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The tension in the line between A and B

  1. Nov 5, 2007 #1
    1. The problem statement, all variables and given/known data
    THREE IDENTICAL BLOCKS (A,B AND C) ARE BEING PULLED BY AN IDEAL STRING BY A HORIZONTAL FORCE F( PULLES THEM TO THE RIGHT). IT THE TENSION BETWEEN BLOCK B AND C IS T=3N, mass of each block= 0.4 kg ,I DONT KNOW HOW TO UPLOAD A PICTURE BUT THE BLOCKS ARE IN THE ORDER OF
    A B C pUllED BY F IN THIS DIRECTION >


    2. Relevant equations

    A) FIND THE FORCE F
    B) FIND THE TENSION BETWEEN BLOCK A AND B


    3. The attempt at a solution

    I TRIED F=MASS OF C* acc. + tension but i got a wrong answer
    and for part b idont hav a clue plz guide me!!!
     
    Last edited: Nov 5, 2007
  2. jcsd
  3. Nov 5, 2007 #2

    Astronuc

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    Staff: Mentor

    Are the blocks moving at constant velocity, i.e. the acceleration is zero, or are they accelerating? What is the coefficient of friction?

    The tension in the line between A and B will be one half of that in B and C, the then tension in the line attached to C must be 1.5 times that between B and C.

    The friction force on each of A, B and C is [itex]\mu[/itex] mg, where m is the mass of A which is also the masses of B and C, and the friction force must = tension between A and B if the blocks are not accelerating.
     
  4. Nov 5, 2007 #3
    thanks for ur responce

    there is no friction

    the blocks are at rest but when the force pulls them they starts to move
    any thing else??
     
  5. Nov 5, 2007 #4

    Astronuc

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    Staff: Mentor

    Ah, OK, then we can assume that the blocks are accelerating.

    So the tension between B and C is related strictly to the resistance to motion of blocks A and B. The tension, 3 N, is due to the masses of A and B (0.4 + 0.4 kg) resisting the acceleration a. Now to find the acceleration, take the tension 3 N and divide by the masses of A and B.

    The tension between A and B is still half that of the string between B and C, since the tension there is only due to the acceleration of mass A.

    Then the tension or force pulling C must be mass (A+B+C)*a or the sum of the tensions between AB and BC.
     
  6. Nov 5, 2007 #5
    thnx man
     
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