In simple words, Stokes law discusses the active force applied on a body when it is dropped in a liquid. Initially, because of low viscous force, this velocity of the falling body remains low. But, as the spherical body falls with its effective weight, it gains acceleration, and this velocity of the body increases gradually.
This also makes the liquid in contact move with a velocity which is the same as that of this body. The movement of an object in this fluid with increasing velocity causes motion in liquid layers resulting in the development of viscous force. This force increases with the increasing velocity until the point of time when it matches this effective force with which this body moves.
In such a situation, the net force on the body becomes zero, and it attains a constant velocity named as terminal velocity (Vt). This idea helped state Stokes law and this derivation of frictional force or Stokes’ drag applies on the interface between the body and fluid.
According to Sir George Stokes, “the force acting between the liquid and falling body interface is proportional to velocity and radius of the spherical object and viscosity of fluid”.
Stokes came up with this formula in 1851 to calculate this drag force or frictional force of spherical objects immersed in viscous fluids. Here, look at the formula mentioned below.
F = 6 * πηrv
F is the drag force or frictional force at the interface.
η is viscosity of a liquid.
r is radius of the spherical body.
V is velocity of flow.
Stokes’ proposition regarding this immersion of the spherical body in a viscous fluid can be mathematically represented as,
F ∝ ηa rb vc
By solving this proportional expression, we can get Stokes law equation. To change this proportionality sign into equality, we must add a constant to the equation. Let us consider that constant as ‘K’, so the transformed equation becomes
F = K * ηa rb vc …………….. (1)
Now, Let us write down the dimensions of this equation on both sides. Please note, K is a constant which has no dimension.
[MLT-2] = [[ML-1T-1]a . [L]b . [LT-1]c ]
We need to simplify this expression by separating each variable as follows
[MLT-2] = [Ma ] [L-a+b+c] [T-a-c]
Comparing the value of length, mass and time, following equations can be found.
a = +1 …………….. (2)
-a+b+c = +1 …………….. (3)
-a-c= -2 …………….. (4)
On solving equation 2,3, and 4, we get a =1, b=1, c=1. On substituting these values in equation 1, we have
F = K * η1 r1 v1 = K ηrv
Further, the value of K was found to be 6π for spheres through experimental observation. The above calculations helped in Stokes equation derivation along with its fundamental formula.
Terminal Velocity Formula
As explained earlier, terminal velocity is attained at an equilibrium position when the net force acting upon the spherical body and acceleration becomes zero. Here is the formula for terminal velocity derived from Stokes law definition.
Vt = 2a2 (ρ−σ) g / 9η
ρ is mass density of a spherical object.
σ is mass density of a fluid.
Question to Solve
Two identical spherical water droplets travelling through the air have a steady velocity of 20 cm/s. Calculate the terminal velocity of a newly formed drop if the two drops combine to become one.
You will be able to solve such numerical questions better by learning the topic in greater detail. In this pursuit, Vedantu’s comprehensive study notes can be your most prominent companion. You can also download our Vedantu app and begin learning complex concepts with ease.
1. Define Stokes Law with an Example.
Ans. According to Stokes law, if a spherical body falls into a viscous liquid then the force acting at the interface is proportional to – Radius of the spherical body, velocity of the sphere, and viscosity of this given fluid.
2. How Can I Derive Stokes Formula?
Ans. You can derive this formula by writing the proportionality equation of Stokes law first. Then, you need to write down the proportionality constant and represent this equation with its dimensional formula. Calculate the value of constants and put them in the original equation to get the formula.
3. How Does Viscosity of Fluid Affect the Travel of Spherical Objects?
Ans. If the fluid is more viscous, then the speed of the spherical object with which it travels will be low and vice versa. For instance, air has low viscosity, and hence the spherical objects can move much faster in this case.