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Conservation of energy of a pulley problem

  1. Feb 26, 2009 #1
    1. The problem statement, all variables and given/known data

    A 75.0 kg mass sits on an inclined plane, and a rope passing over a pulley at the top connects it to a hanging 125 kg mass. The pulley is frictionless and its mass is negligible. The coefficient of friction between the 75.0 kg block and the plane is 0.143. The system is released from rest, and after dropping 8.75 m, the 125 kg mass is moving at a speed of 8.00 m/s. What is the angle of inclination of the plane from the horizontal?

    2. Relevant equations

    Not entirely sure if these are correct...

    μ = F /N
    KE = ½ mv2
    PE = mgh

    3. The attempt at a solution[/b

    Okay...apparently this is a transcendental equation????
    Ive tried using the formula μ = F /N above and I'm coming out with two missing variables. I'm just completely stumped. Any help would be greatly appreciated.
     
  2. jcsd
  3. Feb 26, 2009 #2

    Delphi51

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    Drop a force vector from the block and label it mg.
    Replace this vector with two new ones, one along the ramp, the other straight into the ramp, so that the combination of these two ends where the mg one did.
    Use trigonometry to write expressions for each of the two vectors in terms of mg and the angle of the ramp.
    Write an expression for the friction force in terms of mg and the angle.

    Consider the block, the rope and the hanging mass as one system. Write that the sum of the forces on it equals ma.

    You can find the acceleration from the information in the last sentence, so only the angle should remain unknown and you can use that ma equation to find it.
     
  4. Feb 26, 2009 #3
    thanks for responding so quickly.
    from what i can understand, i've drawn three vectors, the vertical one with a value of mg, being opposite of theta. The other two vectors touch each end of the mg vector and form a triangle. Would these vectors have a designated value?? if so, is it given in the equation? If not, how would I write expressions for the other two vectors?

    Cheers
     
  5. Feb 26, 2009 #4

    Delphi51

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    ramp.jpg

    It is important to realize that the two blue vectors are equal to the red one and take its place.
    You are expressing the force of gravity as a vector along the ramp (F parallel) plus a vector into the ramp (Normal force).

    Carefully figure out which angle in the small triangle is theta.
    Use sine and cosine to find the magnitudes of the blue vectors.
     
  6. Feb 26, 2009 #5
    Ok.
    This graph is very helpful.
    First off, the angle opposite of Fparallel would be theta?

    assuming so...

    Fparallel = mg sin(theta)
    N = mg cos(theta)

    then, even though the equations have 2 undefined variables, I can write a formula for friction? Would I use μ = F /N ?
     
  7. Feb 26, 2009 #6

    Delphi51

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    Good work - carry on!
    Only the angle is unknown.
     
  8. Feb 26, 2009 #7
    would it be...

    μ = (mg sin(theta)) / (mg cos(theta))

    ?
     
  9. Feb 26, 2009 #8
    or would the F component of μ = F/N be (mass)(acceleration) of the hanging object?
     
  10. Feb 26, 2009 #9

    Delphi51

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    Friction = uN = u*mg cos(theta)
     
  11. Feb 26, 2009 #10
    okay so...

    Friction = 105.105 cos(theta)

    and then i have to set the net force on the object equal to ma?
    Fg+Fparallel+N+Friction = ma ??????
     
  12. Feb 26, 2009 #11

    Delphi51

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    Not Fg - you were supposed to forget that and use N and F// in place of Fg.
    Oh, of course you DO include the Fg due to the hanging mass - right.
    Not N - it is perpendicular to the direction of motion and we are only doing the forces that cause the acceleration in the direction things move.
     
  13. Feb 26, 2009 #12
    okay. I did the calculations and I get:

    735 sin(theta) + 105.105cos(theta) = 446.45

    honestly, I'm not really sure how to get theta by itself
     
  14. Feb 26, 2009 #13

    Delphi51

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    That 446.45 doesn't look right. mg = 125*9.81
    And you are missing the ma!
    Should be
    - 735 sin(theta) - 105.105cos(theta) + mg = Ma
    since the friction and the Fparallel are to the left while the mg is to the right.
    The acceleration will come out positive if it goes to the right.
    You can't solve this yet because you have two unknowns.
    Have to work on the
    end of it. Can you deduce the acceleration from this?

    Yes, we are going to have trouble finding the angle even after we get acceleration.
    Can you use a graphing calculator to find it?
     
  15. Feb 26, 2009 #14
    yes, the acceleration should be 3.66, correct?
     
  16. Feb 26, 2009 #15

    Delphi51

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    Yes, 3.66.
     
  17. Feb 26, 2009 #16
    ok, so then the equation would be...

    950.5 = 105.105 cos(theta) + 735 sin(theta) ?
     
  18. Feb 26, 2009 #17

    Delphi51

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    I'm getting 495 = 105 cos(theta) + 736 sin(theta) ?
    For the m in ma, you must use the total mass of all the moving parts - 200 kg.

    You can solve for the angle using a graphing calculator, a spreadsheet, or trial & error.
    Trial & error works very well on a spreadsheet; a bit slow on a calculator unless you can program it.
     
  19. Feb 26, 2009 #18
    I can't figure it out.
    The calculator that I'm using is giving me a ton of different answers and the program I'm doing my homework on says they're all wrong.
     
  20. Feb 26, 2009 #19
    this is killing me, I'm so close.
     
  21. Feb 26, 2009 #20

    Delphi51

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    Let's try the spreadsheet.
    In cell A1 put "degrees". In B1 put "radians". In C1 put "should be 495".
    In A2 put 10. In B2 put =A2*3.14159/180.
    In C2 put =105*cos(B2) + 736*sin(B2).
    You should now have 231 in C2.
    Try different angles in A2 until you get 495 in C2.
     
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