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Get Full Access to University Physics - 13 Edition - Chapter 11 - Problem 67p
Get Full Access to University Physics - 13 Edition - Chapter 11 - Problem 67p

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BIO Downward-Facing Dog. The yoga exercise

ISBN: 9780321675460 31

Solution for problem 67P Chapter 11

University Physics | 13th Edition

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Problem 67P

BIO Downward-Facing Dog. The yoga exercise “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 750-N person as shown in Fig. P11.63?. When he bends his body at the hip to a 90° angle between his legs and trunk, his legs, trunk, head, and arms have the dimensions indicated. Furthermore, his legs and feet weigh a total of 277 N, and their center of mass is 41 cm from his hip, measured along his legs. The person’s trunk, head, and arms weigh 473 N, and their center of gravity is 65 cm from his hip, measured along the upper body. (a) Find the normal force that the floor exerts on each foot and on each hand, assuming that the person does not favor 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 on both hands (but not necessarily the same on the feet as on the hands). [?Hint: First treat his entire body as a system; then isolate his legs (or his upper body).]

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Solution 67P Here, we need to consider the person’s legs and upper body separately and then find the moments of the weight about those portions center of gravity. Let us have a look at the simplified figure below. Now, applying Pythagoras theorem, we can calculate the distance between his feet and hands. Let, d be the distance. Therefore, d = .9 + 0.135 m 2 d = 1.62 m Let the horizontal distance of his center of mass of legs from the feet is =1x Let the horizontal distance of his center of mass of the upper body is = x2 Let 1e the angle made by his feet with the horizontal. Therefore, cos =1 0.9 = 0.55 1.62 = 56.6 0 1 If 2s the angle made by his hands with the horizontal then, 0 0 0 2 90 56.6 = 33.4 Therefore, x = (0.90 0.41) cos 56.6 0 1 x = 0.27 m 1 0 Similarly, x 2 1.62 m (0.1 + 0.6)m × cos 33.4 x 2 1.62 m 0.58 m x 2 1.04 m Now, the net torque about the person’s hand is, x w + x w = d × R 1 1 2 2 2 0.27 m × 277 N + 1.04 m × 473 N = 1.62 m × R 2 74.8 N.m + 491.9 N.m = 1.62 m × R 2 R =2350 N (a) Therefore, his both hands will experience normal force of 350 N. On each hand, the normal force will be = 350/2 N = 175 N Now, the weight of the person is given to be = 750 N Therefore, the normal reaction on his feet R = 150 350 N = 400 N So, on each foot, the normal reaction is = 400/2 N = 200 N (b) Friction force acts as shown in the figure above. The normal reaction is 400 N. Therefore, the component of normal reaction along the feet 0 = 400 N × cos 33.4 = 334 N Therefore, the component of this force along the horizontal is 0 = 334 N × cos 56.6 = 183.8 N This component will be equal in magnitude to the friction force on his feet. Therefore, the friction force on his each foot = 183.8 N/2 = 92 N .

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