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Torque of a woman leaning against a wall

  1. Dec 22, 2008 #1

    A woman who weighs 496 N is leaning against a smooth vertical wall. Find

    Fx= 0= -F Wall = Gx
    Fy= 0= Gy- Fgwoman

    (a) the force FN (directed perpendicular to the wall) exerted on her shoulder by the wall
    T= 0= -Fgw*sin30* ((1.1+.4)/2)- Fgw sin 30 (1.1) + F wall cos 30 (1.5)

    F wall= 353.1844

    (b) horizontal and

    (c) vertical components of the force exerted on her shoes by the ground.

    I understand that the force of the horizontal is equal to the force of the wall, and the force of the vertical components is equal to the woman's weight, but I am not sure what is wrong with my equation, because F wall is incorrect.
  2. jcsd
  3. Dec 22, 2008 #2
    Do you have to find the torque on the woman? Torque about which axis? Wall is smooth. Is the floor smooth too?
    Could you explain the meanings of symbols you have used in the equations?
    Is the woman at rest? If yes, then don't you think that the net torque is zero?
  4. Dec 23, 2008 #3
    yeah, the net torque is zero. I need to find the force of the wall on the woman. -Fgw is the downward force of the woman (496 N)
  5. Dec 23, 2008 #4

    Doc Al

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    Staff: Mentor

    I understand the 2nd term (torque due to her weight) and the 3rd term (torque due to the wall force), but what is the meaning of your 1st term?
  6. Dec 23, 2008 #5
    You have to calculate Fn. Right?
    Take torques about the point where her feet touch the floor. Torque of force by floor = 0
    Take counterclockwise as positive.
    Torque by Fn = Fn * (1.10 + 0.400) * sin(60deg)
    Torque by weight = - 496 * 1.10 * cos(60 deg)
    Net torque = 0
    How much do you get for Fn?
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