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Drag force = Mass * deacceleration?

  1. Oct 21, 2011 #1
    For the motion of a spherical ball under water, can its drag force be calculated by knowing only its mass and deceleration?
  2. jcsd
  3. Oct 21, 2011 #2
    Yes. F=ma. Deceleration and acceleration are essentially the same thing.
  4. Oct 21, 2011 #3
    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).
  5. Oct 21, 2011 #4

    Doc Al

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    Realize that whenever you calculate mass*acceleration for an object, you are calculating the net force on the object.
  6. Oct 21, 2011 #5
    ok awesome. I thought F=ma was only valid for terminal velocity so I was wondering about the drag equation.Thanks again guys!
  7. Oct 21, 2011 #6


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    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.
  8. Oct 21, 2011 #7


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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.
  9. Oct 21, 2011 #8
    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.
  10. Oct 21, 2011 #9


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

    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.
  11. Oct 21, 2011 #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.
  12. Oct 21, 2011 #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?
  13. Oct 21, 2011 #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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