John 8:32

> And you shall know the truth, and the truth shall make you free.

I've been invited?

@KnightadmiresChappo what is your relation to the user Knight? Is it the Knight who has appeared in my chat room connected to math.SE?

@CalvinKhor There I don’t admire Chappo publicly.

What is Chappo?

Chappo’s account was destroyed by some people because he supported Monica.

It’s peaceful here.

fair enough

I did not read it

How about this one:

> When you pray, don’t be like the hypocrites who love to pray publicly on street corners and in the synagogues where everyone can see them. I tell you the truth, that is all the reward they will ever get.

@CalvinKhor You busy right now?

yes, and not usually interested in bible phrases

@CalvinKhor I want to do some physics.

Problem: A box of 10 kg is suspended by the rope of length $2$m from the ceiling. A force $F$ is applied horizontally at the mid point of the rope such that the top half of the rope makes an angle of $45^{\circ}$ with the vertical. Then $F$ equals ? (Take $g= 10 m/s^2$ and rope to be massless)

Did someone call?

Yes

@JohnRennie It was me

@KnightadmiresChappo hi :-)

Good morning sir

Another new room?

I had no other option sir.

Let’s do some Physics?

I cannot see how the length of the string/rope is relevant here.

@JohnRennie

That diagram is wrong ...

hmmm...

where's the mistake?

Sir, I thought the Lower part of the rope will also get stretched

That's not how I interpret the question.

Okay.

How to solve it further?

Consider the tension in the lower part of the string.

The tension has to be equal to mg because the upwards force due to the tension balances the downwards force due to the weight. Yes?

YES

yes

I've marked the tension in red. The only two forces acting on the mass are the tension and the weight, so this should make it clear that mg + T = 0

Yes and for a massless rope, the tension is the same all along the rope

Yes.

And we have to calculate F.

The way we do this is to take moments about the point where the string is attached to the ceiling.

Since the system is stationary the moment due to the weight must be equal and opposite to the moment due to F. Yes?

Moment due to weight is $100/\sqrt{2}$

$\cos 45 = x/1 \implies x = \frac{1}{\sqrt 2}$

Yes, the moment due to the weight is mgx i.e. $mg/\sqrt{2}$

$\sin 45 = y / 1 \implies y = 1/\sqrt 2$

The moment due to F is Fy and that's $F/ \sqrt{2}$

Sir, you're missing the ${\Large / }$

And the moments have to be equal and opposite otherwise there would be a net moment at the midpoint of the string and it would rotate.

Oops :-)

:-)

So, I should note that whenever there are two forces : one horiz. and the other vert. we have to work with the moments not with the translational equilibrium.

You can use translational equilibrium, but the problem is we don't know the tension in the top part of the string.

Thank you so much sir.

There is a tension T' in the top part of the string that isn't equal to T

Isn't tension same all along the massless rope?

If we use the condition that the net vertical force must be zero we find $T' = mg\sqrt{2}$. Yes?

@KnightadmiresChappo no, because the point where F pulls on the rope isn't a pulley.

Okay

The rope isn't free to move there because the force F is acing at that point.

I got it, I got it.

Yes

And then we can use the condition that the net horizontal force must be zero.

Yes

And that will give $T' \cos45 = F$

And F= mg

So $mg\sqrt{2} / \sqrt{2} = F$

Thank you sir :)

:-)

I need to drop out for an hour to do some stuff ...

:)

@JohnRennie Sir, I always make the mistake between $$E = c ~B \\ B = c ~E$$

How to avoid it? Do you know some trick sir?