Why the force exerted by a fluid on an object submerged in it is always perpendicular to it's surface? [duplicate]

When an object is immersed in a fluid, say or example water, then the force exerted by the fluid is always perpendicular to the surface. I'm unable to understand why is this so?

My books writes :- (NCERT Class 11 th Physics Part 2)

When an object is submerged in a fluid at rest, the fluid exerts a force on its surface. This force always normal to the object's surface. This 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. This will cause the fluid to flow parallel to the surface. Since the fluid is at rest therefore it cannot happen. Hence, the force exerted by the fluid has to be perpendicular to the surface.

I have some counter-arguments for this explanation.

As you can see from the picture that if the fluid exerts a force F1 then the block would exert a reactionary force -F1 and this should cause the fluid to move upward in a beaker but why don't we observe it in real life?

F3 on block would compel the block to exert -F3 on the fluid which should cause the fluid to flow in the beaker.

I tried to satisfy myself that reactions of F3 and F4 would cancel each other but we also know that pressure i.e. force (if area is constant) varies with depth and therefore I couldn't account for the absence of this upward motion. For sideway motion it is satisfactory that reactions of F3 and F4 would cancel each other out (although I'm not sure over here too).

So, I request to please solve my misconception. Books like these tend to be logical but always ends up being illogical.

At last I want to make another request: Please give a plausible explanation why the fluid force is always normal to the surface?

My question is addressing Newton's third law in explaining the "normal" nature of the fluid pressure.

marked as duplicate by Aaron Stevens, John Rennie newtonian-mechanics StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Sep 6 at 6:44

• @AaronStevens The duplicate pointed out by you matches my question just a little: when the OP of that question asks why a vertical plate experiences a normal force. – Knight Sep 5 at 13:53

This is a rather flawed explanation by the book. A fluid is never at rest. It can be in a state of equilibrium but not rest. A fluid is an ensemble of molecules continuously moving about at non-zero temperatures. In the absence of convection or any other mean flow, their motion will cause collisions with the object which on average will exert a force normal to the surface. An average; however, is merely that. There is a spread of off normal forces defined by the variance. There is parallel motion of the fluid; it just averages to zero.

As to why the mean force is normal, the simplest explanation is symmetry. From the normal to a surface, there is just as much chance of having a molecule impact at a certain angle as there is for the same angle spun around the normal 180 degrees. Thus on average the off axis components cancel.

• Can you please explain why the fluid doesn’t move due to reactionary forces? Thank you, your answer is very nice and making a complete sense. – Knight Sep 5 at 14:39
• Fluid dynamics is commonly studied through the notion of dividing the fluid into many small 'fluid elements' and developing partial differential equations for the motion of these fluid elements. Any one fluid element is large compared to a molecule but small compared to the fluid as a whole and to structures in the fluid. In equilibrium, all these fluid elements are at rest. – Andrew Steane Sep 5 at 17:33
• At rest is a really poor term to use. It might be fine in statics, but not fluidics. At rest implies no molecular motion which implies no pressure. – Paul Childs Sep 6 at 20:30
• I think Andrew Steane has explained well enough why the fluid doesn't move due to reactionary forces that there is no need to repeat. – Paul Childs Sep 6 at 20:33

You are right to bring in Newton's third law which certainly helps here. At each place on the boundary where the solid body and liquid meet, there is a force on the body and and equal and opposite force on the liquid. The layer of liquid experiencing this force would indeed begin to accelerate if this were the only force on it, but there is also a further force from the next layer of liquid, and so on through all the liquid.

The liquid keeps moving around and re-arranging itself until all these forces are balanced and everything is in equilibrium.

If the liquid is slippery (low viscosity) then in equilibrium each part of the liquid layer can only exert forces on any neighbouring part in the direction perpendicular to the boundary between them, because any other component of force will immediately cause the slippery fluid to accelerate. In short, such a liquid only supports forces that 'squeeze' it (pressure forces), not ones that 'slide' it. Because the molecules repel each other when they are close, the liquid can manage to provide pressure. But because the molecules readily slip past one another, the liquid cannot provide any force in opposition to a force in the direction tending to make one layer slide over another; in the presense of such a force the molecules accelerate until they come up against a pressure force. This rearranges the positions of the molecules, and consequently whatever forces they were experiencing are also rearranged. In particular, the non-squeezing part of the force must have fallen away to zero once the molecules reach equilibrium and stop moving.

• Your answer is really helpful to me. I request you please explain your last para with some visuals or more easy to understand. Thank you. – Knight Sep 5 at 13:58