Exam review problem. Weird answer 
The question asks us to calculate the holding force of the triangular mounting if the rock weighs exactly $2kg$ and the mounting supports the rod at exactly $\frac{1}{4}$ of its length. ($g=9.81 \frac{m}{s^2}).$
This is what I tried:
If I remove the mass and $\frac{2}{4}$ of the rod from the right then the system would still be in equilibrium.

Therefore i know that $\frac{1}{2}m_{rod}=m_{rock}$ $\iff$ $2kg=\frac{1}{2}m_{rod} \iff m_{rod}=4kg$
So the total holding force $=4kg\cdot9,81\frac{m}{s^2}+2kg\cdot 9,81 \frac{m}{s^2}=58,86 N$
Here is the problem: In our solutions, the same calculation is done with 3,0kg but the answer still remains $58,86N$. How can that be? Am I making a mistake somewhere?
 A: Ok, I don't want to give away the farm here, so I will intentionally be a little vague.
First thing to recognize is that in this situation your system is in static equilibrium.  For a system to be in equilibrium, two conditions need to be met:


*

*The net force on the system is 0, which means there is no acceleration.  This is expressed in equation form as $$\Sigma F=0$$

*The net torque on the system is 0, which means there is no _angular acceleration.  This could be expressed in analogous form to equation (1) as $\Sigma \tau =0$, but for whatever reason most textbooks choose instead to state it as $$\Sigma \tau_{cc}= \Sigma \tau_{ccw}$$ where $\tau_{cc}$ are torques that would rotate your system clockwise with respect to your chose pivot, and $\tau_{ccw}$ are torques that would rotate your system counterclockwise with respect to your pivot.


As an extra, because your system is in static equilibrium (not moving), not only do we know that your net torques and net forces are zero, but we have the additional knowledge that your velocity and angular velocity are also zero.  In other words, your system is not moving or rotating.
So what I would do to set up your problem is use your net force and net torque equations.  If we assume the mass of the rod is non-negligible, you should have three individual forces acting on your system.  Since you don't know the mass of your rod, I would choose to just express it as a generic $m$ for now.  Then, for your net torque equation, I would choose the point where the rod is mounted as your pivot, so you should have only one clockwise torque and one counterclockwise torque.  Since they don't tell you otherwise, assume the rod has a uniform density.
