Objects of equal mass, but different terminal velocities?

[TelusPig]

Why do different objects having the same mass but different projected surface areas have different terminal velocities? On wikipedia, the formula for terminal velocity is Vt = sqrt (2mg/pAC) http://en.wikipedia.org/wiki/Terminal_ve ...

I understand from this formula that the terminal velocities would be different since objects with a higher surface area would have lower terminal velocities.

What I don't understand is that if both objects have the same mass, they have they same gravitational force Fg. Terminal velocity means that the net force of the drag force (Fd) and Fg is 0. So in order to reach the terminal velocity, wouldn't the drag have to be the same for each object for Fg to equal -Fd, thus making the terminal velocities the same in each case ?

So... I don't know it seems contradictory depending on how I look at this problem? (I'm not good with physics )

 

[LowlyPion]

Why do different objects having the same mass but different projected surface areas have different terminal velocities? On wikipedia, the formula for terminal velocity is Vt = sqrt (2mg/pAC) http://en.wikipedia.org/wiki/Terminal_ve ...

I understand from this formula that the terminal velocities would be different since objects with a higher surface area would have lower terminal velocities.

What I don't understand is that if both objects have the same mass, they have they same gravitational force Fg. Terminal velocity means that the net force of the drag force (Fd) and Fg is 0. So in order to reach the terminal velocity, wouldn't the drag have to be the same for each object for Fg to equal -Fd, thus making the terminal velocities the same in each case ?

So... I don't know it seems contradictory depending on how I look at this problem? (I'm not good with physics )

What is the "A" in your equation?

 

[TelusPig]

Sorry:
m = mass,
g = 9.81m/s²,
p = density,
A = projected surface area,
C = drag coefficient

 

[LowlyPion]

Sorry:
m = mass,
g = 9.81m/s²,
p = density,
A = projected surface area,
C = drag coefficient

Thanks. I know what A means. What I was getting at was what does its effects mean to you in the equation if the m*g is the same?

 

[D H]

Why do different objects having the same mass but different projected surface areas have different terminal velocities? On wikipedia, the formula for terminal velocity is Vt = sqrt (2mg/pAC)

The answer to your question is in that equation. Think about it.

 

[TelusPig]

Well I know that having a bigger area wold mean that the denominator increases, so that the overall result is smaller, as I said in my first post "I understand from this formula that the terminal velocities would be different since objects with a higher surface area would have lower terminal velocities."

Where I seem to find a contradiction (seem because maybe my reasoning is wrong in the following paragraph) is when I reason that the terminal velocity means that there is no net force acting on the object. Both objects have the same gravitational force since they have the same mass. For the net force to be zero, the upwad drag force would have to be equal to mg. Since both mg's are the same, aren't the drag forces the same? Both objects accelerate until the drag force equals the pull of gravity... so wouldn't both objects have the same terminal velocity?

Maybe my reasoning in the 2nd paragraph is not correct, which is what I'm trying to figure out

 

[D H]

Your reasoning is not correct. Yes, the net force is zero at terminal velocity. You are implicitly assuming that the thing contribution to the drag force is velocity. That is obviously incorrect. Think about it this way: Why do skydivers use parachutes?