A glass full of water has a bottom of area `20 cm^(2)`, top of area `20 cm`, height `20 cm` and volume half a litre.

a. Find the force exerted by the water on the bottom.
b. Considering the equilibrium of the `v`, after. find the resultant force exerted by the side, of the glass on the water. Atmospheric pressure `= 1.0 xx 10^(5) N//m^(2)`. Density of water `= 1000 kg//m^(-3)` and `g = 10 m//s^(2)`

Explanation

(a) Pressure at bottom
=`P_0 + rho gh = 10^5 + 1000 xx 10 xx 0.2`
=`(10^5 + 2000) N//m^2`
Area of bottom surface
=`20 cm^2 = 20 xx 10^-4 m^2`
For at bottom
=`(10^5 + 2 xx 10^3) xx 20 xx 10^-4 = 20 + 4 = 204 N`
(b) Force on water due to sir top surface
`F_1 = P_(0) xx` area at top `= 10^(5) xx 20 xx 10^(-4)`
=`200 N (downward)`
weight of water
`W = rho_omega Vg = 1000 xx (1)/(2) xx 10^-3 xx 10 = 5 N (downward)`
Force on water due to bottom surface `= 204 N (upward)`
Total downward force `= 200 + 5 = 205 N`
Total upward force `= 204 N`
Since water is in equilibrium, force extered by the sides of the glass on the water `= 205 - 204 = 1 N (upward)`.