Question

There is a jar that is 74 cm tall and has a diameter of 29.6 cm, the jar stands vertically and is half loaded up with water (ρwater = 1 g/cm3). Discover the gauge pressure close to the base of the jar. What is the force on the jar's base because of the water?

P1 = Unit in kPa

F1 = Unit in lb

Presently, some oil with ρoil = 0.8 g/cm3 is poured on head of the water so the total level of two liquids together is h2 = 57.7 cm (the liquids don't blend and don't get compacted). What is the gauge pressure and the force on the base of the jar for this situation?

P2 = Unit kPa

F2 = Unit lb

Answer 1

1) So gauge pressure in the jar coloumn is

$P =\rho_w gh$

Here rho is water density = 1gm/cm^3 = 1000kg/m^3

g is gravitational acceleration = 9.8 m/s^2

h = 74cm/2 = 37 cm = 0.37 m

So we get,

So Force on the base of jar

$F= P.A= P.(\pi r^2)= 3.63 kPa(\pi (\frac{0.296}{2})^2)= 249.7 N$in lb unit

$F= \frac{ 249.7 N}{4.45} = 56.11 lb$

2) So oil density is lower than water that's why it floats on water and oil height is then

h2 = 57.7 - 37 = 20.7 cm= 0.207 m

So, gauge pressure is

So total gauge pressure.

Poil Poigh = 800kg/m (9.8)(0.207) 1622 Pa = 1.62k Pa

So net force on Bottom is