# why don't we consider atmospheric pressure or weight of liquid while deriving the equation of thrust in a fluid?

More specifically , when we try and prove that up thrust experienced by an liquid column inside the liquid is equal to the weight of the liquid column , I have seen it being derived like this:

thrust = $(P_2-P_1 )dS$ (where $P_1$ and $P_2$ are the pressures indicated in the image and $\rho$ is the density of the fluid) followed by :

in this entire derivation why don't we consider adding the pressure exerted by air or the weight of the liquid column to the equation of thrust ?

• Could you please write out the equations in MathMode (use \$...\$ and \$\$...\$\$.? They are not easy to see from your photos. Also, what is $dS$ in your thrust equation? And where does the rho enter, which you mention in the text but doesn't include in the expression? – Steeven Nov 26 '15 at 12:53
• sorry about that , I'm actually still familiarizing myself with the mathmode . dS is the sectional area of the column. Also rho is used in the derivation (in its symbolic form). – Ishita Gupta Nov 26 '15 at 13:14

$P_1$ and $P_2$ are defined in terms of a reference pressure and a contribution due to the weight of the liquid column: $$P_1= P_0 + \rho g h_1 \quad P_2=P_0+\rho g h_2$$
Note that the $\rho g h = \rho g V/A = F/A$ is the pressure exerted by the weight of the liquid column. Taking the difference $P_1-P_2=\rho g (h_1-h_2)$ cancels the reference pressure (which could be atmospheric or otherwise).
• @IshitaGupta - There is a $P_1$; in that case $h_1=0$ and the pressure $P_1=P_0$ is atmospheric. Unless your in a perfect vaccuum there is always a pressure to be considered. The buoyancy force comes from the fact that the object experience forces at its surfaces which want to push it in the direction of decreasing pressure. Whether the object moves or not depends on the balance between its weight in the fluid vs the bouyancy force it experiences. – nluigi Nov 26 '15 at 13:31