Statics Problem: Finding Center of Mass

It is convenient to perform two separate integrations for the areas A and B. First we compute the 'whole mass'...

$M= \int \int_{A} \rho_{A}\cdot dx\cdot dy + \int \int_{B} \rho_{B}\cdot dx\cdot dy=$

$= \int_{0}^{.1} \int _{.2}^{.7} dx\cdot dy + \frac{1}{2}\cdot \int_{.1}^{.5} \int _{.2}^{.3} dx\cdot dy + \frac{1}{2}\cdot \int_{.5}^{.6} \int _{0}^{.3} dx\cdot dy= .085$ (1)

... and then we compute $x_{m}$ and $y_{m}$ ...

$x_{m}= \frac{1}{M} \cdot \{ \int \int_{A} \rho_{A}\cdot x \cdot dx\cdot dy + \int \int_{B} \rho_{B}\cdot x\cdot dx\cdot dy \}=$

$=\frac{1}{.085}\cdot \{ \int_{0}^{.1} \int _{.2}^{.7} x\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.1}^{.5} \int _{.2}^{.3} x\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.5}^{.6} \int _{0}^{.3} x\cdot dx\cdot dy \}=$

$=\frac{1}{.085}\cdot \{ \int_{0}^{.1} .5\cdot x\cdot dx + \frac{1}{2}\cdot \int_{.1}^{.5} .1\cdot x\cdot dx + \frac{1}{2}\cdot \int_{.5}^{.6} .3\cdot x\cdot dx \}=$

$=\frac{1}{.085}\cdot \{ .5\cdot |\frac{x^{2}}{2}|_{0}^{.1} + .05\cdot |\frac{x^{2}}{2}|_{.1}^{.5} + .15\cdot |\frac{x^{2}}{2}|_{.5}^{.6} \} = .1970588...$ (2)

$y_{m}= \frac{1}{M} \cdot \{ \int \int_{A} \rho_{A}\cdot y \cdot dx\cdot dy + \int \int_{B} \rho_{B}\cdot y\cdot dx\cdot dy \}=$

$=\frac{1}{.085}\cdot \{ \int_{0}^{.1} \int _{.2}^{.7} y\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.1}^{.5} \int _{.2}^{.3} y\cdot dx\cdot dy + \frac{1}{2}\cdot \int_{.5}^{.6} \int _{0}^{.3} y\cdot dx\cdot dy \}=$

$=\frac{1}{.085}\cdot \{ \int_{.2}^{.7} .1\cdot y\cdot dy + \frac{1}{2}\cdot \int_{.2}^{.3} .4\cdot y\cdot dy + \frac{1}{2}\cdot \int_{0}^{.3} .1\cdot y\cdot dy \}=$

$=\frac{1}{.085}\cdot \{ .1\cdot |\frac{y^{2}}{2}|_{.2}^{.7} + .2\cdot |\frac{y^{2}}{2}|_{.2}^{.3} + .05\cdot |\frac{y^{2}}{2}|_{.0}^{.3} \} = .39235294...$ (3)

The results are shown in the figure...

http://digilander.libero.it/luposabatini/MHF31.bmp

Kind regards

$\chi$ $\sigma$