1. The problem statement, all variables and given/known data 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? 2. Relevant equations Net torque = Iα 3. 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?