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Torque/rotational motion question

  1. Nov 7, 2007 #1
    1. The problem statement, all variables and given/known data
    A person standing on tiptoe, the position of the foot is as shown. *The total gravitational force on the body Fg is supported by the force n exerted by the floor on the toes of one foot. A mechanical model for the situation is shown in Figure P10.70b (http://www.piercecollege.edu/faculty/meyermd/Physics 66/P66 Homework/Chapter 10 HW PDF.pdf Problem 70) where T is the force exerted by the Achilles tendon on the foot and R is the force exerted by the tibia on the foot. Find the values of T,R, and θ when Fg = 700N

    2. Relevant equations
    torque(tau) = r x F (that's r cross F)

    3. The attempt at a solution
    I assumed that the Fg and the normal force cancel out from the 2nd line in the problem statement. I also tried moving the Fg to different positions, but the answers dont seem correct. Using hte [tex]\sum[/tex]of torques, whereas the axis of rotation is at the toes(normal force), i got torque = 0 = -.18 * R * sin(θ) + .25 * T, I dont think that's R sin(θ) because R is not the hypotenuse. I also tried assuming that θis 15 degrees, and that R was perpendicular to the Foot. Other than the sum of torques, I don't know what other formulas to use, or what kind of approach to take for this problem.
    Another method I used was to move the axis of rotation around, and use 4 equations for 4 unknowns, but that proved to be an incorrect method (the results didnt seem correct). I also think that the Fg = 700 N is supposed to come into the equation somewhere, and not cancel out with the normal force, but I have no idea what to do with that if it's not canceled. Can anyone help me with an approach I can take to solve this problem? Thanks
  2. jcsd
  3. Nov 7, 2007 #2


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    Homework Helper

    the sin(theta) shouldn't be there. torque is the force times the perpendicular distance from the pivot to the line of the force... you had the T part right. so it should be:

    that's one equation.

    you get 2 more equations by setting sum of forces in x-direction = 0 and sum of forces in y-direction = 0.
  4. Nov 9, 2007 #3

    Thanks for the help, can't believe i forgot the basics of sum of the forces lol
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