Drag is the force that is exerted on a solid body moving with respect to a fluid. It is due to the movement of the fluid. Therefore a drag is the resistance force caused by the motion of a body through a fluid like water or air. These acts are opposite to the direction of the oncoming flow velocity. So, it is the relative velocity between the body and the fluid. In this topic, we will discuss the concept and drag formula with relevant examples. Let us learn the concept!
Source:en.wikipedia.org
Drag Formula
Concept of Drag
Drag is the force, whose magnitude depends on factors such as the density of the air, the square of the velocity, the viscosity of air, the size and shape of the object and the inclination of the object to the flow.
Therefore, the dependence on the shape, inclination, air viscosity, and compressibility are complex but important factors.
One way to deal with these complex dependencies is to characterize the dependence by a single variable. Here, in this case, this variable is known as the drag coefficient and designated as \(C_d\). It allows us to collect all the effects, simple and complex, into a single equation.
This drag equation states that drag D will be equal to the drag coefficient \(C_d\) times the density r times half of the squared velocity V times the area A.
Formula
Mathematically,
D = \(C_d \times A \times 0.5 \times \rho \times V^2\)
Where,
D | Drag Force |
\(C_d\) | It is the drag coefficient |
\(\rho\) | It is the density of the medium in kg \(m^{-3}\) |
V | It is the velocity of the body in \(ms^{-1}\) |
A | It is the cross-sectional area in m² |
For some given air conditions, shape, and inclination of the object, we can determine the value for \(C_d\) to compute drag. It must be noted that the area A, given in the drag equation is given as the reference area. The drag depends directly on the size of the object.
Since we have to deal with the aerodynamic forces, the dependence can be characterized by some area. In the report, the aerodynamicist has to specify the area used. When using the data, the user may have to convert the drag coefficient using the ratio of the areas.
In the equation density is designated by the letter \(\rho\), instead of “d” or “r”. The combination of terms “density times the squared velocity divided by two” is known as the dynamic pressure and it also appears in Bernoulli’s pressure equation.
Solved Examples
Q.1: A car travels with a speed of 90 km per h, with a drag coefficient of 0.32. If the cross-sectional area is 5 square meters, then find out the drag force. Consider The air density is 1.2 Kg \(m^{-3}\).
Solution:
Given parameters:
Velocity, V = 90 km per h = 90 \(\times \frac {5}{18}\) m per sec = 25 m per sec.
Drag coefficient, \(C_d\) = 0.32
Cross-sectional area, A =5 square meter
Density of fluid, \(\rho\) =1.2 kg per cubic m
The drag force formula is:
D =\( C_d \times A \times 0.5 \times \rho \times V^2 \)
Substituting the values,
D= \(0.32 \times 5 \times 0.5 \times 1.2 \times 25 ^{2} \)
D = \(600 \; N\)
Therefore the drag force will be \(600 \; N\).
Typo Error>
Speed of Light, C = 299,792,458 m/s in vacuum
So U s/b C = 3 x 10^8 m/s
Not that C = 3 x 108 m/s
to imply C = 324 m/s
A bullet is faster than 324m/s
I have realy intrested to to this topic
m=f/a correct this
Interesting studies
It is already correct f= ma by second newton formula…