Can any of you brailets represent (a) in a more simplified equation?

>implying that's not shit bait to make him do it

I'm in med school
well i just need a little push in the right direction, math isn't really my thing

>uses Newtonian mechanics for an atwood machine
>calls us brainlets

Hahahahaha

Use virtual work

WTF is that?

If anyone wants to try and solve this differential equation, then go ahead.

But Euler-Lagrange equation should be much easier.

Now for Euler-Lagrange

You done fucked up, mate. So, you have two different equations for the same motion? What's this, Schrodinger's pulley?

Fuck it's gonna be hard to ague against quads but I'll try.
The first one is a differential equation and isn't finished. But if you were to solve the differential equation for x** (or a) , you should get the equation in the second image

If math isn't really your things, that equation doesn't need to be simplified. Its plug and chug for a correct answer on something theoretical.

You did fine.

But you won't get the same equation. There are no terms with square roots in first equation. Also your second equation has dimensionless term [math]\frac{m_2}{m_1+m_2}[/math].

quick question, why can you leave out the acceleration of the ssecond mass in the dissipation term of the potential?

Thanks

what does T stand for?

both masses have the same acceleration.

Have you ever solved a differential equation m8?

Dude, don't try to be smug with me. [math]\ddot{x}=a[/math]. Your second equation is differntial equation too. You have two different differential equations that decribe the same motion. So, the block has two different accelerations at the same time? How does that work?

tension

>mechanics

kek, i remember when i was 12

>both masses have the same acceleration.
Do they? isn't there a cos(alpha) needed somewhere?
But the normal force of m1 remains the same, so the dissipation is obviously mu.m1.g.x
While the upward force on m1 by m2 depends on the acceleration of m2 so why isn't that term mu.m2.x.(g-a).sin(alpha)?

sorry for no latex and possibly a stupid question

If you consider the tension of the string constant they will have the same acceleration. Its basically a rigid body with that hypothesis.

>you
Nope, just look at the geometry, when on body travels a unit distance, the other ones travels different distance. Trigonometry, nigga.

That doesn't mean the acceleration is different?

The acceleration in the generalized coordinate system is the same you brainlet. Come back when you have studied Lagrangian mechanics

that means exactly that, you are describing only one bodies acceleration. When the rope is horizontal, the acceleration is same. Imagine the very steep rope angle, the m1 body would have to cover a much larger distance, thus higher acceleration.

No

Listen you faggot, using priciple of virtual work, at first, when you are describing the system, you evaluate different distances and shit, eventually you can get a "generalised" acceleration, but different bodies will accelerate different.
Now, the op fag, I think his problem is not supposed to go so deep, so the full problem description would be helpful.

No, seeing that you are a fucking retard, I won't wait for another answer from you.
I present you with a more visual example.
Generalise my dick, you cun't

Define x1 to be in the direction of alpha and x2 to be downwards.
Find the Lagrangian and solve the Euler Lagrangian equations
Use the constraint x1+x2=l where l is the length of the string and add the constraint force.
I haven't checked the guys working it could be wrong but they do have the same acceleration ( in the new coordinates) considering they are literally attached by in in-extensible string

Yes this is a double pulley where 2x1+x2=l
so 2x1''=-x1''... I fail to see the point

Yes, let's say we solved the system and the body m1 moves at 1 m/s2 upwards. The body m2 is going down at 2 m/s2.
Are you all such fucking retards?
The best part is that parts of the string are not even moving...
You do realise, that geometry plays a part, of how bodies move?

The point is that the component in the direction of the string must be the same, I think the confusion lies in the way we are defining our coordinate system

The confusion lies in the way that op faggot described the problem. There is no description except the drawing.
I guess, that as op states, it is not his main course, the problem should be very simple in nature, now the most interesting thing for me is what does the problem ask?

This is a lower division physics problem, you get a better understanding of how to account for that sort of thing in classes where it matters

>This is why nothing ever works right. We are training a bunch of engineers to always assume that
Nope, they are trained to see the essence or problem. Maybe we should include air resistance here? How about time dilation?
Yeah, rope and pulleys... The situations there this would matter are basically nonexistant.