# Fluid mechanics concept help please -- Pressure versus depth

• jonny997
So then the force from pressure on the bottom of the slab acts downwards, and its given by p_bottom*A.And the force due to the weight of the slab acts downwards, and its given by dw.So putting it all together, the sum of these forces must equal zero for the slab to be at rest within the fluid. Therefore, p_top*A - p_bottom*A - dw = 0. In summary, the drawing shown depicts a rectangular slab of fluid with height dy and top and bottom faces of equal area A. The top of the slab experiences a pressure of p+dp, while the bottom experiences a pressure of p. The force from pressure on the top of the slab acts upwards, while the force from

#### jonny997

Homework Statement
I've been trying to make understand the equation for the relationship between the pressure and elevation within a fluid for some time now, I was wondering if someone could help me out...

I don't understand why the force exerted on the top of the fluid element is (p+dp)A and the bottom is pA? The textbook doesn't really provide any explanation for why this is the case... Why is it not the other way around? Wouldn't there be an additional force at the bottom of the element due to there being more fluid above..?
Relevant Equations
dp/dy =-rho g v

In the drawing shown, ##\frac{dp}{dy}## is negative. So ##p+dp## is indeed less then ##p##.

• vanhees71
jbriggs444 said:
In the drawing shown, ##\frac{dp}{dy}## is negative. So ##p+dp## is indeed less then ##p##.
I'm still kinda confused... I'm not sure where the (p+dp)A came from... How do you work out that p+dp is less than p from the drawing?

jonny997 said:
I'm still kinda confused... I'm not sure where the (p+dp)A came from... How do you work out that p+dp is less than p from the drawing?
Well, let us look at the drawing.

We have this rectangular slab. The top and bottom faces are equal and have area ##A##. The slab has height ##dy##. Its weight is ##dw##.

We know that the pressure difference (##dp##) between top and bottom must yield a force large just large enough to support the weight of the slab.

Can you see how that assertion yields the first equation in the drawing.

• vanhees71
No, not exactly. :/

jonny997 said:
No, not exactly. :/
Let us break down that equation then:$$\sum F = pA - (p+dp)A - dw = 0$$Can you go through the three terms there (##pA##, ##-(p+dp)A## and ##-dw##), explain in your own words what you think they might denote and give a justification for why they must total to zero?

You might spend a sentence or two on sign conventions. Explain why each term has the sign it does.

• jonny997
jbriggs444 said:
Let us break down that equation then:$$\sum F = pA - (p+dp)A - dw = 0$$Can you go through the three terms there (##pA##, ##-(p+dp)A## and ##-dw##), explain in your own words what you think they might denote and give a justification for why they must total to zero?

You might spend a sentence or two on sign conventions. Explain why each term has the sign it does.

dw is the weight of the slab of fluid and the negative sign implies that it acts downwards.

(p+dp)A is the force on the slab due to the fluid above it, it also acts downwards, hence the negative sign.

pA is the force acting upwards on the slab and so its positive...

The net force acting on the slab must equal 0 because its at rest within the fluid, so (p+dp)A + dw = pA?

I just wouldve thought that (p+dp)A would be acting upwards, instead of downwards but then I tried that and i end up with dp/dy = rho g, without the negative sign...

jbriggs444 said:
Let us break down that equation then:$$\sum F = pA - (p+dp)A - dw = 0$$Can you go through the three terms there (##pA##, ##-(p+dp)A## and ##-dw##), explain in your own words what you think they might denote and give a justification for why they must total to zero?

You might spend a sentence or two on sign conventions. Explain why each term has the sign it does.
I think its mainly the (p+dp)A force I'm not understanding. Specifically the dp part, I'm not sure exactly what that is supposed to represent or why its acting downwards. I thought there would be more pressure at the bottom?

jonny997 said:
I think its mainly the (p+dp)A force I'm not understanding. Specifically the dp part, I'm not sure exactly what that is supposed to represent or why its acting downwards. I thought there would be more pressure at the bottom?
Yes, there is more pressure at the bottom. Pay attention to sign conventions!

##A## is the area of the top of the slab. Also the area of the bottom, but that need not concern us just now.

##p## is the pressure on the bottom of the slab. One might alternately label it ##p_\text{bottom}##.

##dp## is the difference in pressure between top and bottom of the slab. i.e. it is ##p_\text{top} - p_\text{bottom}##

##p+dp##, accordingly, is equal to the pressure on the top of the slab.

The force from pressure acting on the top of the slab acts in what direction?

The force on the top of the slab arising from the pressure on the top of the slab is accordingly given by...?

• jonny997
jbriggs444 said:
Yes, there is more pressure at the bottom. Pay attention to sign conventions!

##A## is the area of the top of the slab. Also the area of the bottom, but that need not concern us just now.

##p## is the pressure on the bottom of the slab. One might alternately label it ##p_\text{bottom}##.

##dp## is the difference in pressure between top and bottom of the slab. i.e. it is ##p_\text{top} - p_\text{bottom}##

##p+dp##, accordingly, is equal to the pressure on the top of the slab.

The force from pressure acting on the top of the slab acts in what direction?

The force on the top of the slab arising from the pressure on the top of the slab is accordingly given by...?
Ohhh okay, that makes everything a lot clearer... So dp is the difference in pressure between the top and bottom portions of the slap, therefore p_top = p_bottom + dp

So the force from the pressure at the top of slab acts downwards and is given by (p+dp)A

• jbriggs444
In the drawing, they could have put the pressure ##p## on top acting downward and the pressure ##p+dp## at the bottom acting upward. You would have ended up with:
$$\sum F = (p+dp)A - pA - dw = 0$$
Reducing to:
$$\frac{dp}{dy} = \rho g$$
Where ##dp## is now positive, which would feel more intuitive (although the value of ##p## is now different).

It just goes to say that no matter what assumption you start with (i.e. which force is larger and in what direction it acts), as long as you respect the sign convention, you will end up with the correct answer.

jack action said:
In the drawing, they could have put the pressure ##p## on top acting downward and the pressure ##p+dp## at the bottom acting upward. You would have ended up with:
$$\sum F = (p+dp)A - pA - dw = 0$$
Reducing to:
$$\frac{dp}{dy} = \rho g$$
Where ##dp## is now positive, which would feel more intuitive (although the value of ##p## is now different).

It just goes to say that no matter what assumption you start with (i.e. which force is larger and in what direction it acts), as long as you respect the sign convention, you will end up with the correct answer.
Did you switch from y increasing upward to y increasing downward there?

jack action said:
In the drawing, they could have put the pressure ##p## on top acting downward and the pressure ##p+dp## at the bottom acting upward. You would have ended up with:
$$\sum F = (p+dp)A - pA - dw = 0$$
Reducing to:
$$\frac{dp}{dy} = \rho g$$
Where ##dp## is now positive, which would feel more intuitive (although the value of ##p## is now different).

It just goes to say that no matter what assumption you start with (i.e. which force is larger and in what direction it acts), as long as you respect the sign convention, you will end up with the correct answer.
So in that case we would be taking the surface of the fluid to be the origin and measuring downwards?

• jbriggs444
jbriggs444 said:
Did you switch from y increasing upward to y increasing downward there?
jonny997 said:
So in that case we would be taking the surface of the fluid to be the origin and measuring downwards?
Either that or ##dy## becomes negative.

Although not clearly stated in the problem, I realize now that ##p## is assumed to be the pressure at the bottom. I'm used to the more traditional problem where the pressure at the surface is known (atmospheric pressure).

• jonny997
Thanks, I appreciate all the help

• Lnewqban