# Drag Force Formula

The force exerted on a solid body moving in relation to a fluid by the fluid’s movement is known as a drag force. For instance, drag on a moving ship or drag on a flying jet. As a result, a drag force is a resistance created by a body moving through a fluid such as water or air. This drag force opposes the oncoming flow velocity. As a result, this is the body-to-fluid velocity. We’ll go over the concept and formula for drag force with examples in this article. Let’s take a look at the idea.

### What is a Drag Force?

The drag force (D) is the force that opposes a body’s motion through a fluid.

The resisting force of a fluid is called drag force. This force opposes the motion of a submerged object in a liquid. Drag force is thus defined as the force that opposes a body’s motion through a fluid. When a body moves in a fluid-like environment, aerodynamic drag arises. When the fluid is water, it’s also a hydrodynamic drag. It has a natural inclination to act in the opposite direction of the velocity flow. Air resistance frequently limits the maximum speed at which a falling body can travel. Air resistance is a good example of drag force, which is the force that objects feel when traveling through a fluid.

The drag force, like kinetic friction, is a reactive force that arises only when an object is moving and points in the opposite direction of the object’s motion through the fluid. Shape drag and skin drag are two types of this force. Form drag is the resistance of fluids to being moved out of the way by a moving object. As a result, shape drag is comparable to the force generated by solids’ resistance to deformation. The fluid slipping over the moving object’s surface causes skin drag, which is essentially a mechanical frictional force.

### Formula for Drag Force

The drag force is calculated using the following formula:

D = (C_{d}× ρ × V^{2}× A) / 2where,

- D is the Drag Force (N),
- C
_{d}is the Drag coefficient,- ρ is the Density of the medium (kg/m
^{3}),- V is the Velocity of the body (m/s), and
- A is the Cross-sectional area (m
^{2}).

### Sample Questions

**Question 1: What is Drag?**

**Answer** :

Drag is the force exerted by a fluid stream on an object moving in its direction, or the force experienced by an object travelling through a fluid. Moving vehicles, ships, suspension bridges, cooling towers, and other structures must all evaluate the size of the structure and how it might be reduced. Drag forces are traditionally represented by a drag coefficient that is determined regardless of the body’s shape.

The drag coefficient is proportional to the Reynolds number, as shown by dimensional analysis; the exact relationship must be determined experimentally, but it can be used to approximate the drag forces experienced by other bodies in other fluids at other speeds. The idea of dynamic similarity is used by engineers when they use the effects of a model structure to anticipate the behaviour of other structures.

**Question 2: A car is traveling at 83 km/h and has a drag coefficient of 0.30. Determine the drag force if the cross-sectional area is 12 square meters.**

**Answer:**

Given that,

V = 83 km/h = 83 × 5/18 m/sec = 23.05 m/sec,

C

_{d}= 0.30, A = 12 m^{2},ρ = 1.2 kg/m

^{3}Since,

D = (C

_{d}× ρ × V^{2}× A) / 2= (0.30 × 1.2 × 23.05

^{2}× 12) / 2= 2295.21 / 2

= 1147.60 N

**Question 3: With a drag coefficient of 0.25, a plane moves at 610 km/h. Calculate the drag force if the plane’s cross-sectional area is 70 m ^{2}.**

**Answer:**

Given that,

V = 610 km/h = 610 × 5/18 m/sec = 169.44 m/sec,

C

_{d}= 0.25,ρ = 1.2 kg/m

^{3},A=70 m

^{2}Since,

D = (C

_{d}× ρ × V^{2}× A) / 2= (0.25 × 1.2 × 169.44

^{2}× 70) / 2= 602908.11 / 2

= 301454.05 N

**Question 4: With a drag coefficient of 0.14, a bicycle goes at an average speed of 83 km per hour. Calculate the drag force given a cross-sectional area of 5 m ^{2}.**

**Answer:**

Given that,

V = 83 km/h = 83 × 5/18 m/sec = 23.05 m/sec,

C

_{d}= 0.14,A = 5 m

^{2},ρ = 1.2 kg/m

^{3}Since,

D = (C

_{d}× ρ × V^{2}× A) / 2= (0.14 × 1.2 × 23.05

^{2}× 5) / 2= 446.292 / 2

= 223.146 N

**Question 5: What is the drag coefficient of a moving object in the water that experiences a 520 N drag force? The sectional area is 3 m ^{2} and the speed is 20 m/s.**

**Answer:**

Given that,

D = 520 N,

A = 3 m

^{2},V = 20 m/s,

ρ = 1.2 kg/m

^{3}Since,

D = (C

_{d}× ρ × V^{2}× A) / 2= 2 × D / ρ × V

^{2}× A= 2 × 520 / 1.2 × 20

^{2}× 3= 1040 / 1440

= 0.722