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Newton's law of motion problem

  1. Feb 23, 2006 #1
    Here is the problem. I think I set it up right but I am stumped now.

    An object with mass m1=5.00 kg rests on a frictionless horizontal table and is connected to a cable that passes over a pulley and is then fastened to a hanging object with mass m2=10.0 kg. Find the acceleration of each object and the tension in the cable.

    Here is what I came up with:

    m1y: N-m1*g=0

    m1x: T=m1*ax

    m2: T-m2*g=m2*a

    Help please!!! This is due tomorrow and I'm stuck. I'm really bad at solving or finding the right step to get to only 1 unknown in equations.
  2. jcsd
  3. Feb 24, 2006 #2


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    Gold Member

    First of all - the N is positive = mg.
    Now with each of the blocks you have an equation with two unknowns, the acceleration which is the same with both blocks and the tension which is also the same in both equations. Solving this is just simple algebra. You could do it by using the first equation to find T as a function of a (T=m1*a) and then inserting that (m1*a) into the second equation. Then you will have just one unknown.
  4. Feb 24, 2006 #3
    You know that the magnitude of acceleration is the same for both masses?

    [tex] | {\vec a} | = \frac{{| {m_2 \vec g - \vec T} |}}{{m_2 }} = \frac{{ | {\vec T} | }}{{m_1 }} \Rightarrow[/tex]

    [tex]| \vec T | = \frac{{m_1 m_2 }}{{m_1 + m_2 }}| \vec g |[/tex]

    [tex]| \vec a | = \frac{{m_2 }}{{m_1 + m_2 }} | \vec g |[/tex]

    Now just plug in that m1=5.00 kg and m2=10.0 kg //
    (and that [itex]| \vec g |[/itex] = 9.81 m/s2 :smile:)
    Last edited: Feb 24, 2006
  5. Feb 24, 2006 #4
    Thanks very much guys.
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