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Homework Help: Vertical Dynamics Problem

  1. Jul 8, 2008 #1
    1. The problem statement, all variables and given/known data
    Two masses, 1.5 KG and 3.0 KG, are tied together by a light string and looped over a frictionless pully. They are allowed to hang freely. Determine the magnitude of the acceleration for each mass. [Answer: 3.3 m/s2)

    2. Relevant equations
    Fnet = ma
    Fg = mg

    3. The attempt at a solution
    This is what i did..
    [tex]\Sigma[/tex]F = Fg = m1g + m2g
    [tex]\Sigma[/tex]F = (1.5 KG [tex]\bullet[/tex] 9.81 m/s2) + (3.0 KG [tex]\bullet[/tex] 9.81 m/s2)
    [tex]\Sigma[/tex]F = 44.145 N

    [tex]\Sigma[/tex]F = ma
    44.145 N = 4.5 KG [tex]\bullet[/tex] a
    a = 9.81

    The answer is 3.3 m/s2 and to be honest, i dont get how to do any questions like these. Could someone explain how its done please?
  2. jcsd
  3. Jul 8, 2008 #2


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    Hi mykoolspam,

    You should start by treating the masses separately. Draw a force diagram for each block. What two forces are acting on each of the blocks?

    You can get one equation from each force diagram. So from the force diagrams, you can get two equations in two unknonws, and you can solve for a. What do you get?
  4. Jul 8, 2008 #3
    The two forces are Fg and Fn, right?
    So..two equations?
    mass 1 = 1.5 kg
    - Fg = mg = 1.5kg x 9.81
    mass 2 - 3.0 kg
    - Fg = 3.0 x 9.81

    hmm im so lost.
  5. Jul 8, 2008 #4


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    Let's look at mass 1. One of the forces is Fg=m g, because gravity is pulling it down. The other force is not Fn (normal force) because it is not in contact with a surface, it is the tension force T from the rope.

    So draw the force diagram. Which way are the two forces going? Which way is m1 (the lighter mass) accelerating? Once you have those three directions, write out:

    (sum of the forces in the y directions) = m (acceleration in the y direction)

    What do you get for the equation for m1?
  6. Jul 8, 2008 #5
    Oops. Yeah i often use Fn and T interchangeably. Are they the same equations?

    Well, mass one is accelerating upwards while mass two is being pulled down by gravity.
    Fg = 3.0kg * 9.81 = 29.43 N [MASS 2]
    And since mass one is being pulled up:
    T = ma?
    T = 1.5 kg * a

    [Sum of Forces] = 29.43 N + (1.5kg * a)

    And thats as far as i get.
  7. Jul 8, 2008 #6


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    One of the reasons why the force diagram approach is so useful is because it let's us focus on one mass at a time. In your equation for m1, you included the weight of m2 (29.43 N).

    But the 29.43 N is not acting on m1 at all, and so will not appear when you write down the m1 equation; only forces that actually act on m1 appear in the m1 equation.

    The starting point for a 1D problem is:

    [sum of forces] = m a

    On the right hand side, you only have the mass and acceleration, not any of the forces. So for m1:

    [sum of forces] = (1.5) a

    The forces all go on the left side, and in this problem there are only two forces acting on m1: the gravity force [itex]m_1 g[/itex], and then tension force T.

    What do you get for this equation for m1? Remember the signs of the force components tell you the direction of the forces.
  8. Jul 8, 2008 #7
    [Sum of forces] = T - Fg = T - (mg) = T - 14.715
    T - 14.715 = 1.5 * a

    hhmmm..what Am I Missing?
  9. Jul 8, 2008 #8


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    Nothing at all! That is exactly the correct equation for m1.

    Of course it has two unknowns, so you'll need another equation, so now do the exact same thing to find the corresponding equation for m2. (Then you'll have two equations in two unknowns, and it will be just a matter of algebra to solve for a.) What do you get for the m2 equation?
  10. Jul 8, 2008 #9
    m2 =
    T1 = 1.5a + 14.715
    T2 = 3.0a + 29.43

    so..i made T1=T2, but it doesnt work out. I thought about it and i figured T1 cant equal T2 anyways. ha.
    Do i multiply the two equations or..?
  11. Jul 8, 2008 #10


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    That's so close! but there is a sign error. The left side is okay (the tension is positive because the tension is upwards; the weight is downwards and so need to be negative). However, on the right side, you need a negative sign for a, because m2 is accelerating downwards.

    If the pulley is massless and frictionless, like it is here, then the tensions will be the same.
  12. Jul 8, 2008 #11
    ok so...
    -14.715 = 1.5a
    -29.43 = 3.0 * -a

    i get -9.81 and 9.81 = a
    What ami missing this time? x.x
  13. Jul 9, 2008 #12


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    It looks like you set the tensions equal to zero, intead of setting them to each other. Let's look back at one of your earlier posts:

    We said these two equations were correct, except that the (3 a ) should have had a negative a. So they should be:

    T1 = 1.5a + 14.715
    T2 = -3.0a + 29.43

    So since the left hand sides are equal (because T1=T2), the right hand sides must be equal to each other. Set them equal to each other and solve for a. What do you get? I'm not getting 9.8, so if you still get that please post the details of your algebraic manipulations.
  14. Jul 9, 2008 #13
    Ohh haha nevermind i got it! Thanks ALLOTT for your help!
  15. Jul 9, 2008 #14


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    Glad to help!
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