8.3-bo'lim : Massa markazi
3. \((0, 0)\), \((-4, 2)\) va \((0,6)\) uchlari bo‘lgan uchburchakning massa markazini toping.
Yechim
Keling, nuqta bilan ko'rsatilgan massa markazi bilan mintaqaning qisqacha chizmasidan boshlaylik (bu nuqtaning koordinatalari, albatta, oxirgi bosqichda aniqlanishi kerak ....).
Biz buni sizga uchburchakning yuqori va pastki oyoqlarining tenglamalarini tekshirishni qoldiramiz.
Bizga ushbu hududning maydoni ham kerak bo'ladi, shuning uchun avval uni topamiz.
\[A = \int_{{ - 4}}^{0}{{\left( {x + 6} \right) - \left( { - \frac{1}{2}x} \right)\,dx}} = \int_{{ - 4}}^{0}{{\frac{3}{2}x + 6\,dx}} = \left. {\left( {\frac{3}{4}{x^2} + 6x} \right)} \right|_{ - 4}^0 = 12\]2-qadam
Keyinchalik, ikkita momentni hisoblashimiz kerak. Quyidagi hisob-kitoblarga zichlikni kiritmadik, chunki u faqat oxirgi bosqichda bekor qilinadi.
\[\begin{align*}{M_x} & = \int_{{ - 4}}^{0}{{\frac{1}{2}\left[ {{{\left( {x + 6} \right)}^2} - {{\left( { - \frac{1}{2}x} \right)}^2}} \right]\,dx}} = \int_{{ - 4}}^{0}{{\frac{3}{8}{x^2} + 6x + 18\,dx}} = \left. {\left( {\frac{1}{8}{x^3} + 3{x^2} + 18x} \right)} \right|_{ - 4}^0 = 32\\ {M_y} & = \int_{{ - 4}}^{0}{{x\left( {\left( {x + 6} \right) - \left( { - \frac{1}{2}x} \right)} \right)\,dx}} = \int_{{ - 4}}^{0}{{\frac{3}{2}{x^2} + 6x\,dx}} = \left. {\left( {\frac{1}{2}{x^3} + 3{x^2}} \right)} \right|_{ - 4}^0 = - 16\end{align*}\]3-qadam
Nihoyat, massa markazining koordinatalari,
\[\overline{x} = \frac{{{M_y}}}{M} = \frac{{\rho \left( { - 16} \right)}}{{\rho \left( {12} \right)}} = - \frac{4}{3}\hspace{0.25in}\hspace{0.25in}\overline{y} = \frac{{{M_x}}}{M} = \frac{{\rho \left( {32} \right)}}{{\rho \left( {12} \right)}} = \frac{8}{3}\]U holda massa markazi: \(\require{bbox} \bbox[2pt,border:1px solid black]{{\left( { - \frac{4}{3},\frac{8}{3}} \right)}}\).