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Mass Sliding at An Angle; find mass of one

  1. Sep 20, 2011 #1
    1. The problem statement, all variables and given/known data
    M1 and M2 are two masses connected as shown.
    (M1 on hypotenuse of triangle, M2 on right side, (adjacent to right angle)) (Pulley at corner of them, Angle Theta is on the opposite corner)
    The pulley is light and frictionless. Find the mass M1, given that M2 (4.50 kg) is moving downwards and accelerates downwards at 3.09 m/s2, that θ is 20.0°, and that μk is 0.450.



    2. Relevant equations
    The total force (which is the total mass times the acceleration) is equal to the sum of the forces. The normal force is NOT simply "m1g"--WHY?



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 20, 2011 #2
    It would be better if you could show us a diagram of what you are talking about. :smile:
     
  4. Sep 20, 2011 #3

    PeterO

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    I couldn't see the "as shown"?
     
  5. Sep 20, 2011 #4

    PeterO

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    OK, I was expecting "as shown" to be a diagram.

    The Normal force is perpendicular to the slope [that is what the "Normal" means - as in Tangent and normal ... Tangent, parallel to surface; Normal, perpendicular to surface.]
    mg is vertically down, so you need to use some components to calculate the Normal force.
     
  6. Sep 20, 2011 #5

    PeterO

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    Without the string, M2 would accelerate down at g - either 9.8, 9.81 or 10 depending which value you have been advised to use.
    The fact it accelerates at only 3.09 means the string is providing a force trying to prevent the acceleration. This force is due to two contributions.
    Part of it will be friction between M1 and the slope.
    Part of it will be the component of the weight force of M1 acting down the slope.
     
  7. Sep 20, 2011 #6
    yes, but what components?
     
  8. Sep 20, 2011 #7
  9. Sep 20, 2011 #8
    Last edited by a moderator: Apr 26, 2017
  10. Sep 20, 2011 #9

    PeterO

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    We always take components parallel to and perpendicular to something convenient,
    Here that is parallel to the slope and perpendicular to the slope.

    You want those components of M1g.
     
  11. Sep 20, 2011 #10
    The picture should be attached now.
     

    Attached Files:

  12. Sep 20, 2011 #11

    PeterO

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    Last edited by a moderator: Apr 26, 2017
  13. Sep 20, 2011 #12
    Ok, first lets make the equation for the M2 block.
    What are the forces acting on the M2 block? :smile:
     
  14. Sep 20, 2011 #13
    On the M2 block there is gravity and normal force ?
     
  15. Sep 20, 2011 #14

    PeterO

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    Read post 5 and think about it.
     
  16. Sep 20, 2011 #15
    You are forgetting one force "Tension". :smile:
    Now form the equation for M2 block.
     
  17. Sep 20, 2011 #16
    M2 : T - mg = ma
    ?
     
  18. Sep 20, 2011 #17
    Yep, that's right.
    Now plug in the values and find tension. :)
     
  19. Sep 20, 2011 #18
    is the acceleration negative because the box is falling down?
     
  20. Sep 20, 2011 #19
    No. :)
     
  21. Sep 20, 2011 #20
    k, so T = 58.05N
     
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