Question

Three masses are placed along the x-axis: 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. system of masses 1) What is the location of the center of mass of the system?

Answer 1

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:

$\overline x = \frac{\Sigma_{i = 1}^n (m_{i} \cdot x_{i})}{\Sigma_{i = 1}^n m_{i}}$

The current inputs are presented below (It is assumed that distances are positive for all number to the right of the origin:

$x_{1} = - 0.43 m, m_{1} = 3.2 kg\\x_{2} = + 0.31 m, m_{2} = 3.6 kg\\x_{3} = + 0.47 m, m_{3} = 4.1 kg$

Then, the location is finally found:

$\overline x = \frac{(-0.43 m)\cdot (3.2 kg)+(0.31 m)\cdot (3.6 kg) + (0.47 m)\cdot (4.1 kg)}{3.2 kg + 3.6 kg + 4.1 kg}$

$\overline x = + 0.153 m$