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Bob of mass

  1. Jul 3, 2012 #1
    1. The problem statement, all variables and given/known data
    A bob of mass 0.35 kg is attached by a string to the roof of a train that is accelerating rightwards and the mass is inclined at an angle of 15° to the left. Show that the acceleration of the train is 2.63 ms^-2.


    3. The attempt at a solution
    I don't know how to exactly start doing this because I don't know what concepts I should use here. Please help! :)
     
  2. jcsd
  3. Jul 3, 2012 #2

    Doc Al

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    Use Newton's 2nd law. What forces act on the bob? Analyze horizontal and vertical components.
     
  4. Jul 3, 2012 #3
    How do you know what to use?? I was thinking of using mg sin θ in the first place.
     
  5. Jul 3, 2012 #4

    Doc Al

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    Start by drawing a free body diagram of the bob showing all forces acting on it. What forces act on the bob?
     
  6. Jul 3, 2012 #5
    Weight and tension of the string!
     
  7. Jul 3, 2012 #6

    Doc Al

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    Right!

    Now apply Newton's 2nd law. I suggest analyzing horizontal and vertical components separately.
     
  8. Jul 3, 2012 #7
    Alright I'll try! Can I ask you another question over here??
     
  9. Jul 3, 2012 #8

    Doc Al

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    Ask away.
     
  10. Jul 3, 2012 #9
    http://via.me/-2ostlh4
    http://via.me/-2ostlh4 [Broken]
    Here's the question and I did part of it and I don't know how to continue!
     
    Last edited by a moderator: May 6, 2017
  11. Jul 3, 2012 #10

    Doc Al

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    You took moments about the bottom end of the ladder, which is good. What about the other conditions for equilibrium?

    (It's a bit difficult to read your work from the diagram, so write it out here if you want further critique.)
     
  12. Jul 3, 2012 #11
    Other condition will be the upward forces = downward forces?? Am I correct??
    My working was W*0.5L cos θ = Y*L sin θ + f* L cos θ and that frictional force cause a moment too??
     
  13. Jul 3, 2012 #12

    Doc Al

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    Yes.
    Looks good.
     
  14. Jul 3, 2012 #13
    so that means W= X+f?? And I just subt w into the above equation I did??
     
  15. Jul 3, 2012 #14

    Doc Al

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    Yep.
     
  16. Jul 3, 2012 #15
    X=W
    X=W/2 + W/2
    Find value of W/2 in term of Y.
     
  17. Jul 3, 2012 #16

    Doc Al

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    Don't neglect the friction force from the wall.
     
  18. Jul 3, 2012 #17
    Taking as No friction on the wall,
    If we take moment at the base,
    (L/2)WCosθ=LYSinθ
    W/2=YTanθ.

    Then we can proof the equation.
     
  19. Jul 3, 2012 #18

    Doc Al

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    But that's not the problem being discussed here. The wall is rough.
     
  20. Jul 3, 2012 #19
    Yes i make a mistake there.
    With or without friction the equation still valid.
     
  21. Jul 3, 2012 #20

    Doc Al

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    How can you say that?
     
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