@PhysicsBoi1908 I'm very glad we got the right answer. And thank you for posting the question. It really is one of the better ones we have encountered. At first sight, it almost looked like it was a problem with a very trivial solution of ## \frac{m}{M} \geq 1 ##, but it turned out to be very interesting.PhysicsBoi1908 said:
Using energy equations, I think you can write an equation for the velocity of each mass as a function of y . The equation for each mass would have the form ## \frac{dy}{dt}=f(y) ##, which could be written as ## \frac{dy}{f(y)}=dt ##. Perhaps it even has a closed form solution. I haven't worked it yet to see. ## \\ ## Edit: Letting ## y ## the the distance that ## m ## moves, and ## y_1 ## the distance that ## M ## moves, and ## s ## the cable length from ## M ## to the pulley, we have ## s=3L-\sqrt{L^2+y^2} ##, and ## y_1=\sqrt{3}L-\sqrt{s^2-L^2} ##. ## \\ ## Finally, ## K.E.=\frac{1}{2}m(\frac{dy}{dt})^2+\frac{1}{2}M(\frac{dy_1}{dt})^2=mgy-Mgy_1 ##. Let me work it a couple more steps to see if I can get a single equation of ## \frac{dy}{dt}=f(y) ##. ## \\ ## Additional edit: Yes, ## y_1 ## can be expressed completely in terms of ## y ##, but it's not a simple expression. Thereby, we have ## g(\frac{dy}{dt})=f(y) ##, but ## f(y) ## is not a simple function, and ## g(\frac{dy}{dt}) ## is also not simple. ## \\ ## Additional edit: If anything at all useful comes out of this last bit of algebra and calculus, perhaps it is the almost obvious result that if ## \frac{dy}{dt}=0 ##, then ## \frac{dy_1}{dt}=0 ## as well.Delta² said: