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Respected experts, I am in need of your help. This is a question which I can not determine how to approach, how to proceed, what concepts are involved,...nothing! I would be very grateful if you could guide me to the correct method to solve this problem. Here goes the question:
"A particle is placed on a rough plane inclined at an angle theta, where tan θ = μ = coefficient of friction(both static an dynamic). A string attached to the particle passes through a small hole in the plane. The string is pulled so slowly that you may consider the particle to be in static equilibrium at all times. Find the path of the particle on the inclined plane."
Comments (16)
rite... thats the case when the coeff of friction is good enuff to balance gravity. The other( and i mean..the general ) case is rather complicated.. agreed the string is being pulled down at negligible rate..but that does not imply that the particle wont slide... hence, in this case..the trajectory will be a combination of 2 linear motions ..one accelerating and other non-accelerating. ONE MORE POINT : had the stable equilibrium thing wudnt be there...khair chaddo ..any takers ?
Looks like now people are beginning to take interest.
But the answer's given as a circle, in this form : x^2 + y^2 = cy.
Anyway, Lakshya, how did you get it as a parabola? The reasoning's fine, I'd also got to that, but how did you conclude that it is a parabolic path? Waiting for you to shed some more light on it, may be it will be enlightening for us all.
Sorry to say this, sir, but I don't think this solution is completely correct.
This is because you have taken T as a constant, which is not true. In fact, T changes such that the system is in static equilibrium.
You can verify this by just looking at your answer, the reported equation is x^2 + (1-k)y^2 = 0, and this equation does not give any curve if 1-k is positive, and gives 2 straight lines if 1-k is negative. So, you see, this is not the correct answer.
I think the discussions should START AGAIN!
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