# Stoke’s Law

Stoke studied and concluded that the viscous force acting on the object depends upon the coefficient of the viscosity of the object and gave Stoke’s Law.

​When an object moves inside a stationery viscous fluid, then, the layer of the fluid which is in contact with the object moves with the velocity of the object. The velocity of the layers farther from the object goes on decreasing with increase in the distance from the object. Due to this relative motion between the layers of the fluid, viscous force comes into play which opposes the motion of the object. This viscous force increases with the increase in velocity. When this force becomes equal with the weight of the object, the object starts to move with a constant velocity known as terminal velocity.

​Let a spherical object be moving inside a viscous fluid. Stoke found that the viscous force $(F)$ acting on the object depends upon;

1. Coefficient of viscosity $(η)$ of the fluid $\text{i.e. } F ∝η^x$ 2. Terminal velocity $(v)$ of the object $\text{i.e. } F ∝v^y$ 3. Radius $(r)$ of the object $\text{i.e. } F ∝r^z$ Thus, $F ∝η^x v^y r^z$ where $x$, $y$ and $z$ are the dimensions of $F$ in terms of $η$, $v$ and $r$. and, $F=k η^x v^y r^z \text{ ____(1)}$ where, $k =$ proportionality contant.

Writing dimensions of each terms of $(1)$, $\left[MLT^{-2}\right] = \left[L\right]^x \left[ML^{-1}T^{-1}\right]^y \left[LT^{-1}\right]^z$ $\left[MLT^{-2}\right] = \left[M^yL^{x-y+z}T^{-y-z}\right]$ equating, $∴y=1$ $∴-y-z=-2$ $-1-z=-2$ $z=1$ $∴x-y+z=1$ $x-1+1=1$ $x=1$ putting the values of $x$, $y$ and $z$ in $(1)$, $F=k η^1 v^1 r^1$ Experimentally, the value of $k$ is found to be $6π$. $F=6πηrv$ This is called Stoke’s law of viscosity.

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