lets say we have 2 balls inside a jar with radius 1.5, I am trying to calculate the torque with respect to the point of contact of the left ball with the jar.
so far I have that the weight of the object is at a distance of 1.5R with the pivot(moving the weight in his line of action such that is perpendicular with the pivot point).
however I dont have the "straight line" distance of the normal force_f "point of contact of the right ball with the floor" and the normal force_j "point of contact of ball with the jar
$$\Sigma \tau = -1.5RMG+"?"N_f +"?"N_j$$
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Since the free body diagram is as follows:
the four reactions can be calculated by imposing the respective equilibrium on the translation:
$$ \begin{cases} R_A - \frac{1}{2}R_B = 0 \\ \frac{\sqrt{3}}{2}R_B - P_1 = 0 \\ \frac{1}{2}R_B - R_D = 0 \\ -\frac{\sqrt{3}}{2}R_B + R_C - P_2 = 0 \\ \end{cases} \quad \quad \Leftrightarrow \quad \quad \begin{cases} R_A = \frac{\sqrt{3}}{3}P_1 \\ R_B = \frac{2\sqrt{3}}{3}P_1 \\ R_C = P_1 + P_2 \\ R_D = \frac{\sqrt{3}}{3}P_1 \\ \end{cases}\,. $$
If for verification we want to impose equilibrium on the rotation with respect to $B$:
$$ -R_A\frac{\sqrt{3}}{2}r + P_1\frac{1}{2}r + R_C\frac{1}{2}r - R_D\frac{\sqrt{3}}{2}r - P_2\frac{1}{2}r = 0 $$
i.e.
$$ -\frac{\sqrt{3}}{3}P_1\frac{\sqrt{3}}{2}r + P_1\frac{1}{2}r + (P_1+P_2)\frac{1}{2}r - \frac{\sqrt{3}}{3}P_1\frac{\sqrt{3}}{2}r - P_2\frac{1}{2}r = 0 $$
which result in an identity $0=0$, the check is satisfied.
By imposing equilibrium on the translation and rotation of the system of the two balls:
$$ \begin{cases} R_A - R_D = 0 \\ R_C - P_1 - P_2 = 0 \\ -R_A\sqrt{3}\,r - R_C\,r + P_1\,2\,r + P_2\,r = 0 \end{cases} \quad \quad \Leftrightarrow \quad \quad \begin{cases} R_A = \frac{\sqrt{3}}{3}P_1 \\ R_C = P_1 + P_2 \\ R_D = \frac{\sqrt{3}}{3}P_1 \\ \end{cases}\,. $$
Naturally, if $R_B$ is needed, the body system must be exploded as above.