\[ \left\{\begin{array}{cccc} m_1=m_2=m_3=100\,\rm{kg} \\ r_1=r_2=r_3=20\,\rm{cm} \\ \theta_1=0^\circ, \ \theta_2=120^\circ, \ \theta_3=240^\circ\\ r=30\,\rm{cm} \end{array}\right. \] 力のつりあい \[ \Sigma mr\omega^2 = 0 \] \[ 100\,\rm{kg} \times 20 \,\rm{cm} \times \omega^2 \times (\cos0^\circ + \cos120^\circ + \cos240^\circ)+30\,\rm{cm} \times\omega^2 \times (M_A \cos\theta_A+M_B\cos\theta_B)=0 \\ 100 \times 20 \times \left(1-\frac{1}{2}-\frac{1}{2}\right)+30\left( M_A\cos\theta_A+M_B\cos\theta_B \right)=0 \] \[ \therefore M_A\cos\theta_A+M_B\cos\theta_B=0 \ \longrightarrow M_B\cos\theta_B=-M_A\cos\theta_A \] \[ 100\,\rm{kg} \times 20 \,\rm{cm} \times \omega^2 \times (\sin0^\circ + \sin120^\circ + \sin240^\circ)+30\,\rm{cm} \times\omega^2 \times (M_A \sin\theta_A+M_B\sin\theta_B)=0 \\ 100 \times 20 \times \left(0+\frac{\sqrt3}{2}-\frac{\sqrt3}{2}\right)+30\left( M_A\sin\theta_A+M_B\sin\theta_B \right)=0 \] \[ \therefore M_A\sin\theta_A+M_B\sin\theta_B=0 \ \longrightarrow M_B\sin\theta_B=-M_A\sin\theta_A \] モーメントのつりあい \[ 100\,\rm{kg} \times 20\,\rm{cm}\times \omega^2 \times \left( -20\,\rm{cm} \times \cos0^\circ +20\,\rm{cm} \times \cos120^\circ +60\,\rm{cm} \times \cos240^\circ \right)\\ +M_B \times 30\,\rm{cm} \times \omega^2 \times 40\,\rm{cm} \times \ \cos\theta_B=0 \\ \] \[\begin{align} M_B \cos\theta_B &=\frac{-100 \times 20 \times \left( -20-20 \times\frac{1}{2} -60\times \frac{1}{2} \right)}{30 \times 40} \\ &=100\,\rm{kg} \end{align} \] \[ 100\,\rm{kg} \times 20\,\rm{cm}\times \omega^2 \times \left( -20\,\rm{cm} \times \sin0^\circ +20\,\rm{cm} \times \sin120^\circ +60\,\rm{cm} \times \sin240^\circ \right)\\ +M_B \times 30\,\rm{cm} \times \omega^2 \times 40\,\rm{cm} \times \sin\theta_B=0 \\ \] \[\begin{align} M_B\sin\theta_B &=\frac{-100 \times 20 \times \left( -0+20 \times\frac{\sqrt3}{2} -60\times \frac{\sqrt3}{2} \right)}{30 \times 40} \\ &=57.73\,\rm{kg} \end{align} \] 以上より \[ \begin{align} M_B &=\sqrt{\left( M_B \cos\theta_B \right)^2+ \left( M_B \sin\theta_B \right)^2 } \\ &=\sqrt{\left( 100\,\rm{kg} \right)^2+ \left( 57.73\,\rm{kg} \right)^2 } \\ &=115.4\,\rm{kg} \\ \\ &=\sqrt{\left( -M_A \cos\theta_A \right)^2+ \left( -M_A \sin\theta_A \right)^2 } \\ &=M_A \end{align} \] \[ \therefore M_A = M_B = 115\,\rm{kg} \] \[ \frac{M_B\sin\theta_B}{M_B\cos\theta_B}=\tan\theta_B \ より \\ \theta_B= \tan^{-1}\left( \frac{57.73\,\rm{kg}}{100\,\rm{kg}}\right) =29.99^\circ \\ \\ \] \[ \therefore \theta_B=30.0^\circ \\ \] \[ \frac{M_B\sin\theta_B}{M_B\cos\theta_B} = \frac{-M_A\sin\theta_A}{-M_A\cos\theta_A} \ より \\ \theta_A=29.99^\circ +180^\circ = 209.9^\circ \] \[ \therefore \theta_A=210^\circ \]