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?
at a constant speed, there is no acceleration, so no net force. The upward force = weight = 9.8*70
First calculate C_2 = (mg / (v^2 * r^2) where r is unknown, then calculate F = C_2 * r^2 * v^2
I presume this question is from Walter Lewin's Physics course
To find the force due to air drag on the man, we need to use the formula for air drag force. The air drag force can be described using the equation:
F_drag = 1/2 * rho * A * C * v^2
where F_drag is the air drag force, rho is the density of air, A is the cross-sectional area of the object, C is the drag coefficient, and v is the velocity of the object.
In this case, the drag coefficient and cross-sectional area are not provided, and we are assuming that rho is constant. Therefore, we need to approximate the drag force using the terminal velocity.
At terminal velocity, the air drag force is equal to the gravitational force acting on the object, resulting in no net acceleration. So we have:
F_drag = F_gravity
The gravitational force can be calculated using the formula:
F_gravity = m * g
where m is the mass of the object and g is the acceleration due to gravity.
Given that the mass of the man is 70 kg and g is approximately 9.8 m/s², we can calculate the force due to air drag.
F_drag = F_gravity = m * g = 70 kg * 9.8 m/s² = 686 N
Therefore, the force due to air drag on the man is approximately 686 Newtons.