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Newton Laws of motion question

  1. Jun 19, 2012 #1
    1. The problem statement, all variables and given/known data
    A rod AB is shown in figure. End A of the rod is fixed on the ground. Block is moving with velocity √3 m/s towards right. The velocity of end B when rod makes an angle of 60o with the ground.
    [​IMG]



    2. Relevant equations



    3. The attempt at a solution
    I assumed that at any instant the distance of A from block is x and distance of B from ground is y. The length of rod is l.
    l^2=x^2+y^2
    Differentiating with respect to time.
    0=2x(dx/dt)+2y(dy/dt)
    y=xtan(60o)
    dx/dt=√3 m/s
    Solving, i get

    (dy/dt)=-3 m/s
    But that's not correct, the answer says its 2 m/s. I don't seem to get where i am wrong.
    Any help is appreciated. :smile:

    Thanks!
     
  2. jcsd
  3. Jun 19, 2012 #2
    Hi Pranav! :smile:

    Are you sure about that step?

    Recheck your solving,

    [tex]0 = 2x\sqrt{3} + 2\sqrt{3} x \frac{dy}{dt}[/tex]
     
    Last edited by a moderator: Jun 19, 2012
  4. Jun 19, 2012 #3
    Oh yes, sorry, my mistake, i wrote tan60=1/√3. :tongue:

    But now i get (dy/dt)=-1.
     
  5. Jun 19, 2012 #4
    Yes. That's correct. So, whats the velocity of the end B of the rod now?
     
  6. Jun 19, 2012 #5
    Isn't (dy/dt) the velocity of the end B?
     
  7. Jun 19, 2012 #6
    That is the vertical component of B's velocity :wink:
     
  8. Jun 19, 2012 #7
    Ah i get it, thanks for the help Infinitum!
    I shouldn't be studying Physics at midnight.
     
  9. Jun 19, 2012 #8
    :smile:
    Good night, then :wink:
     
  10. Jun 19, 2012 #9
    Hehe, no, i am posting one more question and then i will go to sleep. :D
     
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