Question

A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 458 N and have a center of gravity that is 1.34 m above the floor. His upper legs weigh 120 N and have a center of gravity that is 0.766 m above the floor. Finally, his lower legs and feet together weigh 89.8 N and have a center of gravity that is 0.204 m above the floor. Relative to the floor, find the location of the center of gravity for the entire body.

Answer 1

Answer:

the location of the center of gravity for the entire body is 1.08 m

Explanation:

Given the data in the question;

w1 = 458 N, y1 = 1.34 m

w2 = 120 N, y2 = 0.766 m

w3 = 89.8 N, y2 = 0.204 m

The location arrangement of the body part is vertical, locate the overall centre of gravity by simply replacing the horizontal position x by the vertical position y as measured relative to the floor.

so,

$Y_{centre of gravity}$ = (w1y1 + w2y2 + w3y3 ) / ( w1 + w2 + w3 )

so we substitute in our values

$Y_{centre of gravity}$ = (458×1.34 + 120×0.766 + 89.8×0.204 ) / ( 458 + 120 + 89.8 )

$Y_{centre of gravity}$ = 723.9592 / 667.8

$Y_{centre of gravity}$ = 1.08 m

Therefore, the location of the center of gravity for the entire body is 1.08 m