A metal piece of mass 160 g lies in equilibrium inside a glass of water. The piece touches the bottom of the glass at a small number of points. If the density of the metal is 8000 kg/m3, find the normal force exerted by the bottom of the glass on the metal piece.
Solution
Given:
Mass of the metal piece, m =160 gm = 160 × 10−3 kg
Density of the metal piece, ρm = 8000 kg/m3
Density of the water, ρw = 1000 kg/m3
Let R be the normal reaction and U be the upward thrust.
From the diagram, we have:
mg = U + R
⇒ R = mg −Vρwg [U = Vρwg]
\[\Rightarrow \text{R = mg }- \frac{\text{m}}{\rho_\text{m}} \times \rho_\text{w} \times \text{g}\]
\[= 160 \times {10}^{- 3} \times \left( 10 - \frac{{10}^3 \times 10}{8000} \right)\]
\[ = 160 \times {10}^{- 3} \times 10\left( 1 - \frac{1}{8} \right)\]
\[ = 1 . 4 \text{N }\]
