Question

A stainless steel sphere drops through a graduated cylinder filled .245 meters high with water

knowing that the mass of the ball is 028236 kg, it took .175 seconds to reach the bottom, and its radius is 0.00955 meters

Find the drag force acting

Answer 1

Mass of ball = 0.028236Kg

radius of ball= 0.00955m

Gravity force acting on bal (F1) = m g = 9.81 X 0.028236 = 0.2770N (downwards)

Density of water = 1000 Kg/m³

Buoyancy force on ball (F2)= density X volume of ball X g

= 1000 X ((4/3) X 3.14 X (0.00955³)) = 0.0036 N (upwards)

Acceleration of ball be a m/s^s

applying equation of motion

u=0 m/s t=0.175s

s= ut + 0.5 X a X (t^2)

=> a= (2 s)/ (t²)

=> a= 16

Now F= m *a

F1-F2-F_drag = 0.02836*16

Fdrag= 0.2770-0.0036-0.4538

Fdrag= -0.1804 N (Drag force acts along the water)