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Does the acceleration cancel out?

  1. Jan 29, 2014 #1
    1. The problem statement, all variables and given/known data

    Two masses and a pulley are arranged as shown. m1 has a mass of 13 kg. The coefficent of static friction between ,1 and the table in 0.25. How big will m2 need to be to get m1 to begin to slide?

    2. Relevant equations

    Fnet=ma

    3. The attempt at a solution

    I am having an extraordinarily difficult time trying to derive an equation for this.

    Fnet=ma
    m2g-f=m1+m2a
    I tried rearranging to get m2=kg+f, but for one this doesn't sound right, and for two it seems impossible to solve because there are two variables. Does the acceleration somehow cancel out in this problem, or is there something else I'm missing?

    Thanks so much! :smile:
     

    Attached Files:

  2. jcsd
  3. Jan 29, 2014 #2
    The problem really wants you to find the minimal mass for ##m_2## that causes ##m_1## to slide. The minimal mass causes minimal acceleration - but what is this minimal acceleration?
     
  4. Jan 29, 2014 #3

    lightgrav

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    if you write f in terms of m_1 , you'll find something to put on the right-hand side.
     
  5. Feb 3, 2014 #4
    Wow, so sorry for the errors. The mass of m1 is actually 12 kg, and there's at least one grammar error.

    voko, that's one of the equations I'm stuck on. I tried deriving an equation for acceleration, but it didn't come together, either. lightgrav, thanks! I've tried a bunch of different ways and I'm still stuck. Could we walk through the steps?

    The total mass is m[itex]_{}1[/itex]+m[itex]_{}2[/itex], correct?

    f=m[itex]_{}1[/itex]+m[itex]_{}2[/itex] * a

    Is that correct so far?

    Thanks again!
     
  6. Feb 3, 2014 #5

    BvU

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    Yes. On two counts. Don't forget the brackets....
    However, you haven't answered voko and you haven 't written out f in terms of m1, so lightgrav is also deeply unhappy...
    And I think you can easily make them happier!
     
    Last edited: Feb 3, 2014
  7. Feb 3, 2014 #6
    Okay, thanks! Do you have tips on how to do brackets on LaTex? I tried before but it messed up my entire equation.
     
  8. Feb 3, 2014 #7

    lightgrav

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    what two Forces (|| the rope) cause (m1 + m2)a ?
    ... do you know a formula for friction?
     
  9. Feb 3, 2014 #8
    Yes. f=[itex]M[/itex][itex]_{}N[/itex]
     
  10. Feb 3, 2014 #9
    (With "M" being mu and "N" being Newtons)
     
  11. Feb 3, 2014 #10

    lightgrav

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    friction Force F ≤ μ N
    N is the "Normal" component of the Force from the table surface
    (pushing straight thru the contact Area into the block).
    Why does the table need to push up?

    Is there tension in the rope? that requires contact Force at _both_ ends!
     
    Last edited: Feb 3, 2014
  12. Feb 3, 2014 #11

    BvU

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    ##\TeX## help: ( or [ or | brackets can simply be typed :
    [itex] [ [/itex] itex [itex] ] [/itex] |[(a^2_\epsilon)]|^4 [itex] [ [/itex] /itex [itex] ] [/itex] gives ##|[(a^2_\epsilon)]|^4 ##
    If you want nested stuff to be extra clear you can use \bigl ( and \bigr ), \Bigl (, \biggl, \Biggl :
    [itex] [ [/itex] itex [itex] ] [/itex] \Biggl ( \biggl( \Bigl( \bigl( () \bigr ) \Bigr) \biggr) \Biggr)^4 [itex] [ [/itex] /itex [itex] ] [/itex] gives ##\Biggl ( \biggl( \Bigl( \bigl( () \bigr ) \Bigr) \biggr) \Biggr)^4 ##

    { has a special meaning as grouping symbol in TeX, but then you can use \{ or \lbrace, again with the extra sizing stuff if so desired.
     
  13. Feb 3, 2014 #12
    Ah, right. I was mixing up two formulas, thanks. The table doesn't need to push up, it's a stationary object. But it's exerting forces on the masses, since the masses are exerting forces on it.

    There's tension in the pulley. Where are we going with this? Do we need to calculate the tension?

    Thanks!
     
  14. Feb 3, 2014 #13
    Thank you!
     
  15. Feb 3, 2014 #14

    lightgrav

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    (yes, sliding friction does have an equal sign)

    If the table did not push up, the block would sink through it.
    in fact, as the block does sink into it, the contact Pressure increases until the sinking stops.

    ignore the pulley for now ... it does push on the rope, but its push is perp to the rope.
    and we know that the acceleration will be parallel the rope.

    always look at components that are || and _|_ the acceleration (if feasible).
    There IS a Force applied to the right end of the rope.
    We want that one: it contains our desired unknown!
     
  16. Feb 3, 2014 #15
    Thanks for pointing that out, but I responded to voko and lightgrav and I are trying to walk through this step by step. We're good. Thanks, though! :smile:
     
  17. Feb 3, 2014 #16
    I'm not sure what you were responding to when you said sliding friction has an equal sign.

    Ah, I interpreted your question too literally. Yes, the table definitely exerts a force.

    I think you're calling the object in question a rope and I'm calling it a pulley, correct? Or are we talking about different things?

    I'm having a hard time translating your words into a variable we can use. So far we have f=m1+m2 * a
    Could you please explain what we're trying to do? Are we finding the friction and putting that on the left side as the force? I think I have an good understanding of the concepts to some extent, but I can't figure out how to translate it into numbers.

    Thanks again!
     
  18. Feb 3, 2014 #17

    lightgrav

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    a pulley is a round wheel fixed in place, that can spin freely around an axle.
    rope, cord, cable, string are all nearly the same ... string sounds a lot like sPring, so I try to avoid it
    ("ideal" textbook string has constant length until it breaks ; ideal spring stretches with any tension)

    your post 1: m2 g - f = (m1 + m2) a

    all you need to do is plug the formula for friction (my post 3),
    and decide how small the acceleration can be for m1 to start sliding (voko's post 2)
     
  19. Feb 3, 2014 #18
    I don't see a rope in the diagram, maybe that's where the confusion lies.

    Okay, I can't believe I'm overcomplicating physics, for crying out loud.

    m2 g - MN = (m1 + m2) a

    Like that? (With "M" being mu and "N" being the normal force.)

    Thank you!
     
  20. Feb 3, 2014 #19

    lightgrav

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    (what is that thing that keeps the dangling m2 from falling quickly to the floor?)

    What is the Normal Force, in this situation (why isn't it zero?)
    ... how small can "a" be , to produce some Δx?
     
  21. Feb 3, 2014 #20
    Tension?

    I'm not sure, can we break this down a little further?

    Finding "a" is one of the main points I was confused on for this situation. I haven't dealt with a situation like this one, and I'm not sure if I have enough information to figure it out.
     
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