1. The problem statement, all variables and given/known data A 60.0-kg grindstone is a solid disk 0.550m in diameter. You press an ax down on the rim with a normal force of 140N(Figure 1) . The coefficient of kinetic friction between the blade and the stone is 0.60, and there is a constant friction torque of 6.50N⋅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 120 rev/min in 8.00s? I then have the questions: 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? and 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? 2. Relevant equations T = RFsin(theta) 3. The attempt at a solution I set up an equation, shown below, which sets the Torque net to all of the variables that add up to get that net force, which is equal to I * alpha.. I found alpha to be 1.57 using the formula w = wo + alpha * t. I is 1/2MR^2. Tnet = Fr - Taxle - Taxe = I * alpha r being the handle length, Taxle being RF, or (.275) * (60) * (9.8), Taxe being Mk * Fn or (.6) * (140). After doing this out, I got 155.51 Nm. The system is telling me I'm wrong. Help please, I'm dying.