Question

Part 1

Answer 1

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$.

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$.

Determination of the coordinates of the center of gravity

Let suppose that each square has an uniform density, the coordinates of the center of gravity of each square with respect to the origin are, respectively:

$\vec r_{1} = (0.5\cdot a, 0.5\cdot a)$

$\vec r_{2} = (1.5\cdot a, 0.5\cdot a)$

$\vec r_{3} = (0.5\cdot a, 1.5\cdot a)$

$\vec r_{4} = (1.5\cdot a, 1.5\cdot a)$

The center of gravity of the entire system is found by applying definition of weighted averages:

$\vec r_{cg} = \frac{W_{1}\cdot \vec r_{1}+W_{2}\cdot \vec r_{2}+W_{3}\cdot \vec r_{3}+W_{4}\cdot \vec r_{4}}{W_{1}+W_{2}+W_{3}+W_{4}}$

Where $W_{1}$, $W_{2}$, $W_{3}$ and $W_{4}$ are weights of the each square, in newtons.

Now we proceed the coordinates of the center of gravity of the entire system:

$\vec r_{cg} = \frac{(50\,N)\cdot (0.5\cdot a, 0.5\cdot a) + (30\,N)\cdot (1.5\cdot a, 0.5\cdot a)+(30\,N)\cdot (0.5\cdot a, 1.5\cdot a)+(70\,N)\cdot (1.5\cdot a, 1.5\cdot a)}{50\,N+30\,N+30\,N+70\,N}$

$\vec r_{cg} = \frac{5}{18}\cdot (0.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (1.5\cdot a, 0.5\cdot a) +\frac{1}{6}\cdot (0.5\cdot a, 1.5\cdot a) + \frac{7}{18}\cdot (1.5\cdot a, 1.5\cdot a)$

$\vec r_{cg} = \left(\frac{19}{18}\cdot a, \frac{19}{18}\cdot a \right)$ $\blacksquare$

a) The x-coordinate of the center of gravity is $\frac{19}{18}\cdot a$. $\blacksquare$

b) The y-coordinate of the center of gravity is $\frac{19}{18}\cdot a$. $\blacksquare$

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