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Why is the direction of pressure always perpendicular to surface area of a body for fluids ? We also assume that it is an ideal fluid here.

So , pressure acts in all directions because fluid has a tendency to flow.

Now , in book it says that for a block in water. Pressure must only act perpendicular to the surface area of block because otherwise , there is no friction between adjacent layers of the fluid. Also ,pressure must act in perpendicular direction on the sides on the tub if a block is put inside a tub which has water contained in it.

How does pressure being perpendicular does not cause that. How can we prove that pressure always act perpendicular in this way.

Also , in real life. This law must not be valid right since it only for making calculations easy? Is that true so.

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    $\begingroup$ If the pressure isn't perpendicular to the surface then it will be wind and you'll have waves. "Also ,pressure must act in perpendicular direction on the sides on the tub if a block is put into water inside a tub." I think you need to edit this sentence. It doesn't make any sense. $\endgroup$ – Transistor Jan 23 at 17:46
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    $\begingroup$ @Transistor When two opposite perpendicular pressures meet , they can also form waves right . About the sentence to be edited , I have edited it to make it more understandable. It is written in my book about that also. $\endgroup$ – srijan Sri Jan 23 at 18:03
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    $\begingroup$ @srijan Sri - The fluid does exert force in the direction parallel to the surface. We choose to call that comment of the force shear (in fluids, it is the effect of viscosity). The normal force is pressure. $\endgroup$ – Pete W Jan 23 at 23:52
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This is actually something not easily answered because it is part of the definition of pressure. I will instead point you to other answers which hopefully make sense.

The following is an excerpt from Lumen Physics.

*The force exerted on the end of the tank is perpendicular to its inside surface. This direction is because the force is exerted by a static or stationary fluid. We have already seen that fluids cannot withstand shearing (sideways) forces; they cannot exert shearing forces, either. Fluid pressure has no direction, being a scalar quantity. The forces due to pressure have well-defined directions: they are always exerted perpendicular to any surface. (See the tire in Figure , for example.)

enter image description here

Finally, note that pressure is exerted on all surfaces. Swimmers, as well as the tire, feel pressure on all sides. (See Figure 3.)

enter image description here*

Basically, a key part of the above is saying that pressure is always perpendicular it is because if there were a component of force parallel to the surface, then the object will also exert force on the fluid parallel to it as a consequence of Newton's third law.

Additionally, there is a similar question to physics stackexchange.

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  • $\begingroup$ What is the problem if the object also exerts a force on the fluid parallel to it? $\endgroup$ – srijan Sri Jan 23 at 19:52
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    $\begingroup$ when you submerge a solid object in a stationary fluid you don't expect it to move. If forces parallel to the surfaces you would have some translation or rotation. However, the solid body will move only up or down (due to pressure differential from top/bottom sides). $\endgroup$ – NMech Jan 23 at 19:54
  • $\begingroup$ Ok.@NMech right. But this condition is not there in real life scenario right that pressure is always perpendicular to surface. $\endgroup$ – srijan Sri Jan 23 at 19:56
  • $\begingroup$ could you please rephrase the last comment, because I couldn't follow your meaning? $\endgroup$ – NMech Jan 23 at 19:57
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    $\begingroup$ As I said in the post, this is a tricky matter to argue because, this is basically a definition. however, In this particular situation, if you are looking for a genaral definition you are looking for something that encompasses all situations. So you have a case which shows that pressure can only be perpendicular. If you are thinking about, how parallel forces are transmitted/generated then you need to consider the development shear forces. $\endgroup$ – NMech Jan 23 at 20:07

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