**1. Definition of the center of gravity**

A rigid body is composed of many particles, therefore the gravitational force acting on each of these particles. In other words, each particle has its own weight. The center of gravity of an object is a point on the object where the weight of all parts of the object is considered to be centered at that point.

If an object is homogeneous (the density of each part of the object is the same or the object is composed of similar material) and the shape of the object is symmetrical (for example square, rectangular, circle) then the weight of the object coincides with the center of mass of the object located at the center of the object. For triangles, the center of mass is located at 1/3 h (h = height of the triangle).

**2. Equation of the center of gravity**

If the shape of the object is symmetrical and the object is homogeneous then the center of gravity of the object coincides with the center of the mass of the object, where the center of gravity and center of mass is located in the center of the object. Conversely, if the object is homogeneous but not symmetrical then the position of the object’s weight can be determined using the following formula.

The coordinates of the object’s weight on the x-axis:

The coordinates of the object’s weight on the y axis:

x = the midpoint of the object on the x-axis, y = the midpoint of the object on the y axis, A = area. If the object is in three dimensions, besides determining the coordinates of the object’s weight on the x and y axes, it also determines the coordinates of the object’s weight on the z-axis. Area (A) is replaced by volume (V).

Sample problem 1.

Determine the coordinates of the center of gravity of the homogeneous object in the figure on the side!

Solution:

Devide object into three parts.

A_{1 }= (10-0)(10-0) = (10)(10) = 100

A_{2 }= (20-10)(30-0) = (10)(30) = 300

A_{3 }= (30-20)(10-0) = (10)(10) = 100

x_{1} = 1⁄2 (10-0) = 1⁄2 (10) = 5

x_{2} = 1⁄2 (20-10) + 10 = 1⁄2 (10) + 10 = 5 + 10 = 15

x_{3 }= 1⁄2 (30-20) + 20 = 1⁄2 (10) + 20 = 5 + 20 = 25

y_{1} = 1⁄2 (10-0) = 1⁄2 (10) = 5

y_{2} = 1⁄2 (30-0) = 1⁄2 (30) = 15

y_{3 }= 1⁄2 (10-0) = 1⁄2 (10) = 5

The coordinates of the object’s weight on the x-axis:

The coordinates of the object’s weight on the y axis:

The coordinates of the object’s weight are (15; 11)

Sample problem 2.

Determine the coordinate of the object’s weight in the figure!

Solution:

Divide object into two parts, area 1 = square, area 2 = triangle.

A_{1 }= (5-1)(3-0) = (4)(3) = 12

A_{2} = 1⁄2 (6-0)(9-3) = 1⁄2 (6)(6) = (3)(6) = 18

x_{1} = 1⁄2 (5-1) + 1 = 1⁄2 (4) + 1 = 2 + 1 = 3

x_{2} = 1⁄2 (6-0) = 1⁄2 (6) = 3

y_{1 }= 1⁄2 (3-0) = 1⁄2 (3) = 1.5

y_{2} = 1/3(9-3) + 3 = 1/3(6) + 3 = 2 + 3 = 5

The coordinates of the object’s weight on the x-axis:

The coordinates of the object’s weight on the y axis:

The coordinate of the object’s weight is (3; 3.6)