The static coefficient of friction is
The kinetic coefficient of friction is
For our diagram,
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
# 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
# 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
the kinetic coefficient of friction is
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The coefficients of static and kinetic friction between a 516 crate and the warehouse floor are 0.715 and 0.450, respectively. A worker gradually increaseshis horizontal push against this crate until it just begins to move and from then on maintains that same maximum push.What is the acceleration of the crate after it has begun to move? Start with a free-body diagram of the crate.
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