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Centre of gravity in between two points of different wieght

  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

    Im lost and exam in two weeks! I have read about torque and center of gravity but can't find anything that helps me understand this.

    Thanks in advance

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

    kuruman

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    Hi amy84. Welcome to PF. Don't let the existence of one scale confuse you. Suppose the good doctor had two scales placed on each end. Would that change anything? No, because the bed doesn't "know" or care what it is resting on. Scale or floor, it exerts the same force. Once in equilibrium, always in equilibrium. So you have two scales, one at the feet reading 700 N and the other scale at the head reading 800 N. Since the bed is in equilibrium, the sum of all the torques must be zero. Can you calculate the sum of all the torques (using the center of gravity as reference) and say they are equal to zero? Making a drawing that shows where all the forces acting on the system are will be extremely helpful.
     
  4. Sep 3, 2009 #3
    thanks for your post

    I have attached a picture i did on paint which is the same as the one in my question (obviously much worse)

    Again in the last question i just have no idea about , i mean the force at point Y is 800 how can the answer be 7/15 multiplied by the whole length.

    i cant find the valid formulas or equations

    Thanks in advance

    Amy
     

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  5. Sep 3, 2009 #4

    kuruman

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    Have you studied torques? Remind yourself what they are and how they are calculated.
     
  6. Sep 3, 2009 #5
    Yes torque is the force vector multiplied by the moment length, so how does this apply here , what are the angles? I much appreciate your time in reading my questing, but could you explain the last question to me so I can make my own deductions or try and figure it out from the worked answer.

    thanks

    amy
     
  7. Sep 3, 2009 #6

    kuruman

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    There are three forces acting on the system at different points. All three are along the vertical. Two forces you are already given, they are the normal forces exerted by the floor and act at the two ends of the system in the "up" direction. The third force is gravity and acts at the center of gravity in the "down" direction. You know its magnitude, but you don't know its point of application; that's what you are looking for. Say that point is at distance x, measured from the patent's head. You are looking for this x.

    I think I have explained the last question. I suggest that you form a strategy for finding x that involves writing an equation with x as part of it. I will leave it at that, but if you need more help, please ask.
     
  8. Sep 3, 2009 #7
    okay that sheds new light of the answer, never thought of the forces at X and Y to be upwards, so how does these forces related to the center of gravity? i have looked at examples of say a picture frame being held up being in static equilibrium, so the angles of the foces upwards come into play, but there are no angles from the force upwards to the table, i just dont get it, do you know what values the 7 and the 15 relate to? is it 700/1500 (the total force).

    Im really sorry for my ignorance but with my exam coming up im just panicy, what you said makes sense but that doesnt actually help me calculate it, please if you have the time explain how the calculation was done with formulas and how they relate, i wish i could just move on to another section but its really bugging me -
     
  9. Sep 3, 2009 #8

    kuruman

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    You need to draw yourself a diagram of a rectangle (the system) and the three forces acting on the system. Two forces(700 N and 800 N) are up and one force (weight) is down. Starting at the center of gravity draw the moment length to the 800 N force and the 700 N force. Note that the moment length to the 800 N force is x. Now answer the following questions

    1. Can you calculate the torque generated by the 800 N force?
    2. Can you find the moment length to the 700 N force in terms of x and L?
    3. Can you find the torque generated by the 700 N force?

    Note that in each case the moment length and the force are perpendicular to each other. Answer the three questions, and if you still don't see what is going on, ask again. There is plenty of time before your exam, but the important thing is for you to do it yourself and understand what you are doing.
     
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