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henry3369
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Homework Statement
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. P10.57). 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. 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 in 9.00 s?
Homework Equations
Net torque = Iα
The Attempt at a Solution
α = 1.396 rad/s^2
Torque from axle + Torque from friction + Torque from crank handle = Iα
6.50 + (0.6)(.26)(160) - thandle = (1/2)(50)(.26)2(1.396)
thandle = 29.1 N*m = rF
F = 29.1/.26 = 111.926 N
The answer is 67.6 N though. What am I doing wrong?
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