想請問大家~~~~在右圖中，容器底面積20平方公分，內裝高10公分的水，今投入一以線懸吊之鐵球，鐵球質量160公克，密度8公克∕立方公分，當鐵球完全沒入水中，但不與容器底部接觸時，容器底面所承受水的壓力為何？(圖放不上來抱歉拉)?

Volume of iron ball = 160/8 cm^3 = 20 cm^3 = 2 x 10^-5 m^3
Volume of water displaced = 2 x 10^-5 m^3
Weight of water displaced = (2 x 10^-5) x 1000g N = 0.02g N
where g is the acceleration due to gravity, taken to be 9.81 m/s^2

Hence, upthrust on ball = 0.02g N
By Law of Action and Reaction (Newton's 3rd Law), the downward force acts on the water = 0.02g N

Since original water pressure = (10/100) x 1000g N/m^2 = 100g N/m^2
the force acting on the bottom of vessel = 100g x (20 x 10^-4) N
= 0.2g N

The total force = (0.02g + 0.2g) N = 0.22g N = 2.16 N