# Why do hydrostatic fluid exert force on any wall normally with magnitude $ρgh$ always?

I understand the derivation in middle of container, but for a container like an inverted cone the particle very near to the wall exerts pressure on wall of magnitude $$\rho gh$$ normally. I ask “why?” and tried to discover it, but I had some problems.

Just like in normal derivation I assumed that since very small volume of liquid is at rest, it experiences equal and opposite force from both the liquid side and wall side, but from the liquid side in the normal direction to the wall pressure is found to be $$2\rho gh$$ (integrating all the normal components of pressure), then the same amount of pressure (by normal force) is produced by wall, so that liquid is stationary and due to this a reaction force is developed in wall with $$2\rho gh$$ magnitude normally...

Where did I go wrong?

• I don't like the sound of "integrating all the normal components of pressure". What exactly did you do? [If you want a derivation of $p=h \rho g$ I'll supply my favourite one, but I imagine you are seeking your own derivation.] – Philip Wood Jul 19 '17 at 17:08
• @PhilipWood i stated that i know the derivation of ρgh when a small element is considered surrounded by liquid from every side, if u have derivation for why pressure on the wall is ρgh go ahead and tell me your derivation. – user45838 Jul 19 '17 at 18:09
• @PhilipWood As for integration, assume small element of liquid touching the wall on one side liquid on other side, liquid will exert a force on this (ρgh·Area of small element) and this force will be from many directions on this small element since wherever it is surrounded by liquid it will experience a force due to it, if the direction is diagonally down of one of these forces then there will also exist another force vector which will have it as diagonally up, cancelling the net vertical effect but the horizontal component integrates up and points toward the wall normally with magnitude 2ρgh – user45838 Jul 19 '17 at 18:16
• It sounds to me like you are working from the wrong premise. If you apply your method on the surface at the bottom of a container you should get $\rho g h$. If you don't, you have a scaling problem (probably because you are "double counting" contributions). Pressure is defined as the sum of all the forces - regardless of how you account for them, the integral by definition must be the same. And once you have that insight, it will not change when you change the orientation of the surface. – Floris Jul 19 '17 at 18:30
• @user45838 Imagine a piece of liquid in the form of half a small cube, the cube having been bisected by a plane that includes two parallel cube-edges. Orientate the half cube with the new plane and two other adjacent half-side planes vertical. You'll then find that the two full side planes are at 45° angles. Calculate the forces on all five sides of the half cube, using the $h \rho g$ formula for pressure and taking account of the areas and angles of the sides. The forces should add vectorially to zero. If you do this, it'll be quite instructive. – Philip Wood Jul 19 '17 at 22:07

Imagine a piston of small area, A inside an empty cylinder, close to its bottom end, which is open. The cylinder is immersed in liquid, at any angle, so that the piston is at depth h and orientated at any angle. When the piston is pushed down the cylinder by a small distance $\Delta x$, let's say a force $F_\bot$ normal to the piston is needed. Thus the work done by the piston on the liquid is $F_\bot \ \Delta x$. But the displaced liquid (volume $A \Delta x$) makes the liquid level rise; effectively the displaced liquid rises by height $h$ and gains gravitational potential energy $mgh = \rho A \ \Delta x \ gh.$ If the fluid is perfectly non-viscous, or the process is very slow (quasi-static), we can equate the work done to this gain in PE. Thus: $$F_\bot \ \Delta x=\rho A \ \Delta x \ gh.$$ Dividing both sides of the equation by $A \Delta x$ and writing $p=\frac{F_\bot}{A}$ we get $$p=h \rho g.$$
You do not need to do integration here. Let's draw a free body diagram, and lets assume that the small volume is so small that we don't need to worry about pressure changes from top to bottom but the pressure is $$\rho g h$$ everywhere.
Now, let's calculate $$\vec f_1$$, $$\vec f_2$$, and $$\vec f_3$$ using $$\vec f=P \vec A$$ directly since those are in the fluid and we know $$P=\rho g h$$. Then by Newton's 2nd law $$\vec f_0=-(\vec f_1+\vec f_2+\vec f_3)$$. So we have $$\vec f_1=\rho g h (0,d^2)$$ $$\vec f_2=\rho g h (d^2, 0)$$ $$\vec f_3=\rho g h \left(0,- d \left[d-d \tan(\theta) \right] \right)$$ and therefore $$\vec f_0=-(\vec f_1+\vec f_2+\vec f_3)=\left(-\rho g h d^2, -\rho g h d^2 \tan(\theta)\right)$$
So $$||\vec f_0||=f_0=\rho g h d^2 \sec(\theta)$$ $$\hat f_0 = \vec f_0/f_0=\left(-\cos(\theta),-\sin(\theta)\right)$$ And since $$A_0 = d^2/\cos(\theta)$$ we find that the force is directed normal to the area and has a magnitude of $$\rho g h A_0$$, so $$f_0= P \vec A_0$$ as expected, even at a slanted wall.