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Pulley Question

  1. Dec 16, 2009 #1
    I had just asked a question about boats going across a river but that question and this question really have me stumped.
    1. The problem statement, all variables and given/known data
    http://img138.imageshack.us/img138/7910/68512243.png [Broken]
    A)Find the acceleration using newtons laws
    B)Find the velocity using Conservation of Total Mechanical Energy.

    2. Relevant equations
    Idisk=1/2mR^2 (what is R?)
    M1=T+Ff-MG=ma
    M2=1/2mR^2-t+t=ma
    M3=-T-Mg=ma
    I doubt these are right but I tried, I'm terrible at summing forces.


    3. The attempt at a solution
    A)FnMuk-1.1(9.8)+1/2mR^2-t+t-t-Mg=m(a)
    I know I am way off but just so you can see I at least tried.
     
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Dec 16, 2009 #2

    rl.bhat

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    Radius of the disc R is required to find the acceleration.
    The tensions in the two sections of the string are not equal.
    So rewrite the equations for M1, M2 and M3.
     
  4. Dec 16, 2009 #3
    m1=T-MG-Fn-FF ?
    m2= ?? My only guess would be the previous equation.
    m3=T+MGH
     
  5. Dec 16, 2009 #4

    tiny-tim

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    No, you're ignoring what rl.bhat :smile: said …

    there are two tensions, so call them T1 and T2.

    (yes, I know the tension along a string is usually constant, but that's only if the pulley is frictionless … in this case, there is friction at the pulley, otherwise it wouldn't turn, so the tension is different on either side.)
     
  6. Dec 16, 2009 #5
    ohh ok i see i see, thank you for your response.... so
    M1=Muk*M+T1
    M2=M*.5mr^2+T1-T2
    M3=T2-mg
    any better?
     
  7. Dec 16, 2009 #6

    tiny-tim

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    (have a mu: µ and try using the X2 tag just above the Reply box :wink:)

    You're not writing these properly.

    The first and third equations are F = ma.

    The second equation is torque = Iα.

    Start again (and be careful about the ±s). :smile:
     
  8. Dec 16, 2009 #7
    M1=[tex]\mu[/tex]*M-T1=ma
    M2=I[tex]\beta[/tex]
    M3=-T2+MG=ma
    any better?
     
  9. Dec 16, 2009 #8

    tiny-tim

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    (what happened to that µ i gave you? :confused:)

    The third equation is correct.

    In the first equation, the signs are wrong.

    The second equation isn't an equation (and what is ß anyway?)
     
  10. Dec 16, 2009 #9
    I tried to use it!! it came out goofy looking.. By the big B i ment to put[tex]\alpha[/tex].
    M1=-µ*M+T1=ma
    M2= I [tex]\alpha[/tex]= M2G+T2
    M3=-T2+MG=ma
    what about now?
     
  11. Dec 16, 2009 #10

    tiny-tim

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    :rofl: :rofl:
    The first equation is fine now (and easy to read :wink:), except you need a g (why are you writing "G"?)

    In the second equation, why do you have an Mg … that pulley isn't going anywhere, is it? Also, you need to change α into a function of a, you need to use T1, and you need torque on the RHS, not force.
     
  12. Dec 16, 2009 #11
    ok so then
    M2=M2+T1=I[tex]\alpha[/tex]
    I don't really understand what you mean change a to a function of a sorry. Does this look at least a little better
     
  13. Dec 16, 2009 #12

    tiny-tim

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    grrr … not much.

    What is M2 doing there?

    What happened to T2?

    And, I repeat, T is a force, you need a torque

    (and you need a in all three equations, or you won't be able to solve them :redface:, so you must write α in terms of a)
     
  14. Dec 16, 2009 #13
    m2 is just a pulley it is just letting the rope go through it pretty much. So add a -T2 and set it = r x F which is =r x ma ?
     
  15. Dec 16, 2009 #14

    tiny-tim

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    I think you need to get some sleep. :zzz:

    Start again in the morning. :smile:
     
  16. Dec 16, 2009 #15
    I fell asleep at 6 and woke up super early, I am up for the day.Can you please just help me with this I just am not understanding how to do torque..
     
  17. Dec 16, 2009 #16

    tiny-tim

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    Does that mean you've only had about 2 hours sleep?

    You've been making silly mistakes for the last three posts. If you're not normally like this, then you need more sleep.

    torque is force times (perpendicular) distance, in this case tension times radius. :smile:
     
  18. Dec 16, 2009 #17
    no i had about 8 slept from 6-230ish...anyway...I just dont understand it, they may be silly to you but this makes 0 sense to me, thats why I came here, I wouldn't be here if i knew how to do it well.. I do appreciate your attempts at help though.. ok so then M2=t2*radius ?
     
  19. Dec 16, 2009 #18

    tiny-tim

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    No, torque = Iα, so Iα = T2*radius - T1*radius.

    ok, now what is α in terms of a?
     
  20. Dec 16, 2009 #19
    a\R=[tex]\alpha[/tex]?
     
  21. Dec 16, 2009 #20

    tiny-tim

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    Yup! :biggrin:

    ok, now put all the numbers into the three equations, and solve them. :smile:
     
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