As shown in the figure, a metal piece having a mass of 160 g rests in equilibrium within a glass of water. The piece makes contact with the bottom of the glass at only a few points. Given that the metal's density is 8000 kg m ⁻³ , determine the normal force applied by the glass bottom on the metal piece. (Take g = 10 m s ⁻² )

Step-by-step solution

Given: Mass of the metal piece, m = 160 g = 0.16 kg Density of the metal, _m = 8000 kg m ⁻³ Density of water, _w = 1000 kg m ⁻³ Acceleration due to gravity, g = 10 m s ⁻² The volume of the metal piece is: V = m _m = 0.16 8000 = 2 10⁻⁵ m ^3 Since the metal piece makes contact with the bottom at only a few points, water is present under it, and the buoyant force acts on its entire volume. The buoyant force F_B is: F_B = V _w g = (2 10⁻⁵) 1000 10 = 0.2 N The weight of the metal piece W is: W = mg = 0.16 10 = 1.6 N For