Two blocks of equal mass(m) are tied to each other through a light string. One of the blocks is pulled along the line joining them with constant force F. Find the tension between the strings joining the blocks.

Explain me the process

yes but pls draw the FBDs for me

I guess I got them wrong

You know that the two blocks are being pulled with a force F, and their combined mass is 2m

Yes

The blocks are tied together, so they must have the same acceleration.

OK so far?

yes

So the acceleration of the two blocks is:

a = F/(2m)

tension in string?

OK, the trailing block is being pulled by the string, and the force on that block is the tension in the string. Yes?

yup

The acceleration of that block is F/(2m), as we've just worked out above, and the mass of the block is m, so what is the force on the block?

F/2

Correct! :-)

So that's the tension in the string

now if I have to draw an FBD why is it that a foce ma will act on the trailing block in opp direction

opp=opposite

OK, consider the leading block. It has two forces on it. We have the force F pulling it one way, and the tension in the string pulling it the other way. Yes?

yes

now what next

Why unifrom division?

The force F is pulling the leading block, and the leading block is pulling the trailing block (through the string)

uniform*

ok

Now draw FBD thats it

Pls @JohnRennie

I am not getting FBD correct

@gateprep the division of the force depends on the masses of the blocks. In this case the masses are equal so we end up with the force being divided equally.

Why is ma in opposite direction in trailing block

If the masses of the blocks were different the force wouldn't be divided equally.

@gateprep We know both blocks are accelerating to the right with an acceleration F/(2m). Yes?

yes

And both blocks have a mass m, so the net force on each block must be F/2.

Where the net force is the force to the right minus the force to the left.

OK so far?

there @JohnRennie ?

Morning :-)

The way I'd do that is consider it as a sum of capacitors with width $dx$ and integrate to get the total capacitance.

@JohnRennie What to do about the slanted part

If you consider the part of the capacitor between $x$ and $x+dx$ then we can assume the sizes of the two dielectric parts are constant over the infinitesimal region $dx$ so we can easily calculate the capacitance of this element. Yes?

yes

So we can write an expression for the capacitance of the element at distance $x$, $C(x)$

@JohnRennie do we have to treat the slanted part as straight for dx element.

@Abcd yes

@JohnRennie How can we do that? x_x

@Abcd suppose I asked you to integrate the line $y = kx$

@JohnRennie yes

You would do it by considering rectnagular elements of height $y$ and width $dx$, so the area is $ydx = kxdx$

i.e. even though the line is slanted you'd consider it constant within the infinitesimal element of width $dx$.

This is just basic integration.

Oh yes, thats the first example of integration.

Exactly.

Okay thanks.

So within our element here of width $dx$ we assume the dividing line between the two dielectrics to be constant.

@JohnRennie constant = horizontal?

Calculate the capacitance of the element of width $dx$ at a position $x$ then just integrate from $x=0$ to $x=a$

@Abcd you seem to be making this more complicated than it is.

@JohnRennie Are we treating the blue line as horizontal or not?

At a distance $x$ the thickness of the lower dielectric is $dx/a$, and the thinkness of the upper dielectric is $d - dx/a$

@Abcd what do you think?

@JohnRennie yes probably

Yes, we are treating the blue line as horizontal. Just as we would when integrating $y=kx$

We take tension in a string as always a pulling force but in case of rod it can be pulling or pushing? Any logic?

@Jasmine well, a string can't push ...

@JohnRennie Oh yeah we can only pull something towards us by a string and not push but by a rod which is very solid and not flexible we can both pull or push. Am I right?

So while solving problems in rod in which direction do we take the force of rod?

@Jasmine Yes

@Jasmine when you are doing any problem with forces you need to decide what direction is positive and what directing is negative. Forces are vectors remember, so forcs pointing in opposite directions have opposite signs.

It doesn't really matter what direction you define as positive and what direction you define as negative as long as you are consistent.

The forces in a rod are no different to any other forces.

@JohnRennie consider a horizontal circular motion of a string tied to a stone we replace it by a rod so will the rod push or pull?

Well, what direction does the force on the stone point?

@JohnRennie are you talking of centripetal?

@JohnRennie inwards

@Jasmine yes, so the rod is behaving just like a string would

@JohnRennie any example when the rod would push?

@Jasmine Suppose I tie one end of the rod to my car, and I fix the stone to the other end of the rod. Then I jump in the car and floor the accelerator.

@JohnRennie with respect to what do you integrate ? Is it $V$ ?

@NehalSamee we're differentiating not integrating. And we are differentiating wrt $V$ i.e. we are calculating $dU/dV$.

@JohnRennie the rod would push the stone

@Jasmine yes, so there's an example where the rod is pushing rather than pulling. If I replaced the rod by a string then the stone wouldn't move (until my car crashed into it :-)

@JohnRennie oops ... Man ... 'twas a typing mistake

@JohnRennie yeah got it through the example.

@NehalSamee what we're doing is finding the value of $V$ at which the total energy is a minimum i.e. the value of $V$ that we will get in the real system, because the system will always have the lowest energy state.

@JohnRennie yes. .. got that yesterday ... Just curious regarding the method of differentiation ...

you ppl not having problems with limits continuity differentiability

john rennie did you see the iit paper?

Those who have given 12th or are in 12th- how did/do you all manage English? It's so boring plus I dint understand anything.