Question

A person practicing a yoga pose (look at the diagram below) is in static equilibrium. The diagram shows normal forces acting on the hands and feet of this person. There are also horizontal static friction forces on the hands and feet not shown in the diagram. Assume the lower body has a magnitude of 450 N and the weight of the upper body has a magnitude of 600N.

A. Consider the system to be the entire body and choose the rotation axis through the point at which the feet makes contact with the floor. Use the torque about this axis to find the total normal force and the floor exerts on the hands.

B. Now use the vertical component of the net force on the entire body to find the normal force of the floor on the feet.

C. What are the directions of the static friction forces on the hands and feet? Find an equation (don't calculate)  relating magnitudes for these friction forces.

Lower body Upper body 56° 34° Feet Hands 28 cm 72 cm 58 cm

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Answer #1

1450N 56 600N 3:07 0 2- 72cm 28 cm < s8cm To=4500-28 + 60001-N₂ X 1058 = 0 Al Normal force hand N2 = 459.490 exerts on floor

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