Question

What is the total force on the bottom of a tank of dimensions 4m x 2m x 2m when it is full of water. Density of water= 1000 kg/cubic m. (g= 10 m/sq. s)​

Answer 1

Given :

• Dimensions of the tank :

⠀⠀⠀⠀⠀⠀⠀⠀⠀Length = 4 m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀Breadth = 2 m.

⠀⠀⠀⠀⠀⠀⠀⠀⠀Height = 2m.

• Density of water = 1000 kg/m³.

• Acceleration due to gravity = 10 m/s².

To find :

Force acting on the bottom of the tank.

Solution :

Pressure on the tank :

We know the formula for Pressure i.e,

$\boxed{\bf{P = h \rho g}}$

Where :

• P = Pressure.
• h = Height.
• ρ = Density of the liquid.

Now using the Formula for Pressure and substituting the values in it, we get :

$:\implies \bf{P = h \rho g}$

$:\implies \bf{P = 2 \times 1000 \times 10}$

$:\implies \bf{P = 20000}$

$:\implies \bf{P = 2 \times 10000}$

$:\implies \bf{P = 2 \times 10^{4}}$

$\boxed{\therefore \bf{P = 2 \times 10^{4}\:Pa}}$

Hence the Pressure of the tank is 2 × 10⁴ Pa.

Force acting on the tank :

We know the Alternative Formula for Pressure i.e,

$\boxed{\bf{P = \dfrac{F}{A}}}$

Where :

• P = Pressure
• F = Force
• A = Area

Now using the above formula and substituting the values in it, we get :

$:\implies \bf{P = \dfrac{F}{A}}$

$:\implies \bf{2 \times 10^{4} = \dfrac{F}{4 \times 2}}$

$:\implies \bf{2 \times 10^{4} = \dfrac{F}{8}}$

$:\implies \bf{2 \times 10^{4} \times 8 = F}$

$:\implies \bf{16 \times 10^{4} = F}$

$:\implies \bf{F = 16 \times 10^{4}\:N}$

Hence the force acting on the tank is 16 × 10⁴ N.