Answer:
0.153 meters to the right of the origin.
Explanation:
The location of the center of mass of a particle system is given by this formula:
[tex]\overline x = \frac{\Sigma_{i = 1}^n (m_{i} \cdot x_{i})}{\Sigma_{i = 1}^n m_{i}}[/tex]
The current inputs are presented below (It is assumed that distances are positive for all number to the right of the origin:
[tex]x_{1} = - 0.43 m, m_{1} = 3.2 kg\\x_{2} = + 0.31 m, m_{2} = 3.6 kg\\x_{3} = + 0.47 m, m_{3} = 4.1 kg\\[/tex]
Then, the location is finally found:
[tex]\overline x = \frac{(-0.43 m)\cdot (3.2 kg)+(0.31 m)\cdot (3.6 kg) + (0.47 m)\cdot (4.1 kg)}{3.2 kg + 3.6 kg + 4.1 kg}[/tex]
[tex]\overline x = + 0.153 m[/tex]
1. the location of the center of mass of the system is 0.153 meters to the right of the origin.
Since m1 = 3.2 kg is a distance x1 = 0.43 m to the left of the origin, m2 = 3.6 kg is a distance x2 = 0.31 m to the right of the origin, and m3 = 4.1 kg is a distance x3 = 0.47 m to the right of the origin
Now
the location should be
= ((0.43m)*(3.2kg) + (0.31 m) * (3.6kg) + (0.47m) * (4.1 kg)) / 3.2kg + 3.6 kg + 4.1 kg
= +0.153m
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