Drag force = Mass * deacceleration?

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Discussion Overview

The discussion revolves around the calculation of drag force on a spherical ball moving underwater, specifically whether it can be determined solely from the mass and deceleration of the ball. The scope includes theoretical considerations, practical applications, and the relationship between different forces acting on the object.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that drag force can be calculated using the formula F=ma, where deceleration is treated as negative acceleration.
  • Others argue that the mass times acceleration yields the net force, which does not account for weight, thus requiring additional calculations to find drag force.
  • A participant expresses confusion about the applicability of F=ma outside of terminal velocity, questioning the need for the more complex drag equation.
  • One participant suggests that the drag force can be simplified to mass times deceleration plus weight, while another emphasizes the importance of considering changing deceleration due to varying velocities.
  • There is a discussion about comparing two equations for drag force, with questions about whether to use average or instantaneous velocity in the calculations.
  • A later reply introduces the idea that buoyancy may also play a significant role in the forces acting on the ball, depending on its material density.

Areas of Agreement / Disagreement

Participants express differing views on the relationship between mass, deceleration, and drag force, with no consensus reached on whether drag can be calculated solely from mass and deceleration. The discussion includes competing models and interpretations of the forces involved.

Contextual Notes

Participants highlight limitations in their calculations, such as the need to account for weight and buoyancy, and the variability of drag force depending on velocity and time. There are unresolved questions regarding the applicability of different equations for drag force.

Who May Find This Useful

This discussion may be useful for those interested in fluid dynamics, physics of motion, and the complexities of force calculations in experimental settings.

Xenekaro
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For the motion of a spherical ball under water, can its drag force be calculated by knowing only its mass and deceleration?
 
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Yes. F=ma. Deceleration and acceleration are essentially the same thing.
F=Force
m=mass
a=acceleration.
 
Yes. Deceleration is just a negative acceleration, so it will make F negative, and thus will be the opposing, unbalanced force (in this case the drag).
 
Realize that whenever you calculate mass*acceleration for an object, you are calculating the net force on the object.
 
ok awesome. I thought F=ma was only valid for terminal velocity so I was wondering about the drag equation.Thanks again guys!
 
No. The mass times the acceleration in this case will give you drag minus weight (draw a free body diagram). You will have to add the weight to the net force to get the drag.
 
Xenekaro said:
ok awesome. I thought F=ma was only valid for terminal velocity so I was wondering about the drag equation.Thanks again guys!

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.
 
boneh3ad said:
No. The mass times the acceleration in this case will give you drag minus weight (draw a free body diagram). You will have to add the weight to the net force to get the drag.

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.
 
You have the weight acting downward, which is -mg and the drag acting upward opposing the motion, which is F_d If you sum the forces, you get:
\sum F = F_d - mg = ma
Solving that for F_d gives you:
F_d = m(a+g)

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.
 
  • #10
Ok thanks for the explanations! I just have 1 more question. Say if I want to compare:
Fd=m(a+g) and
Fd = 0.5pv2acd

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 Fd of the two equations. The second equation refers to the theoretical force of drag in my experiment and the first one is the experimental Fd.
 
  • #11
In both equations you have instantaneous values. Instantaneous acceleration in the first one and instantaneous velocity in the second one. And they provide the instantaneous value of the drag force, which is variable.
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?
 
  • #12
I would suggest you likely have 3 forces equate to the deceleration, not 2.

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

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