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Find the acceleration of the smaller mass. Pulley physics

  1. Sep 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A 2kg mass and a 3kg mass are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. Choose coordinate systems for the two masses with the positive direction up for the lighter weight and down for the heavier weight.
    Find the acceleration of the smaller mass.


    2. Relevant equations
    F=ma


    3. The attempt at a solution
    Fnet = Ft-Fg = 3g-2g = g the tension force on the smaller mass is equivalent to the weight of the heavier mass, so the net force acting on the object is g

    (m1+m2)a=g
    a=1.96 m/s² I know this is the answer, but I'm not sure why. If we already know that the force on the object is g, why don't we just say g=2a and get a=4.9 m/s²?
     
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  3. Sep 19, 2009 #2

    kuruman

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    Re: Pulley?

    We don't know that the force on the object is g, in fact g is an acceleration not a force. There is tension T that acts on each mass as well. Draw a complete free body diagram for each mass, then write Newton's Second Law in each case. This should give you two relations linking the masses, the acceleration and the tension.
     
    Last edited: Sep 19, 2009
  4. Sep 19, 2009 #3
    Re: Pulley?

    The force would be 9.8 newtons. 3kg*9.8m/s²-2kg*9.8m/s² = 9.8 newtons. The tension force would just be the gravity of the other mass since it's frictionless, and it's a pulley wouldn't it?
     
  5. Sep 19, 2009 #4

    kuruman

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    Re: Pulley?

    The tension acting on each mass is the same, I agree. However, if the tension on the other mass is the same as the weight but in the opposite direction, then the net force on the mass will be zero and the mass will not accelerate, don't you think? Like I said, draw a free body diagram for each mass.
     
  6. Sep 19, 2009 #5
    Re: Pulley?

    The tension acting on each force would be the weight of the other mass, so the net force acting on each mass would be the same, 9.8 newtons right? I drew a free body diagram for each mass. On the smaller weight, tension is greater than gravity for the, and on the greater weight, gravity is greater than the tension force.
     
  7. Sep 19, 2009 #6

    kuruman

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    Re: Pulley?

    Let's consider the smaller weight for the time being. You say that tension is greater than gravity and that is absolutely correct. So if you wanted to write an expression for the sum of all the forces (=net force) on that mass, how would you write this sum? (fill in the blank)

    Fnet on smaller mass = ___________ ?
     
  8. Sep 19, 2009 #7
    Re: Pulley?

    The weight of the bigger mass - the weight of the smaller mass = Net force. The only upward pull (which I'm using as tension here) on the smaller mass is the weight of the larger mass and vice versa. 3*9.8-2*9.8 = 9.8 newtons.
     
  9. Sep 19, 2009 #8

    kuruman

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    Re: Pulley?

    So you are saying that the net force on the 2 kg mass is 9.8 N up and the tension on the 3 kg mass is 9.8 N down. Hmm... let's see what that implies. The net force is mass times acceleration so acceleration is net force divided by mass. According to you then

    Acceleration of 2 kg mass = 9.8/2 = 4.9 m/s2

    Acceleration of 3 kg mass = 9.8/3 = 3.3 m/s2

    The acceleration cannot be different otherwise the rope connecting the two mass will stretch, which is not the case here. So what do you think? Will you draw free body diagrams for each mass separately or will you not? You might wish to check out this reference

    http://en.wikipedia.org/wiki/Atwood_machine
     
  10. Sep 20, 2009 #9
    Re: Pulley?

    If we just look at the thing as a whole, this makes sense: the net force on the entire system is 9.8 newtons, and the acceleration for both of them is the same in the same direction: clockwise. I thought that a free body diagram was for the entire system; how can I draw it for just one? Am I wrong that the tension force for one mass would just be the weight of the other?
     
    Last edited: Sep 20, 2009
  11. Sep 20, 2009 #10

    kuruman

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    Re: Pulley?

    The net force on a mass hanging from a string m is always

    Fnet = T - mg

    If the mass accelerates up, then from Newton's Second Law

    Fnet = T - mg = +ma (T > mg).

    If the mass accelerates down,

    Fnet = T - mg = - ma (T < mg).

    If the mass moves with constant velocity or is at rest,

    Fnet = T - mg = 0 (T = mg).

    You need to write such equations for each of the hanging masses separately and you will end up with two equations and two unknowns, the acceleration a and the tension T.
     
  12. Sep 21, 2009 #11
    Re: Pulley?

    Small:
    Fnet = T-mg = ma
    Fnet = T-19.6 = 2a
    T - 19.6 = 2a

    Big:
    Fnet = T-mg = -ma
    Fnet = T- 29.4 = -3a
    T - 29.4 = -3a

    A few questions: since it's a frictionless pulley, wouldn't the tension force be equivalent to the weight of the other? In the last equation, shouldn't the ma be positive because "Choose coordinate systems for the two masses with the positive direction up for the lighter weight and down for the heavier weight"? I can't solve it like a system of equations because the tension forces are different right?
     
  13. Sep 21, 2009 #12

    kuruman

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    Re: Pulley?

    This is an ideal pulley. Ideal pulleys change the direction of the tension but do not affect its magnitude. So the symbol "T" in the above equations stands for the same number. Go ahead and solve the system.

    ** Edit **
    If you take "down" to be positive for the heavier mass, then on the other side of the second equation you need to make the sign of the tension negative (because is is up) and the sign of the weight positive (because it is down). Essentially, this amounts to multiplying both sides of the second equation that you have by -1. It does not change anything.
     
    Last edited: Sep 21, 2009
  14. Sep 21, 2009 #13
    Re: Pulley?

    29.4-T=3a

    T-19.6=2a

    T=29.4-3a

    29.4-3a-19.6=2a
    9.8=5a
    a=1.96 m/s²

    T-19.6=2(1.96)
    T=23.52N

    So the answer is 1.96 m/s²

    Thanks very much! However, I still have a problem: what exactly is tension? I was defining it as the entire upward force on the mass which is obviously incorrect.
     
  15. Sep 22, 2009 #14

    kuruman

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    Re: Pulley?

    Very good. Here, tension is the force exerted by the string on each mass. You can view it as the "entire upward force" but don't confuse it with the "Normal force" which is also an upward force. Tension is exerted by a string on a mass and points away from the mass along the string. Normal force is exerted by a surface on a mass and points away from the surface. You can't have tension without a string or normal force without a surface.

    I am curious, why do you say that your way of looking at tension is obviously incorrect?
     
  16. Sep 22, 2009 #15
    Re: Pulley?

    I solved for T = 23.52N, but it was for both masses. I thought that it was the upward force due to the mass of the other weight. For instance, I thought that for the smaller mass, the tension would be the weight of the bigger mass and vice versa.
     
  17. Sep 22, 2009 #16

    kuruman

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    Re: Pulley?

    I think you have sorted things out now. You may to practice some tension problems or read solve examples in your textbook involving tension.
     
  18. Sep 22, 2009 #17
    Re: Pulley?

    Just making sure: I was wrong?
     
  19. Sep 22, 2009 #18

    kuruman

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    Re: Pulley?

    The tension is 23.5 N acting "up" on both masses.
     
  20. Sep 22, 2009 #19
    Re: Pulley?

    So the force is just from whatever the pulley is attached to?
     
  21. Sep 22, 2009 #20

    kuruman

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    Re: Pulley?

    No. The forces acting on each mass are from whatever is touching or interacting with each mass. Here you have the rope and the invisible force of gravity acting on each mass. The rope force is the same for both masses.
     
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