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Homework Help: Force and friction problem

  1. Mar 8, 2010 #1
    1. The problem statement, all variables and given/known data
    Three boxes of masses 25 15 and 30 kg are connected:
    link to my picture:
    http://i36.photobucket.com/albums/e47/jo860/force.jpg

    The tension on the connecting strings are T1 and T2. The assembly is realeased from rest at t = 0. There is no friction between the box and table.
    a) find acceleration of the boxes
    b) find T1
    c) find T2
    d) If there is friction between the box and the table and the coef. of friction is .15, How much is the acceleration of the boxes?


    2. Relevant equations

    F = ma
    W = mg
    F = friction x normal force


    3. The attempt at a solution

    i feel like i'm overthinking the problem. Is it right to add all of the vectors going in the same direction to each other? Maybe I'm under thinking the problem. I never got clarification on what's what.
    I just need someone to explain the problem in simpler terms. I've already asked alot of people (tutor, prof, friends) this is my last resort sooo...
    I hope someone can explain the mechanics of the problem

    The major thing i know to do is that f = ma can be applied to two masses at the same time by adding the masses...and that's all. my answers
     
  2. jcsd
  3. Mar 8, 2010 #2

    rl.bhat

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    1) All the masses must have the same acceleration.
    2)A string must have the same tension in all its segments.
    3) Apply Newton's second law to individual masses.
    To start with,
    For 30 kg mass
    m1g - T1 = m1a ......(1)
    Proceed in the same manner for other two masses. And solve equations to find a.
     
  4. Mar 8, 2010 #3
    could you explain how you got

    m1g - T1 = m1a ......(1)? this whole equation represents the net force right?

    are you subtracting T1 because it's pulling up?
    so does this equation sort of assume that the system will move up?
    and which way is the f = m1a force pointing?

    sorry i have so many questions...this is why i think im over thinking it...
     
  5. Mar 8, 2010 #4

    rl.bhat

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    could you explain how you got

    m1g - T1 = m1a ......(1)? this whole equation represents the net force right?
    Yes.

    are you subtracting T1 because it's pulling up?

    Yes
    so does this equation sort of assume that the system will move up?

    No. It is moving down.

    and which way is the f = m1a force pointing?

    In the down ward direction.
     
  6. Mar 8, 2010 #5
    Could you explain why it's pointing down?
    is it because we assume that the system is moving down?
     
  7. Mar 8, 2010 #6

    rl.bhat

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    Not the assumption but it is the fact. T2 pulls the 25 kg mass on the table. 15 kg and 30 kg ,asses move in the downward direction due to gravitational force.
    Why are you not proceeding to the other two equations for 15kg and 25 kg?
     
  8. Mar 9, 2010 #7
    oh...right, i should have looked at the picture again.

    ok so since it's the same acceleration throughout the entire system
    would this be right?
    Ms = mass of entire system
    Ms *a = (m1 + m2) g
    but i can't explain what force the Ms *a is, or where it's pointing

    and for T1 is it T1 = mg - m1a ?

    and for T2, I ignore all the other stuff pretty much right becuz it's on another axis
    and the 25 kg block is movingrightward and since tension is same for all the rope, i can just pay attention to this part
    which is T2 = 25a right?

    and for the friction, the normal force is eqaul to W
    so it must be Fn = 25 g
    and since friction always opposes teh motion, motion being T2 (righ?)
    so the eqation for finding the a with friction would be
    T2- FrictionForce = Msa?
    again...i can't explain where the extra Msa comes from, but i just have a feeling it should be there like the previous one...
     
  9. Mar 9, 2010 #8

    rl.bhat

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    It is stated that there is no friction between the box and table
    I have given the first equation for m1.
    The second equation is
    T1 - T2 = m2a.
    And the third equation is
    T2 = m3a
    Now solve these equations to find T1, T2 and acceleration.
     
  10. Mar 9, 2010 #9
    m1 = 30
    m2 = 15
    m3 = 25

    so for a i got a = mg /( m1 + m2 + m3)

    then i just plaug that into the T1 and T2 equation to to get those values
    but can u explain why its
    T1 - T2 = 15a ?
    why is it only the m2? Is this equation saying that the T1 is pulling down on m2 while T2 is pulling up on it and that the F = 15a force added to T2 = T1 because its also pointing down?
     
  11. Mar 9, 2010 #10

    rl.bhat

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    so for a i got a = mg /( m1 + m2 + m3)
    It should be
    a = m1g /( m1 + m2 + m3).
    T1 - T2 = 15a ?
    why is it only the m2? Is this equation saying that the T1 is pulling down on m2 while T2 is pulling up on it

    Yes.
    the F = 15a force added to T2 = T1 because its also pointing down?
    I didn't understand what you mean by this statement. T2 is not equal to T1.
     
  12. Mar 9, 2010 #11


    i was kinda asking if the T1 - T2 = 15a eqatuion was arbitrarily picked, but now i see taht it is the only eqatuion to relate the Two forces.

    Just to clarify the Force, F = 15a is extra force that's throwing the thing off of equilibrium and pulling the system down right?

    its T1 - T2 = 15 a because we present up is negative and down is positive?
    or will it always be the one pulling down minus the one pulling up?
     
  13. Mar 10, 2010 #12

    rl.bhat

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    The positive and negative directions are arbitrarily selected. In this problem, the 30 kg and 15 kg masses are moving in the down ward direction, which also the direction of g. So 15a is taken as positive.
     
  14. Mar 30, 2010 #13
    ok i understand it completely now
    thank you
    took a little while to answer back...and doo more problems
    but i got it.
    it depends on the direction and common sense i guess
    now that i understand it, i don't understand how i didn't get it before hah
    thanks tho, sorry for late reply
     
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