91影视

Given lamina is a composition of rectangles.

density is constant.

Density is constant so substitute $蚁(x,y)=k$in the formula of the x coordinate of the center of mass $\overline{x}.$

role="math" localid="1650343334637" $\begin{array}{rl}\stackrel{炉}{x}& =\frac{{鈭�}_{\mathrm{惟}}x蚁(x,y)dA}{{鈭�}_{\mathrm{惟}}蚁(x,y)dA}\\ \stackrel{炉}{x}& =\frac{{鈭�}_{h_a}^{h_b}{鈭�}_{b_a}^{b_b}xkdxdy}{{鈭�}_{h_a}^{h_b}{鈭�}_{b_a}^{b_b}kdxdy}\\ \stackrel{炉}{x}& =\frac{k\frac{\left({b_b}^2-{b_a}^2\right)}{2}{鈭�}_{h_a}^{h_b}dy}{k\left(b_b-b_a\right){鈭�}_{h_a}^{h_b}dy}\\ \stackrel{炉}{x}& =\frac{k\frac{\left({b_b}^2-{b_a}^2\right)}{2}\left(h_b-h_a\right)}{k\left(b_b-b_a\right)\left(h_b-h_a\right)}\\ \stackrel{炉}{x}& =\frac{\left({b_b}^2-{b_a}^2\right)\left(h_b-h_a\right)}{2\left(b_b-b_a\right)\left(h_b-h_a\right)}\end{array}$

Similarly, substitute $蚁(x,y)=k$formula of the y coordinate of the center of mass $\overline{y}$

role="math" localid="1650343375260" $\begin{array}{rl}\stackrel{炉}{y}& =\frac{{鈭�}_{\mathrm{惟}}y蚁(x,y)dA}{{鈭�}_{\mathrm{惟}}蚁(x,y)dA}\\ \stackrel{炉}{y}& =\frac{{鈭�}_{h_a}^{h_b}{鈭�}_{b_a}^{b_b}ykdxdy}{{鈭�}_{h_a}^{h_b}{鈭�}_{b_a}^{b_b}kdxdy}\\ \stackrel{炉}{y}& =\frac{k\frac{\left({h_b}^2-{h_a}^2\right)}{2}{鈭�}_{b_a}^{b_b}dx}{k\left(h_b-h_a\right){鈭�}_{b_a}^{b_b}dx}\\ \stackrel{炉}{y}& =\frac{k\frac{\left({h_b}^2-{h_a}^2\right)}{2}\left(b_b-b_a\right)}{k\left(h_b-h_a\right)\left(b_b-b_a\right)}\\ \stackrel{炉}{y}& =\frac{\left({h_b}^2-{h_a}^2\right)\left(b_b-b_a\right)}{\left(h_b-h_a\right)\left(b_b-b_a\right)}\end{array}$

So the center of mass of rectangular lumina of constant density isrole="math" localid="1650343386806" $\left(\overline{x},\overline{y}\right)=\left(\frac{\left({b_b}^2-{b_a}^2\right)\left(h_b-h_a\right)}{2\left(b_b-b_a\right)\left(h_b-h_a\right)},\frac{\left({h_b}^2-{h_a}^2\right)\left(b_b-b_a\right)}{\left(h_b-h_a\right)\left(b_b-b_a\right)}\right).$

Consider lumina ${惟}_1,{惟}_2,\&{惟}_3.$

As the center of mass of each rectangle is at$\left(\overline{x},\overline{y}\right)=\left(\frac{\left({b_b}^2-{b_a}^2\right)\left(h_b-h_a\right)}{2\left(b_b-b_a\right)\left(h_b-h_a\right)},\frac{\left({h_b}^2-{h_a}^2\right)\left(b_b-b_a\right)}{\left(h_b-h_a\right)\left(b_b-b_a\right)}\right).$

The graph state that the center of mass of ${惟}_1$is atrole="math" localid="1650343946155" $\left(\overline{x_1},{\overline{y}}_1\right)=\left(\frac{{b_1}^2\left(h_2-h_1\right)}{2b_1\left(h_2-h_1\right)},\frac{\left({h_2}^2-{h_1}^2\right)b_1}{\left(h_2-h_1\right)b_1}\right).$

Similarly center of mass of ${惟}_2$is atrole="math" localid="1650343984563" $\left(\overline{x_2},{\overline{y}}_2\right)=\left(\frac{{b_1}^2{h}_1}{2b_1{h}_1},\frac{{h_1}^2{b}_1}{h_1{b}_1}\right).$

center of mass of ${惟}_3$is atrole="math" localid="1650344007160" $\left(\overline{x_3},{\overline{y}}_3\right)=\left(\frac{\left({b_2}^2-{b_1}^2\right)h_1}{2\left(b_2-b_1\right)h_1},\frac{{h_1}^2\left(b_2-b_1\right)}{h_1\left(b_2-b_1\right)}\right).$

The Center of mass is at the sum of all centers of mass.

$$\begin{gathered} \left(\overline{X},\overline{Y}\right)=\left(\overline{x_1},{\overline{y}}_1\right)+\left(\overline{x_2},{\overline{y}}_2\right)+\left(\overline{x_3},{\overline{y}}_3\right) \\ =\left(\frac{{b_1}^2\left(h_2-h_1\right)}{2b_1\left(h_2-h_1\right)},\frac{\left({h_2}^2-{h_1}^2\right)b_1}{\left(h_2-h_1\right)b_1}\right)+\left(\frac{{b_1}^2{h}_1}{2b_1{h}_1},\frac{{h_1}^2{b}_1}{h_1{b}_1}\right)+\left(\frac{\left({b_2}^2-{b_1}^2\right)h_1}{2\left(b_2-b_1\right)h_1},\frac{{h_1}^2\left(b_2-b_1\right)}{h_1\left(b_2-b_1\right)}\right) \\ =\left(\frac{{b_1}^2\left(h_2-h_1\right)+{b_1}^2{h}_1+\left({b_2}^2-{b_1}^2\right)h_1}{2b_1\left(h_2-h_1\right)+2b_1{h}_1+2\left(b_2-b_1\right)h_1},\frac{\left({h_2}^2-{h_1}^2\right)b_1+{h_1}^2{b}_1+{h_1}^2\left(b_2-b_1\right)}{\left(h_2-h_1\right)b_1+h_1{b}_1+h_1\left(b_2-b_1\right)}\right) \\ (\stackrel{炉}{X},\stackrel{炉}{Y})=\left(\frac{b_1^2{h}_2+b_2^2{h}_1鈭�b_1^2{h}_1}{2(b_1{h}_2+b_2{h}_1鈭�b_1{h}_1)},\frac{h_2^2{b}_1+h_1^2{b}_2鈭�h_1^2{b}_1}{2(b_1{h}_2+b_2{h}_1鈭�b_1{h}_1)}\right) \end{gathered}$$