When an object moves through a fluid, the fluid exerts a viscous force vector F

Question

When an object moves through a fluid, the fluid exerts a viscous force vector F on the object that tends to slow it down. For a small sphere of radius R, moving slowly with a speed v, the magnitude of the viscous force is given by Stokes' law, F = 6??Rv, where ? is the viscosity of the fluid. What is the viscous force on a sphere of radius R = 4.90 times 10-4 m that is falling through water (? = 1.00 times 10-3 Pa · s) when the sphere has a speed of 3.1 m/s? The speed of the falling sphere increases until the viscous force balances the weight of the sphere. Thereafter, no net force acts on the sphere, and it falls with a constant speed called the "terminal speed." If the sphere has a mass of 1.03 10-5 kg, what is its terminal speed?

Explanation / Answer

F = 6??Rv

putting values,,

F = 6*3.14*10^-3 * 4.9*10^-4 * 3.1 = 286.17 *10^-6 N

2) 1.03*10^-5 *9.81= 6*3.14*10^-3 * 4.9*10^-4* V

V= 1.09 m/s

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