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Time it takes for block to slide down an incline in elevator

  1. Jan 31, 2016 #1
    1. The problem statement, all variables and given/known data

    MECHANICS:

    Given Theta, L, M, and acceleration of elevator relative to ground. Find the time it takes for the block to reach the end of the incline.

    Here is a diagram: http://k-elahian.com/tmp/nip.PNG [Broken]

    2. Relevant equations
    f=ma
    kinematics
    relative acceleration

    3. The attempt at a solution
    I've gotten 3 different answers so far. Basically I do f=ma for the m mass and then split it into x and y (slanted system) and then I relate the accelerations of the the bodes Amg=Ame+Aeg (ground is g and elevator is e).
    After finding Ame I use that in my kinematics equation.
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Jan 31, 2016 #2

    Student100

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    Your notation is a bit odd, what does Amg=Ame+Aeg mean exactly? What is the A? Can you explain what exactly you mean?
     
  4. Jan 31, 2016 #3

    Suraj M

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    I'm not sure if I understand what you've tried to do
    But start by calculating the pseudo force on the block due to the elevators Motion and also the other forces on it. I think you can proceed from there.
    You say you've got 3 answers
    Unless you state what were your answers, how should we judge if it's right or wrong?
     
  5. Jan 31, 2016 #4

    Suraj M

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    What does Aeg mean? Is that relative acceleration?
     
  6. Jan 31, 2016 #5
    Yes,
    Amg is acceleration of m relative to g
    Aeg is acceleration of elevator relative to ground.
    Since I cannot use fictitious forces I need to use relative accelerations.

    Amg=Ame+Aeg just shows that the a of the mass relative to the ground is the sum of the 2 relative accelerations vectors
     
  7. Jan 31, 2016 #6

    Student100

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    Okay, so what equation did you find pre-substitution?
     
  8. Jan 31, 2016 #7

    -mgsinΘ = mamgx for the x and -mgcosΘ+Fn=mamgy for the y
    are the x and y components of f=ma for the block, but I am using a slanted coordinate frame
     
  9. Jan 31, 2016 #8

    Student100

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    Okay, so I don't understand what you're trying to write.

    Maybe it will help if we just look at an incline plane first without the added acceleration, If we rotate the coordinate axis ##\theta##, what's the acceleration down the plane in this case?
     
  10. Jan 31, 2016 #9
    yeah I did that in the equation above. for the y, -mgcosΘ+Fn are the 2 forces acting on the block. (slanted frame). amgy just means the acceleration of the mass relative to the ground in the y direction
     
  11. Jan 31, 2016 #10

    Student100

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    Why is there an acceleration in the y direction in the rotated coordinate system?

    Hang with me for a second I'll get you to the answer, ignore the elevator for a second, what's the sum of the forces in the x and y looking only at an inclined plane?
     
  12. Jan 31, 2016 #11
    Ok, It's just Fg and Fn. Those are the only 2 forces acting on the block.

    for X: -mgsinΘ , no fn here and for Y: -mgcosΘ+Fn
     
  13. Jan 31, 2016 #12

    Student100

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    So ##mgsin(\theta)=ma## and ##N - mgcos(\theta) = 0##, correct?

    You know why the y direction is in equilibrium when we rotate the system, correct?
     
  14. Jan 31, 2016 #13
    Yes, if we were not in a elevator yes. But that would be a non-inertial frame so we can't do that, according to my teacher
     
  15. Jan 31, 2016 #14

    Student100

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    Yes, so if you now look back at the accelerating system relative to the plane, this time don't rotate your coordinate system and solve for ##a_{rel}## relative down the plane.
     
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