Recent content by beetlegauss

Are you Alex, by any chance? Because my physics teacher's name is Alex...

Alright, coolio, I think I've got both of them now. lol, thank you both.

Yes, actually, I did the problem two different ways, both of which ended up with the same answer. The first way, which i thought was wrong, was this. Ff=2m (so now I'm accelerating, and there is no force-of-train) 9.8mu=2m u=2m/9.8m u still equals .2

Ft is, um, Force of the train. I didn't know what else to label it as. The only thing I was confused about is whether or not the Force of the train on me is it's acceleration times my mass.

Oh yeah, thanks. So the coefficient is .2, and the force on the person is his mass times the trains acceleration, then?

I have a question for homework, and this is a high school physics class. The question reads: Suppose that you are standing on a train accelerating at 2.0m/s^2. What minimum coefficient of static friction must exist between your feet and the floor if you are not to slide? I've drawn a free...