Asked by Rachel

A 70 kg person is having a tug-of-war. The rope is horizontal and 1.3 m above the
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).
Answered by bobpursley
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.
Answered by Jessica
I don't get this question still! please help
Answered by Aaron
I don't understand how you got that solution. Could someone please elaborate? PLEASE
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