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Help with Tension

  1. Sep 8, 2003 #1
    Okay this is a problem I can't figure out:

    59. A pulley is essentially weightless and frictionless. If m1 weighs 100N and the m2 weighs 300 N, and someone holds on to m1 so that the system is motionless, what is the tension in the rope and the acceleration of m1? How much force must the eprson exert and in what direction?

    There is a drawing of the pulley system attached to this post.

    Anyway, this is what I did:

    Ft = tension
    m1g = 100 N
    m2g = 300 N
    so with some calculations, with g = 9.81 m/s^2, m1 = 10.2 kg and m2 = 30.6 kg

    Then I said m1a = m1g - Ft --> m1a = m1g - Ft
    And m2a = Ft - m2g --> Ft - m2g

    Then I said Ft = m1g - m1a and Ft = m2a + m2g
    Then m1g - m1a = m2a + m2g
    So plugging everything in and solving for a:

    (100 N) - (10.2 kg)a = (30.6 kg)a + 300 N
    a = -4.90 m/s^2

    OK I wasn't sure if I was suppose to calculate with the negative sign, but anyway... trying to find Ft:

    Ft = m1g - m1a
    Ft = 100 N - (10.2 kg)(-4.90kg)
    Ft = 150 N

    Did I do this correctly? Because the answer in the back of my Physics book says the answer is suppose to 300 N, but I can't figure out how they got this answer. Any help would be appreciated.
     
  2. jcsd
  3. Sep 9, 2003 #2

    dduardo

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    I get an answer of T = 300 N

    Take each side of the pulley separately

    the first equation would be

    [sum] F(yW1) = T - F - W1 = 0

    the second equation would be

    [sum] F(yW2) = T - W2 = 0

    Equating the two equation you get

    T - F - W1 = T - W2

    F = W2 - W1 = 300 N - 100 N = 200 N

    Plug F into the first equation and solve for T

    T = F + W1 = 200 N + 100 N = 300 N

    So T = 300 N
     
  4. Sep 9, 2003 #3

    HallsofIvy

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    I have a question: the problem says "someone holds on to m1 so that the system is motionless". Doesn't that MEAN that the acceleration of m1 is 0?
     
  5. Sep 9, 2003 #4

    dduardo

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    Yes the acceleration of m1 and m2 is zero.
     
  6. Sep 9, 2003 #5
    What does F represent in your equations dduardo?
     
  7. Sep 9, 2003 #6

    dduardo

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    "someone holds on to m1 so that the system is motionless"

    That implies that the person is applying a force to the system to put it in equilibrium
     
  8. Sep 9, 2003 #7
    Wow, there are too many equations flying around for this question.

    Since the system is in equilibrium, all acceleration is 0. So is all force, so

    m1g - m2g + F = 0, where F is the force of the person.

    Put in the values

    100 - 300 + F = 0
    F = 200
     
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