Recent content by trdewitt

Well Ill try it. I have two trys left so Ill start with your answer. It makes sense but at the same time my experience says 6.5 - 23.1 and then adding that to the other side is legal. I got 66.32 Edit: You were right, guess I'm abnormally tired since I made such a stupid mistake. Thank you for...

These are my calculations: F(.5) - 6.5 - (.6)(140)(.275) = 1/2(60)(.275)^2(1.57) f(.5) -16.6 = 3.56194 f(.5) = 20.1619 f = 40.3239

23.1 What radius did you use for the axe? I used the grindstone, .275.

I have no idea then, I feel like everything is correct, I don't know what's wrong. Am I close to yours at least?

I see what you mean, I'm starting to get it a little more. I put in the radius of the wheel into that part of the equation and it looks like my final answer became 40.32

Ok, I'll add in the variables where I felt they should go. T_net = F(.5) - 6.5 - (.6)(140) = 1/2(60)(.275)^2(1.57) To this my answer was: 162.124 Thoughts?

I thought I already did, what else do you need to know?

My answers were 162.52, 155.51, and 142.51. All wrong so far. I only have two chances left, could you share your answer?

So, taking the piece of information, my recalculated answer comes to 162.124. What do you think? I figured I was going to be wrong somewhere related to that piece of information, but I didn't know exactly what it meant so i didn't know where to put it. Btw, before I put this into my hmwk, since...

Homework Statement 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...