Question

Consider the following distribution of objects: a 5.00-kg object with its center of gravity at (0, 0) m, a 3.00-kg object at (0, 4.00) m, and a 4.00-kg object at (3.00, 0) m. Where should a fourth object of mass 8.00 kg be placed so that the center of gravity of the four-object arrangement will be at (0, 0)?

Answer 1

Answer:

(-1.5,-1.5)m

Explanation:

we know that:

$X_{cm} = \frac{m_1x_1+m_2x_2....m_nX_n}{m_1+m_2...m_n}$

where $X_{cm}$ is the location of the center of gravity in the axis x, $m_i$ is the mass of the object i and $x_i$ the first coordinate of center of gravity of object i.

so:

$0 = \frac{(5kg)(0)+(3kg)(0)+(4kg)(3)+(8kg)x_4}{5kg+3kg+4kg+8kg}$

Where $x_4$ is the first coordinate of the center of gravity for the fourth object.

Therefore, solving for $x_4$, we get:

$x_4 = -1.5m$

At the same way:

$Y_{cm} = \frac{m_1y_1+m_2y_2....m_ny_n}{m_1+m_2...m_n}$

where $Y_{cm}$ is the location of the center of gravity in the axis y, $m_i$ is the mass of the object i and $y_i$ the second coordinate of center of gravity of object i. replacing values we get:

$Y_{cm} = \frac{(5kg)(0)+(3kg)(4)+(4kg)(0)+(8kg)y_4}{5+3+4+8}$

Where $y_4$ is the second coordinate of the center of gravity for the fourth object.

solving for $y_4$:

$y_4 = -1.5m$

It means that the object of mass 8kg have to be placed in the  

coordinates (-1.5,-1.5) m.