A 70 kg person is having a tug-of-war. The rope is horizontal and 1.3 m above the

Question

ground. This person’s center of mass is 1.0 m from the ground when he is in an
upright standing position. If the force his opponent exerts on the rope is 400 N,
what angle must the person be from the horizontal to be in equilibrium? (Assume
the knees and hips are not bent).

Answer 1

write the sum of moments about his feet.

assume angle theta ..

400*1.3*sinTheta=70g*1.0*cosTheta

tan Theta=70(9.8)/400*1.3

theta= you do it.

Answer 2

I don't get this question still! please help

Answer 3

I don't understand how you got that solution. Could someone please elaborate? PLEASE