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24: A ball is falling at terminal speed in still air. The forces acting on it are upthrust, viscous drag and weight. What is the order of increasing magnitude of these 3 forces?

Answer :upthrust  < viscous drag  < weight

Could some one please explain why in still air upthrust of a ball is smaller than the viscous drag ?

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  • $\begingroup$ I second John Rennie's answer. "Buoyant force" is what "upthrust" evokes for me in this context, but it still is not altogether clear. I see from the Google search it's often used to mean that in connexion with ships, although I haven't heard the word used thus before. I'm guessing it's probably more precisely defined somewhere in supporting text for your question set. $\endgroup$ Jul 9, 2015 at 1:30

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Although it isn't absolutely clear from the question I would guess that the up thrust is the force due to Archimedes' principle, so the up thrust is equal to the weight of air displaced by the object. Since the object is not accelerating, the net force on it must be zero so:

$$ mg = \text{drag} + \text{up thrust} $$

Obvious both the drag and up thrust must be less than or equal to $mg$, but their relative size will depend on the density of the ball. You would have to do the calculation to determine which was bigger. Unless your ball is something exotic, like a hot air balloon, a back of the envelope calculation should make it pretty obvious that the up thrust is a tiny force.

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Up thrust is approximately constant: $$U=\rho\cdot g\cdot H$$

  1. At $H$ height, air density low = up thrust
  2. At $H-x$ (when object travel certain height), air density increase=up thrust

Comparing statements 1 and 2, up thrust is constant at $H$ height and $H-x$ BUT drag or viscous force increases as velocity of an object increases, so viscous drag is greater than up thrust.

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  • $\begingroup$ Welcome to Physics! Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. I've added it in for you, but you may want to check the mark-up to ensure I've correctly interpreted your statements. $\endgroup$
    – Kyle Kanos
    Jul 8, 2015 at 17:24
  • $\begingroup$ @user85417 Please do not reply to over-a-year-old questions which already have correct answers with totally incorrect answers invoking irrelevant physical principles. In this case it is sufficient to notice that a beach ball is a good counterexample where upthrust > viscous drag, proving both you and the quoted-answer wrong, and John Rennie above right. The air density has nothing to do with anything here. $\endgroup$
    – CR Drost
    Jul 8, 2015 at 18:00

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