One yoga exercise, known as the \"Downward-Facing Dog,\" requires stretching you
ID: 2155137 • 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 person, as shown in the simplified model in the figure below (Figure 1) . When he bends his body at the hip to a 90.0 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 250 , and their center of mass is 41.0 from his hip, measured along his legs. The person's trunk, head, and arms weigh 485 , and their center of gravity is 65.0 from his hip, measured along the upper body.A) Find the normal force that the floor exerts on each foot, assuming that the person does not favor either hand or either foot.
B) Find the normal force that the floor exertson each hand, assuming that the person does not favor either hand or either foot.
C) Find the friction force 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).
D) Find the friction on each foot, 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).
Explanation / Answer
From the figure, Distance between the feet and hands is v( 90^2 + 135^2) = 162.3 cm His legs and feet weigh a total of 270N, and their center of mass is 41.0cm from his hip, i.e 49 cm from the feet. The person's trunk, head, and arms weigh 480N, and their center of gravity is 65.0cm from his hip, measured along the upper bodyline 70 cm from the floor. Angle between the leg and floor ? = tan inverse of 75/90 = 39.8° Angle between the hands and floor ? = tan inverse of 90 - 39.8 = 50.2° The reaction force along the leg is R1 The reaction force along the hand is R2 Upward forces = downward forces. R1 sin 39.8 (on the feet) + R2 sin50.2 (on the hands) = 270 +480 = 750 Taking moment about the feet of the person. 270 * 49 cos 39.8+ 480*[162.3 - 70 cos 50.2] – R2 * 162.3 = 0 The upward force acting on the hands is R2 sin 50.2 = 410 .1 N R2 = 533.8 N If f 2 is the frictional force on the hands, f2 = R2 cos 50.2 = 533.8 cos 50.2= 341.7 N --------------------------------------… Answer for the second question : R1 sin 39.8 + R2 sin 50.2 = = 750 R1 sin 39.8 + 410 .1 = = 750 R1 = 531.N If f 1 is the frictional force on the feet , f2 = R1 cos 39.8 = = 531 cos 39.8 = 408 N ======================================…
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