two small sized objects are placed on a uniform 9.00 kg plastic beam 3.00 m long. assume that m1 = 7.75 kg and m2 = 7.60 kg. find the location of the center of mass of this system by setting x = 0 to the left end of the beam. express your answer in meters.

[tex]\boxed{ \begin{array}{ccc} \text{\underline{Center of Mass of a System of Particles:}} \\\\ \vec{R}_{\text{cm}} = \frac{1}{M} \sum_{i=1}^n m_i \vec{r}_i \\\\ \text{Where:} \\ \bullet \ \vec{R}_{\text{cm}} \ \text{is the center of mass} \\ \bullet \ M \ \text{is the total mass of the system} \\ \bullet \ m_i \ \text{are the masses of the particles} \\ \bullet \ \vec{r}_i \ \text{are the position vectors of the particles} \end{array}}[/tex]

[tex]\boxed{ \begin{array}{ccc} \text{\underline{X-Coordinate of Center of Mass:}} \\\\ x_{\text{cm}} = \dfrac{\sum (m_i x_i)}{\sum m_i} \\\\ \text{Where:} \\ \bullet \ x_{\text{cm}} \ \text{is the x-coordinate of the center of mass} \\ \bullet \ m_i \ \text{are the masses of the particles} \\ \bullet \ x_i \ \text{are the x-coordinates of the particles} \end{array}}[/tex]

Given:

m_b = 9.00 kg
m₁ = 7.75 kg
m₂ = 7.60 kg
L = 3.00 m

Assuming m₁ is attached to the beam at the left end or the origin, and m₂ is attached to the right end or (3.00, 0) m, we can calculate the center of mass:

[tex]\Longrightarrow x_{\text{cm}}=\dfrac{m_1x_1+\dfrac{1}{2}m_bL+m_2x_2}{m_1+m_b+m_2}[/tex]

[tex]\Longrightarrow x_{\text{cm}}=\dfrac{(7.75 \text{ kg})(0 \text{ m})+\dfrac{1}{2}(9.00 \text{ kg})(3.00 \text{ m})+(7.60 \text{ kg})(3.00 \text{ m})}{7.75 \text{ kg}+9.00 \text{ kg}+7.60 \text{ kg}}[/tex]

[tex]\therefore x_\text{cm}\approx\boxed{1.49 \text{ m}}[/tex]

Therefore, the location of the center of mass of the system is approximately 1.49 m from the left end of the beam.