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Questions about principle of moments

  1. Jun 21, 2007 #1
    1. The problem statement, all variables and given/known data

    Hi
    I have a physics exam tmorrow and I was hoping I could get a little help with the moments problem. I attached the drawing and the question asks:
    a-1. State the principle of moments
    b-1.Write down expressions including W, Rc and FA for
    (i)the total counter clockwise moments about pivot B
    (ii)the total clockwise moments about pivot B
    (iii)what is the value of Rc when the first boom loses contact with its cradle
    2.Use the results in 1. (i),(ii) and (iii) above to find the mass of the counterweight that must be used if FA=25N is just enough to start raising the boom (g=10ms-2)

    2. Relevant equations



    3. The attempt at a solution
    a-1 the principle of moments states that for a system in equilirium the sum of the clockwise moments is equal to the sum of the anitclockwis moments
    b-1(i)600N X 6m=3600Nm

    And we were never atught how to answer the rest so from there i'm lost.Help plz
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     

    Attached Files:

  2. jcsd
  3. Jun 21, 2007 #2

    berkeman

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    Staff: Mentor

    But what is a moment defined as, in terms of a force and a lever arm length?

    Starting with the sum of the moments being zero in equilibrium is good, but then you need to use that to help you calculate the forces that are spaced at different distances from the pivot....
     
  4. Jun 22, 2007 #3

    andrevdh

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    Homework Helper

    In this problems the moments are easy to calculate since the moment arms are just the horizontal distances to the line of action of the involved forces.

    For instance for the weight of the arm of the boom, 600 N, which rotates clockwise about point B the moment will be

    [tex]600 \times 3 = 1800 \ Nm[/tex]

    clockwise since the moment arm for this force is three meters.

    The normal reaction force from the cradle,[tex]R_C[/tex], rotates the boom anticlockwise about the point B so its moment will be

    [tex]R_C \times 6 = 6R_C \ Nm[/tex]

    anticlockwise about B.

    To answer part (iii) you need to consider what happens as the applied force, [tex]F_A[/tex], is increased. What will happen is as the applied force is increased the reaction force of the cradle will decrease while the boom is still in equilibrium. When the reaction force becomes zero the boom is at the point of lifting up, but the clockwise anticlockwise moments will still balance out at this point. Only when the applied force is increased even further will the boom begin to lift.
     
    Last edited: Jun 22, 2007
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