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# Stoke’s Law

• Difficulty Level : Medium
• Last Updated : 06 Jun, 2021

Have you ever asked that raindrops dropped from heights do not harm people due to the force of gravity? Stoke’s law explains why this is the case. This is a fascinating example of a retarding force that is proportional to velocity. George Gabriel Stokes developed an equation for the frictional force, also known as the drag force, in 1851. Stokes Law, named after the scientist George Gabriel Stokes, describes the relationship between the frictional force of a sphere moving in a liquid and other quantities (such as particle radius and velocity of the particle). If a sphere or a body moves through a fluid, a friction force must be overcome. In this article, let us look at what is Stoke’s law and its derivation.

### Stoke’s Law

Stoke’s Law is an equation that expresses the settling velocities of small spherical particles in a fluid medium. The law is established by taking into account the forces acting on a certain particle as it falls through a liquid column under the influence of gravity.

The force that retards a sphere passing through a viscous fluid is directly proportional to the sphere’s velocity, radius, and fluid viscosity. Sir George G. Stokes, an English physicist, stated the viscous drag force F as:

F = 6 × πηrv

where F is the frictional force, η is the viscosity of the liquid, r is the radius and v is the velocity of the flow.

Stokes law, in a brief, discusses the active force applied to a body as it is dropped into a substance. Because of the low viscous impact, the falling body’s velocity initially stays low. However, when the spherical body sinks with its effective weight, it achieves acceleration, and the body’s velocity steadily increases.

The force that retards a sphere passing through a viscous fluid is directly proportional to the sphere’s velocity, radius, and fluid viscosity. Sir George G. Stokes, an English physicist, stated the viscous drag force F as Stokes’ law has applications in many fields, including sediment settlement in freshwater and determining the viscosity of fluids.

In this case, the net force on the body is zero, and it achieves a fixed velocity known as terminal velocity. This definition aided in the formulation of Stokes’ law, and this derivation of frictional force or Stokes’ drag applies at the boundary between the body and the fluid. Sir George Stokes said that “the force acting between the liquid and falling body interface is proportional to velocity, the radius of the spherical object, and fluid viscosity.”

### Importance of Stoke’s Law

Following are the importance of Stoke’s Law:

• Millikan employs this rule in his oil-drop experiment to determine the electronic charge.
• This law protects a person who is falling from a great height using a parachute.
• Cloud formation is explained by this rule.

### Stoke’s Law Derivation

The viscous force acting on a sphere is directly proportional to the following factors:

1. Coefficient of viscosity (η).
2. The radius of the sphere (r).
3. The velocity of the object (v).

Mathematically, this can be written as:

F ∝ ηa rb vc

Here, the powers a, b and c are some unknowns.

Therefore, lets determine these a, b, and c by substituting the proportionality sign with an equality sign as:

F = k ηa rb vc                                                                                                                                                                                                          ……(1)

where, k is the constant of proportionality.

Now, writing the dimensions of parameters on either side of equation (1):

[MLT–2] = [ML–1T–1]a [L]b [LT-1]c

Simplifying the above equation:

[MLT–2] = [Ma ⋅ L–a+b+c ⋅ T–a–c]                                                                                                                                                                            ……(2)

According to classical mechanics, mass, length and time are independent entities.

Equating the superscripts of mass, length and time respectively from equation (2), we get

a = 1                                                                                                                                                                                                                 ……(3)

–a + b + c = 1                                                                                                                                                                                                       ……(4)

–a –c = 2

or

a + c = 2                                                                                                                                                                                                               ……(5)

Substituting (3) in (5), we get

1 + c = 2

⇒ c = 1                                                                                                                                                                                                                 ……(6)

Substituting the value of (3) & (6) in (4), we get

–1 + b + 1 = 1

⇒ b = 1                                                                                                                                                                                                                ……(7)

Substituting the value of (3), (6) and (7) in (1), we get

F = k η r v

The value of k for a spherical body was experimentally obtained as 6π. Therefore, the viscous force on a spherical body falling through a liquid is given by the equation:

F = 6 π η r v

Hence, this is the required relation of Stoke’s Law.

### Limitations of the Stoke’s Law

1. Negative density difference in Stoke’s equation: Stokes’ equation is invalid if the density difference in the equation is negative that is when the particles are lighter than the dispersion medium. This results in floatation or creaming most commonly seen in emulsion systems.
2. High content of dispersed solids: When the solid content of a suspension is high, Stokes’ equation may not show the real sedimentation rate. High solid content imparts additional viscosity to the system, which must be taken into consideration if the correct rate of settling is to be determined. The equation contains only the viscosity of the medium.
3. Dielectric Constant: The dielectric constant, which is not used in Stokes’ equation, is a significant parameter in many contexts. The electrical potential between two charges is inversely proportional to the medium’s dielectric constant. As a result, the zeta potential is affected by the medium’s dielectric constant. The assumption is that if an automobile has a low dielectric constant, the double layer is several times thicker than in an aqueous medium, resulting in different zeta potential and, therefore, a different setting. These considerations are critical for using non-aqueous vehicles such as sesame oil, corn oil, and chlorofluorocarbon propellant (in aerosol suspension).
4. Brownian movement: Another aspect that can affect the precision of Stokes’ equation results is Brownian movement, which is a spontaneous (zigzag) movement of particles floating in a fluid caused by collisions with fast-moving atoms or molecules in the gas or liquid. Sedimentation is mitigated to some degree by Brownian migration. This results in a significant difference between the real rate of sedimentation and the rate measured using Stokes’ equation.

