Finding the acceleration of two masses on a pulley system

[Jpyhsics]

So I guess my question is which of these two equations do we use in the tension force equations (if either is correct...and if my tension force equations are correct...):
a₂= -a_surface + a₁
a₂= -2a₁

 

[haruspex]

a2= -a_surface + a1 ?

Which way are you defining as positive for a_surface?

 

[haruspex]

it seems as though the acceleration of the surface is not needed

You are to find the accelerations of the masses in terms of the acceleration of the surface.

 

[FriedChicken885]

So I went through the entire question with +x as left and +y as up, I got to the two equations a(2) = 2a(1) and -a(2) = a - a(1) and solved for a(1) = -a. Yet I got the incorrect answer, what did I do wrong?

 

[haruspex]

-a(2) = a - a(1)

If you are taking a₂ as positive up and a₁ positive left then you have a sign error there.

 

[FriedChicken885]

If you are taking a₂ as positive up and a₁ positive left then you have a sign error there.

I also tried it for:
a₂ = 2a₁ and a₂ = a - a₁ and solved for a₁ = ⅓a, but that's still the incorrect answer. Also I wonder if I'm missing something since I don't use the given value of m₁.

 

[haruspex]

a₂ = a - a₁

still the wrong combination of signs.
The easiest way to get this right is to imagine holding one acceleration at zero and thinking how the other two interact.
Since a₁ and a₂ are both positive towards the pulley, if a were zero their sum would be constant. So we have a₁+a₂= f(a).
If we imagine a₂ held at zero but the string remaining taut then m₁ must be at constant distance from the pulley. Since a is positive right but (bizarrely) a₁ is positive left then, again, their sum must be constant. This leads to a₁+a₂= -a.

 

[Jpyhsics]

So does this solution make sense to you?

 

[FriedChicken885]

a₁+a₂= -a.

So does this solution make sense to you?

I again tried solving that solution with the system of equations:

a₁+a₂= -a and a₂ = 2a₁

and I got a₁ = (-1/3)a which is still incorrect despite every step seeming to be correct.

Also do we need to use the numerical value for m₁ at all?

 

[haruspex]

and I got a₁ = (-1/3)a which is still incorrect despite every step seeming to be correct.

hmm.. that is the answer I get.

Also do we need to use the numerical value for m₁ at all?

No. Only the ratio of the masses can be relevant.

 

[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.