\overline F \parallel плоскости A_{yz}
\overline S_1, \overline S_2 \parallel плоскости A_{xy}
\overline T_1 \parallel T_2 \parallel плоскости \overline{A_{xy}}
H_2 \cdot T_1 = T_2 = 2 \cdot 100 = 200H
S_1 = H_1 \cdot S_2 = 3 \cdot S_2;

Дано:

R = 0.5m
AB = 1.2m
G = 300H
T_1 = 100H
F = 100H
M = 10\cdot H_2 Hm
H_1 = 3.0
H_2 = 2.0
\alpha = 30^0
\beta = 60^0
\gamma = 30^0

A O_1 = O_1 O_2 = \frac{AB}{H_1} = \frac{1.2}{3} = 0.4m
r = (\frac{H_2}{H_1}) R = \frac{2}{3} 0.5 = 0.33m

Определить:

\overline R_A (X_A,Y_A,Z_A) = ?
\overline R_B (X_B,Y_B,0) = ?
\vert \overline {S_1} \vert = ?

Решение:

Система сил F_j принадлежит \{ G, \overline F, \overline T_1, \overline T_2, \overline S_1, \overline S_2, \overline R_A, \overline R_B \}, \{ \overline M \}

(1)   \begin{equation*} \sum^{u=8}_{j=1} F_{jx} = 0 ; \end{equation*}

    \[ -S_1 \cos\beta - S_2 \cos\gamma + \vert X_A \vert + \vert X_B \vert = 0 \]

(2)   \begin{equation*} \sum^{u=8}_{j=1} F_{jy} = 0 ; \end{equation*}

    \[ -F \cos\alpha + S_1 \sin\beta + S_2 \sin\gamma + \vert Y_A \vert + \vert Y_B \vert - T_1 - T_2 = 0 \]

(3)   \begin{equation*} \sum^{u=8}_{j=1} F_{jz} = 0 ; \end{equation*}

    \[ -G -F \sin\alpha + \vert Z_A \vert = 0 \]

(4)   \begin{equation*} \sum^{u=8}_{j=1} M_{AX} (\overline F_j) + M_x = 0; \end{equation*}

    \[ \vert AO_2 \vert F \cdot \cos\alpha - F \sin\alpha \cdot r \cos\gamma + (T_1+T_2) \cdot \vert AO_2 \vert - S_1 \sin\beta \cdot \vert AO_1 \vert - S_2 \sin\gamma \cdot \vert AO_1 \vert - \vert Y_B \vert \cdot \vert AB \vert = 0 \]

(5)   \begin{equation*} \sum^{u=8}_{j=1} M_{AY} (\overline F_j) + M_y = 0; \end{equation*}

    \[ F \sin\alpha \cdot r \cdot \sin\gamma - S_1 \cos\beta \vert AO_1 \vert - S_2 \cos\gamma \vert AO_1 \vert + \vert X_B \vert \cdot \vert AB \vert = 0 \]

(6)   \begin{equation*} \sum^{u=8}_{j=1} M_{AZ} (\overline F_j) + M_z = 0; \end{equation*}

    \[ -F \cos\alpha \cdot r \cdot \sin\gamma + (T_1 - T_2) \cdot r + (S_2-S_1) \cdot R - M = 0 \]

Из (6) \Longrightarrow S_1 = ?
Из (5) \Longrightarrow \vert X_B \vert = ?
Из (4) \Longrightarrow \vert Y_B \vert = ?
Из (3) \Longrightarrow \vert Z_A \vert = ?
Из (2) \Longrightarrow \vert Y_A \vert = ?
Из (1) \Longrightarrow \vert X_A \vert = ?

Ответ:

\overline R_A (X_A,Y_A,Z_A) =
\overline R_B (X_B,Y_B,0) =
\vert \overline {S_1} \vert =