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Forces pulley and bricks Problem

  1. Sep 29, 2016 #1
    1. The problem statement, all variables and given/known data
    I have a man who is 100kg and a barrel of bricks weighing at 220kg. The man is using a pulley to lower the bricks onto the ground however since they are much higher in weight he rapidly travels up. Then the bricks fall onto the ground and the barrel weighs 60kg since it's lighter than the man, he then rapidly goes down now. I have to find the acceleration of when the man is traveling up and the bricks are going down and when the man is going down and the barrel is going up.

    2. Relevant equations
    Fnet = ma
    Fg = mg

    3. The attempt at a solution

    For the man it would be:
    Fg=mg
    Fg= (100)(9.81)
    Fg=981N

    For the barrel full of bricks it would be:
    Fg= mg
    Fg= (220)(9.81)
    Fg= 2158.2N

    For the barrel that's empty it would be:
    Fg= mg
    Fg= (60)(9.81)
    Fg= 588.6N

    I am not quite sure what to do next. All I know is that I have to find acceleration of the barrel and man separately. I thought that I would have to use Fnet = Fg but then there is tension which I do not know. So then I tried to do:
    Fnet = ma
    Fg = ma
    but it does not work of course.








     
  2. jcsd
  3. Sep 29, 2016 #2

    andrewkirk

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    Let M be the mass on the barrel end of the rope and m be the mass of the man, ##a## be the acceleration (which has the same magnitude for both m and M) and T the tension in the rope. Let g be the acceleration due to gravity.

    We have five variables, of which only two are unknown: ##a## and T.

    This can be solved by writing two equations, one that applies the ##F=ma## law to the barrel and one that applies it to the man. Note that the values of F we use here are the net forces on the accelerated objects, arising from gravity and the tension in the rope.

    With two equations and two unknown variables, you should be able to solve the problem.
     
  4. Sep 29, 2016 #3
    I got 4.0m/s2 for the first acceleration and for the second 2.5m/s2. Still gonna find tension.
     
  5. Sep 29, 2016 #4

    andrewkirk

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    Each of your two equations will have ##a## and ##T## in them. Substitute the value you got for ##a## into either of the equations and you'll be able to solve for ##T##.

    But first check that calc that gave you 4.0 ##ms^{-2}##. I agree with your second answer but for the first one I get a lower answer.
     
  6. Sep 29, 2016 #5
    Was it 3.6m/s2?
     
  7. Sep 29, 2016 #6

    andrewkirk

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    I get 3.67, which rounds to 3.7, not 3.6.
     
  8. Sep 29, 2016 #7
    I apologize for that. For Ft I got 1348.875N for my first tension. As for my second 735.750N.
     
  9. Sep 29, 2016 #8
    Is the velocity of the man relative to the barrel of bricks 21m/s?
     
  10. Sep 29, 2016 #9

    andrewkirk

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    Since both are accelerating in opposite directions, the relative velocity changes over time. A time needs to be specified in order to calculate a relative velocity.
     
  11. Sep 29, 2016 #10
    I did relative velocity using formula VBC = VBA + VAC
     
  12. Sep 30, 2016 #11

    haruspex

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    I do not understand how velocities came into it. Everything up to post #7 concerned forces and accelerations.
     
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