A glass full of water has a bottom of area 20 cm^{2}, top of area 20 cm^{2}, 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 water, find the resultant force exerted by the sides of the glass on the water. Atmospheric pressure = 1.0 × 10^{5} N/m^{2}. Density of water 1000 kg/m^{3} and g = 10 m/s^{2}. Take all numbers

to be exact.

#### Solution

Given:

Atmospheric pressure, pa = 1.0 ×10^{5}N/m^{2 }

Density of water, ρw =10^{3}kg/m^{3}

Acceleration due to gravity, g =10m/s^{2}

Volume of water , V = 500 mL ≈ 500g ≈ 0.5 kg

Area of the top of the glass, A = 20 m^{2}

Height of the glass, h = 20 cm

(a) Force exerted on the bottom of the glass = Atmospheric force + Force due to cylindrical water column or glass

=p_{a} × A + A × h × ρ_{w} × g

=A(ρ_{a }+ hρ_{w}g)

=20×10^{-4}(10^{5}+20 ×10^{-2} ×10^{3} ×10)

=204 N

(b) Let F_{s }be the force exerted by the sides of the glass. Now, from the free body diagram of water inside the glass, we can find out the resultant force exerted by the sides of the glass.

Thus, we have:

Pa × A + mg = A × h × ρ_{w }× g + F_{s }+ P_{a} × A

⇒ mg = A × h × ρ_{w }× g + F_{s}

⇒ 0.5 × 1 = 20 × 10^{-4 }× 20 × 10^{-2 }× 10^{-3 }× 10 + F_{s}

⇒ F_{s }= 5 - 4 = 1N (upward)