Finding the acceleration of two masses on a pulley system

[haruspex]

So does this solution make sense to you?

That is the same as the general equation I obtained.

 

[Jpyhsics]

That is the same as the general equation I obtained.

So I plugged in my values and it was wrong, so I don't really know what else to do.

 

[FriedChicken885]

hmm.. that is the answer I get

Okay, thank you very very very much for all of your help!

 

[kuruman]

Should one not consider that the tension supporting the hanging mass is no longer parallel to the vertical side of the table but at an angle?
Also, I think that $g=9.81 ~m/s^2$ in the expression posted by OP in #62 should be replaced by an effective acceleration. This problem can be solved quite simply in the accelerated frame with additional fictitious horizontal forces $-m_i\vec a_0$ acting on masses $m_i$. Here, $\vec a_0$ is the given horizontal acceleration of the table.

 

[haruspex]

Should one not consider that the tension supporting the hanging mass is no longer parallel to the vertical side of the table but at an angle?

we are asked for the accelerations immediately after the surface has started to accelerate. The string is still vertical.

Also, I think that $g=9.81 ~m/s^2$ in the expression posted by OP in #62 should be replaced by an effective acceleration.

I don't see why. Do you get a different answer? If so, please post your solution.

 

[kuruman]

I don't get a different answer, but I am bothered by the fact that one writes the same equation for the hanging mass whether the pulley supporting it accelerates or not. When the pulley accelerates, the hanging mass must have a horizontal acceleration at t = 0 because its horizontal position changes a moment later. Should this not affect the tension? I need to think about this some more.

On Edit: OK, I got it. At t = 0 the horizontal acceleration is instantaneously zero and then changes as it reaches a constant value in the steady state.

 

[haruspex]

At t = 0 the horizontal acceleration is instantaneously zero

Right.

then changes as it reaches a constant value in the steady state.

Or maybe oscillates.