Help with Drag.

  • Thread starter xieon
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  • #1
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Alright so far I've got:

Force of Drag = (.5)(A)(p)(V^2)(Cd)


The only thing is I keep getting a Fd that is larger than Fg.

Could someone explain to me how to figure out the acceleration an object falls at when experienced to a certain drag?

Ex: 9.8 m/s/s is a freefalling object, when exposed to drag of D the acceleration becomes...
 

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  • #2
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Is your V terminal?
 
  • #3
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Is your V terminal?
I was thinking that the V would be the speed when the drag was introduced (such as a parachute)
 
  • #4
PhanthomJay
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I was thinking that the V would be the speed when the drag was introduced (such as a parachute)
Do you have a specific example in mind? As cesiumfrog noted, are you talking terminal velocity? At terminal velocity, the drag force equals the gravity force, and the object moves at constant speed (no acceleration). An object falling from rest thru (say) air will immediately experience a drag force calculated from ther formula you noted. At first, since v is small, the drag force will be small, and the object will continue to accelerate downward at an acceleration less than g. Nonetheless, it will still be accelerating, and thus will gain speed. As it gains more and more speed, the drag force becomes larger and larger, and thus the acceleration of the object becomes smaller and smaller, until such time as the drag force equals the objects weight, at which point there is no net force acting on the object, and the acceleration becomes zero, and the object moves downward at constant speed (terminal velocity). Terminal velocity is calculated by setting the objects weight equal to the drag force.The drag force thus cannot exceed the objects weight in this situation. But supposing at this terminal speed a parachute is opened. Now, since the Area of the objects exposure is now greatly increased, then the drag force will exceed the objects weight, and serve to decelerate (slow) the object, until such point as the drag force , now reducing due to the decreasing speed, again ultimately equals the objects weight, and it will reach a terminal a velocity again, but this time at a velocity much slower than the terminal velocity prior to the chute being opened. The actual calculation of the objects acceleration during its downward fall is solved by a very complex (to me, not being skilled in the calculus) differential equation, which shows that the object approaches, but never quite reaches, the calcualted terminal velocity.
 
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