A man of mass 70 kg is in free fall in the air. He reaches a terminal speed of about v_term = 54 m/s. (Note that v_term >> v_crit and we are in the pressure dominated regime where F_fric is directly propotional to v^2 ; we have g = 9.8 m/s )
What is the force (in N) due to air-drag on the man?
To calculate the force due to air drag on the man, we can use the equation for air resistance:
F_drag = 0.5 * (density * Cd * A) * v^2
Where:
- F_drag is the force due to air drag
- density is the density of the air (typically around 1.2 kg/m^3)
- Cd is the drag coefficient (which depends on the shape of the object)
- A is the cross-sectional area of the object
- v is the velocity of the object
Since the man is in free fall and has reached terminal velocity, we can assume that the force due to air drag is equal to the force due to gravity. Therefore:
F_drag = m * g
Where:
- m is the mass of the man (70 kg)
- g is the acceleration due to gravity (9.8 m/s^2)
Substituting the values:
F_drag = 70 kg * 9.8 m/s^2
F_drag = 686 N
So, the force due to air drag on the man is approximately 686 Newtons.