@Abcd just that "Indians are spamming (Chem.SE) now" is a very very vague allegation, I know you didn't mean to make any allegation, but be careful on the internet
what you wrote exactly sounded like the former
that said, SE is very great at fighting spam
you won't run into much trouble; i've seen worse forums with worse spam problems
@Abcd in those accelerating elevator type problems the easiest approach is just to note that the gravitational acceleration is changed. Just add the acceleration of the elevator to $g$ and do the problem as normal.
(I've just scribbled my way through the calculation and I get the same result)
When the elevator is accelerating upwards at $w_0$ in effect it increases $g$ to $g' = g + w_0$
So the acceleration is now: $$ a = \frac{g'(m_1 - m_2)}{m_1 + m_2} = \frac{(g + w_0)(m_1 - m_2)}{m_1 + m_2} $$
That's taking $w_0$ as a positive number. I suppose if we're doing our sign conventions properly it would be: $$ a = \frac{(g - w_0)(m_1 - m_2)}{m_1 + m_2} $$
That's the acceleration relative to the elevator car
I'm going to write $A = g + w_0$ so our equation becomes: $$ a_1 = A + \frac{T}{m_1} $$
@Abcd the elevator car is accelerating upwards while gravity accelerates downwards. There are two ways you can handle this. If you insist that the values of $g$ and $w_0$ are both positive numbers then you have to remember to add $g$ and subtract $w_0$.
But down that route lies the potential for making sign errors.
My preferred approach is that if we have two forces $F_1$ and $F_2$ acting on a body then the net force is always $F_1 + F_2$.
And we make the values of $F_1$ and $F_2$ positive or negative depending on their direction.
Signs are hard and easy to get confused about. With experience you can be careless with signs because you're used to knowing intuitively which way the forces act.
If you don't have that experience you need to be extra careful.
Yes. What I've done is work in the stationary inertial frame of the shaft and simply add up all the forces on the mass. That gives me the equation you've just linked.
Suppose $w_0 \gg g$ i.e. the acceleration is dominated by $w_0$
Oh, hang on ...
Better idea, suppose we make $g$ negligibly small e.g. we're doing the experiment in space
As soon as the elevator starts accelerating away from us both masses will accelerate away from us as well i.e. the acceleration of both masses has the same sign (negative if we're taking upwards to be negative). OK so far?
@Abcd suppose you're in the elevator so you're measuring the accelerations relative to the elevator floor. In this frame we must have $a_1 = -a_2$ otherwise as you say the thread would go slack.
But we're not in the elevator, we're in the lift shaft, and the accelerations we measure have $w_0$ added on to them i.e. $a'_1 = a_1 + w_0$ and $a'_2 = a_2 + w_0$, where $a'$ are the accelerations measured in the lift shaft.
(hmm, I missed that factor of 2 in my previous comment ... hmm ...)
Ah OK, yes, that must be correct. Suppose $m_1 = m_2$ then in the elevator frame both are stationary so in the shaft frame both have $a' = w_0$ so $a'_1 + a'_2$ is indeed $2w_0$.
You've gone quiet. You're either stunned by my genius or you're wondering if I've gone insane :-)
@Abcd let's try the elevator frame and see what happens. If you're standing in the elevator looking at the masses and they are equal then they will just remain motionless. Yes?
@Abcd let's try the elevator frame and see what happens. If you're standing in the elevator looking at the masses and they are equal then they will just remain motionless. Yes?
Abcd, when the pulley m goes down by let say X, there will be two free lengths created and as the left end of the string is fixed, that string will be stretched from the man end which is movable
How can I post images in this chat room so that I can post the diagram of what I am saying?
@Abcd that's a risky method because when you do that you are working in a non-inertial frame and it's easy to make mistakes.
As a general rule I would work in the lab frame. Write down the net force on all the bodies and you should end up with a set of simultaneous equations that you can solve to get the accelerations of all the bodies.
$m_1$ has only two forces acting on it, $m_1 g$ and $T$.
We don't know what $T$ is at the moment, but when we do the force balance for all the masses we'll get a set of equations from which we can calculate te tensions in all the strings.
Basically the only things we don't know are the tensions in the strings.
When we write the net forces we get a set of simultaneous equations involving the tensions, and from those we can calculate the tensions then the accelerations.
@JohnRennie Even in this problem, I agree that the string will not slack in the pulley's frame, but what about the lab frame? How would the motion seem to us then?
The acceleration of the pulley will be added, I know.
