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Homework Help: 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?



    F=ma


    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.
     
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