Question

The mass and coordinates of three objects are given below: m1 = 6.0 kg at (0.0, 0.0) m, m2 = 1.5 kg at (0.0, 4.1) m, and m3 = 4.0 kg at (1.9, 0.0) m. Determine where we should place a fourth object with a mass m4 = 7.9 kg so that the center of gravity of the four-object arrangement will be at (0.0, 0.0) m

Answer 1

Answer:

The location of the center of gravity of the fourth mass is $\vec r_{4} = (-0.961\,m,-0.779\,m)$.

Explanation:

Vectorially speaking, the center of gravity with respect to origin ($\vec r_{cg}$), measured in meters, is defined by the following formula:

$\vec r_{cg} = \frac{m_{1}\cdot \vec r_{1}+m_{2}\cdot \vec r_{2}+m_{3}\cdot \vec r_{3}+m_{4}\cdot \vec r_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}$ (1)

Where:

$m_{1}$, $m_{2}$, $m_{3}$, $m_{4}$ - Masses of the objects, measured in kilograms.

$\vec r_{1}$, $\vec r_{2}$, $\vec r_{3}$, $\vec r_{4}$ - Location of the center of mass of each object with respect to origin, measured in meters.

If we know that $\vec r_{cg} = (0,0)\,[m]$, $\vec r_{1} = (0,0)\,[m]$, $\vec r_{2} = (0, 4.1)\,[m]$, $\vec r_{3} = (1.9,0.0)\,[m]$, $m_{1} = 6\,kg$, $m_{2} = 1.5\,kg$, $m_{3} = 4\,kg$ and $m_{4} = 7.9\,kg$, then the equation is reduced into this:

$(0,0) = \frac{(6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4.0\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4}}{6\,kg+1.5\,kg+4\,kg+7.9\,kg}$

$(6\,kg)\cdot (0,0)\,[m]+(1.5\,kg)\cdot (0,4.1)\,[m]+(4\,kg)\cdot (1.9,0)\,[m]+(7.9\,kg)\cdot \vec r_{4} = (0,0)\,[kg\cdot m]$

$(7.9\,kg)\cdot \vec r_{4} = -(6\,kg)\cdot (0,0)\,[m]-(1.5\,kg)\cdot (0,4.1)\,[m]-(4\,kg)\cdot (1.9,0)\,[m]$

$\vec r_{4} = -0.759\cdot (0,0)\,[m]-0.190\cdot (0,4.1)\,[m]-0.506\cdot (1.9,0)\,[m]$

$\vec r_{4} = (0, 0)\,[m] -(0, 0.779)\,[m]-(0.961,0)\,[m]$

$\vec r_{4} = (-0.961\,m,-0.779\,m)$

The location of the center of gravity of the fourth mass is $\vec r_{4} = (-0.961\,m,-0.779\,m)$.