Conditions under which Stoke’s law is valid are:

• The fluid through which the body moves must have infinite extension.
• The body is perfectly rigid and smooth.
• There is no slip between the body and the fluid. The motion of the body does not give rise to turbulent motion. Hence, the motion is streamlined.
• The size of the body is small, but it is larger than the distance between the molecules of the liquid. Thus, the medium is homogeneous and continuous for such a body.

Applications of Stoke’s Law:-

• Velocity of Raindrops: Raindrops do not reach extremely high speeds during their free fall. If this does not occur, a person walking in the rain will be injured. This is because the viscous drag in air opposes the velocity of raindrops as they descend due to gravity. The drop reaches a terminal velocity when the viscous force equals the force of gravity. As a result, raindrops reaching the earth have a low kinetic energy.
• Parachute: When we jump out of an airplane, a parachute assists us in landing safely on the ground. Because in this case, the person falls with g acceleration due to gravity, but the acceleration decreases due to viscous drag in air until the person attains terminal velocity. The person then drops at a constant speed and opens his parachute close to the ground at a predetermined period, allowing him to land safely near his destination.
• It is used to calculate the charge on an electron. (The oil drop method of Millikan)

### Terminal Velocity

When a body falls through a viscous fluid, it produces relative motion between its different layers. As a result, the body experiences a viscous force that tends to retard its motion. As the velocity of the body increase, the viscous force (F = 6 π η r v) also increases. A stage is reached when the weight of the body becomes just equal to the sum of the upthrust and viscous force. Then no net force acts on the body, and it begins to move with a constant velocity. So, here comes the concept of Terminal Velocity.

The maximum constant velocity acquired by a body while falling through a viscous medium is called its Terminal Velocity.

As the speed of an object increases, so does the drag force acting on it, which also depends on the substance it is passing through (for example air or water). At some speed, the drag or force of resistance will equal the gravitational pull on the object (buoyancy is considered below). At this point, the object stops accelerating and continues falling at a constant speed called the terminal velocity (also called settling velocity).

The formula for terminal velocity (vt) is,

vt = 2a2 × (ρ−σ)g / 9η

where ρ and σ are mass densities of sphere and fluid respectively.

### Sample Problems

Problem 1: A raindrop of radius 0.3 mm falls through the air with a terminal velocity of 1 m/s. The viscosity of air is 18 × 10-5 Poise. Find the viscous force on the raindrop.

Solution:

Given that,

The radius of the raindrop, r = 0.3mm = 0.03 cm.

The terminal velocity, v = 1 m/s =100 cm/s.

The viscosity of air, η = 18 × 10-5 Poise.

According to Stokes law, force of viscosity on rain drop is

F = 6π η r v

= 6 × 3.142 × 18 × 10-5 Poise × 0.03 cm × 100 cm/s

= 1.018 × 10-2 dyne

Problem 2: Consider a spherical object is flowing through water. The velocity of the body at some instant is 4 m/s. Find the drag force on the object due to the fluid. Assume that Stokes law is valid. (Given: viscosity of water = 0.001 Kg m−1 s−1, radius of spherical object = 4 mm).

Solution:

According to the formula, the drag force on an object is,

F = 6π η r v

Putting the given values in the above expression,

F = 6 × 3.14 × 0.001 Kg m−1 s−1 × 4 × 10−3 m × 4 m/s

= 3.014 × 10-4 N

Problem 3: The speed of water in a water stream is 18 km/h near the surface. If the stream is 10 m deep, then determine the Shearing stress between the surface layer and the bottom layer (coefficient of viscosity of water, η=10−3 Pa s).

Solution:

Since, the velocity of water at the bottom of the river is 0 m/s,

Therefore, dv = 18 km/h = (18 × 5/18) m/s = 5 m/s

Also, dx = 10 m and η=10−3 Pa s

The Force of viscosity, F = η A (dv/dx)

Therefore, the Shearing stress = F / A = η (dv/dx)

= 10−3 Pa s × (5 m/s / 10 m)

= 5 ×10−2 N m-2

Problem 4: What will be the net force on the water drop of radius r which is falling through the air of coefficient of viscosity η with a constant velocity of v?

Solution:

When a drop falls at a constant velocity, the viscous force is balanced by its weight.

Therefore, the resulting force is zero.

Problem 5: What is the velocity of a rain drop called and why?

Solution:

A rain drop falls with a velocity called terminal velocity. Because of the viscosity of the air, raindrops fall with terminal velocity. Terminal velocity is the speed of an object when its net force is zero. In the case of raindrops, the mass of the drop acts downward while the viscous force of the air acts upward.

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