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Newton's law problem

  1. Aug 16, 2015 #1

    Titan97

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    1. The problem statement, all variables and given/known data
    Two Blocks A and B, each of mass M are attached by a thin inextensible string through a frictionless pulley. Block B is released from rest. Block A hits the pullty in time tA and block B hits the vertical surface at time tB. Find the relation between tA and tB.
    Untitled.png
    2. Relevant equations
    ##F=ma##

    3. The attempt at a solution
    Length of the string from pulley to block B increases as it moves down. If the string does not become slack, and since its inextensible, I think tA=tB since acceleration of both ends of string should have equal acceleration. But I think its wrong. (I have attached the solution given by the author, but I want to solve it myself)
    Iet block B descend by a distance ##y## and cover an angle ##\theta##. Let the length of string be ##R## which can be expressed as a function of ##y## and ##\theta##.
    ##R=ycosec\theta##.
    For block A, ##T=Ma##.
    For block B, along y-direction, ##Mg-Tsin\theta=Ma_y## and ##Tcos\theta=Ma_x##
    From this, substituting value of T with Ma, ##a_x=acos\theta## and ##a_y=g-asin\theta##
    Then, net acceleration of Block B is ##a_B=\sqrt{a^2+g^2-2agsin\theta}##
    ##|2agsin\theta|\le 2ag##.
    But now, acceleration of both ends are different.
    Here is the solution given in the textbook.
    Capture.PNG
     
  2. jcsd
  3. Aug 16, 2015 #2
    What is the relation between the acceleration of the block B along the string with that of block A's acceleration ?
     
  4. Aug 16, 2015 #3

    Titan97

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    They must be equal...
     
  5. Aug 16, 2015 #4
    And block B has acceleration along string , and perpendicular to string , so - ?
     
  6. Aug 16, 2015 #5

    Titan97

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    It wont move along a circle.
     
  7. Aug 16, 2015 #6
    And acceleration of block B with respect to block A would be - ?
     
  8. Aug 16, 2015 #7

    Titan97

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    Why should i do that? I already found acceleration of block B
     
  9. Aug 16, 2015 #8
    I have been trying to establish that acceleration of block A is lesser than that of block B , from -
     
  10. Aug 16, 2015 #9
    As your analysis stands now, your kinematic relationships between the accelerations of blocks A and block B are incorrect.

    Have you been learning about using cylindrical coordinates yet? If so, this problem is much easier to solve using cylindrical (r-θ) coordinates, with the origin of the coordinate system for block B located at the pulley. Do you know the equations for the radial acceleration and the angular acceleration if both r and theta are both changing with time?

    Let r be the radial distance of block B from the pulley and let theta be the same theta you have been using.

    Chet
     
  11. Aug 16, 2015 #10

    PeroK

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    That solution makes no sense. Instead, you could go back to your initial approach. Hint: You can solve this without any calculation.

    Perhaps think about the cases where block B is much heavier or much lighter than block A. What happens in those cases?

    Then you might see what happens when the two blocks are of equal mass.
     
    Last edited: Aug 16, 2015
  12. Aug 16, 2015 #11

    Titan97

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    Firstly @Chestermiller , I only know the definition of cylindrical coordinates. But I will try to use it.
    Secondly, @PeroK , If block B is much heavier than block A, the time it takes to fall down wont change right? A peice of feather and a block of iron take the same time to fall in vacuum. If block A is heavier, since there is no friction, I still don't think there will be any change.
     
  13. Aug 16, 2015 #12

    PeroK

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    If block B is much heavier then it will fall straight down and pull block A into the pulley.

    If block B is much lighter, it will swing into the wall like a pendulum and A will hardly move.

    The question is, if they are the same mass, which one gets pulled more, or do they get pulled equally?
     
  14. Aug 17, 2015 #13

    Titan97

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    @PeroK , here is how I tried to solve it. The force acting along horizontal for block B is ##Tsin\theta## and the force acting along horizontal on block A is ##T##.
    ##Tsin\theta\le T##. Hence, the acceleration along horizontal direction for block A is more and hence it hits the pulley first or takes lesser time.
     
  15. Aug 17, 2015 #14
    I'm embarrassed to admit that I tried a much more complicated approach to this problem and was not able to reach a conclusion. That's because I violated my own first three most important rules of modeling: 1. keep it simple, 2. keep it simple, 3. keep it simple.

    Nice job Titan97.

    Chet
     
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