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Homework Help: Forces and tensions between masses check and help

  1. Sep 15, 2010 #1
    1. The problem statement, all variables and given/known data
    Two small blocks, each of mass m, are connected by a string of constant length (4h) and negligable mass. Block A is placed on a smooth tabletop as shown, 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 time t=0. Express your algerbraic answers in terms of h, m, and g




    2. Relevant equations
    ∑Fb = ma = Fw - T
    ∑Fa = ma = T


    3. The attempt at a solution

    a) Determine the acceleration of Block B as it descends.
    - I came up with acceleration = gravity/2 because after finding the sum of the system of equations above I ended up with a(ma + mb) = mg(Fw) which simplifies to a=g/2

    b) Block B strikes the floor and does not bounce. Determine the time it takes for block B to hit the floor.
    - I used the equation 2h/g = t² to come up with 4h/(g/2) = t² which simplifies to
    2h/√g = t

    e) Determine the distance between the landing points of the two blocks.
    - I used the kinematics equations X - Xo = Vot + .5at2 and Y - Yo = Vot + .5gt2
    and found that the time from Block A leaving the table to hitting the floor was 2h/√(g) and by using that in the equation for movement along the x-axis I found that the distance between the two blocks after landing = h


    How does all of this look to you? I feel fairly confident about it im just not 100% positive.
    Thanks!
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Sep 15, 2010 #2

    collinsmark

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    Hello Foonis

    Welcome to Physics Forums!
    Sounds reasonable :approve:
    Something is not right with the above, but I'm not quite sure what. How did the h get out from under the square root? Also, don't forget, Block B is not dropped from a height of 2h, it is dropped from a height of h.
    Once again, how did h get out from under the square root?
    I came up with a somewhat different answer.
     
  4. Sep 15, 2010 #3
    Oh, you're right, I didnt notice that and ran with it. So I plugged it all in again and I got b) 2h/g=t^2 as the base equation to get 2h/(g/2)=t^2 which should simplify to
    2√(h/g)=t.
    e) I fixed it to end up with the time between the block(a) leaving the table and hitting the floor as 2√(h/g) and by plugging that in my final answer for the distance between the blocks is now √h. How does all of this look now?
     
  5. Sep 15, 2010 #4

    PhanthomJay

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    good.
    you have a math error in this equation, but the equation itself is not correct. Why did you use g when you already showed that a=g/2?
    This is the wrong approach. When the first block hits the floor, the tension force becomes 0 (the string goes slack), but the block on the table continues to move, and then this becomes a parabolic motion problem as the 2nd block leaves the table and lands some distance away from the 1st block.
     
  6. Sep 15, 2010 #5

    collinsmark

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    Your equation for the time agrees with mine. :approve:
    so far so good. :smile:
    Oooh. Something doesn't seem right to me. :frown: You'll need to show your work for this one. How did you find the x-component of Block A's velocity?
     
  7. Sep 15, 2010 #6
    Wow this is a tough one haha hmm..

    Well I used a kinematics equation to find the velocity as the block slides along the table after the other block hits the ground, which I set up like:
    X=Xo+VoT+.5at^2, which i substituded the values to get:
    X=0+(0)(2√(h/g))+.5(g/2)(2√(h/g))^2
    so
    X=.5(g/2)(2√(h/g))^2
    so
    X=.5(g/2)(4h/g)
    so
    X=(g/2)(2h/g)
    so
    X=2h/2
    so
    X=h?

    Im extremely confused about this last one..

    Thanks for all the help though, it is really saving me :)
     
  8. Sep 15, 2010 #7
    wait could I use the equation Vf = Vi + at ?
    im going to try that and see what i get
     
  9. Sep 15, 2010 #8
    with that equation I got the velocity of Block A to be (g√h)/(√g)

    Am I any closer?
     
  10. Sep 15, 2010 #9

    collinsmark

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    Yes, that's the one. :approve:
    'Looks good :approve:
    Now you know the velocity in the x-direction (which is constant when the block flies off the table) and the amount of time it's in the air. Solve for distance. :wink:
     
  11. Sep 15, 2010 #10
    so then i could just use distance=velocity x time
    so
    x=((g√h)/(√g))(2√(h/g))
    so
    x=2gh/g
    so
    x=2h

    So the distance between the blocks after they land is 2h?
     
  12. Sep 15, 2010 #11

    collinsmark

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    'Looks right to me. :approve:
     
  13. Sep 16, 2010 #12
    Thanks you helped a ton :D
     
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