Center of Gravity Example Problem

Lecture 28
Example Problem

Center of Gravity

Given:
the plate shown in the diagram has a weight of 1.0#/in² of horizontal surface.
Determine:
the center of gravity of the plate knowing that it is symmetrical about the X-X axis.

Solution:
The principle of moments states that the total weight about an axis is equal to the sum of the moments of the component weights about that same axis. Thus, the first thing to do is to divide the plate into several simple parts. Then, determine the area and the center of gravity (or centroid) for each of the component parts. After this is completed, take the moments of each of the parts around a convenient axis (in this case select the Z-Z axis about which to take these moments). Lecture 27

Sum M_{A_total} = M_A₁ + M_A₂ + M_A₃

This simple equation can be rewritten as follows in which each of the component parts are described:

(A_total)(distance from reference axis to centroidal axis) = (A₁)(distance from centroid of A₁ to reference axis) + (A₂)(distance from centroid of A₂ to reference axis) + (A₃)(distance from centroid of A₃ to reference axis)

[(6in)(4in)](2in) + [(6in)(3in)](7in) + [(15in)(2in)](11in) = (24 +18+30)(y)

48 in³ + 126 in^3 + 330 in³ = 72 in² y
504 in³ = 72 in² y

and then solving for y ... the centroidal axis is 7 inches from the reference axis.

The actual center of gravity occurs midway through the depth of the plate at the point calculated above. As the plate thickness is reduced the line of action of the center of gravity will remain while the center of gravity moves proportionally along this line of action always effective at the midpoint of the depth of the plate. If the plate thickness is reduced to zero it has no weight and the former center of gravity position is now referred to as the centroid of the area.

Lecture 28 Example Problem

Lecture 28
Example Problem