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2 Blocks 1 Pulley - First Time Poster

  1. Sep 29, 2011 #1
    Hi! First Time Poster.

    Thank you for helping me with my problem!

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

    A 28.3-kg block (m1) is on a horizontal surface, connected to a 5.5-kg block (m2) by a massless string as shown in the Figure. The pulley is massless and frictionless. A force of 208.9 N acts on m1 at an angle of 29.7o. The coefficient of kinetic friction between m1 and the surface is 0.201. Determine the upward acceleration of m2.

    \
    \
    \
    M1-------O
    |
    |
    M2

    I'm not sure how to post pictures within my message just yet.

    2. Relevant equations

    X and Y components of the force on F1

    Sum F = M * A

    Kf = uK * Fn

    g = 9.8m/s


    3. The attempt at a solution

    **Y+ is upwards and X+ is to the right**

    I found the X and Y components of the Force on M1

    I added M1 and M2... not sure if I need this but just in case.

    And I drew Free Body Diagrams for both M1 and M2

    - M1 has Normal Force acting Y+, which equals its Weight (m*g)... M1 also has Friction and Tension acting X+... the x component of the force acting X-

    - M2 only has Tension acting Y+, and Weight (mg)... I suspect the tension to be equal to the tension... but I'm not sure.

    Thanks! I hope I gave you enough information!
     

    Attached Files:

  2. jcsd
  3. Sep 29, 2011 #2
    I'm not really sure where to go from here. Any direction would be appreciated.

    Thank you!

    Also, can anyone see this post?? I'm not sure what the icon next to the post means.
     
  4. Sep 29, 2011 #3
    Can someone please message me? I've been awake and waiting for quite some time now...
     
  5. Sep 29, 2011 #4

    ehild

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    Show your equations. The figure is very nice.

    ehild
     
  6. Sep 29, 2011 #5
    Ok thank you for writing back!!! One second while i type it up!!!
     
  7. Sep 29, 2011 #6
    Known
    m1 = 28.3
    m2 = 5.5 kg
    theta = 29.7
    f = 208.9
    uK = 0.201

    Want


    Acceleration of M2 in the Y direction...


    ________________

    Fx = 208.9 cos (29.7) = 181.457
    Fy = 208.9 sin (29.7) = 203.501

    M1 + M2 = 28.3 + 5.5 = 33.8 (M3) (I'm not sure if I'll need this)



    _____________________

    For M2

    Sum Fy = M*Ay

    T - m*g = M* Ay


    _____________________

    For M1

    F - T - Friction = M * A

    F - T - uK * Fn = M * A




    _________________

    Thats all I have at the moment... :/
     
  8. Sep 29, 2011 #7
    Does this suffice?
     
  9. Sep 29, 2011 #8

    ehild

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    Use M1 and M2, respectively, in the equations for M1 and M2. They are not the same, and what is m in the equation for the hanging mass?
    The length of the string does not change, so the magnitude of the vertical acceleration of M2 is the same as the horizontal acceleration of M1.
    When you corrected the equations, add them: T will cancel and you can solve for the common acceleration a. Show it.

    ehild
     
  10. Sep 29, 2011 #9
    M1: F-T-ukFn = m a

    (208.9cos(29.7)) - T - 0.201*277.34 = 28.3*a

    M2: T - mg = m a

    T - 5.5*9.8 = 5.5 a

    (T - (5.5*9.8) ) / 5.5 = a

    ___________________________________

    How would I solve for something like this? Two variables?
     
  11. Sep 29, 2011 #10

    ehild

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    You assumed that M1 moves on the left and M2 moves upward. It is not sure that true, but try. Then write:

    M1:
    F cosθ - T - μ Fn =M1a,

    Fn=M1g-F sinθ, so

    F cosθ - T-μ(M1g-F sinθ)=M1a.

    M2:

    T-M2g=M2a.

    You can plug-in the numerical data at the end.

    Add the equations in bold, or express T from the second one and substitute in the first one.

    ehild
     
  12. Sep 29, 2011 #11
    208.9 cos 29.7 - (5.5 a + 28.3 * -9.8 ) - 0.201 ( 208.9 - sin (29.7) ) = 28.3 - a

    a = 86.357????
     
  13. Sep 29, 2011 #12
    Yeah, this is not making sense... sorry. Can you please explain it again?
     
  14. Sep 29, 2011 #13
    Yeah, this is not making sense... sorry. Can you please explain it again?
     
  15. Sep 29, 2011 #14
    nerd, do as ehild said. "Add the equations in bold" it will cancel out T, then solve for a

    EDIT: i can't emphasize the importance of a free body diagram enough. Always draw a FBD for different bodies in question.
    believe me, life becomes easier when one does
     
  16. Sep 29, 2011 #15

    ehild

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    Correct the red parts of the equation. g=9.8 m/s2, its direction was taken into account in the second equation. I can not imagine, why did you make 28.3-a from M2a. Isolate a, it is present at both sides. Take care of the parentheses.
    It is better to solve a problem symbolically, and plug in data at the end.

    ehild
     
    Last edited: Sep 29, 2011
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