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Magnitude of static friction force

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  1. Feb 20, 2016 #1
    1. The problem statement, all variables and given/known data
    A 500 kg ox runs and speeds up by 3 m/s2, but is held back by a 120 kg plow that is attached to its back. Assuming the coefficient of static friction between the horse's hooves and the soil is 0.73, what is the magnitude of the static friction force the soil exerts on the horse's hooves (not maximum static friction) which explains the horse's acceleration?

    2. Relevant equations
    Fsfr ≤ μFN
    FN=mg
    3. The attempt at a solution

    FN = 500(9.8) = 4900 N

    Fsfr ≤ 0.73(4900)
    max static friction = 3600 N

    ^I got up until here. I'm confused about finding the magnitude of the static friction force...do I have to find the force the plow exerts on the horse and subtract that from 3600 N? If so, how would I find that force?
     
    Last edited: Feb 20, 2016
  2. jcsd
  3. Feb 20, 2016 #2

    PhanthomJay

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    You could find the force between the ox and plow, but it is not necessary to solve this problem.
    You have a typo...the acceleration is 3 m/s^2.
    You are missing a very important relevant equation. Which one?
     
  4. Feb 20, 2016 #3
    I have acceleration and mass, so would I use Newton's second law a = ∑F/m?
     
  5. Feb 20, 2016 #4
    3 = (- Fplow on ox + Fsfr )/500
    1500 = (- Fplow on ox + Fsfr)
    1500N would be the net horizontal force right? Would that be the magnitude of Fsfr ?
     
  6. Feb 20, 2016 #5

    PhanthomJay

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    No.
    You have calculated the net horizontal force acting on the ox correctly, but the net horizontal force acting on the ox includes the sum of both the static friction force and the force between the ox and plow. So you would have to calculate the force between the ox and plow to find the friction force. But instead, apply newtons law to the entire ox-plow system. Then you don't need that force between ox and plow because it is internal to the system.
     
  7. Feb 20, 2016 #6
    If I combine the masses so it's one system and then apply N2L:
    3 = F / 620
    so 1860 = Fsfr ?
     
  8. Feb 20, 2016 #7

    PhanthomJay

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    Yes. But don't forget units. The static friction force is 1860 N. (If you wish, for extra credit, you might want to calculate the force between that ox and plow).
     
  9. Feb 20, 2016 #8
    I got it, thanks :)
     
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