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Pulley Question, mechanics question

  1. Oct 12, 2016 #1

    Kajan thana

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    Gold Member

    1. The problem statement, all variables and given/known data
    Two particles A and B of mass 4kg and 3kg respectively are connected by a light inextensible string which passes over a small smooth pulley. The particles are released from rest with taut.

    When A has travelled a distance of 2m it strikes the ground and immediately comes to rest.
    Question: Assuming the B does hit the pulley find the greatest height that B reaches above its initial position.

    2. Relevant equations
    Kinematic equations

    3. The attempt at a solution

    I found out the the time it takes for the particle A to go down by 2m using this equation s=ut+0.5at^2.
    The time is 1.69 rounded to two decimal place so if Particle A go down by 1.69 second thant the Particle B will go up by 1.69 second.

    Now I know that for Particle B, U=0 A=-1.4 T=1.69 so using kinematics I can find the displace which will be -1.183m so the distance it goes by 1.183m. Which is a incorrect answer.
     
  2. jcsd
  3. Oct 12, 2016 #2
    Why is the distance that particle B goes up different from the distance that particle A goes down in the 1.69 sec? Let's see your calculation of the acceleration.
     
  4. Oct 12, 2016 #3

    gneill

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    Staff: Mentor

    Particle B is rising and accelerating upwards so long as the string is taut (particle A hasn't hit the ground). So when A hits the ground and the string goes slack, particle B will still have whatever velocity it achieved until that moment. It becomes a projectile launched with that initial speed upwards from that point. So when you wrote:

    You were not considering the new projectile status of particle B. At the instant particle B becomes a free projectile, the motion from its travels while moving due to A's influence are behind it. You need to consider the new situation for B from that point on.

    Edit: Ah, Chestermiller got there before me with a much more subtle approach :smile:
     
  5. Oct 12, 2016 #4

    Kajan thana

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    4g-t=4a and t-3g=3a solving it with elimination will get you a=g/7 same as 9.8/7
     
  6. Oct 12, 2016 #5
    Good. So, since their accelerations are the same and their initial velocities are the same, how come they travel different distances in the same amount of time?
     
  7. Oct 12, 2016 #6

    Kajan thana

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    Thanks Mentor It made so much sense now after you clear explanation so just to make sure I understood the concept.
    The acceleration I found is constant until it touches the ground after that it will change.
     
  8. Oct 12, 2016 #7
    So, are you OK now, and have you achieved the answer you desire?
     
  9. Oct 12, 2016 #8

    Kajan thana

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    Gold Member

    Yes I think I did mentor.

    So there will be 2m distance caused by the particle A. Then when the particle A touches the ground the string no longer will taut so the acceleration can not be applied so I will apply a=-9.8 with the initial velocity as same as the instantaneous final velocity of particle A. and the final velocity will be 0. Using kinematics the answer will be 16/7.



    Thanks again
     
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