Question

easy bio A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 438 N and have a center of gravity that is 1.28 m above the floor. His upper legs weigh 144 N and have a center of gravity that is 0.760 m above the floor. Finally, his lower legs and feet together weigh 87 N and have a center of gravity that is 0.250 m above the floor. Relative to the floor, find the location of the center of gravity for his entire body.

Answer 1

Answer:

1.034 m above the floor

Explanation:

The location of center of body for a compound body, when the weights are given is calculated as:

$\bar x = \frac{W_1x_1+W_2x_2+W_3x_3+W_4x_4+W_5x_5+.......+Wnx_n}{W_1+W_2W_3W_4W_5+.....+W_n}$

where,

$\bar x$ is the center of gravity of the entire body

W = weight of the individual body

x = center of gravity of the individual body

Thus on substituting the values we get,

$\bar x = \frac{438\times 1.28+144\times 0.760+87\times 0.250}{438+144+87}$

or

$\bar x = \frac{691.83}{669}$

or

$\bar x =1.034m$

Hence, the center of gravity of the entire body lies 1.034 m above the floor