1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Force of air resistance

  1. Jul 31, 2010 #1
    If F is the force of air resistance on an object with mass m moving at a constant velocity, which of the following best describes the acceleration of the object when the force of air resistance is reduced by a factor of 4?


    It says the answer is equal to (3/4)F/m = a

    Can someone explain why its the 3/4?
  2. jcsd
  3. Jul 31, 2010 #2
    It is an application of Newton's 2nd law.

    Write down the sum of all forces in the case where an object is moving at constant velocity:

    [tex] \sum F = F_{\mathrm{thrust}} - F_{\mathrm{res}} = 0 [/tex],

    i.e. the object is not accelerating or decelerating, thus the thrust must equal the wind resistance. Now write down the sum of all forces when the resistance force suddenly drops by 1/4:

    [tex] \sum F = ma = F_{\mathrm{thrust}} - F_{\mathrm{res}}/4 [/tex].

    The thrust force remains the same as before which was equal to the old wind resistance, however the resistance has now dropped a quarter meaning the thrust is more powerful than the new wind resistance which results in an acceleration. From the first part you have that [tex]F_{\mathrm{thrust}} = F_{\mathrm{res}}[/tex], so plugging this into the second part yields the result.
  4. Jul 31, 2010 #3
    So, qualitatively speaking:
    resisting force = driving force since its at constant velocity, right?
    so if the resisting dropped from F to 1/4F... then the net force would be 3/4F in favour of the driving force. ...which means the object is accelerating.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook