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alexgmcm
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Homework Statement
I have dutifully reproduced the problem in OODraw please forgive the poor aesthetic quality but I believe all the important information is visible.
Homework Equations
[tex]mgL-Lsin(\theta)m'g = \frac{mv^2}{2} + \frac{m'v'^2}{2} [/tex]
[tex] T=T'[/tex]
[tex] v=v'[/tex]
[tex] \therefore v^2 = \frac{2gl(m-m'sin(\theta)}{m+m'}[/tex]
[tex] v = \sqrt{\frac{2gl(m-m'sin(\theta)}{m+m'}}[/tex]
Also:
[tex]m'a = T-m'gsin(\theta)[/tex]
[tex]ma = mg - T[/tex]
[tex]\therefore m(g-a)=m'(a+gsin(\theta))[/tex]
The Attempt at a Solution
I assumed that T must be equal to T' as it is the same rope and hence can only have one tension? It just seemed to make more sense. The Lsin(theta) bit comes from the assumption that the rope length remains constant and hence m' is dragged parallel to the slope the same distance that m falls. I also assumed that v' must equal v as otherwise the rope would break, I just don't see how it could make physical sense if they weren't equal.
I need an expression for V in terms of L and theta and possibly g. That makes the second part of the question solvable. But I cannot see how to cancel the expressions for the masses out of my equations.
Any help?
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