Pascal's Law 1

Linda Fahlberg-Stojanovska

Pascal’s Law – is the principle of transmission of fluid-pressure It says that "a pressure exerted anywhere in a confined incompressible fluid is transmitted equally in all directions throughout the fluid” That is: “pressure acts isotropically at a point" (isotropic means equally in all directions)

Here we prove that the pressure at a point M in a static fluid acts equally in all directions. Start with

• static fluid (a fluid at rest so only normal and body (no shear) forces) • a point M in the fluid

Question: What is the pressure at M? Remember:

We cannot talk about pressure without a surface. A point and a normal vector n determine a surface S. Pressure forces act normal to a surface S and thus in the direction of n .

• Draw an arbitrary unit vector n to M. • Draw "smaller and smaller nice", flat surfaces S , S , S , … each containing M such than n is

normal to each of S , S , S , … at M.

• The pressure on each of S , S , S , … is F/A so: iAreaof S

iiS

SpF

.

• Stating i means Area of Si 0 . When this happens, the surfaces Si the point M. o Now, remember that the surfaces Si are all normal to n and thus depend on n . o This means that when Area of Si 0 , the pressure iSp goes to a number depending

both on the point M and the vector n . o We call this number the pressure pn,M with respect to n at M.

• That is, ,0 i

limAreaof S

i

i

SAreaof S

n MpF

where SF is the pressure force acting on S.

Pascal's Law states that ,n Mp does NOT depend on n and we can write pM = pressure at M.

To prove our statement, we must show that ,n Mp does NOT depend on the choice of n .

We show that , ,, ,x M yn zM M Mp pp p where px,M is the pressure at M with respect to the

positive x-axis, py,M is the pressure at M with respect to the positive y-axis and pz,M is pressure at M with respect to the positive z-axis.

Since these 3 equalities are true for an aribtrary n , we can conclude that pn,M does NOT depend on our choice of n and this proves Pascal's law. Proof of , ,, ,x M yn zM M Mp pp p

• Draw a point O "behind and below" M that is "close" to M and draw a cartesian system at O.

Pascal's Law 2

Linda Fahlberg-Stojanovska

• Draw the "small" flat triangular surface with area A that lies on S with vertices ( x,0,0), (0, y,0) and (0,0, z) on the cartesian axes (below left). The sides of this triangle are parallel to the cartesian plane and have area Ax, Ay and Az. This makes: Ax= y z/2, Ay= x z/2, Az= x y/2 and Volume= x y z/3 .

Notice that we are being a bit sloppy with notation so as not to crowd the picture. Pressures are NOT vectors; the arrow indicates the normal direction along which the corresponding pressure acts.

• The (unit) directional vector of n is , ,cocos ,cos ,sny zn xnu yy z nnnn

Then

co

cos

cs

osy ny y

z n

nx x

z z

x

A A

A A nA A n

A

n

A

A

Let Mx=(0,y,z), that is Mx is the projection of M right onto the y-z plane (above right). Let My=(x,0,z), that is My is the projection of M left onto the x-z plane Let Mz=(x,y,0), that is Mz is the projection of M down onto the x-y plane We will show (1) ,,n M y Mypp , (2) ,,n M x Mxpp and (3) ,,n M z Mzpp

Remember that the pressure at M requires A 0. As A 0, we have Mx,My,Mz M and thus , ,, ,x M yn zM M Mp pp p , which was what we

wanted. 1. We show ,,n M y Mypp . (The projection with respect to y is the "easiest" to see.)

To show ,,n M y Mypp , we compute the forces in the positive y-direction:

Components of forces in the y-direction coming from all 4 sides of the tetrahedron. x. Component on x-side(back right): Fyx = 0 (normal to y-direction) z. Component on z-side (bottom ) Fyz = 0 (normal to y-direction) y. Component on y-side (back left)

F

yy = p

y,My Ay

Pascal's Law 3

Linda Fahlberg-Stojanovska

s. Component on surface (top) Fys = purple· A = - pn,M cos( ny )· A = - pn,M Ay Fluid at rest is in equilibrium sum of all forces is 0 F

yx + F

yz + Fyy

+ Fys

= 0

0 + 0 + py,My Ay+ 0 + (- pn,M

Ay) = 0

py,My = pn,M

2. Analogously ,,n M x Mxpp

We include picture since that is the "hardest part". 3. We show that ,,n M z Mzpp .

This proof is NOT quite analogous to (1) since we must ALSO consider the body force of gravity. To show ,,n M y Mypp , we compute the forces in the positive z-direction:

Components of forces in the z-direction coming from all 4 sides of the tetrahedron. x. Component on x-side(back right): Fzx = 0 (normal to z-direction) y. Component on y-side (back left): Fzy = 0 (normal to z-direction) z. Component on z-side (back) Fzz = pz,Mz Az – weight

weight = density g volume = - g 1/3 x y z = - g 1/3 2 Az z

Notice the EXTRA infintesimal z. So this term is much smaller than other terms. So we can consider it 0.

s. Component on surface (top) Fzs = purple· A = - pn,M cos( nz )· A = - pn,M Az Fluid at rest is in equilibrium sum of all forces is 0

Fzx + Fzy + Fzz + Fzs = 0

0 + 0+ pz,Mz Az – 0 + (- pn,M Az) = 0

pz,Mz = pn,M

QED

nz,n Mp

M

,n Mp

M

nx

ny,n Mp

M