Basic Thermodynamics ~ J. Pohl © | www.THERMOspokenhere.com |
Parachutist's Drag Force
A soldier (m = 90 kg) parachutes from a
C-130 Hercules at an altitude of 2000 meters moving horizontal at 120 m/s. Some moments later, in free fall, the soldier experienced an acceleration of 6.8 m/s². Calculate the frictional drag force of the surrounding air at that instance of the fall?
Solution: We will solve this problem somewhat "at length" by looking at phases of the jump.
i) Before the jump: For this phase we set our time period to be indefinite, to start at t = 0+ and to continue to unspecified times later, t. While standing in the C-130, the acceleration of the soldier is zero and his velocity is 350 m/s. The momentum equation for this event is:
![]() | (1)(1) |
Solution: The term "force" is a signal that Newton's 2'nd Law applied to some "BODY" is the consideration. The physical property of the 2'nd Law is the momentum of the BODY.
![]() | (1)(1) |
Acceleration, (not a property but a spatial characteristic) of a BODY is relevant for systems having constant mass. The 2'nd Law for the parachutist (as a mass) in this case has two forces acting: that of gravity and fluid resistance or drag.
![]() | (2)(2) |
A system sketch should have been drawn with the person modeled as a BODY and the forces acting. Acceleration and forces are vectors; coordinates and a vector basis are required to state them. The sketch ( not done) would show the coordinates 0XZ with unit vectors I - J - K (with K upward or in the positive Z direction). The acceleration and gravity force are in the negative K direction. In the vector equation below, the drag force, its magnitude and direction, are the unknowns.
![]() | (3)(3) |
The above equation has only one unknown entity. So do algebra and calculator.
![]() | (4)(4) |
So, Newton's 2'nd Law, being a vector equation, states the force to have a magnitude of 270 Newtons and to be directed upward.
Parachutist's Drag Force
At the precise instance of leap, the acceleration is Earthward at 9.8m/s². A
few moments later, in free fall, the acceleration of a parachutist becomes 6.8 m/s². The mass of the person is 90 kilograms. Calculate the frictional drag force of the surrounding air at that instance of the fall?