Fluid mechanics problem, need explanation please

[Bassel]

Hi please check the highlighted question in the attachment and the solution and see the problem, i need help please. I have two questions about it:

1- why didn't we take the plug as a vertical wall and thus the force would be integral of pressure x area?

2-WHERE DID THE FORCE DUE TO ATMOSPHERIC PRESSURE GO?

Thnx for your help.

 

[tiny-tim]

Hi Bassel! Welcome to PF!

1- why didn't we take the plug as a vertical wall and thus the force would be integral of pressure x area?

You did.

You did ∫ ρgA(H+h) dh, but the plug is so small that you can take H+h to be a constant, H …

so that integral is just ∫ ρgAH dh

2-WHERE DID THE FORCE DUE TO ATMOSPHERIC PRESSURE GO?

it nipped round the other side when you weren't looking, and started pushing back

(so it's the same on both sides, and you can ignore it)

 

[Bassel]

by the other side you mean from outside ??

 

[tiny-tim]

yes, the plug has two sides

 

[PJC01]

I am happy to provide an explanation for the fluid mechanics problem you have presented. First, let's address your first question: why didn't we take the plug as a vertical wall and thus the force would be integral of pressure x area?

In fluid mechanics, we often assume that the forces acting on a surface are normal (perpendicular) to that surface. This means that if we consider the plug as a vertical wall, the force acting on it would be normal to the wall, not parallel to it. This is because the pressure in a fluid acts in all directions, not just horizontally. Therefore, we cannot simply use the integral of pressure x area to calculate the force on the plug.

Now, let's address your second question: where did the force due to atmospheric pressure go?

In the solution provided, the force due to atmospheric pressure was taken into account by including it in the calculation of the net force on the plug. The force due to atmospheric pressure is acting on the top surface of the plug, pushing it down. This is balanced by the force due to the fluid pressure acting on the bottom surface of the plug, pushing it up. Therefore, the force due to atmospheric pressure is already included in the calculation and does not need to be considered separately.

I hope this helps to clarify the problem and solution for you. If you have any further questions or need more explanation, please don't hesitate to ask. Science is all about asking questions and seeking understanding, so keep asking and learning!