What will be center of gravity of the given L-section shown in figure?

 

Concept:

Centre of gravity:

\(\bar y = \frac{{{A_1}{y_1} + {A_2}{y_2}}}{{{A_1} + {A_2}}} \)

where, A1, A2 = Area of section 1 and section 2 respectively, y1, y2 = Centre of gravity with respect to the reference axis.

Calculation:

Given:

A1 = 4 × 2 = 8 cm2, A2 = 10 × 2 = 20 cm2,

Taking X-axis as the reference axis,

Centre of gravity:

\(\bar y = \frac{{{A_1}{y_1} + {A_2}{y_2}}}{{{A_1} + {A_2}}} = \frac{4~\times~2 \times 1~+~10 ~\times ~2\times 5}{4\times 2~+~10\times 2}=3.857~cm\)

and 

\(\bar x = \frac{{{A_1}{x_1} + {A_2}{x_2}}}{{{A_1} + {A_2}}} = \frac{4~\times~2 \times 4~+~2 ~\times ~10\times 1}{4\times 2~+~10\times 2}=1.85~cm\)