Question

Problem 3. (20 points) Consider the following simplified model of a person in static equilibrium in a yoga pose. The diagram below shows the upward normal forces acting on the hands and feet, as well as the downward weight forces on the upper and lower body. There are also horizontal static friction forces on the hands and feet not shown in this diagram. Assume the weight of the lower body has magnitude 250 N and the weight of the upper body has magnitude 450 N. Lower body Upper body 56° 34 Feet Hands 28 cm 72 cm 58 cm (a) (8 points) Take the system to be the entire body and choose the rotation axis through the point at which the feet make contact with the floor. Use the net torque about this axis to find the total normal force the floor exerts on the hands. (b) (8 points) 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) (4 points) What are the directions of the static friction forces on the hands and feet? Find an equation relating the magnitudes of these friction forces. (You don't need to calculate these magnitudes.)

Answer 1

Pr-3 lower body upperbody E A28 cm off 72cm fita 250N 58cm USON MA20 FH X 158 - 450x100 - 250x28 20 Fit - 329411 {fy-o ff + FH = 250+450 =ff=370089N Fit I fe= uff & H = ff = M Ft ff fit Pru MA = mBam = = 2 kg A gf AN TA OF B 1 mg a F - I- ma 50-T= za T-mg-ma T- 2x9.8=29 50- 2x9.8= 4a &a=7.6m/s² b 2x908+ axa = 34.8N T of & 50 mg 50-FrT=ma Now when po ama T-mg 69.6N (To the left Jong a=-g 50-fa-mg - fl= (50 +249.8). Manta Oxford