Question

A glass full of water has a bottom of area 20 $c{m}^2$, 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 =$1.0\times {10}^{5}NM-2$. Density of water = 1000 kg $m^{-3}$ and g = 10 m $s^{-2}$. Take all numbers to be exact.

Answer 1

Pa = $1.0\times {10}^{5}N/m^2$,

Pw = ${10}^{3}kg/m^3$

g= $10m/s^2$

v = 500 ml = 500 g = 0.5 kg

(a) Force exerted at the bottom

= Force due to Cylindrical water colum + atm. force

= $A\times h\times P_w\times g+P_a\times A$

= $A(h{p}_{w}g+P_a)$

= $20\times {10}^{-4}(20\times {10}^{-2}\times {10}^3\times 10+{10}^5)$

= 204 N

(b) To find out the resultant force exerted by the sides of the glass, from the free body diagram of water inside the glass,

$P_a\times A+mg=A\times h\times {\rho }_w\times g+F_s+P_a\times A$

$\Rightarrow mg=A\times h\times p_w\times g+F_s$

$\Rightarrow 0.5\times 1=20\times {10}^{-}4\times 20\times {10}^{-2}\times {10}^{-3}\times 10+F_s$

$\Rightarrow F_s$ = 5-4 = 1N (upward)

This force 1 N is provided by the sides of the glass.