I'm at Hyderabad now :D

I can type at the rate of 120 WPM on my keyboard but I can't type any faster than 20 WPM on mobile :P

How is $T.x_1$ = $2T.x_2$ (wrote x1 by mistake)?

← 2 messages moved from The h Bar

@JohnRennie ^

We assume the pulley is massless

Yes

I know the reason for development of T and 2T in the different strings

The acceleration of the pulley is given by $a = F/m$, and since for the pulley $m=0$ that means the net force must also be zero otherwise the pulley would have infinite acacleration.

Yes

Ah, sorry, I didn't read your diagram carefully. Yes, I think there's a mistake.

@JohnRennie If F=0, then it would be undefined not zero

@Abcd Yes, I'm being careless with my description though you can take limits of $F \rightarrow 0$ and $m \rightarrow 0$ if you want to be rigorous.

@JohnRennie Okay

The point is that the net force on the pulley is zero, but anyway that wasn't what you were asking. I'm not sure what $T_{x_1} = 2T_{x_1}$ means in your notes.

@JohnRennie I mean $T.x_1 = T.x_2$ (Dot product of Tension and displacement) (I wrote second x1 by mistake)

Presumably you mean $T_{x_1} = 2T_{x_2}$, which is just that net work is zero we were talking about yesterday.

Oh.. hang on.

Ah, $x_1$ is the change in length of the string not the vertical displacement. i.e. the vertical displacement is half of $x_1$

@JohnRennie Not clear

$x_1$ is the distance moved by the mass on the slope.

So it's the same as we were talking about yesterday. The distance moved by the $2M$ mass is $x_2 = x_1/2$.

@JohnRennie I understood that. I didn't get the dot product thing

You don't need to worry about the dot product in this case because the force and displacement are collinear.

ok, so how are they equal?

In general the work done is $\mathbf F \cdot \mathbf x$ not just $Fx$, but they are only different if the force and displacement are in different directions.

@Abcd Well, what is the work done on the mass $M$?

T.x1

And what is the work done on the mass $2M$?

2T.x1/2

And the zero work principle just tells us that the two works have to be equal. Which they obviously are because $T x_1 = 2T x_1/2$.

I didn't know the zero work principle.

That's what we talked about yesterday. The string can't do any net work, so the work done at one end must be equal and opposite to the work done at the other end so the total work adds up to zero.

ohkay

@Abcd it is virtual work theorem .

$\sumT_i\cdotx_i=0$

@Abcd this question is from h c verma, I can bet it. Isn't it?

@Blue I has been there before for the interview.

It is small. IIRC 70 ha

I have a classmate there

Wait

I meant acres

even IITKGP has just 2000 acres

No

He joined last year

How did u get 7000?

Nope

Having dinner brb

@blue, hi"

@kenshin : The physics Q and A website seems to have gone AWOL again ... Sorry, it is back now.