m = 50.0 kg F= 160 N

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A 50.0 kg grindstone is a solid disk 0.520 m in diameter.
You press an ax down on the rim with a normal force of 160 N
(Fig. ). The coefficient of kinetic friction between the blade and
the stone is 0.60, and there is a constant friction torque of 6.50 N . m
between the axle of the stone and its bearings. (a) How much force must
be applied tangentially at the end of a crank handle 0.500 m long to
bring the stone from rest to 120 rev/min in 9.00 s? (b) After the grindstone
attains an angular speed of 120 rev/min, what tangential force at
the end of the handle is needed to maintain a constant angular speed of
120 rev/min? (c) How much time does it take the grindstone to come
from 120 rev/min to rest if it is acted on by the axle friction alone? 

m = 50.0 kg
F= 160 N
Transcribed Image Text:m = 50.0 kg F= 160 N
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