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An object with a mass of #12 kg# is on a plane with an incline of # -(3 pi)/8 #. If it takes # 4 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?

An object with a mass of #12 kg# is on a plane with an incline of # -(3 pi)/8 #. If it takes # 4 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?
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The static coefficient of friction is #2.5031# (4dp)
The kinetic coefficient of friction is #2.4587# (4dp)

Explanation:

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For our diagram, #m=12kg#, #theta=(3pi)/8#

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

#R-mgcostheta=0#
#:. R=12gcos((3pi)/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+12gsin ((3pi)/8) \ \ N#

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

# F = mu R => 4+12gsin ((3pi)/8) = mu (12gcos((3pi)/8)) #
# :. mu = (4+12gsin ((3pi)/8))/(12gcos((3pi)/8)) #
# :. mu = 2.5030953 ... #

Once the object is moving the driving force is reduced from #4N# to #2N#. Now #D=3#, reapply Newton's Second Law down parallel to the plane and we get:

# D+mgsin theta -F = 0 #
# :. F = 2+12gsin ((3pi)/8) \ \ N#

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

# F = mu R => 2+12gsin ((3pi)/8) = mu (12gcos((3pi)/8) #
# :. mu = (2+12gsin ((3pi)/8))/(12gcos((3pi)/8)) #
# :. mu = 2.4586544 ... #

So the static coefficient of friction is #2.5031# (4dp)
the kinetic coefficient of friction is #2.4587# (4dp)

answered by: Steve M
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An object with a mass of #12 kg# is on a plane with an incline of # -(3 pi)/8 #. If it takes # 4 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?
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