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3 Object Tension

  1. Oct 16, 2004 #1
    This should be a simple problem, but I'm just missing one step somewhere:

    A 0.840kg glider on a level air track is joined by 2 strings (with negligible mass) to 2 hanging masses and pass over light, frictionless pulleys. (a) Find the acceleartion and (b) the tension in the strings.


    So far what I've been able to figure out is that the tension in the system should be equal throughout and some force diagrams, but I can't figure out the next step

    Masses:
    /\
    | Tension force from string
    |
    -----
    |
    |
    \/ Weight of mass

    Glider:
    /\
    | Normal Force
    |​
    <----|----> (string tensions)
    |
    |
    \/ Weight of mass 1 and mass 2​

    Can anyone help my mind-block?
     
  2. jcsd
  3. Oct 16, 2004 #2
    i dont understand the question , could u re-phrase it ? pls
     
  4. Oct 16, 2004 #3

    Pyrrhus

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    Homework Helper

    Use Newton's 2nd Law on each of the bodies, to make the equations.

    [tex] \sum_{i=1}^{n} \vec{F}_{i} = m \vec{a} [/tex]
     
  5. Oct 16, 2004 #4
    Okay, so from that I get 4 formulas, but no where to substitute:

    Glider:
    Fnet(x)=F(t) - F(t) =0

    Fnet(y)= F(N) - a (m1+m2)
    a=F(N) / (m1+m2)​

    Masses:
    Fnet(x)=0

    Fnet(y)= F(t) - mg
     
  6. Oct 16, 2004 #5

    Pyrrhus

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    Homework Helper

    Are the bodies hanging of same mass?
     
  7. Oct 16, 2004 #6

    Doc Al

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    Staff: Mentor

    All you care about are the forces in the direction of motion for each body. So you should have 3 equations, one for each body.
    If you mean by this that the tension in both strings is equal, that is wrong. (If the tension were equal, how could the glider accelerate?)

    Label the two tensions [itex]T_1[/itex] and [itex]T_2[/itex], then rewrite your equations for the net force for each body.
     
  8. Oct 16, 2004 #7
    Okay, I see somewhat clearer now (hopefully I'm not going in the wrong direction here).

    Glider:
    Fnet = F(T1)+F(T2) - a(m1+m2)

    Masses:
    Fnet = F(T1) - m1g and Fnet = F(T2) - m2g

    Using substitution, a=[m1g + m2g] / [m1 + m2] or g (9.8m/s^2)

    It seems to me the acceleration being gravity alone is too simple. Even past that, I don't know how I'd figure out the tension not knowing the mass of the either of the hanging weights (or a formula/substitution) making the masses unimportant.
     
  9. Oct 16, 2004 #8

    Doc Al

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    Staff: Mentor

    Fnet (on glider) = T1 - T2 (I assumed a coordinate system with T1 in the positive direction. Note that the only forces on the glider are the two tensions, which pull in different directions.)

    This is OK.
    Before you go any further, you must find the connection between the three objects: they all have the same acceleration, but one mass goes up while the other goes down.

    Here's what I'd do. Pick the direction of the resulting acceleration. I will ASSUME that the setup will accelerate so that the m1 mass goes down, and m2 goes up. (If I guess wrong, the answer will be negative.) Now I can assign accelerations to each object: m1 has a down, m2 has a up, glider has a towards m1. With that in mind, you can write Newton's 2nd law for each object:
    glider: Fnet = m(glider)a ==> T1 - T2 = m(glider)a
    m1: T1 - m1g = - m1a

    Got the idea? Your turn.
     
  10. Oct 16, 2004 #9
    Ok, I got that, but now I don't have enough values to get the actual answer. Here's what I've got so far (signs may be different as my the direction of the 2 masses may be different.

    -m(2)a = -m(1)g + m(glider)a + T(2)
    -m(2)a = -m(1)g + m(glider)a + m(1)a + m(2)g
    -m(glider)a - m(2)a-m(1)a = m(2)g - m(1)g
    m(glider)a + m(2)a + m(1)a = m(1)g - m(2)g
    a(m(glider)+ m(2) + m(1)) = g(m(1) - m(2))
    a = g ( m(1)-m(2) / m(glider) + m(2) + m(1))

    Then
    T(1) = m(glider)a + T(2)
    T(1) = m(glider)a + m(1)a + m(2)g
    T(1) = a(m(glider)+m(1)) + m(2)g

    T(2) = T(1) - m(glider)a
    T(2) = -m(1)g - m(2)a - m(glider)a
    T(2) = -m(1) - a(m(2) + m(g))

    So those are the equations I come out with, but without knowing the mass of at least one of the hanging weights, I'm not sure how to calculate an numeric answer.
     
  11. Oct 17, 2004 #10

    Doc Al

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    Staff: Mentor

    If you don't know the 2 masses, you can't get a numeric answer for the acceleration or the two tensions.


    Correct. a = g (m1 - m2)/(m(glider) + m1 + m2)
    I'm not exactly sure what you are calculating here.

    You can't.

    Just for the record, here's how I would set this one up. The equations for each body are:
    glider: [itex]T_1 - T_2 = Ma[/itex], where M = mass of glider
    m1: [itex]m_1g - T_1 = m_1a[/itex]
    m2: [itex]T_2 - m_2g = m_2a[/itex]

    To get the acceleration, add the three equations. To find the tensions, plug the acceleration into the last two equations.
     
  12. Oct 17, 2004 #11
    Okay, I think I've got it then!

    Thank you so much for all of your help!
     
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