Balance a cart on an edge

The points $A$ , $B$ and $C$ each form, together with the origin $O$ , right-angled triangles whose sides are in the ratio $5:4:3$ .

In each of the points $A$ , $B$ and $C$ a force acts on the car. In point $B$ this is the weight with $\vec F_B = (0, - m g)$ . The force at point $A$ is exerted by the edge on the wheel. Since the wheel does not transmit torque to its axis (the wheel is rotatably mounted), the force acts along the connecting line between $A$ and the origin. Therefore, the force vector can written as $\vec F_A = (- \frac{4}{5} , \frac{3}{5} ) F_A$ . The force $\vec F = \vec F_C$ force at point $C$ has an unknown angle $\alpha$ to the horizontal. Thus, the vector yields $\vec F_C = (\cos \alpha, \sin \alpha) F_C$ .

The sum of all forces must be zero:

$\begin{aligned} & & 0 = \sum_i \vec F_i &= \vec F_A + \vec F_B + \vec F_C \\ & & &= \left( \begin{array}{c} - \tfrac{4}{5} F_A \\ \tfrac{3}{5} F_A \end{array} \right) + \left( \begin{array}{c} 0 \\ - m g \end{array} \right) + \left( \begin{array}{c} \cos \alpha F_C \\ \sin \alpha F_C \end{array} \right) \\ \Rightarrow & & 0 &= - \frac{4}{5} F_A + \cos \alpha F_C && (I)\\ & & m g &= \frac{3}{5} F_A + \sin \alpha F_C && (II) \end{aligned}$ By multiplying the first line by the factor 3/4 and adding the two lines, the unknown $F_A$ can be eliminated: $\frac{3}{4} (I) + (II) = m g = \left(\frac{3}{4} \cos \alpha + \sin \alpha \right) F_C \qquad (III)$

In addition, the total torque must be zero $\begin{aligned} & & 0 = \sum_i \vec T_i &= \vec A \times \vec F_A + \vec B \times \vec F_B + \vec C \times \vec F_C \\ & & &= \left( \begin{array}{c} 4 \\ -3 \\ 0 \end{array} \right) \times \left( \begin{array}{c} - \tfrac{4}{5} F_A \\ \tfrac{3}{5} F_A \\ 0 \end{array} \right) + \left( \begin{array}{c} -6 \\ 8 \\ 0 \end{array} \right) \times \left( \begin{array}{c} 0 \\ - m g \\ 0 \end{array} \right) + \left( \begin{array}{c} 12 \\ 9 \\ 0 \end{array} \right) \times \left( \begin{array}{c} \cos \alpha\, F_C \\ \sin \alpha \, F_C \\ 0 \end{array} \right) \\ & & &= \left( \begin{array}{c} 0 \\ 0 \\ 0 \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ 6 m g \end{array} \right) + \left( \begin{array}{c} 0 \\ 0 \\ (12 \sin \alpha - 9 \cos \alpha) F_C \end{array} \right)\\ \Rightarrow & & m g &= \left( \frac{3}{2} \cos \alpha - 2 \sin \alpha \right) F_C & & (IV)\end{aligned}$ By comparing equations (III) and (IV) we get $\begin{aligned} & & \frac{3}{4} \cos \alpha + \sin \alpha &= \frac{3}{2} \cos \alpha - 2 \sin \alpha \\ \Rightarrow & & 3 \sin \alpha &= \frac{3}{4} \cos \alpha \\ \Rightarrow & & \tan \alpha &= \frac{\sin \alpha}{\cos \alpha} = \frac{1}{4} \\ \Rightarrow & & \alpha &= \arctan \frac{1}{4} \approx 14^\circ \end{aligned}$ Inserting the result for $\alpha$ in equation (III) or (IV) results $F_C = \frac{m g}{\frac{3}{4} \cos \alpha + \sin \alpha} = \frac{m g}{\frac{3}{2} \cos \alpha - 2 \sin \alpha} \approx 1.03 \cdot mg$