Answer:
81.1 cm
Step-by-step explanation:
The sum of moments about the hip joint is ...
(8.60 kg)(92 cm/2) + (5.30 kg)(92 cm +92 cm/2))
= 395.6 kg·cm +731.4 kg·cm = 1127 kg·cm
The mass of the leg is 8.60 +5.30 kg = 13.9 kg, so the above moment is equivalent to that mass at a distance of ...
(1127 kg·cm)/(13.9 kg) = 81.08 cm
The center of mass of the stretched-out leg is about 81.1 cm from the hip.
Answer:
Step-by-step explanation:
We know that:
The length of the leg is: [tex]l=92cm[/tex]
The problem says that each part of the leg has the same length, that is:
[tex]l_{upper}=\frac{92}{2} = 46cm= l_{lower}[/tex]
Also, the leg is uniform, so the center of mass of each part is in the middle, this is the position (r):
[tex]r_{upper} = \frac{46}{2}= 23cm\\ r_{lower} = 46 + 23 = 69cm[/tex]
This means that the center of mass of the upper leg is 23cm from the hip, and the lower leg if 69c m from the hip.
In addition, we know each mass:
[tex]m_{upper} = 8.60kg\\m_{lower}=5.30kg[/tex]
Now, we have all values needed, we use the proper equation to calculate the center of mass of the leg:
[tex]r \ _{CM} = \frac{m_{upper}r_{upper} + m_{lower}r_{lower} }{m_{upper}+m_{lower}} \\r \ _{CM} =\frac{(8.60)(23)+(5.30)(69)}{8.60+5.30} =\frac{197.8+365.7}{13.9} \\r \ _{CM} =40.5cm[/tex]
Therefore, the center of mass of the leg is 40.5m from the hip in a horizontal direction because it's stretched out horizontally Specifically, the coordinates are (40.5;0) cm.
