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Wize University Physics Textbook (Master) > Newton's Laws of Motion: Forces and Dynamics

Drag Force

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Drag Force


Drag is a resistive force that occurs when an object moves through a fluid. This force is due to the collisions with fluid particles and is extremely complicated. However, the drag force in air can be simplified to:



D=14.v2. A\boxed{D=\frac{1}{4}.v^2.\ A}



  • Where:
  • AA = surface area of the object experiencing drag
  • vv = speed of the object
  • DD = drag force (also shown by FDF_D)



Watch Out!
Be careful in choosing the surface involved in drag force! This area is usually the effective area of the surface which is the size of its shadow!


Exam Tip
Drag force is always applied against direction of motion!





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Drag Force


Drag is a resistive force that occurs when an object moves through a fluid. This force is due to the collisions with fluid particles and is extremely complicated. However, the force can be simplified to:



D=12ρCDAv2\boxed{D=\frac{1}{2}\rho C_DAv^2}


  • Where:
  • ρ\rho = density of the fluid
  • AA = surface area of the object experiencing drag
  • vv = speed of the object
  • CDC_D = drag coefficient
  • DD = drag force (also shown by FDF_D)




Watch Out!
Be careful in choosing the surface involved in drag force! This area is usually the effective area of the surface which is the size of its shadow!


Exam Tip
Drag force is always applied against direction of motion!












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Terminal Velocity


  • Say, a body is freely falling in vacuum. Then the force acting on it F=mg.
  • In presence of air or other fluids, it will experience a drag force (resistive force due to fluid) against its motion.
  • This resistive force is usually increasing with the speed of the object
  • On some cases it may happen that the drag force will become equal to the force of gravity. Hence the body will experience no net force; hence it will fall with a constant velocity, known as the terminal velocity.



Exam Tip
Terminal velocity could be found by making gravitational force and drag force equal to each other! (You might need to consider other forces as well if there are more forces exerting on the object)

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Example: A Skydiver Terminal Velocity


A 60-kg skydiver jumps out of airplane. After a while, he reaches a final terminal velocity of 250 km/h. What is his effective cross section surface area of the airplane?

Solution:
When a falling object reaches terminal velocity, force of gravity is equal to drag force. Thus:

D=14.v2. A     mg=14v2. A →A=4mgv2=4×60×9.81(69.4)2=0.488m2 \begin{aligned} D=\frac{1}{4}.v^2.\ A\ \end{aligned} \\ \begin{aligned} \ \ \ \\\ {mg=\frac{1}{4}v^2.\ A\ \rightarrow A=\frac{4mg}{v^2}=\frac{4\times60\times9.81}{\left(69.4\right)^2}=0.488m^2} \end{aligned}



Effective cross section surface area is 0.488m20.488m^2.







A 4×4×4 cm  4\times4\times4\ cm^{\ \ }1 kg1\ kg cube is initially falling through the air at 32 m/s32\ m/s. It is falling with one face flat towards the ground. Find the initial drag force on this cube. Find the terminal velocity of the cube. You may consider the density of air to be 1.3 [kgm3]1.3\ \left[\frac{kg}{m^3}\right]and CD=1C_D=1.