An object with a mass of #5 kg# is on a plane with an incline of # - pi/8 #. If it takes #8 N# to start pushing the object down the plane and #3 N# to keep pushing it, what are the coefficients of static and kinetic friction?

Answer 1

The static coefficient of friction is #0.5909# (4dp)
The kinetic coefficient of friction is #0.4804# (4dp)

For our diagram, #m=5kg#, #theta=pi/8#

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

#R-mgcostheta=0#
#:. R=5gcos(pi/8) \ \ N#

Initially it takes #8N# to start the object moving, so #D=8#. If we Apply Newton's Second Law down parallel to the plane we get:

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

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

# F = mu R => 8+5gsin (pi/8) = mu (5gcos(pi/8)) #
# :. mu = (8+5gsin (pi/8))/(5gcos(pi/8)) #
# :. mu = 0.5909306 ... #

Once the object is moving the driving force is reduced from #8N# to #3N#. Now #D=3#, reapply Newton's Second Law down parallel to the plane and we get:

# D+mgsin theta -F = 0 #
# :. F = 3+5gsin (pi/8) \ \ N#

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

# F = mu R => 3+5gsin (pi/8) = mu (5gcos(pi/8)) #
# :. mu = (3+5gsin (pi/8))/(5gcos(pi/8)) #
# :. mu = 0.4804482 ... #

So the static coefficient of friction is #0.5909# (4dp)
the kinetic coefficient of friction is #0.4804# (4dp)

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Answer 2

Nicholas Agrusa

The coefficient of static friction is $\mu_s = \tan(\theta)$ , where $\theta$ is the angle of the incline. Given $\theta = -\frac{\pi}{8}$ , $\mu_s = \tan\left(-\frac{\pi}{8}\right)$ . The coefficient of kinetic friction is $\mu_k = \tan(\alpha)$ , where $\alpha$ is the angle whose tangent is the ratio of the force needed to keep the object moving to the force pressing the object against the incline. Therefore, $\mu_k = \tan\left(\arctan\left(\frac{3}{5}\right)\right)$ .

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Answer from HIX Tutor

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