- #1
hangainlover
- 83
- 0
Homework Statement
I know this tipping a block question has been asked many times and I know how to find the critical angle at which a block tips over.
I was just wondering how one would go about analyzing the torque of the system. We all know that anywhere from 0 to the critical angle, the block won't tip over around the left bottom vertex. I wanted to show the block won't rotate by showing the net torque of the system is zero. I can easily show it for zero angle slope but for some arbitrary angle, it doesn't go to zero...
It makes intuitive sense that the block won't rotate because the block won't wade into the slope...
but I want to see the net torque going to zero..
(just to let you guys know, I know that at the critical angle, the only source of torque is the gravitational weight because the normal force and the friction all pivot at the left end of the base, so r = 0)
Homework Equations
Torque = r x F
The Attempt at a Solution
(there is no friction)
I have attached my work.
So, according to my diagram, at the critical angle, Alpha = Theta. when the CG is right above the left end of the base, in which case, the block starts tipping over.
I guess my work is valid only when Theta =0 (when the block is just resting on a horizontal surface, in which case Fn= Fg and r x (Fn + Fg) = 0, which I can show algebraically )
Attachments
Last edited: