694# (4dp)
the kinetic coefficient of friction is #0.4280# (4dp)