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Alborosie
3 years ago
9

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
Physics
1 answer:
sergij07 [2.7K]3 years ago
4 0

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).

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