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

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

Step-by-step 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) = (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) = ((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)`.