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Static equilibruim and centre of gravity - HELP!

  1. Sep 3, 2009 #1
    1. The problem statement, all variables and given/known data

    Hi, Imagine a patient lying down on a bed , you want to weigh her without moving her, but the scale can only support the pair of legs of the bed at one end - or the pair at the other end , but not simultaneously. So he puts the scale at point X (the pairs of bed legs either side of the her feet ) at it was 700N , at point Y (the pair of legs either side of her head) the scale read 800N , The Dr knows from previous determination the bed weighs 90kg.


    2. Relevant equations

    i)the total force exerted by the floor on the bed A) 1500 - this i understand
    ii) the mass of the patient is ? A) 60kg - again i understand this

    here comes the brain teaser

    III) Suppose the length of the bed , the horizontal distance between X and Y is L.

    At what horizontal distance from Y is the center of gravity of the patient , bed and bedding considered a single body located?

    Answer = (7/15)L

    PLEASE HELP!!

    3. The attempt at a solution

    i now understand that the forces upwards are 700n and 800n so the total force of gravity must be equal to that ? but how do i locate it what equation do i need, is the total magnitude of the point of gravity 1500N and the center of gravity from point Y is the direction??

    Thanks in advance

    Amy
     
  2. jcsd
  3. Sep 3, 2009 #2

    tiny-tim

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    Welcome to PF!

    Hi Amy! Welcome to PF! :wink:

    Hint: use moments :smile:
     
  4. Sep 3, 2009 #3
    please guys im begging you ! Just explain this to me i havent done physics in 7 years!

    I have read about torque moment arms etc , but my problem is i dont know how to apply it to this question. a previous forum member explain the forces upwards are 700n and 800n which i understand but how do i tie this with vectors and torque, someone please go through the calculations with me.
     
  5. Sep 3, 2009 #4

    tiny-tim

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    Hi amy84! :smile:

    Moment (or torque) = force times distance …

    it's like a seesaw: if you sit twice as far away, then you exert twice as much "turning force".

    If this was a seesaw, where would you put the pivot to balance the bed?

    and i'm going to bed now … g'night! :zzz:​
     
  6. Sep 3, 2009 #5
    well on a see saw you can move up if you are heavier to balance it, in this problem you can not, and in a see saw you can not change the pivot point it is stuck to the ground,

    thanks for you help but just going to bed when you could have just explained the answer a bit better is a bit harsh - what a power trip! so i will spend another night looking for this answer while you got have explained it in a bit more detail?

    Anyone else kind enough to shed some light
     
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