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3 masses attached by rope on table with 2 pulleys.

  1. Feb 12, 2009 #1
    1. The problem statement, all variables and given/known data

    You have 3 masses. M1 = 4 kg. M2 = 1kg. M3 = 2kg, M1 is attached to M2 by a rope. M2 is attached to M3 by a rope. M1 is hanging from a pulley attached to the left end of a table, M2 is on the table, and M3 is hanging from a pulley attaced to the right end of the table.

    The coefficient of friction for the table is 0.48

    Find the downward acceleration of M1, the leftward acceleration of M2 and the upward acceleration of M3.

    Also find the Tensions of the left rope(the one attaching M1 and M2) and the right rope(the one attaching M2 and M3)

    KNOWN:
    • Acceleration for M1, M2 and M3 are the same, right?
    • The masses are moving, therefore the net force /= 0, right?


    2. Relevant equations

    Ff = mu k * FN

    F = ma


    3. The attempt at a solution

    I dont understand this at all.

    Ff = (0.48) (9.8) (1kg) = 4.704

    FnetM1 = 19.6 - 4.704 = 15.096 N (to the left).

    Acceleration = 15.096 / mass(1kg) = 15.1 m/2

    This is wrong. I cant figure anything out...for Gods sake someone please help me.
     
  2. jcsd
  3. Feb 12, 2009 #2

    Delphi51

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    This looks correct, now just find the acceleration.
    It would look a little nicer this way:

    The sum of the forces = Ma. Goes left so that's positive. Friction force opposes.
    m1g - 4.704 - m3*g = Ma
     
  4. Feb 12, 2009 #3
    Ok calm down MaximaMan :D. Ok so firstly have you draw a force diagram with all the relevant forces and any resultant accelerations? it is vitally important that you get a good diagram draw as it really helps to visualize the problem.

    Now in order to solve this problem you will end up with a system of equations, it actually is as simple as writing out the equations to describe the forces and any resultant forces on all of the masses. You have to also realize something important about the acceleration on all the masses, and this line "Find the downward acceleration of M1, the leftward acceleration of M2 and the upward acceleration of M3" is almost there to throw you off.

    What you currently missing in you equations it the tensions (which is why a diagram will help) draw what is depicted from the question, draw the weights acting on the masses, and also the relevant tensions in the strings.
     
  5. Feb 12, 2009 #4
    Yeah I got that. But which mass do I use? The sum of M1,2,3? Or M1?

    Ive tried a couple of different answers and none of them were right(though I dont have the right one).

    Galadirith, the problem actually asks for what direction each mass is moving. I just filled it in here for simplicity sake(m1 down, m2 left etc etc). I have drawn this problem out, like 4 different ways. Each mass as its own free body diagram, and also as the whole diagram. I still cant sort it out to get this...
     
  6. Feb 12, 2009 #5

    Delphi51

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    The M in Ma is the total of all the moving masses.
    Are you getting around 2 m/s^2?
     
  7. Feb 12, 2009 #6
    I put in 1.9 and it told me my answer was not within 10% of the correct answer.
     
  8. Feb 12, 2009 #7
    Wait. Okay, so somehow, I managed to get 1.9 something, but you're correct the answer is 2.1xxxx. I put it in and got it right.


    Now onto the tension. For the left rope, isnt it simply T1 = m1*g and T2 = m2*g?


    Galadirith, my professor said for the sake of life, that we are always to consider the string to be massless and rigid(aka, it cant stretch).
     
  9. Feb 12, 2009 #8
    Oh MaximaMan of course, I didnt mean to imply anything about the properties of the ropes themselves. With you equations of the tensions in the strings, remember that M1 and M3 are also accelerating.
     
  10. Feb 12, 2009 #9

    Delphi51

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    T1 is the force needed to overcome the friction plus the gravity on M3 plus the ma needed to accelerate m2 and m3.

    Galadarith, I would be most interested in seeing the solution with the tensions in there from the start. It seems unnecessary to me, but there would be a certain elegance to having all the forces in the first line of the solution. Would you have an equation for each mass - a system of 3 equations to solve to find the acceleration and the two tensions?
     
  11. Feb 12, 2009 #10
    Delphi51, yes basically, and I want to retract my previous comment, you were absolutely right I totally confused what was going on.
     
  12. Feb 12, 2009 #11
    T1 = 4.704 Ff + 2(9.8) + (1kg)(2kg)(2.1 m/s2) = 30.6 N right? - correct answer, thanks guys!

    However! I calculated T2 the same way, and didnt get the right answer. I got 54.4 N and that is incorrect. Any ideas?
     
    Last edited: Feb 12, 2009
  13. Feb 12, 2009 #12

    Delphi51

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    T2 is just the force of the rope on m3, causing it to accelerate at 2.13. Just use F=ma.

    Hey, Galadarith, thanks for the tutoring on tension forces! I think I understand how to incorporate them at the beginning now. I'm a retired high school teacher trying to figure these questions out just for brain exercise, and of course I really enjoy helping kids.
     
  14. Feb 12, 2009 #13
    Well I ran out of chances to get the right answer for T2.

    I did f=ma and the answer was wrong. I swear...lol. Anyhow, I did it F = (2kg) (2.1 m/s2) and I tried F = (2kg) (9.8 - 2.1) thinking that M3 had the acceleration of gravity and the upward acceleration. (I dont know why I thought this, just trying to use my head! Though, half the time I over think things, thinks IB!)

    I am still interested in the answer for learning purposes though.
     
  15. Feb 12, 2009 #14
    Since there is an acceleration, the net force cannot be zero. Don't forget that the only time an object can be in motion with a net force of zero is if there is no acceleration.
     
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