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Dynamics problem(pulley)

  1. Apr 21, 2008 #1
    1. The problem statement, all variables and given/known data

    http://14.photobucket.com/albums/a322/guitaristx/phy.jpg

    Just add the http for the image above..

    2. Relevant equations

    Please let me know how the FBD looks and I think im having trouble with my coordinate system( direction and signs)

    The answer for acceleration is 5.21m/s^2...and the tension is 215N.....I cant seem to get the equation to work out to the correct answer...any help and direction would be appreciated...

    3. The attempt at a solution

    http://i14.photobucket.com/albums/a322/guitaristx/Scan10005.jpg

    just add the http for the image above...

    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
    Last edited by a moderator: Apr 21, 2008
  2. jcsd
  3. Apr 21, 2008 #2

    Doc Al

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    Staff: Mentor

    Your FBD looks OK. Pick a direction parallel to the plane as positive and stick to it, such as: Down = positive; up = negative.
     
  4. Apr 21, 2008 #3
    I chose up as positive and down as negative and for the equations that I have for the first FBD...and the only way I come up with the correct Tension answer is by making the acceleration negative....and for the second FBD my calculations came up with a completly different tension....when the tensions should obviously be the same
     
  5. Apr 21, 2008 #4

    Doc Al

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    Staff: Mentor

    Then why is tension marked -T?
     
  6. Apr 21, 2008 #5
    The tension is upward....but the problem states that P is a force pulling block A downward...so I figured Tension would be negative because block A is being pulled downward(left) hence the Negative...
     
  7. Apr 21, 2008 #6

    Doc Al

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    Staff: Mentor

    I don't quite understand that reasoning. The tension is upward and thus positive. (Ropes can't push--only pull.)
     
  8. Apr 21, 2008 #7
    Last edited by a moderator: Apr 21, 2008
  9. Apr 21, 2008 #8

    Doc Al

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    Staff: Mentor

    Only because they chose down to be positive.
     
  10. Apr 21, 2008 #9
    ok...I think i see now what you mean...let me try some new calculations...
     
  11. Apr 21, 2008 #10
    i changed the T to positive....but when I plug in 5.21 for the acceleration I get a -215 for the Tension??
     
  12. Apr 21, 2008 #11

    Doc Al

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    Staff: Mentor

    Show me the equations you used.

    Don't forget that the acceleration will be negative (down the incline).
     
    Last edited: Apr 21, 2008
  13. Apr 21, 2008 #12
    heres the equ.

    T-P+Fr-mgSin(30) = ma

    I tried the neg. accel. and it gave me an answer even further off...
     
  14. Apr 22, 2008 #13

    Doc Al

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    Staff: Mentor

    As long as you have the acceleration negative, that equation should work out OK. What numbers did you plug in? What did you use for Fr?
     
  15. Apr 22, 2008 #14
    Thanks for the help on the first one, it finally worked out.....

    But for the second FBD I came up with

    T-Fr-mgSin30 = ma

    But for Friction there are two surfaces. So I have to find both normal forces acting on block A and block B

    heres the numbers I came up with

    Friction for B = .10(15)(9.8)Cos(30)= 12.7N

    Friction for A = .10(30)(9.8)Cos(30)= 25.4N

    So

    T-12.7N-25.4-73.5N = 15 (5.21)

    the acceleration wouldnt be negative now would it?

    but now T is not coming out to 215 as it should....do you see any errors in my calculations?
     
  16. Apr 22, 2008 #15

    Doc Al

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    Staff: Mentor

    Looks good.
    Careful: What's the normal force of the bottom block against the incline?
     
  17. Apr 22, 2008 #16
    ohh.. I did miss that, the total mass of both blocks acting as N on the lower block.

    The numbers worked out.

    So to solve the system I am trying this

    T-372 = 30(-a)
    T-177 = 15 (a)


    So can I cancel the acceleration to find T?..or is there a better way....
     
  18. Apr 22, 2008 #17
    *T-137 = 15a
     
  19. Apr 22, 2008 #18
    Got it!...thanks for all the help:approve:
     
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