# How to calculate the angle of water in a moving beaker

How to calculate the angle of water in a moving beaker?

Say the beaker is moving to the right direction. The water inside the beaker will be tilted to the left. An angle will be formed and how am i supposed to calculate the angle?

Assumed parameters:

F = force applied to move the beaker
a = accelerations of the beaker
m = mass of the beaker

Thanks!!

## Answers and Replies

The water surface is determined by the forces that are acting on it. Also, you can assume that the fluid is static (it does not move, and this is valid when the water surface reaches the final position). For static fluids you have only one equation:
grad P=acceleration =>
dP/dz=-g*rho
dP/dx=a*rho=F/m, where rho is the density

Since P is a function of x,z you can write the following expression
dP=(dp/dx)*dx + (dP/dz)*dz
integrate this equation to find the slope

Last edited:
hi masterx,

thanks for your reply. however i still don't quite understand how to get the slope. In your reply, you assumed that grad of P = acceleration. What is P? also, a*rho=F/m. a is acceleration? i though a = f/m? for the 2 eqns i obtained, must i set boundary conditions to find the 2 constants? Thanks agn.

Regards
Siu Kwok

P is the pressure
I made a typo, a*rho=F*rho/m

You could set b.c., but it would not change the slope.

Remember, that the pressure at the water surface is constant, and equal to 1atm.

tiny-tim
Science Advisor
Homework Helper
welcome to pf!

hi siukwok! welcome to pf! (please don't post the same problem more than once )

here's two more ways to do it …

imagine that the water is ice … at that acceleration, at what angle would the ice have to be for nothing to slide up or down on it?

use the beaker as a frame of reference: that means you must introduce a fictional horizontal "gravitational" force, and you can then apply the usual inertial laws Hi all,

taking over to ask some questions regarding this question.

MasterX mentioned that:
dP/dz=-g*rho, if i am not wrong, it was derived from bernoulli's equation? By taking beakers frame of reference, the velocity term drops off and P = -rho*gh+Const.

Differentiating, we get dP/dz=-g*rho.

Question here is:
1) What is the physical meaning of dP/dz=-g*rho?
2) How do you know that you should solve the equation in this manner? Taking differentiation of P.
Edit: added should.

Any help will be great. Thanks.

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Hi Delzac,

I did not understand your second question.

First of all, the equation dP/dz=-g*rho is not derived from Bernoulli's equation. Bernoullis' equation states that P/rho+g*z+v*2=const.

You may consider the equation dP/dz=-g*rho as a constitutive equation. It simply says that as we move downwards the pressure increases. In the problem with the beaker, the pressure at the surface is equal to the atmospheric pressure. As we move towards the bottom of the beaker the pressure increases. If a person was sitting at the surface of the water, he would only feel the atmospheric pressure. If, instead, he is at the bottom of the beaker, he would feel the atmospheric pressure and the weight of the fluid. The question here is, how do you calculate the "weight of the fluid"? Well, you do not!! You, instead, calculate the pressure.

By the way, if you use Bernoullis' equation for the beaker, you would find that P_bottom=P_surface+g*rho*h. If you solve the dP/dz=-g*rho you will get exactly the same answer

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I missed out the word 'should', added it back into the post.

So the second question is:
2)How do you know that you should solve the equation in this manner? Taking differentiation of P.

But we know you didn't. Nonetheless, how did you arrive at dP/dz=-g*rho, or know that the pressure gradient is g mutiply by density?

For me, i was thinking on the line of balancing the forces, much like those static equilibrium question. And for this question, pressure on one side is higher than the other, so the water should move to equalize it. But it isn't since there is a fictitious force acting on it. And can bernoulli's equation be applied here?

As I said, this equation is not derived from Bernoullis' equation. So you do not need to differentiate with respect to P.

Actually, you are right, You need to take a force balance. Are you familiar with the Navier-Stokes equation? This equation is an extension to Bernoullis' equation, when viscosity is important. In the case of static fluid, the Navier-Stokes equation simplifies to dP/dz+rho*g=0 ( in the z-direction).

The water does not move!!! It is static. If the water was moving, I could not use this eqn and I had to either use the Navier-Stokes equation or the Bernoullis' equation.

Nope, never heard of Navier-Stokes Equation, and after checking out the wikipedia page, i am quite sure its beyond me.

I am quite sure that i can derive it from bernoulli's equation though:

$$\frac{1}{2}\rho v^2 + \rho gh + P = Const$$

Since fluid is static, v=0.

$$P = -\rho gh + Const$$
$$\frac{dP}{dz} = -\rho g$$

In this case, how should be go about balancing the force? Its a bit strange. And is the force exerted on the fluid a constant regardless of where the fluid is? If yes/no, how do you show that? As of yet, i can think of any equation that can do this.

