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Acceleration of masses

  1. Jan 3, 2005 #1
    Hi. Can anybody help me with this problem, please? I don't know what to do.

    Problem: A 2.0-kg mass and a 3.0-kg mass are attached to a lightweight cord that passes over a frictionless pulley. The hanging masses are free to move. The coordinate systems is with positive direction up for 2.0-kg mass and negative direction down for 3.0-kg mass. Find the acceleration of the smaller mass.

    Is there any different equation for this? Or do i somehow change the F=ma to fit?
  2. jcsd
  3. Jan 3, 2005 #2


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    It is done with F=ma. You already know the mass m so all you need to have the acceleration of the mass is the total force acting on it.

    For one there's the force of gravity, which acts downward (which means negative sign!), and there's also a force upward which is the result of the 3kg wanting to go downward as well due also to the force of gravity acting on it. This upward force is the tension in the rope.

    There are two things to be aware of: 1) The tension acting on the small mass is the same as the tension on the bigger mass. 2) The acceleration of both masses is the same.

    I let you work on figuring out how to solve the thing from there.

    Hint: you will need the equation of motion (F=ma) for the 3kg mass too.
  4. Jan 3, 2005 #3
    One Important thing here is remember that this is a system of equations and it is easier to solve it that way!

    Good Luck
  5. Jan 3, 2005 #4
    okay so i got this far...with the previous help.

    for the smaller mass - F(g) = -2g = -2(9.8) = -19.6N (acting downward)
    - F = 3g = 3(9.8) = 29.4N (acting upward)

    so the net force is: F(net) = 29.4 - 19.6 = 9.8 N
    using the equation of motion: F(net) = ma ... i can say that a = F(net)/m

    a = 9.8/5 (now i think im supposed to add the two masses (2kg and 3kg) together, right?)

    a = 1.96 = 2.0 m/s2

    well i think its right...but i can be always wrong....
  6. Jan 4, 2005 #5


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    It is the right answer, but it's almost luck because the reasoning is at most dubious. :tongue2:

    I think you supposed that the tension in the rope (the upward force) was equal to the weight of the big mass, but unfortunately, this is not so. Your equations of motion for the 2kg mass should be

    [tex]-F_{earth \rightarrow small} + \tau = m_{small}a \Rightarrow -19.6 + \tau = 2a[/tex]
    [tex]\Rightarrow \tau = 2a+19.6 \ \ (*)[/tex]

    where, you will have guessed, [itex]\tau[/itex] is the tension in the rope.

    As you can see, you need to know what the tension is in order to solve for a. The trick is to write down the equation of motion of the 3kg mass too. That way you will have 2 equations with 2 unknown. Remember this: Every time you have n equations of degree 1 with n unknown, you can always find a way to rearange your equations to find what the unknown are. That's what we're gonna do.

    The equation of motion for the 3kg mass is

    [tex]F_{earth \rightarrow big} - \tau = m_{big}a \Rightarrow 29.4 - \tau = 3a[/tex]
    [tex]\Rightarrow \tau = 29.4 - 3a \ \ (**)[/tex]

    Now try equating (*) with (**): you get a single equation with a as the only unknown.

    N.B. substituting a by 1.94 in (*) or (**), you can also find the tension in the rope, which is 23.52 N; less than the weight of the 3kg mass!
    Last edited: Jan 4, 2005
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