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2 blocks on a ramp with a pulley PLS.HELP

  1. Jun 14, 2009 #1
    1. The problem statement, all variables and given/known data

    Two blocks,connected by a string over a pulley, slide along the frictionless ramp shown in the diagram. What is the tension in the string?

    2. Relevant equations

    [tex]\sum[/tex]F=ma

    3. The attempt at a solution
    so i set up an equation for each mass
    m1=3kg
    [tex]\frac{Fg1+Ft}{m1}[/tex]=a
    m1=5kg
    [tex]\frac{Fg2+Ft}{m2}[/tex]=a

    i equated the two since the a would be the same to solve for Ft
    m2[(m1*g*cos30deg)+T1]=m1[(m2*g*cos60deg]+T1]

    when i solve the equation and simplify for T i get 26.9N when i should be getting around 25 N..pls help
     

    Attached Files:

  2. jcsd
  3. Jun 15, 2009 #2

    tiny-tim

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    Hi brunettegurl! :smile:

    (i can't see the diagram yet, but …)

    if I'm guessing right, the accelerations aren't both a …

    one is minus a :wink:
     
  4. Jun 15, 2009 #3

    Doc Al

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    Staff: Mentor

    Also: Gravity acts down the incline while tension acts up--they must have different signs.
     
  5. Jun 15, 2009 #4
    m1=3kg
    [tex]\frac{-Fg1+Ft}{m1}[/tex]=a
    m1=5kg
    [tex]\frac{Fg2-Ft}{m2}[/tex]=a

    i equated the two since the a would be the same to solve for Ft
    m2[(-m1*g*cos30deg)+T1]=m1[(m2*g*cos60deg]-T1]
    does that look better??
     
  6. Jun 15, 2009 #5

    tiny-tim

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    Hi brunettegurl! :smile:
    Same thing applies to the other block. :wink:
     
  7. Jun 15, 2009 #6
    so is my new equation wrong??
     
  8. Jun 15, 2009 #7

    tiny-tim

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    oops!

    oops! must have misread it :redface:

    yes, the first two equations are correct …

    but shouldn't the final one have sin rather than cos?
     
  9. Jun 15, 2009 #8
    why would it be sin instead of cos??
     
    Last edited: Jun 15, 2009
  10. Jun 15, 2009 #9

    ideasrule

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    Try splitting the force of gravity, a downward vector, into a component parallel to and a component perpendicular to the ramp. You'll see that the parallel component is mgsin(a), where a is the angle of inclination of the ramp.
     
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