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Friction incline with spring

  1. Mar 31, 2015 #1
    1. The problem statement, all variables and given/known data
    80kg box is on a 20 degree incline against a spring and is released from rest. Coefficient of static friction is .25. Determine max and in values of the intial compression force in the spring so that block will not slip on release.

    Then calculate the magnitude and direction of friction acting on the block if the spring compression force is 200N

    2. Relevant equations

    FBD: http://i57.tinypic.com/2hs7orr.png

    (80kg)*(9.81m/s^2) = 784.8N
    Fnormal = (784.8 N) *(cos20) = 737.47 N
    Fwparallel = 784.8sin20 = 268.42 N
    Fwperpendicular = 784.8 * cos20 = 737.47 N

    Ffriction = (268.42)*(.25) = 67.11N



    3. The attempt at a solution
    See 2



    I'm not sure where to go with the spring. Don't we need a spring constant for this?
     
  2. jcsd
  3. Apr 1, 2015 #2

    Svein

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    Say again?
     
  4. Apr 1, 2015 #3
    Whoops.

    Ffriction = (737.47)*(.25) = 184.37N.
     
  5. Apr 1, 2015 #4

    Svein

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    So, how much is left for the spring to do?
     
  6. Apr 1, 2015 #5
    268.42 - 184.37 = 84.05N
     
  7. Apr 1, 2015 #6

    haruspex

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    No. You are only asked for the force, not the compression length.
     
  8. Apr 1, 2015 #7
    I'm not understanding what it's asking for "min and max values of intial compression force for the spring" at all. I just can't turn the words into something meaningful to me.
     
  9. Apr 1, 2015 #8

    SammyS

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    If the spring is compressed enough it may exert enough force to push the block up the incline. What is the maximum force the spring can exert, without pushing the block up the incline?

    If the incline is steep enough, the block will need an external force applied so that it doesn't slide down the incline. In that case, what minimum force must the spring exert?
     
  10. Apr 1, 2015 #9
    Wouldn't the maximum force it can exert be what I solved above since that is enough to stop the box from moving down further?
     
  11. Apr 1, 2015 #10

    haruspex

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    No, there are two different constraints. If the force is too small, the block will slide down. If too great, it will slide up. Think about which way friction acts in each case.
     
  12. Apr 1, 2015 #11
    Friction will be opposite the direction of motion.
     
  13. Apr 1, 2015 #12

    haruspex

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    ... or, for static friction, opposes the motion that would occur were there no friction.
    So what equations do you get for the two extremes?
     
  14. Apr 1, 2015 #13
    Ok, I'm still confused how to even go about this.

    The only thing I'm familiar with in regards to springs is hooke's law.
     
  15. Apr 1, 2015 #14

    haruspex

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    The fact that it is a spring is irrelevant here. There is a block of known mass on a slope with known angle and known friction. A force is applied to it in the up slope direction. What is the minimum value of the force to prevent its slipping down the slope? What is the maximum to avoid pushing it up the slope?
     
  16. Apr 1, 2015 #15
    So it's sliding down the slope with a force of 268.42 N, right?

    Because of friction that force is reduced to 84.05 N so the spring would have to at least support that to prevent it from slipping down further.

    But in that case wouldn't the minimum have to be the maximum be also equal to the minimum? :S
     
  17. Apr 1, 2015 #16

    haruspex

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    Think again about the direction of friction when the force from the spring is very large.
     
  18. Apr 1, 2015 #17
    If the force is large the spring pushes it up so the friction will be moving down the slow while the block moves up.
     
  19. Apr 1, 2015 #18

    haruspex

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    Right, so what equation do you get for the forces in that case?
     
  20. Apr 1, 2015 #19
    Fspring - Ffriction - Fwparallel

    Fspring - 184.37 - 268.42
    Fspring = 452.79 N
     
  21. Apr 1, 2015 #20

    haruspex

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    Good. All ok now?
     
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