Problem Solving Strategies

sanya

06:00

Hi sir @JohnRennie

John Rennie

Hi :-)

@sanya Hello?

sanya

hi

Theres this question.... Find the relation by coordinates(x,y) in a container containing fluid at which pressure is P/2. The density of the fluid is ρ and the container is accelerating rightwards. The pressure at origin is P.

I get a straight line equation y=P/2ρg - xa/g

John Rennie

OK ... ?

sanya

Theres a conclusion made from this relation that pressure at each point on this line is P/2, can you tell me how we can see that?

At (x,y) it is P/2 i get that, but on the other points....

John Rennie

06:23

Hi, sorry, I had to drop out for a few mins.

sanya

Its okay

John Rennie

In the accelerating container the surface of the fluid is at an angle. Yes?

sanya

@JohnRennie Can you explain?

John Rennie

Imagine you have a glass of water sitting still on the table. The surface of the water is level.

sanya

Yes

John Rennie

06:27

Now start pushing the glass away from you so it accelerates. The water moves in the glass so its higher on the side nearest you and lower on the side further from you. That is, the surface of the water is now at an angle.

Yes?

sanya

Okay

John Rennie

Do you want to go through the derivation of why this is?

sanya

Sure

John Rennie

Give me a few moments to draw a diagram

This shows a tank of water being accelerated to the right with acceleration 𝑎, and there is also a vertical acceleration 𝑔 due to gravity.

OK so far?

sanya

Yes

But the point x,y is outside the fluid?

John Rennie

06:35

The points A and B are at the same pressure because they are both in contact with the air. So both will be at 1atm.

sanya

@JohnRennie Yes

John Rennie

@sanya Outside the fluid the pressure is just 1 atm everywhere, so I think the question only makes sense if the point x,y is in the fluid.

sanya

@JohnRennie Yes

John Rennie

Suppose we go from A to C then to B. Since A and B are at the same pressure then the pressure change AC must be the same as the pressure change CB.
OK so far?

sanya

A to C pressure increases and the same decreases from C to B?

John Rennie

06:40

Yes

sanya

Ok

John Rennie

The pressure change from A to C is just ρgy. Yes?

sanya

Yes

John Rennie

I'll just state the pressure change as we move horizontally. We can go through the proof if you want.

The acceleration 𝑎 works like a horizontal version of gravity, and the change in pressure as we move a distance 𝑥 horizontally is just ρax.

That is, it's just like moving vertically but we replace 𝑔 by 𝑎.

So for the pressure changes to be equal we need:
ρgy = -ρax

OK so far?

sanya

Okay

John Rennie

06:44

So y/x = a/g

sanya

yes

John Rennie

But y/x is just the gradient of the surface, so what we've found is that the gradient of the surface depends on the acceleration.

If a = 0 the gradient is zero i.e. the surface is flat, which is of course correct.

As 𝑎 increases the gradient of the surface gets steeper and steeper.

Does this make sense so far?

sanya

Yes

John Rennie

This is why the surface of a liquid becomes sloped if we accelerate the container that the liquid is in, and that is what is being done in this question.

35 mins ago, by sanya

I get a straight line equation y=P/2ρg - xa/g

Notice here that the gradient of the line is -a/g

Just as we found above.

Yes?

sanya

Oh. Yes

John Rennie

06:49

38 mins ago, by sanya

Theres this question.... Find the relation by coordinates(x,y) in a container containing fluid at which pressure is P/2. The density of the fluid is ρ and the container is accelerating rightwards. The pressure at origin is P.

I'm not sure how the coordinates work here i.e. exactly where they are taking the origin.

Is there a diagram to go with the question?

sanya

Im referencing your diagram the origin is at the bottom-most leftmost point

John Rennie

Suppose the depth of the fluid on the left side is ℎ

We are told that the pressure at the bottom left is P so we know:
P = ρgh
Yes?

sanya

yes

John Rennie

So h = P/(ρg)

So the equation for the surface is:
y = P/(ρg) - (a/g)x

i.e. the gradient is -a/g as we found earlier and the y intercept is P/(ρg)

Yes?

sanya

Yes

John Rennie

06:58

We found the equation for the surface, but the question asks for the equation of the line where the prressure is P/2.

And that's the red line I've drawn on the diagram. Yes?

sanya

Yes

John Rennie

i.e. on the left it start halfway down where the pressure is P/2.

sanya

Yes

John Rennie

So the red line has the same gradient at the surface. The only difference is that the y intercept is now h/2 instead of h i.e. ¹⁄₂P/(ρh)

OK so far?

sanya

Yes

John Rennie

07:01

So the equation of the red line is:
y = P/(2ρg) - (a/g)x

sanya

yes

John Rennie

And that's the answer to the question :-)

Can you see how this works?

Aayush Sethia

Had a doubt

Can you help

sanya

Yes but
Theres a conclusion made from this relation that pressure at each point on this line is P/2 , how is this true?

@JohnRennie Yes very well:)

John Rennie

@sanya Every point on the red line is at a height h/2 below the surface. Yes?

@AayushSethia Hi :-) Yes, but let me finish Sanya's question.

sanya

07:05

@JohnRennie Oh yes its tilted

Aayush Sethia

John please help in irodov question 1.43

sanya

Yes thankyou its very clear @JohnRennie

John Rennie

@sanya OK :-)

Aayush Sethia

mech section

John Rennie

@sanya You'll see many questions like this and they all work in basically the same way.

The horizontal acceleration changes the pressure in the same way that the gravitational acceleration does.

sanya

07:07

I never knew how to see it....this will help a lot:)

@JohnRennie Yes

Aayush Sethia

In the question I am not able to understand why in the acceleration part we cannot do (2wr)^2 /2r

John Rennie

Aayush Sethia

Exactly

See I have taken velocity with respect to point O and it coming out to be 2wr

Please help only in the acceleration part

John Rennie

Aayush Sethia

I successfully found velocity

John Rennie

07:13

What did you get as the velocity?

@AayushSethia Hello ... ?

John Rennie

07:32

@AayushSethia I guess you've had to drop out so I'll summarise how I would do it.

If we take the centre of the circle as the origin then the position of the point A is:
x = R cosφ, y = R sinφ

And φ = 2θ

And θ = ωt, if we take t = 0 when A is at the same level as O.

So we get:
x = R cos(2ωt), y = R sin(2ωt)

Then to get the velocity we just differentiate with respect to time:
v_x = -2ωR sin(2ωt), v_y = 2ωR cos(2ωt)

And to get the acceleration differentiate again:
a_x = -4ω²R cos(2ωt), a_y =-4ω²R sin(2ωt)

sudarsan

14:10

hi @JohnRennie sir

sir can u teach me the concepts local maxima and local minima

please :-)

Transcript for

Problem Solving Strategies