How Do You Calculate Tan A Using Slide Times t1 and t2 on a Smooth Wedge?

[fterh]

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 t_{1}\sqrt{1-\frac{t^{2}_{1}}{(n+1)(t^{2}_{1}+t^{2}_{2})}}

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, S_{BC} = \frac{g}{2}cos(\vartheta)t^{2}_{2}, while S_{AC} = \frac{g}{2}cos(90-\vartheta)t^{2}_{1} which is also S_{AC} = \frac{g}{2}sin(\vartheta)t^{2}_{1}.

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

 

[fterh]

How do I solve the 2nd part?

 

[fterh]

Bump.

 

[tiny-tim]

hi fterh!

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 …

hint: the centre of mass is stationary (no horizontal external forces), so the ratio of the speeds is … ?

 

[fterh]

hi fterh!


hint: the centre of mass is stationary (no horizontal external forces), so the ratio of the speeds is … ?

inversely related to the ratio of the masses? I'm thinking, due to Newton's third law, the horizontal component of the normal force acting on the particle is equal to the horizontal force acting on the wedge in the opposite direction.

So F=ma, m_particle * a_particle = m_wedge * a_wedge, and since v=at, at any time the ratio of the speeds will be inversely related to the ratio of the masses. :D Am I right?

 

[tiny-tim]

hi fterh!

So F=ma, m_particle * a_particle = m_wedge * a_wedge, and since v=at, at any time the ratio of the speeds will be inversely related to the ratio of the masses. :D Am I right?

yes and no

there is no horizontal external force (there is a vertical external force), so only the horizontal components of velocity will be inversely related to the ratio of the masses …

you'll need to adjust the particle's horizontal component to find the particle's total speed

 

[fterh]

hi fterh!


yes and no

there is no horizontal external force (there is a vertical external force), so only the horizontal components of velocity will be inversely related to the ratio of the masses …

you'll need to adjust the particle's horizontal component to find the particle's total speed

What after that? :/

 

[tiny-tim]

then you find the particle's acceleration down the slope, and from that you find the time that the question asks for