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A person is standing on a level floor. His head, upper
torso,arms, and hands together weigh 438 N and have a center of
gravitythat is 1.28 m above the floor. His upper legs weigh 144 N
and havea center of gravity that is 0.760 m above the floor.
Finally, hislower legs and feet together weigh 87 N and have a
center ofgravity that is 0.250 m above the floor. Relative to the
floor,find the location of the center of gravity for the
entirebody.

Respuesta :

Answer:

[tex]y_{cg} = 1.03 m[/tex]

Explanation:

Given data:

weigh (head+arms + head) w_1 = 438 N

centre of gravity y_1= 1.28 m

weigh (upper leg) w_2 = 144 N

Center of gravity y_2 = 0.760 m

weigh ( lower leg + feet) = 87 N

centre of gravity = y_3 = 0.250 m

location of center of gravity [tex]= \frac{w_1 y_1 + w_2 y_2 + w_3 y_3}{w_1 +W_2 +w_3}[/tex]

[tex]y_{cg} = \frac{438 \times 1.28 + 144\times 0.760 + 87 \times 0.250}{438+144+87}[/tex]

[tex]y_{cg} = 1.03 m[/tex]

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