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fterh
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Homework Statement
A smooth wedge, whose central cross-section is a triangle ABC, right-angled at C, rests with the face containing AB on a smooth horizontal plane. When the wedge is held fixed, a particle released from rest, takes a time t1 to slide the full length of CA. The corresponding time for CB is t2. Show that tan A = t2/t1, and find AB in terms of t1 and t2. If the mass of the wedge is n times that of the particle and the wedge is free to move, show that the time of sliding down CA becomes [tex]t_{1}\sqrt{1-\frac{t^{2}_{1}}{(n+1)(t^{2}_{1}+t^{2}_{2})}}[/tex]
The Attempt at a Solution
I take "Show than tan A = t2/t1" to mean that the distance of BC/AC is t2/t1, which is correct, right? I let angle A be theta.
So in this case, [tex]S_{BC} = \frac{g}{2}cos(\vartheta)t^{2}_{2}[/tex], while [tex]S_{AC} = \frac{g}{2}cos(90-\vartheta)t^{2}_{1}[/tex] which is also [tex]S_{AC} = \frac{g}{2}sin(\vartheta)t^{2}_{1}[/tex].
But how do I simplify it to t2/t1, without the square on both numerator and denominator?
Edit: Okay I got it, thanks for looking! :D
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