Question

A glass full of water has a bottom of area 20 cm², top of area 20 cm², 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 × 10⁵ N m⁻². Density of water 1000 kg m⁻² and g = 10 m s⁻².

Answer 1

Given:
$$\begin{gathered} \mathrm{Atmospheric}\mathrm{pressure},p_{\mathrm{a}}=1.0\times {10}^5\mathrm{N}/{\mathrm{m}}^2 \\ \mathrm{Density}\mathrm{of}\mathrm{water},{\rho }_W={10}^3\mathrm{kg}/{\mathrm{m}}^3 \\ \mathrm{Acceleration}\mathrm{due}\mathrm{to}\mathrm{gravity},g=10\mathrm{m}/{\mathrm{s}}^2 \\ \mathrm{Volume}\mathrm{of}\mathrm{water},V=500\mathrm{mL}\approx 500\mathrm{g}\approx 0.5\mathrm{kg} \end{gathered}$$
Area of the top of the glass, A = 20 m²
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
$$\begin{gathered} =p_{\mathrm{a}}\times A+A\times h\times {\rho }_{\mathrm{w}}\times g \\ =A\left({\rho }_a+h{\rho }_{w}g\right) \\ =20\times {10}^{-4}\left({10}^5+20\times {10}^{-2}\times {10}^3\times 10\right) \\ =204\mathrm{N} \end{gathered}$$

(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:
$$\begin{gathered} {\mathrm{p}}_{\mathrm{a}}\times A+mg=A\times h\times {\rho }_{\mathrm{w}}\times g+F_{\mathrm{s}}+{\mathrm{p}}_{\mathrm{a}}\times A \\ \Rightarrow mg=A\times h\times {\rho }_{\mathrm{w}}\times g+F_{\mathrm{s}} \\ \Rightarrow 0.5\times 1=20\times {10}^{-4}\times 20\times {10}^{-2}\times {10}^{-3}\times 10+F_{\mathrm{s}} \\ \Rightarrow F_s=5-4=1\mathrm{N}\left(\mathrm{upward}\right) \end{gathered}$$