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Find the torque about the knee

  1. Nov 17, 2008 #1
    1. The problem statement, all variables and given/known data
    I need to solve a problem where a 7.5 Kg object is attached to an ankle and leg lifts are done. There is a picture that shows the leg being at rest where the angle is 0, then legs its lift to 30 degree angle, 60 and 90 degree angle

    I am suppose to find the torque about the knee due to this weight for the 4 positions (each angle)


    2. Relevant equations

    torque = rF sin (angle)

    3. The attempt at a solution

    So i used the equaiton above to solve this problem which gave me

    at 0 degrees = 0 N.m
    at 30 degrees = 15 N.m
    at 60 degrees = 25 N.m
    at 90 degrees = 30 N.m

    is this right????
     
  2. jcsd
  3. Nov 17, 2008 #2

    LowlyPion

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    Re: torque

    How many Newtons is 7.5 kg ?

    Do you have access to any Sine tables to look up the values of Sine for those angles?

    How long is it again from the ankle to the knee? I missed seeing that in your statement.
     
  4. Nov 17, 2008 #3
    Re: torque

    ok 7.5 kg = 73.5 N
    also the lenth of the leg is 40 cm so .4 m
     
  5. Nov 17, 2008 #4

    LowlyPion

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    Re: torque

    And what is that as torque over that distance - expressed as N-m?
     
  6. Nov 17, 2008 #5
    Re: torque

    .4 m x 73.5 N = 29.4
     
  7. Nov 17, 2008 #6

    LowlyPion

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    Re: torque

    OK, so is that one of your answers and if so for what angle?
     
  8. Nov 17, 2008 #7
    Re: torque

    it would be for angle 0 right??
     
  9. Nov 17, 2008 #8

    LowlyPion

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    Re: torque

    If the leg is laying flat when just lifted then yes. 0 degrees relative to x-axis. 1 down.

    Now what is the distance of the weight from the pivot when the leg is at 30 degrees?

    Which trig function gives you that distance the force is acting straight down through to the pivot?
     
  10. Nov 17, 2008 #9
    Re: torque

    cos
    I will have to use the .4 m as the side of the triagle the 30 degree angle makes this way i can find the distance of the 30 degree angle by using cosine

    so

    .4 (cos 30) = .346 m

    using this i have to plug in to the torque equation do i have to use sin of 30 or no because since the force is not exactly being applied at an angle right??
     
  11. Nov 17, 2008 #10

    LowlyPion

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    Re: torque

    The Cos already accounts for the angle.

    Using .346 m is correct.
     
  12. Nov 17, 2008 #11
    Re: torque

    oh ok in that cas
    .346 m (73.5) = 25.4 N-m for angle 30

    for angle 60 .173 m (73.5) = 12.7 N-m

    for angle 90 will it be 0 since cos of 90 = 0???
     
  13. Nov 17, 2008 #12

    LowlyPion

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    Re: torque

    Not quite. It's Cosine taken over the original .4m not the .346 at 30 degrees.
     
  14. Nov 17, 2008 #13
    Re: torque

    so i would have to use cos of 60 times .4
     
  15. Nov 17, 2008 #14

    LowlyPion

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    Re: torque

    Correct.
     
  16. Nov 17, 2008 #15
    Re: torque

    oh ok but still for the angle of 90 its 0 or no?
     
  17. Nov 17, 2008 #16

    LowlyPion

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    Re: torque

    Cos 90 is zero.

    Multiplying by zero covers up a lot of sins.
     
  18. Nov 17, 2008 #17
    Re: torque

    so at 90 there is no torque
     
  19. Nov 17, 2008 #18

    LowlyPion

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    Re: torque

    All of the weight is acting directly through the pivot when it's straight up isn't it? Hence no moment arm no torque.
     
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