Respuesta :
Answer:
Δy = 7.1 cm
Explanation:
The center of mass of a body is defined
[tex]y_{cm}[/tex] = 1 /M ∑[tex]m_{i} y_{i}[/tex]i
Where M is the total mass of the body, m mass of each part and ‘y’ height
Let's apply this equation to our case
We locate the reference system on the shoulders
The height of the arms is at its midpoint
y = -75/2 = 37.5 cm
With arms down
[tex]y_{cm}[/tex] = 1/70 (63 y₀ - 3.5 37.5 - 3.5 37.5)
[tex]y_{cm}[/tex] = 1/70 (63 y)₀ - 7 37.5)
With arms up
[tex]y_{cm}[/tex]’= 1/70 (63 y₀ + 3.5 y + 3.5 y)
[tex]y_{cm}[/tex]’= 1/70 (63y₀ + 7 35.5)
let's subtract the two equations
[tex]y_{cm}[/tex]’ - [tex]y_{cm}[/tex] = 1/70 2 (7 35.5)
Δy = [tex]y_{cm}[/tex]’ - [tex]y_{cm}[/tex] = 2 7 35.5 / 70
ΔY = 7.1 cm
Given,
The weight of the man is 70 kg.
The mass of the rod is 3.5 kg.
Computation:
Raising the arm, therefore lifts the center of mass of the arm through, firstly 37.5cm to get the arm straight out in front of you, then another 37.5cm to get the arm straight up.
Total is 7cm.
Finally, work out the ratio of the mass of the arms to the total mass of the person
[tex]\frac{7}{70}=0.1[/tex]
So, the raised center of mass is,
[tex]75\times0.1=75cm[/tex]
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