The problem involves two masses, m and M, connected by a string and affected by gravity. The objective is to determine the minimum ratio of m to M when the two blocks cross each other, as depicted in an accompanying image. The scenario requires understanding the forces and constraints acting on the masses as they move.
The discussion is ongoing, with participants sharing their attempts and insights. Some suggest that energy considerations may play a significant role in understanding the problem, while others are exploring the implications of different mass ratios. There is no explicit consensus on the correct approach or interpretation of the problem yet.
Participants note potential ambiguities in the problem statement, such as the definition of "crossing" and whether the pulleys have a specified radius. Additionally, there are constraints related to the geometry of the system and the assumptions made about the motion of the masses.
@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: