Acceleration of U-pipe containing water

  • #1
songoku
2,430
364
Homework Statement
A U-pipe having uniform cross sectional area is filled with water. The picture on the left is when the U-pipe is at rest and picture on the right is when the U-pipe is accelerated with acceleration ##a##. Find ##a## in terms of gravitational acceleration, ##g##

##a) \left(\frac{2h}{L}\right)g##

##b) \left(\frac{h}{L}\right)g##

##c) \left(\frac{h}{2L}\right)g##

##d) \left(\frac{h}{L}\right)^2 g##

##e) \left(\frac{2h}{L}\right)^2 g##
Relevant Equations
ΣF = m.a
P = F/A
P = ρ.g.h
1612597478940.png


I am thinking of using ΣF = m.a to find the acceleration but I don't know what causes the resultant force to the right. I thought it will be the difference in pressure on the left and right leg of the U-pipe (picture on the right), which is ΔP = ρ.g.h but the force (ΔP . A) won't be directed to the right (it will be downwards).

What approach should I use to solve this question? Thanks
 
Physics news on Phys.org
  • #2
songoku said:
Homework Statement:: A U-pipe having uniform cross sectional area is filled with water. The picture on the left is when the U-pipe is at rest and picture on the right is when the U-pipe is accelerated with acceleration ##a##. Find ##a## in terms of gravitational acceleration, ##g##

##a) \left(\frac{2h}{L}\right)g##

##b) \left(\frac{h}{L}\right)g##

##c) \left(\frac{h}{2L}\right)g##

##d) \left(\frac{h}{L}\right)^2 g##

##e) \left(\frac{2h}{L}\right)^2 g##
Relevant Equations:: ΣF = m.a
P = F/A
P = ρ.g.h

View attachment 277505

I am thinking of using ΣF = m.a to find the acceleration but I don't know what causes the resultant force to the right. I thought it will be the difference in pressure on the left and right leg of the U-pipe (picture on the right), which is ΔP = ρ.g.h but the force (ΔP . A) won't be directed to the right (it will be downwards).

What approach should I use to solve this question? Thanks
Pressure has no direction. It is only when you multiply it by an area vector (the vector being normal to the plane of the area) that you get a force vector.
So, what area do you think you need to consider?

I note that you are not given the width of the tube. You will need to assume it is small compared with L, despite the drawing.
 
  • Like
Likes songoku and Lnewqban
  • #3
haruspex said:
Pressure has no direction. It is only when you multiply it by an area vector (the vector being normal to the plane of the area) that you get a force vector.
So, what area do you think you need to consider?

I note that you are not given the width of the tube. You will need to assume it is small compared with L, despite the drawing.
Since the pressure is hydrostatic pressure, I think the area should be the cross sectional area of the U-pipe and area vector will be vertical so the resultant force produced will have vertical direction (downwards)

What will be the force to produce acceleration to the right?

Thanks
 
  • #4
The shape of the recipient is irrelevant regarding the angle that the surface of the fluid under acceleration adopts respect to the horizon.

Underneath the surface of any incompressible fluid, being it accelerated or not, each isobar layer must be parallel to that surface.
They give you L and h, you have a slope or an angle of inclination there.

Please, see:
 
  • Like
Likes songoku
  • #5
.
 
  • #6
Lnewqban said:
The shape of the recipient is irrelevant regarding the angle that the surface of the fluid under acceleration adopts respect to the horizon.

Underneath the surface of any incompressible fluid, being it accelerated or not, each isobar layer must be parallel to that surface.
They give you L and h, you have a slope or an angle of inclination there.

Please, see:

I don't really understand the slides you post, so many things I am not familiar with. I need to re-read it many times and find other sources that explain the basic so I can understand it fully, but what I tried after reading is:

$$\tan \theta = \frac a g$$
$$\frac h L = \frac a g$$
$$a=\left( \frac{h}{L}\right)g$$

Is this correct? Thanks
 
  • #7
songoku said:
What will be the force to produce acceleration to the right?
The U-pipe is being accelerated by an external force, which direction is similar to the resulting acceleration.
Then, the U-pipe tries to accelerate the mass of fluid in the same direction, which resists that impossed increasing velocity from the initial state of repose.

From the point of view of the moving U-tube, there is a mysterious force “pulling the fluid down” in a direction that must be perpendicular to the plane formed by both free surfaces of the fluid.

Please, see:
https://en.m.wikipedia.org/wiki/Fictitious_force#

:cool:
 
  • Like
Likes songoku
  • #8
songoku said:
I don't really understand the slides you post, so many things I am not familiar with...
Please, read example on page 17.
You are welcome. :smile:
 
  • Like
Likes songoku
  • #9
Lnewqban said:
Please, read example on page 17.
You are welcome. :smile:
Based on @haruspex suggestion to assume L >> width of tube, I assume the U-pipe will look like this when being accelerated:
1612624337914.png

##h## = height of water surface when the container is at rest
##\Delta z## = rise in water surface

$$\tan \theta = \frac a g$$
$$\frac{\Delta z}{\frac{1}{2}L}=\frac a g$$

Is this correct? If yes, how to relate ##\Delta z## to ##h##? Thanks
 
  • #10
songoku said:
What will be the force to produce acceleration to the right?
Think about the bottom (horizontal) section of water (length L on your diagram) in isolation: call it 'B' (for 'bottom').

The unbalanced forces acting on B are the horizontal normal reactions of the walls at each end of B, magnitudes ##N_{left}## and ##N_{right}##.

The resultant force on B is ##N_{left} - N_{right}## to the right . This is the 'F' in 'F=ma' when considering B in isolation.

If the cross-sectional area of the pipe is A, the pressures at the ends of B are ##P_{left}=\frac {N_{left}} {A} ## and ##P_{right}=\frac {N_{right}} {A}##.

Since there is no vertical acceleration, the pressure at the left end of B is ##P_{left}=\frac {\rho g h_{left}}{A} ## and at the right end ##P_{right}=\frac {\rho g h_{right}}{A} ##.

Once you understand that, the rest should be straightforward.
 
  • Like
Likes songoku and Delta2
  • #11
songoku said:
Based on @haruspex suggestion to assume L >> width of tube, I assume the U-pipe will look like this when being accelerated:
View attachment 277525
The error in this diagram is the magnitude of h, which should be the height difference between the highest and lowest points of the free surface.
 
  • Like
Likes songoku
  • #12
songoku said:
the area should be the cross sectional area of the U-pipe
Yes, but which part of it? Taking a section low on a vertical part doesn't get you far because the fluid can't go any lower. Try each end of the horizontal section.
 
Last edited:
  • Like
Likes songoku
  • #13
Lnewqban said:
The error in this diagram is the magnitude of h, which should be the height difference between the highest and lowest points of the free surface.
Ah ok, so the result will be: ##a=\left(\frac{\frac{1}{2}h}{\frac{1}{2}L} \right)g=\left(\frac{h}{L} \right)g##

Steve4Physics said:
Think about the bottom (horizontal) section of water (length L on your diagram) in isolation: call it 'B' (for 'bottom').

The unbalanced forces acting on B are the horizontal normal reactions of the walls at each end of B, magnitudes ##N_{left}## and ##N_{right}##.

The resultant force on B is ##N_{left} - N_{right}## to the right . This is the 'F' in 'F=ma' when considering B in isolation.

If the cross-sectional area of the pipe is A, the pressures at the ends of B are ##P_{left}=\frac {N_{left}} {A} ## and ##P_{right}=\frac {N_{right}} {A}##.

Since there is no vertical acceleration, the pressure at the left end of B is ##P_{left}=\frac {\rho g h_{left}}{A} ## and at the right end ##P_{right}=\frac {\rho g h_{right}}{A} ##.

Once you understand that, the rest should be straightforward.

haruspex said:
Yes, but which part of it? Taking a section low on a vertical part doesn't get you far because the fluid can't go any lower. Try each end of the horizontal section.

I think these two suggestions are the same. Considering each end of the horizontal section, the Newton's 2nd law equation will be:
$$N_{left} - N_{right}=m.a$$
$$\rho ghA=\rho ALa$$
$$a=\left(\frac{h}{L}\right)g$$

Thank you very much for all the help haruspex, Lnewqban and Steve4Physics
 
  • Like
Likes Lnewqban and Steve4Physics
  • #14
You got it right.
You are welcome, songoku. :smile:
 
  • Like
Likes songoku
Back
Top