Question

A rectangular tank of base 100 m2has a height of 2m. Calculate the thrust at the bottom of the tank, if it is filled upto the brim with water of density 103 kg m-3 ?

Answer 1

Pressure = force or thrust /Area
ρgh= Thrust /Area 
Thrust = ρgh×Area   = 10³×10×2×100
 =2×106 N

Answer 2

The thrust at the bottom of the tank is 1960 kN

Explanation:

The formula for thrust is,

$\text{Thrust}=\text { Pressure } \times \text{Area}$

Pressure created by water in the bottom of the tank can be found using the below formula,

$\text { Pressure }=\text {Density of liquid} \times \text {Acceleration due to gravity} \times \text {Height of liquid }$

Thus, here density of water is known as $1000 \ \mathrm{kg} \mathrm{m}^{-3}$, acceleration due to gravity is $9.8 \ m s^{-2}$ and height of the liquid is 2 m.

$\text{Pressure}=1000 \times 9.8 \times 2$

$\Rightarrow \text { Pressure }=19600 \ \mathrm{N} / \mathrm{m}^{2}$

So the area of the tank is given as $100 \ m^{2}$

$\therefore \text { Thrust }=19600 \times 100 \ N$

$\therefore \text { Thrust }=1960000 \ N$

$\therefore \text { Thrust }=1960 \ \mathrm{kN}$