Solutions - 0110-01

(1) Compute masses:

$$\begin{array}{c}{m}^{\text{total}}=m^{\text{rect}}+m^{\text{tri}}\\ \\ \gg \gg \rho A^{\text{rect}}+\rho A^{\text{tri}}\\ \\ \gg \gg \rho (4)(3)+\rho \left(\frac{1}{2}\right)(4)(2)\\ \\ \gg \gg 12\rho +4\rho \gg \gg 16\rho \end{array}$$

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(2) Consider symmetries of rectangle:

$${\text{CoM}}^{\text{rect}}=({\bar{x}}^{\text{rect}},{\bar{y}}^{\text{rect}})=(0,3/2)$$

Therefore $M_y^{\text{rect}}=0$ and:

$${\bar{y}}^{\text{rect}}=\frac{M_x^{\text{rect}}}{m^{\text{rect}}}$$

Therefore:

$$\begin{array}{c}{M}_x^{\text{rect}}={\bar{y}}^{\text{rect}}\cdot m^{\text{rect}}\\ \\ \gg \gg (3/2)\cdot 12\rho \gg \gg 18\rho \end{array}$$

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(3) Consider symmetry of triangle:

$${\bar{x}}^{\text{tri}}=0\gg \gg M_y^{\text{tri}}=0$$

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(4) Compute $M_x^{\text{tri}}$ by integration:

$$w(y)=4+\frac{0-4}{2}(y-3)\gg \gg 10-2y$$ $$\begin{array}{c}{M}_x^{\text{tri}}={\int }_3^5\rho yw(y)dy\\ \\ \gg \gg \rho {\int }_3^{5}y(10-2y)dy\gg \gg 44\rho /3\end{array}$$

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(5) Optional step: infer ${\bar{y}}^{\text{tri}}$:

$${\bar{y}}^{\text{tri}}=\frac{M_x^{\text{tri}}}{m^{\text{tri}}}\gg \gg \frac{44\rho /3}{4\rho }\gg \gg 11/3$$

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(6) Additivity of moments:

$$\begin{array}{c}{M}_x^{\text{total}}=M_x^{\text{rect}}+M_x^{\text{tri}}\\ \\ \gg \gg 18\rho +44\rho /3\gg \gg 98\rho /3\end{array}$$ $$\begin{array}{c}{M}_y^{\text{total}}=M_y^{\text{rect}}+M_y^{\text{tri}}=0+0=0\end{array}$$

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(7) Compute CoM:

$$\bar{x}=0,\bar{y}=\frac{M_x^{\text{total}}}{m^{\text{total}}}$$ $$\gg \gg \bar{y}\gg \gg \frac{98\rho /3}{16\rho }\gg \gg 49/24$$

Thus:

$$\text{CoM}=(0,49/24)$$