One yoga exercise, known as the \"Downward-Facing Dog,\" requires stretching you
ID: 1976118 • Letter: O
Question
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 certain 735 N person. When he bends his body at the hip to a 90.0 degree angle between his legs and trunk, His legs are 90 cm long, his trunk is 75 cm and his arms are 60 cm. Furthermore, his legs and feet weigh a total of 250 N, and their center of mass is 41.0 cm from his hip, measured along his legs. The person's trunk, head, and arms weigh 485 N, and their center of gravity is 65.0 cm from his hip, measured along the upper body.Find the normal force that the floor exerts on each foot, assuming that the person does not favor either hand or either foot.
In the free body diagram you should have a right triangle with the right triangle as the top point. the left side of the triangle should be 90 cm (representing his legs); and the right side of the triangle should be 135 cm (representing his arms and trunk).
Explanation / Answer
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.AND 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. AND
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