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Mechanics lift problem

  1. Oct 22, 2012 #1
    1. The problem statement, all variables and given/known data
    a light scale pan is attatched to a vertical inextensible sring. The scale-pan carries two masses A and B. The mass of A is 400g and the mass of B is 600g. A rests on top of B, as shown in the diagram.
    The scale-pan is raised verticaly, using the string with acceleration 0.5 m/s².
    a)find the tension in the string.
    b)find the force exerted on mass B by mass A
    c)find the force exerted on mass B by the scale-pan.

    visual representation of the problem http://i1269.photobucket.com/albums/jj597/bubakazouba/test_zps62ac1992.png [Broken]


    2. Relevant equations
    F=ma
    g=9.8 m/s²

    3. The attempt at a solution
    I got a) correct, I just said
    Tension-(0.4x9.8 +0.6x9.8)=0.4x0.5+0.6x0.5
    Tension-9.8=0.5
    tension =9.8+0.5=10.3N
    --------------------
    b)find the force exerted on mass B by mass A
    I think the answer is mg its the weight, because its the only downward force on A so it should be 0.4x9.8=3.92, but its not right.
    What am i missing here?
    Thanks in advance
     
    Last edited by a moderator: May 6, 2017
  2. jcsd
  3. Oct 22, 2012 #2

    tiny-tim

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    hi bubakazouba! :smile:
    what about the acceleration? :wink:
     
  4. Oct 22, 2012 #3
    what about it?
     
  5. Oct 22, 2012 #4

    tiny-tim

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    you haven't used it
     
  6. Oct 22, 2012 #5
    why should I use it?
     
  7. Oct 22, 2012 #6

    tiny-tim

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    i] it's in the question

    ii] F = ma :wink:
     
  8. Oct 22, 2012 #7
    I dont get it why should I use any equation, isn't it obvious the only force that acts on B is the weight, mg
     
  9. Oct 22, 2012 #8

    tiny-tim

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    that's the LHS of the equation

    what about the RHS?​
     
  10. Oct 22, 2012 #9
    what equation?
     
  11. Oct 22, 2012 #10
    I feel so stupid right now :'(
     
  12. Oct 22, 2012 #11
    Draw a free body diagram on A, showing the forces acting on it. Write a force balance on A, recognizing that A is accelerating upward at a rate of 0.5 m/s2.
     
  13. Oct 23, 2012 #12
    ok I got a force 4.12N upwards what does that mean?
     
  14. Oct 23, 2012 #13
    That's the contact force exerted by mass B on mass A. Now, from Newton's third law of action-reaction, what is the contact force that mass A exerts on mass B?
     
  15. Oct 23, 2012 #14
    4.12?
    but why, why isn't it the weight?
     
  16. Oct 23, 2012 #15
    4.12?
    but why, why isn't it the weight?
     
  17. Oct 23, 2012 #16
    Because mass A is accelerating upward, and so mass B has to exert enough upward contact force not only to support the weight of mass A (which would be the case if A were in equilibrium), but also to accelerate it. Also, by Newton's third law, the contact force A exerts on B is equal in magnitude and opposite in direction to the contact force B exerts on A.
     
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