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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) - t

_{handle}= (1/2)(50)

^{(.26)2}(1.396)

t

_{handle}= 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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