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Downward-Facing Dog Equilibrium Question

  1. Nov 1, 2009 #1
    1. The problem statement, all variables and given/known data
    One yoga exercise, known as the downward-facing dog, requires stretching your hands straight out above your head and bending down to lean against the floor. This exercise is performed by a 750N person. When he bends his body at the hip to a 90 degree angle between his legs and trunk, a right angle triangle is formed with his legs having a length of 90cm from the feet to the hip, his trunk having a length of 75cm from the hip to the top of the head, and his arms having a length of 60cm from the head to his hands. The person's legs and feet weigh 277N, with their centre of mass located 41cm from the hip, measured along the legs. The person's trunk, head, and arms weigh 473N, with their centre of gravity located 65cm from the hip, measured along the upper body.

    a) Find the normal force that the floor exerts on each foot and hand, assuming the person does not favour either hand or either foot.

    b) Find the friction force on each foot and on each hand, assuming that it is the same on both feet and hands (but not necessarily the same on the feet as on the hands).

    Hint: First treat his entire body as a system, then isolate either the legs or upper body.



    2. Relevant equations
    Static equilibrium equations (net force = 0, net torque = 0)



    3. The attempt at a solution
    Sorry for the wall of text there, but I didn't know how to slim it down any way :tongue:. My problem is that I can't seem to get the right answers for this question. What I first did was draw a right triangle, and found the other two angles using the inverse of tan. I then isolated the legs, making the pivot point at the hips. Afterwards, I drew a free body diagram of the legs, noting that the legs made an angle of 56.3099 degrees with the horizontal, and using this information I set up a net torque expression as such:

    net torque = rNFN + rFFF = 0, where the subscript N denotes normal force and F denotes the weight of the feet and legs.

    I then plugged in the following values: 0 = (0.9sin33.69)FN + (0.41sin33.69)(-277N) and solved for the normal force, which came out to be: 126.18889N. Dividing this by 2 got me: 63.1N per foot, which is wrong; this answer is supposed to be 200N per foot. Am I missing a force in my method here, or is there another mistake I am making? If anyone can help me out it would be greatly appreciated, thanks in advance.
     
  2. jcsd
  3. Nov 1, 2009 #2
    Isolating the legs or upper body only comes at part b.

    To find the normal force on the feet, you can use the fact that all the torques around the hands must be 0.
     
  4. Nov 1, 2009 #3
    Ahh I see now how to do part A. I'm still lost on part B however; how would I determine the force of friction on both the hands and feet? I know they will both act horizontally, but i'm not sure how to separate the upper and lower halves of the body to solve for either of them.
     
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