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Friction/Force Problem

  1. Dec 4, 2015 #1

    kid

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    1. The problem statement, all variables and given/known data
    A 15kg crate slides across the floor. The coefficient of kinetic friction between the crate and the floor is 0.28. How much horizontal force is needed to accelerate the crate at 3.0 m/s/s?

    2. Relevant equations
    F=ma?

    3. The attempt at a solution
    F=?
    m= 15 kg
    u= 0.28
    a= 3.0 m/s^2

    That's as far as I really got. I'm pretty lost on this one. Thanks for your help.
     
  2. jcsd
  3. Dec 4, 2015 #2

    PeterDonis

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    2016 Award

    Staff: Mentor

    How is the force of friction determined? What do you need to know to calculate it?

    Also, there are two forces involved: the force applied to the crate, and the force of friction. How do those two forces combine to determine the actual motion of the crate?
     
  4. Dec 4, 2015 #3

    kid

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    I guess to determine the Fk I would need uk (0.28) and Fn (9.8 m/s/s). So Fk=uk*Fn would be =2.744.... is this sounding right so far or is it off? (thanks a lot for your help by the way.)
     
  5. Dec 4, 2015 #4

    kid

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    I found that F= 41.16 N, and I believe that's right, but I don't know if it's relevant in this equation.
     
  6. Dec 4, 2015 #5

    cnh1995

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    That's the force required to 'overcome' the friction and set the crate in motion. Won't you need a bigger force to accelerate the crate at 3m/s2? Draw FBD and write the equation of motion considering this 'bigger' force.
     
  7. Dec 4, 2015 #6

    kid

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    Would V02 = V2 - 2aΔx be the equation of motion? Would that mean I'd have to find the velocity?

    Ok, so I drew a diagram. I calculated 41.16 and for Fk = μkmg, would I have to have a different kinetic value? Or like, treat the first value like a static value?

    I'm sorry if I'm totally wrong on this one, I'm just growing more and more puzzled.
     

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  8. Dec 4, 2015 #7

    cnh1995

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    F(applied)-F(friction)=Fnet,which will be responsible for the 3m/s2 acceleration.
     
  9. Dec 4, 2015 #8

    kid

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    That is very helpful. My only question is how to find F(friction) at this point. Would it be the 41.16 number I got before or would that be Fnet?
     
  10. Dec 4, 2015 #9

    cnh1995

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    That's the magnitude of frictional force.
    Fnet is what causes the acceleration.
     
  11. Dec 5, 2015 #10
    Static friction isn't an issue here, worry about solving the problem once the crate is in motion.
    You could work this problem thus :
    1) calculate the opposing friction force
    2) calculate the force required to accelerate the crate at the required rate
    3) add the forces
     
  12. Dec 5, 2015 #11

    haruspex

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    To be pedantic, those two forces act in opposite directions. To fix this up, another step is needed:
    1a) calculate the force required to overcome the opposing frictional force.
     
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