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- Thread starter Xenekaro
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F=Force

m=mass

a=acceleration.

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Doc Al

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sophiecentaur

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Accceleration will be zero at terminal velocity (it has reached a steady velocity)

The drag force will be equal to the weight force under that condition.

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Ya, I was worried about this. So supposing I know that the (mass*deceleration) value is x and the weight of the object is y. Would the drag force be x+y? Or are there other factors to be considered?

I find the drag force calculations to be too simple to actually warrant the need to formulate the more complex drag equation. I mean why worry about drag coefficient if you can calculate the drag just by knowing the object's mass and deceleration rate.

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[tex]\sum F = F_d - mg = ma[/tex]

Solving that for [itex]F_d[/itex] gives you:

[tex]F_d = m(a+g)[/tex]

Xenekaro said:I find the drag force calculations to be too simple to actually warrant the need to formulate the more complex drag equation. I mean why worry about drag coefficient if you can calculate the drag just by knowing the object's mass and deceleration rate.

Probably because you almost never know the deceleration of a given object. That deceleration will also change depending on how long an object has been falling since the drag is dependent on the velocity of the object. It also isn't simple when you are talking about something with more forces acting on it like an aircraft or a car.

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F

F

Would the velocity in the second equation have to be the average or instantaneous velocity? Average velocity refers to the mean of initial and final velocity which I used to calculate acceleration in the first equation.

Instantaneous velocity would probably work better for equation II but I am trying to compare the F

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nasu

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An average acceleration won't be too useful, in my opinion. anyway, averaged over what? The whole motion or just the portion before reaching terminal velocity?

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Depending on the ball material (density), there will likely be a significant buoyancy term as well.

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