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Finding Tensile Force in Pulley Problem

  1. Nov 25, 2008 #1
    1. The problem statement, all variables and given/known data
    See Picture, The problem gives three boxes(M1, M2, and M3) and a few pulleys, no value for the coefficient of friction is given. None of the masses are known, and we are told to assume M1 and M3 are sliding. The objective is to find Ft

    [​IMG] (if you can't see that well enough i made this) [​IMG]

    2. Relevant equations

    3. The attempt at a solution

    First I started of by drawing free body diagrams for all of the boxes, Using these(and assuming M1 and M2 are sliding towards the middle) I came up with


    M2 Would then have the same equations except M1 would be M2 because it also has

    (I feel like that might be wrong but I'm not sure)
    I'm not sure about the signs on all of those either.

    Then I have for M3

    At this point I'm not sure what to do, (the lack of values is killing me) I would think putting Ft from the M1 equation into the Ft equation would be a good start, but then everything except mass would cancel wouldn't it? So that doesn't seem right. I feel like I'm going in the right direction and I need to use these equations simultaneously some how but I'm not sure what to do for the next step as I don't see how this is going to get me Ft.

    Also I'm starting to think M2 Net force is different
    FNET=Ff-Ft (Since the ropes would be moving in opposite directions) and that would also change the equation for M2 to


    But even if that's right, what would i do with it...?

    Any help is appreciated, I'm currently stumped. :confused:
  2. jcsd
  3. Nov 25, 2008 #2

    Doc Al

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    OK, but be careful to distinguish the accelerations of each mass--they aren't the same. Call this acceleration a1.

    OK. Use a2.

    Consider M3 and the pulley as a single object and apply Newton's 2nd law as before. (Careful with the tension force here.)

    Key point: How does the acceleration of M3 relate to a1 and a2? Figure that out.

    Hint: Set up these equations correctly and you'll have 3 equations and 3 unknowns. (Treat the masses and μ as knowns.)

    (You're on the right track.)
  4. Nov 25, 2008 #3
    If they are both treated as the same object then it would have two tensile forces on it correct?

    Giving you FNET=M3a3-Ft1-Ft2
    and a3=g=9.81m/s^2

    Both of the tensile forces will be equal I assume so,

  5. Nov 25, 2008 #4

    Doc Al

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    Well, I wouldn't use a3 to represent g; a3 is the acceleration of M3.


    Play around with it, or figure it out by considering that the length of the rope doesn't change. Example: If M1 moves 1 m to the right and M2 moves 1 m to the left, how far does M3 move down?
  6. Nov 25, 2008 #5
    I Solved for a3 in the net force equation for M3 and got
    then I substituted the values i got for tension for the first and second box earlier and came up with

    a3=9.81m/s^2 - ((M1a1+µM1a1)/M3)-((M2a2+µM2a2)/M3)

    assuming i did my algebra right...

    Is that what you meant by relating all three accelerations?

    Im looking at what i said and it makes me think, If a3 = 9.81-2Ft/M3
    M1=M2 and a1=a2.
    Although what you said earlier to label the accelerations different makes it seem like they shouldn't be equal.


    I Think this whole post is going in the wrong direction, read below post
    Last edited: Nov 25, 2008
  7. Nov 25, 2008 #6
    Okay, so... If the Left box moves 1 meter right, and the Right box moves 1 meter left, The middle box moves down 1 meter So, (a1+a2)/2=a3
  8. Nov 26, 2008 #7

    Doc Al

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    Exactly! This is the constraint equation relating the motion of the three masses. Now you should have all you need to solve for the tension.

    (It's worth your time to learn how to derive this a bit more rigorously from the fact that the string length is fixed.)
  9. Nov 30, 2008 #8
    If you solve for a1 in the first equation and a2 in the second equation

    Plug the results into (a1+a2)=a3 and then substitute that into the the equation for M3

    Using that solve for Ft

    And you end up getting

    Ft=[2m1m2m3(1+u)] / [m3(m1+m2) + 4m1m2]

    I think i did that right but my teacher said that i will have M1 M2 M3 g and u. when I solve for Ft. My equation is missing the g.
    What could i be doing wrong
  10. Nov 30, 2008 #9

    Doc Al

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    OK, except that a3 = (a1 + a2)/2

    That cannot be correct since the units don't match. The right hand side has units of mass, not force. (And yes, you should have a g in your answer.)

    To find out where you went wrong, show the equations that you used.
  11. Nov 30, 2008 #10
    Okay, for some reason i accidently substituted g for a1 and a2 on the right of the first two equations, im reworking it now with the correct variables there, and i also see one other mistake...

    This time through im using the equations

    M1a1=Ft-µM1a1 (1st equation)
    M2a2=Ft-µM2a2 (2nd equation)
    M3a3=Fg-Ft or M3a3=M3g-Ft (3rd equation)
    a3 = (a1 + a2)/2 (4th equation) that's what I meant last time I just forgot to write the divided by 2 as I was typing it fast sorry

    Ill edit this post when i get an answer with these equations
  12. Nov 30, 2008 #11

    Doc Al

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    Careful: The tension pulls up on M3 twice.
  13. Nov 30, 2008 #12
    Right. thats what i have written down, forgot that two when i typed it in. Other than that do all the equations look good? I'm still working on the algebra to solve it for Ft.
  14. Nov 30, 2008 #13

    Doc Al

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    Yep; looks good.
  15. Nov 30, 2008 #14
    I feel like im really bad at algebra right now, so far i have


    Everything i do to this makes it feel like im going backwards, if i multiply by ones on the right side to get a common denominator it takes that straight back to m3g-2Ft

    Any suggestions on the next step? I'm still trying :cry:
  16. Nov 30, 2008 #15

    Doc Al

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    Not sure how you ended up with Ft in the denominator.

    Why don't you follow the same approach you described in post #8?
  17. Nov 30, 2008 #16
    i had Ft/2m1(1+u) + Ft/2m2(1+u)=g-2Ft/m3 and i tried dividing by Ft

    thats how it got on the bottom

    With that i continued to 1/M1+1/M2=2g(1+u)/Ft - 4(1+u)/M3

    I can multiply by Ft here to get it back on top on both the fractions on the left and the last fraction

    And originally i solved for a1, a2, and substituting that into (a1+a2)/2=a3 and then substituted a3 into m3a3=m3g-2Ft

    Thats where i started trying to get Ft
  18. Nov 30, 2008 #17

    Doc Al

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    Since you want to solve for Ft, don't divide by it. Instead, start by collecting all terms with Ft on the left side of the equation. Then just factor out the Ft and simplify the result.
  19. Nov 30, 2008 #18
    I think i got it :surprised


    Is that correct? god i hope so

    Edit:: I think i can pull out an M3 from the denominator and make it m3(m2-m1)+4m1m2(1+u)
    Last edited: Nov 30, 2008
  20. Nov 30, 2008 #19

    Doc Al

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    You are very close but I think you made an error somewhere. (As a sanity check, you might want to consider the special case where all masses are equal and there's no friction. In that case the tension will be mg/3.)
  21. Nov 30, 2008 #20
    Im trying to set all the masses equal and set my u to 0, and i cant find out where i messed up.

    I substituted like that at the point where i got my last equation and it came to 2gm/5.


    Can you see anything i'm doing wrong?

    from this point i would get the Ft on the left and factor/solve

    Edit:: Wow I feel stupid it does work out to gm/3... For some reason i thought 4+1+1 was equal to 5 >.<. I'm working it out again
    Last edited: Dec 1, 2008
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