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Mechanics, Static friction and springs

  1. Nov 14, 2009 #1
    1. The problem statement, all variables and given/known data
    A block of mass m lies on a rough surface, inclined at angle theta to the horizontal. It is attached to a spring of force constant; the other end of the spring is fixed to a point on the table below the inclined block. The co-eff of static friction is mu-s and kinetic friction is mu-k. friction between the table and spring is negligible.

    a) show that the friction force is 0 when the block is stationary and the spring is compressed by l sin theta where l = mg/k


    2. Relevant equations
    f = kl (spring force)
    Fs = mu-s x Fn (friction force as a product of reaction force and co-eff of friction)



    3. The attempt at a solution

    i first drew a diagram and said that the system is in equilibrium. If the block is not moving, the the compression force from the spring + the friction force (acting against this compression) must = the wait of the block down the inclined plane.

    So Fs = mu-s x Fn

    Fn = mg sin theta, so Fs = mu-s x mg sin theta.

    Therefore sum of forces = 0

    kl (spring force) + mg sin theta + mu-s mg sin theta = 0

    But I do not understand how the friction force can = 0.

    The closest result I have been able to get (after finally assuming friction force = 0) is l = mg sin theta/k

    I don't know what I'm doing wrong because I resolved the forces in the directions they are acting?/
     
  2. jcsd
  3. Nov 14, 2009 #2
    If I assume that the compression force works together with the friction force to oppose the motion of the b lock down the slope, I could assume that force from the spring >>>>> friction force? Therefore it is safe to assume that Friction force is 0 when the spring is in play, although I still can't work out how they got to their expression..

    l sin theta, where l = mg/k. I only get l = mg sin theta / k..

    It would help if I knew what l stood for, at first I thought it was a distance/length but I'm not so sure now...
     
  4. Nov 14, 2009 #3

    tiny-tim

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    Hi Unto! :smile:

    (have a theta: θ :wink:)
    You're making this very complicated! :cry:

    You should have started "Let the distance be x".

    Then you would have got x = mg sinθ / k (instead of your l = mg sinθ / k),

    and therefore x = l sinθ, where l = mg/k. :wink:
     
  5. Nov 14, 2009 #4
    OMG!!!!!

    I literally thought I was stupid and was going out of my head. I kept resolving, I even used F = kx and was like wtf am I doing wrong...

    Thank you!!!!

    There are other parts to this question, I will post them and my attempts if I get stuck if that is ok with you.

    Thank you very much
     
  6. Nov 14, 2009 #5
    The block is now displaced by a distance A, compressing the spring further. For large displacements find the initial acceleration of the block.

    -------------------------------------------------------------------------

    K so the block is going to get pushed up the inclined plane by the force from the spring due to it's compression. I gathered also that the friction is no longer static, but kinetic as it is opposing a now moving object.

    So, SUM of Forces = ma

    kx - (mu-k) mg cos θ - mg sin θ = ma.

    Is this right? The compression force is being opposed by the weight of the block down the plane and the kinetic friction.
     
  7. Nov 15, 2009 #6

    tiny-tim

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    (just got up :zzz: …)
    (have a mu: µ :wink:)

    Yes, that's fine. :smile:

    (but don't forget to write x in terms of A, and to use the first part of the question)
     
  8. Nov 15, 2009 #7
    Ok thank you. 1 Last question.

    Since the acceleration of the block is kA - µk mg cos θ - mg sin θ = ma, I was asked to formulate an expression when acceleration was 0. I obtained:

    x or A = l(µk cos θ + sin θ). This was from re-arranging the above equation once ma = 0.

    I am now asked to find V at this point. If acceleration is 0 at this displacement, should I still use V^2 = U^2 + 2as? Actually no..

    I can't use the others since they have time in them.

    I cam up with V = sqrt(2l(µk cos θ + sin θ)) where l = mg/k from our previous expressions.

    I'm not sure this is correct however.

    When the spring is compressed to a certain displacement and released, the block is accelerated until the spring finishes relaxing (since the force from the spring is a resultant force), and even then, there is still the kinetic friction slowing it down until it stops. So where am I supposed to find V with so many things going on?

    I assumed that S = x when there is no more acceleration. Since there is no more resultant force, and therefore no more acceleration when the spring has finished relaxing. But this would mean I have ignored Kinetic friction

    But hasn't kinetic friction already been included in my above expression for when a = 0?

    Man I'm confused :s. I really love Physics, I just always have to question what I do, I hate just following examples and not knowing why.
     
  9. Nov 15, 2009 #8
    This last question?
     
  10. Nov 16, 2009 #9

    tiny-tim

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    Hi Unto! :smile:
    That's mA'' = kA + constant. :wink:
     
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