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Formula Manipulation

  1. Sep 29, 2005 #1
    I am trying to manipulate a formula to get a known answer — only I am not getting it:

    The necessary information: Two small blocks, each of mass m, are connected by a string of constant length 4h, with negligible mass. Block A is placed on a smooth tabletop (no friction), and block B hangs over the edge of the table. The tabletop is a distance 2h above the floor. Block B is then released from rest at a distance h above the floor at T=0. Express all algebraic answers in terms of h, m, and g.

    On to the problem I am stuck on: Block B strikes the floor and does not bounce. Determine the time t1 at which block B strikes the floor.

    Now, the known answer is [tex]t_1=2\sqrt\frac{h}{g}[/tex]

    My attempted work goes as follows: I begin with the equation [itex]d=\frac{1}{2}gt^2[/itex] (falling body), getting [itex]t^2[/itex] by itself, I end up with the equation:

    [tex]t^2=\frac{h}{.5g}[/tex]

    taking the square root of both sides then, I get

    [tex]t=\sqrt\frac{h}{.5g}[/tex]

    If I got the steps right so far, I am completely at a loss as to what to do now.
     
  2. jcsd
  3. Sep 29, 2005 #2

    hotvette

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    You are forgetting that the falling mass is pulling the mass on the table, thus your equation of motion of the falling mass is missing an upward force.
     
  4. Sep 29, 2005 #3
    What would this upwards force be? Tension? Block B's falling acceleration is [itex]\frac{g}{2}[/itex]

    Instead of [itex]d=\frac{1}{2}gt^2[/itex], the formula would then become [itex]d=\frac{1}{4}gt^2[/itex] and later [itex]t=\frac{h}{.5g}[/itex]right? I still don't know how the 2 ends up outside of the square root.
     
  5. Oct 1, 2005 #4

    hotvette

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    I'm baffled. I came up with different answer than either of the ones you mentioned. Two questions:

    1. How do you know the answer is [itex]t_1=2\sqrt{\frac{h}{g}}[/itex]? I got [itex]t_1=2\sqrt{\frac{2h}{g}}[/itex]. Could there be a typo in the answer?

    2. Since this is a high school section, what concepts are you working with?
    - Free body diagrams?
    - Equations of motion? (e.g. F=ma)
    - Energy equations? (e.g. sum of potential energy + kinetic energy, etc.)

    Just want to know what level of dialog is appropriate.
     
  6. Oct 1, 2005 #5

    VietDao29

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    hotvette, you are wrong. I think you should re-check your calculation. I think there's nothing wrong with the answer.
    Lucretius, yes, block B's acceleration is g / 2. But there's a slight error in your arrangement:
    [tex]\frac{1}{4}gt ^ 2 = h[/tex]
    [tex]\Leftrightarrow t ^ 2 = \frac{4h}{g}[/tex]
    [tex]\Leftrightarrow t = \pm \sqrt{\frac{4h}{g}} = \pm 2\sqrt{\frac{h}{g}}[/tex]
    Since t must be greater than 0. So you just choose:
    [tex]t = 2\sqrt{\frac{h}{g}}[/tex]
    Viet Dao,
     
  7. Oct 1, 2005 #6

    hotvette

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    Don't think so. I also got the acceleration = g/2. The answer is correct if the distance is h, which it isn't. The distance is 2h.

    [tex]t_2 = \frac{4x}{g} = \frac{4*2h}{g} = \frac{8h}{g}[/tex]
     
    Last edited: Oct 1, 2005
  8. Oct 1, 2005 #7

    VietDao29

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    Why can't the distance be h??? You should re-read the problem. :wink:
    So is the answer the book provided right, then? :smile:
    Viet Dao,
     
  9. Oct 1, 2005 #8

    hotvette

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    Absolutely right, I missed it! Too late at night for me I guess. The right answer is indeed:

    [tex]t = 2\sqrt{\frac{h}{g}}[/tex]

    So, what are the lessons here? I can think of 3:

    1. Double check your math
    2. Read the problem carefully
    3. If you can, avoid working problems when you are tired
     
    Last edited: Oct 1, 2005
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