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2 blocks one over the edge with a pulley

  1. Dec 12, 2006 #1
    1. The problem statement, all variables and given/known data
    I have 2 blocks of mass M one on a frictionless horizontal surface the other hanging over the edge of the surface connected by a string of length l. The string goes over a pulley of mass m. Find the acceleration.


    2. Relevant equations
    PE = mgh
    KE = 1/2mv^2


    3. The attempt at a solution

    I know how to do the problem without the pulley having any mass how do you factor in the pulley's mass?
     
  2. jcsd
  3. Dec 12, 2006 #2

    rsk

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    Are you sure this is an energy question? These are usually to do with forces on the blocks.
     
  4. Dec 12, 2006 #3

    rsk

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    And come to think of it, the mass of the pulley isn't usually given or needed - at least in the versions we do....
     
  5. Dec 12, 2006 #4

    berkeman

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    Staff: Mentor

    He can do it with energy, but the two m values will be different.

    Alternately, draw the FBD for the rope -- what is the tension?
     
  6. Dec 12, 2006 #5

    berkeman

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    Staff: Mentor

    Oops, I missed the pulley mass in the question. So you need to factor in something about the energy in the pulley wheel's rotation as well. Let's see some more equations, vivitribal.
     
  7. Dec 12, 2006 #6
    HINT : Torque and the rotational inertia of the pulley (ie a circle)

    marlon
     
  8. Dec 12, 2006 #7
    Ok so the force down from the hanging block is M*g which has to pull the pulley and the other block which would be (m+M)*a? or would it be m*alpha +M*a?
     
  9. Dec 12, 2006 #8
    You need three things here :

    1) Newton's second law for the mass on the table (tension on mass1 T1)
    2) Newton's second law for the mass hanging down (tension on mass2 T2 + gravity)
    3) Newton's second law in angular form for the pulley : F' R = I (F' is the tangential force on the pulley, which is the tension in the wire ofcourse. Keep in mind that you have TWO forces acting on the pulley). To get rid of the alpha, you can use : = a/R

    For 3) you need the rotational inertia I = 1/2(mR^2)of the pulley, which is a solid disk (m is the mass of the pulley, r is the radius).


    ATTENTION : keep an eye on the signs of the force vectors here. i suggest you take the positive direction along the x and y axis to be the ones in which the blocks move (x : to the left, y to the right)

    marlon
     
    Last edited: Dec 13, 2006
  10. Dec 12, 2006 #9

    berkeman

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    I think maybe that's a small typo? If the pulley is on the left edge of the table (say), maybe you meant take +x aimed to the left and +y aimed down?
     
  11. Dec 12, 2006 #10
    Let's try this again sum of the forces = m*a so M1*T1+M2*g+I*(T2-T1)-M2*T2= (M1+M2+I)*a and solve for a. Is that right or am I still missing something
     
  12. Dec 12, 2006 #11

    berkeman

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    No, you are mixing unlike quantities. Tension is a force, not an acceleration. And the moment of inertia I is not the same as a mass.

    Remember F=ma, and review how to use the moment of inertia....
     
  13. Dec 12, 2006 #12
    OK so ignoring the pulley for a second the equation would be T1-T2+M2*g=(M1+M2)*a is that correct at least?
     
  14. Dec 12, 2006 #13

    berkeman

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    Which is the hanging mass (I'm assuming M2)? And if the pully wheel is massless, what can you say about T1-T2?
     
  15. Dec 12, 2006 #14
    * M1-------o * ^ * +y
    * ____T1-> | * | * -x +x
    * _________ | * T2 * -y
    * _________ |
    * _________ M2

    Wouldn't T1-T2 be the acceleration?
     
    Last edited: Dec 12, 2006
  16. Dec 12, 2006 #15
    So I did this question and one thing that is EXTREMELY tricky (i.e. I kept bangning my head against the wall, trying to do this while I was sleepy...), is that on the FBD for the blocks, you have T1 going to the right and T3 going up. For the FBD for the pulley, these are OPPOSITE in direction; i.e. T1 on the pulley is going to the left and T3 on the pulley is going down. :eek:
     
  17. Dec 12, 2006 #16
    Yes I realized that about the tensions and the pullies which you can use to find the pulley's angular accel which has to be the same as the systems accel but how do you find tensions?
     
  18. Dec 13, 2006 #17
    Can someone help please?
     
  19. Dec 13, 2006 #18

    OlderDan

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    There is no solution to this problem as stated. Some people have mentioned the moment of inertia of the pulley, and that is needed to solve the problem. However, the problem statement gives you no way to find it.

    If you assume a radius R for the pulley, and make the usual assumption that the pulley is a disk, then you can solve it. Since you were not asked about the tensions in the string, you could solve this with conservation of energy. If you prefer, you can use the approach outlined by marlon in #8, except for the typo. I = ½mR², not ½m².

    If you use energy, only one object in the problem is changing its gravitational potential energy (GPE). The loss of GPE of that one object is transformed into KE of all three objects. If you use marlon's outline, you should write three separate equations for the three objects before trying to lump everything into one equation. If you write those separate equations, it will be easy for others to check what you have done.
     
  20. Dec 13, 2006 #19
    yep you are right

    marlon
     
  21. Dec 13, 2006 #20
    How did you acquire that ?

    Listen,you ALWAYS need to split up the problem in several distinct directions (here x and y): THIS IS THE VERY FUNDAMENT OF NEWTONIAN PHYSICS. After having drawn a FBD, you need to apply Newton's second law in each direction separately. Just calculate each vector along x and y directions, beware of the signs !!!

    1) mass 1 : only a horizontal tension force T1 : m1a = ?
    2) mass2 : vertical tension force T2 and gravity : m2a = ?
    3) Pulley, TWO FORCES T1 and T2, calculate the torque for both forces

    WATCH THE SIGNS OF THE VECTORS (many students make mistakes with these). In the case of the sign of the torque and alpha : use the right hand rule. Keep in mind that the positive z axis is POINTING DOWN, so clockwise rotation is positive here !!!

    The answer has nearly been given now !!!
     
    Last edited: Dec 13, 2006
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