Question : Using the method of dimensional analysis, derive an expression for the viscous force acting on a small sphere moving through a fluid. The viscous force depends on the radius of the sphere, its velocity, and the coefficient of viscosity of the fluid.
Doubt by Prateek
Solution : Let the viscous force (F) acting on a small sphere depend on its radius (r), velocity (v), and coefficient of viscosity (η).
F ∝ ravbηc
F = kravbηc — (1)
where k is a dimensionless constant.
F = [MLT-2]
r = [L]
v = [LT-1]
η = [ML-1T-1]
Substituting in equation (1)
[MLT-2] = [L]a [LT-1]b [ML-1T-1]c
[MLT-2] = [McLa+b-cT-b-c]
Comparing powers of M both sides
c = 1
Comparing powers of T both sides
−b − c = −2
−b − 1 = −2
b = 1
Comparing powers of L both sides
a + b − c = 1
a + b − 1 = 1
a + b = 2
a + 1 = 2 [∵ b=1]
a = 1
Putting a = 1, b = 1 and c = 1 in equation (1)
F = krvη
For a sphere moving slowly through a viscous fluid, Stokes’ law gives:
F = 6πηrv
Note: Dimensional analysis gives the proportionality relation. The numerical constant 6π is obtained experimentally or from fluid mechanics, not from dimensional analysis alone.