A person is standing on a level floor. His head, upper torso, arms, and hands together weigh 496 N and have a center of gravity that is 1.37 m above the floor. His upper legs weigh 199 N and have a center of gravity that is 0.711 m above the floor. Finally, his lower legs and feet together weigh 81.2 N and have a center of gravity that is 0.284 m above the floor. Relative to the floor, find the location of the center of gravity for the entire body.

1.09 m

Explanation:

Given

Weight of the body, W1 = 496 N

Weight of his upper legs, W2 = 199 N

Weight of his lower legs, W3 = 81.2 N

Center of gravity of the body, y1 = 1.37 m

Center of gravity of the upper legs, y2 = 0.711 m

Center of gravity of the lower legs, y3 = 0.284 m

Center of gravity relative to the floor, y(f) = ?

y(f) = (W1y1 + W2y2 + W3y3) / (W1 + W2 + W3), substituting the values, we have

y(f) = [(496 * 1.37) + (199 * 0.711) + (81.2 * 0.284)] / (496 + 199 + 81.2)

y(f) = (679.52 + 141.489 + 23.0608) / 776.2

y(f) = 844.0698 / 776.2

y(f) = 1.09 m

divide the right side by the left side

0.024÷1.2=0.02 for example

answer: 2×10^-2kg

explanation: density= mass per volume

vol=mass/density

converting to s.i units we have;

3÷1000kg = 0.003

1.5÷100m = 0.15

0.003/0.15 = 0.02kg