And the main question is how we should approach this problem, which i am not very clear with.

Hi,

i don't think the solution below is correct. But i managed to get an answer which is within my desired range. Please take a look.

Assumptions: Fluid is static

Gradient P = acceleration

dP/dz= -gρ

P = -gρz+c

dP/dx= a_x ρ=Fρ/m

P = a_x ρx+c_1

Making use of the air flow gun which gives impulse = 2.86 (given),

Thus, let F = 2.86, m=1kg, ρ of water = 1000 kg/m3, g = 9.81m/s2

dP/dz= -9810
P = -9810z + c ---------------> (1)

dP/dx = 2860
P = 2860x + c1 ---------------> (2)

To simplify the problem, assume c & c1 = 0, <--------i know that this looks wrong but i couldn't find a way to find the constant.

Let (1)÷(2) :
-9810z/2860x= 1

Gradient of water = z/x = -2860/9810 = -0.29
tanθ = -0.29
θ = 16.2ᵒ

You answer is correct. I tried to derive everything from first principles and arrived at

$$\theta=\arctan\frac{a}{g}$$

However, I arrived at this answer with a help from a different problem. What I still don't quite get is why you can just add up these two

dP/dz=-g*rho
dP/dx=a*rho=F*rho/m

to get

dP=(dp/dx)*dx + (dP/dz)*dz

what happened to the negative sign in dP/dz?

You are a little bit confused, and I do not know how to help you.

Lets start with the eqn P=-rho * g * h + const

What is the meaning of this eqn? It simple related the pressure to the depth. Nothing more than that!!

Similarly, dP/dz=-rho*g relates the pressure slope to gravity force.

As I wrote before, this is a constitutive eqn. As such, it gives you the pressure at a certain depth (or height).

When I wrote before that you need to take a force balance, I said that only if you want to prove this eqn.

Here is an example:

Assume that you have a body (cube of area A, height h and density rho_b) submerged in water (density rho) at depth H. The body does not move.
If you do a force balance on the body, you have that gravity is equal to the force exerted by the fluid on the body. Of course, this force is the pressure plus the buoyancy (B).

So, -rho_b*A*h*g+(P_bottom-P_top)*A + B=0 => B=-rho_b*A*h*g-(P_bottom-P_top)*A

and based on the previous eqn, P_top=P_surface-rho*g*H
and P_bottom=P_surface-rho*g*(H+h)

So, B=-rho_b*A*h*g+rho*A*h*g=A*h*g*(rho-rho_b)

You answer is correct. I tried to derive everything from first principles and arrived at

$$\theta=\arctan\frac{a}{g}$$

However, I arrived at this answer with a help from a different problem. What I still don't quite get is why you can just add up these two

dP/dz=-g*rho
dP/dx=a*rho=F*rho/m

to get

dP=(dp/dx)*dx + (dP/dz)*dz

what happened to the negative sign in dP/dz?

Because P is a function of x, z. Therefore, you can write that
dP=(dp/dx)*dx+(dP/dz)*dz

Why does the negative sign bother you?
The above eqn simplifies as follows:

1=(a*rho)dx+(-g*rho)*dz
Therefore, the slope is dz/dx=a/g=tan(theta)

huh? where did the 1 come from?

What is the pressure at the surface equal to? It is usually equal to 1 atm.

Hmmm, i think i roughly get it already. Thanks for your help!

Ah... okay, I get it now. Thanks!

Just another question. Why is it that you can imagine the water as ice when you want to calculate the angle of the water?

tiny-tim
Science Advisor
Homework Helper
Just another question. Why is it that you can imagine the water as ice when you want to calculate the angle of the water?

Hi Delzac! Because the important question is whether an extra drop of water would move up or down the surface, or stay where you put it …

for that purpose, it doesn't matter whether the surface is liquid or solid, so if we pretend it's solid, we can forget about Bernoulli etc, and just concentrate on the acceleration and F = ma Ah, that makes sense! Thanks!

Out of curiosity, how would you solve the problem with the ice. The ice is a solid undeformable material. As the beaker moves, the ice would stay still. So, how would you solve for the angle?

Are you asking me? I hope not. But here's what i think.

I think tiny-tm was saying that we can imagine the water to be ice, and calculate at what angle, if a drop of water is placed on ice, would the drop of water stay still for a given acceleration.

This, i think simplify things a lot. I can even compare it with a wedge and block with frictionless surface kind of problem.

tiny-tim
Science Advisor
Homework Helper
I think tiny-tm was saying that we can imagine the water to be ice, and calculate at what angle, if a drop of water is placed on ice, would the drop of water stay still for a given acceleration.

This, i think simplify things a lot. I can even compare it with a wedge and block with frictionless surface kind of problem.

(just got up :zzz: …)

Yes, that's exactly right! Oh, I see. Interesting point of view!!
Thanks