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Friction and Pulleys

  1. Apr 10, 2015 #1
    d0e251d086df8322da1f6e44416ed214e59ffb2c.png


    1. If flat car is given an acceleration a = 3 m/s^2 starting from rest, compute tension (in N) in the light inextensible string connected to block A of mass 30 kg. Coefficient of friction between block and flat car is = 0.50.

    Neglect mass of pulley and its friction. Take g = 10.



    2. Friction = ## \mu N ##
    F = ma
    ma = mg - T



    3. No idea. I couldn't approach the problem. Please tell me how to approach these types.
     
  2. jcsd
  3. Apr 10, 2015 #2

    pcm

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    Write the newton's laws in the frame of the ground. It is an inertial frame. Write down all the forces(along both axes). Also observe that for every meter the car moves, block A moves by 2 meter.
     
  4. Apr 10, 2015 #3
    I hope the original problem gave some units for g!
     
  5. Apr 10, 2015 #4

    haruspex

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    In case pcm's post is not clear enough, it's the forces on the block that matter. You don't need to analyse forces on the cart since its motion is a given.
     
  6. Apr 11, 2015 #5
    I also want it's answer.Please guide how to start?
     
  7. Apr 11, 2015 #6

    haruspex

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    What are the horizontal forces acting on the block? What is the acceleration of the block? What equation does that give you?
     
  8. Apr 11, 2015 #7
    Friction and tension.
     
  9. Apr 11, 2015 #8
    What is the acceleration of the block?
    Same as acceleration of the car(truck)?
     
  10. Apr 11, 2015 #9
    Net force on the block=mass of the block multiplied by acceleration of the truck.
     
  11. Apr 11, 2015 #10
    if the acceleration of the block = 2 * truck acceleration, then the force required in the string = m * a
    then add the friction force
     
  12. Apr 11, 2015 #11
    why should I add?
     
  13. Apr 11, 2015 #12
    acceleration force is forward (say + ve) and friction force is rearward (say - ve), the tension force is the difference.
    adding gets the same result.
     
  14. Apr 11, 2015 #13

    haruspex

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    Yes, it produces the same equation, but I feel it would be less confusing for gracy to stick to the normal formulation ##\Sigma F_x = m a_x##.
    gracy, do you see why the acceleration of the block is not the same as the acceleration of the truck?
     
  15. Apr 11, 2015 #14
    But according to
    The page number 6 of attached pdf
    the direction of friction force should be forward.
     

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    Last edited: Apr 11, 2015
  16. Apr 11, 2015 #15

    haruspex

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    That's a different situation. In the textbook example, the only force accelerating the block is the friction, so of course it must be acting in the forward direction. In this thread, there is a string pulling on the block.
    Friction always opposes relative motion of the two surfaces in contact. Which way does the block move relative to the truck?
     
  17. Apr 11, 2015 #16
    rearward.
     
  18. Apr 11, 2015 #17
    Why?
     
  19. Apr 11, 2015 #18

    haruspex

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    Yes, sorry, I meant in this thread, which way does it move relative to the "flat car".
     
  20. Apr 11, 2015 #19

    haruspex

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    What is the acceleration of the block relative to the car?
    Does the total length of the string change?
     
  21. Apr 11, 2015 #20
    After Considering pull on block by string?
     
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