Why do we feel pressure from the downside in a liquid

[Frigus]

We know that pressure at a point in liquid is due weight of water couloum above that point but why does we experience force from downside face of object

 

[jbriggs444]

We know that pressure at a point in liquid is due weight of water couloum above that point but why does we experience force from downside face of object

If you make a hole in water, the water will move to fill the hole. If you want a hole to persist, you have to apply a force.

 

[Frigus]

If you make a hole in water, the water will move to fill the hole. If you want a hole to persist, you have to apply a force.

Sir I want to say that when if if I put an object in liquid,that object will also feels an pressure from downside and I want to know why object feels that pressure from downside

 

[jbriggs444]

Sir I want to say that when if if I put an object in liquid,that object will also feels an pressure from downside and I want to know why object feels that pressure from downside

Ever poked a hole in the bottom of a boat?

 

[Frigus]

Ever poked a hole in the bottom of a boat?

Thanks sir,I understand what you want to say, thanks a lot

 

[jbriggs444]

You can trace the behavior back to Pascal's law.

Suppose that you have a body of fluid such as water and suppose that the fluid is at rest in a stable condition. Draw an imaginary lattice dividing the water up into a bunch of tiny cubes. Focus your attention on one such cube of water.

The pressure on the front and back of the cube must match -- otherwise the water in that cube would start flowing. That would violate the assumption that the water is stable. Similarly, the pressure on the left and right of the cube must match.

The pressure on the top and bottom must also match, nearly. The discrepancy is the weight of the fluid in the cube. The pressure on the bottom must be just enough higher than the pressure on the top that it balances the weight of the water. That's where $P=\rho g h$ comes from.

Now trace a path through a chain of cubes, the pressure at the far end of the chain will match the pressure at the near end plus a correction for $P=\rho g h$ where h is the height differential between the two ends. As was discussed above, the pressure at the top and bottom of the final cube will be nearly identical.

The pressure force from the top side of that final cube points upward on whatever object sits above it.