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Mechanics problem

  1. Aug 25, 2012 #1
    Hello everybody ,
    I am stuck at this question :frown:,
    1. The problem statement, all variables and given/known data

    Two blocks of masses m=1kg and M =2kg are connected by a non-deformed light spring .They are lying on a rough horizontal surface . The coefficient of friction is 0.4 ,What minimun constant force has to be applied in the horizontal direction to the block m , in order to shift the order block M .

    2. Relevant equations
    I came up with work energy conservation theorem
    F.x = 1/2 kx^2 +f1.x + 1/2mv^2
    (where F is external force
    f1is force of friction on m. and v is velocity of mass m)
    F-kx-f1=ma1
    and when block M is just about to move kx=f2
    (here f2 is force of friction on M

    , its leading me nowhere .
    Any help?
     
  2. jcsd
  3. Aug 25, 2012 #2

    ehild

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    Work with the forces instead of energy.

    ehild
     
  4. Aug 25, 2012 #3
    After working with forces I get
    F-umg - uMg = ma

    Now how do i find a here?
     
  5. Aug 25, 2012 #4

    ehild

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    A has to be just a bit greater than zero, so practically zero.


    ehild
     
  6. Aug 25, 2012 #5
    But this is block 1 which will be moving , not block 2 which will be stationary till the last moment
     
  7. Aug 25, 2012 #6

    ehild

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    What does "shift" mean? You need to made the second block moving. The force F has to be enough to move the first block and stretch the spring. How the first block will move and what is the maximum stretching you can have with the applied constant force F? That stretching of the spring must be enough to overcome the friction of the second block.

    ehild
     
    Last edited: Aug 25, 2012
  8. Aug 25, 2012 #7
    I am really not getting ehild can you frame some equations ?
     
  9. Aug 25, 2012 #8

    ehild

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    Newton's second law for the first mass?

    ehild
     
  10. Aug 25, 2012 #9
    let me try again .
    Thanks
    Ehild
     
    Last edited: Aug 25, 2012
  11. Aug 25, 2012 #10

    ehild

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    I reconsidered your attempt with work and energy, it might be a better approach as mine one.

    Should the block of mass m have non-zero velocity when the bigger block starts moving?

    But it would be useful to perform an experiment. What happens when you exert some force on mass m, bigger than f1=μmg?

    ehild
     
  12. Aug 25, 2012 #11
    Does anything change in the problem if the spring is replaced with a rigid link?
     
  13. Aug 25, 2012 #12

    ehild

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    Yes, it will.

    ehild
     
  14. Aug 25, 2012 #13
    yes ehild , The block with mass m will be moving and spring will be extending .
    When it reaches the limit ie kx=uMg the block M will move .
    I am not getting how to go about it .which equations to use? confusing!
     
  15. Aug 25, 2012 #14

    ehild

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    Does the mass m need to move at the instant when kx=uMg?

    ehild
     
  16. Aug 25, 2012 #15
    I'll rephrase my question: does anything change with regard to the magnitude of the force if the link is rigid?
     
  17. Aug 25, 2012 #16

    ehild

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    Yes. The smallest force which is enough just to move the bigger mass out of its steady state is less than the force needed to move both blocks together with the same velocity (and acceleration).

    See picture. What happens when you connect the weight?

    ehild
     

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    Last edited: Aug 25, 2012
  18. Aug 25, 2012 #17
    I do not see how. To move the bigger mass, the spring has to pull it with some force F. To make the spring do it, it must be extended till its reaction force is equal to F. How is that different from a rigid link?
     
  19. Aug 25, 2012 #18

    ehild

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    In case of a rigid link both blocks have to move together. In case of the spring the block pulled by F will start to move if F overcomes friction, but the second block stays in rest untill the displacement x of the first block yields enough force to overcome the friction of the second block. The question is what constant force is needed to displace the first block so far.

    ehild
     
  20. Aug 25, 2012 #19
    I Think this is the solution....
    for the body to accelerate horizontally then the forces on m are
    External Force F
    Frictional force -mgf
    Force due to the spring -kx
    Force Due Friction on M Mgf


    Mgf=2mgf
    F-(mgf+kx)>2mgf
    F>3mgf+kx
    F>12+kx If k is negligible then
    F>12 Newtons
     
  21. Aug 25, 2012 #20
    Ehild , when this force f is doing work .
    Will we say F.xo ( where xo is the actual displacement )
    or F.x(x is displacement had there been no friction), because of the work done in overcoming friction ?
     
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