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Question regarding Moments

  1. Sep 8, 2015 #1
    1. The problem statement, all variables and given/known data
    "A child and her father are playing on a seesaw. They are exactly balanced when the girl (mass 46 kg) sits at the end of the seesaw, 2.75m from the pivot. If her father weighs 824 N, how far is he from the pivot?"

    2. Relevant equations
    (sum of) clockwise moment(s) = (sum of) anticlockwise moment(s)
    moment (Nm) = force (N) * perpendicular distance from pivot to line of action of the force (m)

    3. The attempt at a solution
    girl is mass 46kg (kilograms) whilst her father has weight 824 N (newtons)?!
    46 x 2.75 = y x 824
    126.5 = 6.51 * 824
    y = 6.51?
    = 6.51 metres from pivot?

    Please show how you got your answer =]

    Thanks,
    Robm50
     
  2. jcsd
  3. Sep 8, 2015 #2

    Orodruin

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    6.51 * 824 is definitely larger than 126.5 - you have made an arithmetic error.

    Edit: Also, the girl has mass 46 kg - not weight 46 N, which is what you seem to be assuming.
     
  4. Sep 8, 2015 #3
    How would I go about solving this problem, then?
     
  5. Sep 8, 2015 #4

    Orodruin

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    What is the weight corresponding to a mass of 46 kg?
     
  6. Sep 8, 2015 #5
    That's exactly my problem; I have no idea how to find such a weight.
     
  7. Sep 8, 2015 #6

    Orodruin

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    So, what is weight? If you do not know this you need to go back to your course literature and restudy gravitational forces.
     
  8. Sep 8, 2015 #7
    Weight is the result of the gravitational attraction pulling an object's mass particles towards the Earth

    Is that correct?
     
  9. Sep 8, 2015 #8

    Orodruin

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    Weight is the force with which gravitation pulls an object towards the ground. So how do you compute this force based on the mass of an object?
     
  10. Sep 8, 2015 #9
    Would it be as simple as by using the formula:

    "Weight = Mass * Gravitational Field Strength"?
     
  11. Sep 8, 2015 #10

    Orodruin

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    Yes. This will give you the weight of an object.
     
  12. Sep 8, 2015 #11
    Thankyou for your help, Sir.
     
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