Question

Using the method of dimensions derive an expression for the viscous force f acting on a spherical body

Answer 1

This is what stoke's law states that viscous force or viscous dragging force acting on a spherical body.

Answer 2

Concept:

• We have to apply dimension analysis here
• We need to identify the correct dimensions of the physical quantities

Given:

• The viscous force depends on the coefficient of viscosity n, the radius of the spherical body r and the velocity of the body v
• Dimensions of the coefficient of viscosity n = [ML^-1T^-1]
• Dimensions of the radius of the sphere r = [L]
• Dimensions of the velocity of the body v = [LT^-1]
• Dimensions of force F = [MLT^-2]

Find:

• Expression for viscous force

Solution:

F = k n^a* r^b * v^c

F = [MLT^-2]

F = [ML^-1T^-1]^a * [L]^b * [LT^-1]^c

F = [M^aL^-aT^-a] * [L^b] * [L^cT^-c]

F = [M^aL^(-a+b+c)T^(-a-c)]

[MLT^-2] = [M^aL^(-a+b+c)T^(-a-c)]

On comparing, we get

a = 1

b+c-a = 1

b+c -1 = 1

b+c = 2

-a-c = -2

-1-c = -2

c = 1

b = 1

The expression is F = k n^1* r^1 * v^1

F = k n r v

where k = 6π, a constant

The expression for the viscous force acting on a spherical body is 6πnrv.

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