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Moment of force on a door

  1. Aug 16, 2011 #1
    1. The problem statement, all variables and given/known data
    A regular house door has 3 forces acting on it.At the moment the door is open by 35 deg. Lets say two children are pushing on a door and mum is on the other side pushing back
    1/ at 0.05m is 50N acting directly perpendicular to the door
    2/ at 0.07m is 50N acting at right angles to the door frame
    3/ At 0.1m is an opposite direction force (clockwise[mum]from the other side of the door of 100n acting (as in case 2) perpendicular to the door frame or wall.


    2. Relevant equations
    What is the sum of forces??



    3. The attempt at a solution My attempt is as follows
    Summation sign:?::: =(50N) (0.05m) + (50N) (0.07 cos 35Deg) - (100N) (0.1 cos 35Deg)

    I am working of Moavini page 264.

    I am essentially asking if my equation is correct in that i have SUBTRACTED the last force as it opposes the first two - am i right?? I subtract because the is clockwise not anti clockwise

    Thanks ahead every body... keen engineering student... Peter
     
  2. jcsd
  3. Aug 16, 2011 #2

    PhanthomJay

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    Hi Peter welcome to the Forums!

    Are you looking for sum of forces or sum of moments? If moments about the hinged size of the door, and your distances are measured from that side, then your answer for sum of moments looks good. Your subtraction of the last term is also correct. I don't have a copy of Moavani page 264 handy:smile:
     
  4. Aug 17, 2011 #3
    Right thanks Phan, My teacher answered me to say that

    Anti clockwise forces are given a negative notation and
    Clockwise forces a positive notation. Although you can swap those values around, i think it is a good bookeeping excercise - as opposing forces are negative.

    which is where i went wrong.

    thanks ..Peter
     
    Last edited: Aug 17, 2011
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