Rido12 said:
https://www.physicsforums.com/attachments/3352
I've seen someone's solution in which they did the following:
$$600(200)-R_{AH} (600) = 0$$
How did they arrive at that conclusion by taking the moment about $A$?
Hey Rido! (Wave)
The third equilibrium condition is that the sum of the torques must be zero.
The torque of the vertical 600 N force with respect to A is
$$+600\text{ N} \cdot \frac{400\text{ mm}}{2} = +600\text{ N} \cdot 200\text{ mm}\tag 1$$
Apparently clockwise has been chosen to be positive, so this torque is positive.
In A and C there will be vertical force components and horizontal force components.
The vertical force components have a lever-arm of zero with respect to A, so the corresponding torques are zero.
The horizontal force component at A also has a lever-arm of zero.
So that leaves only the horizontal force component at C.
Let's use the symbol $H_C$ for the horizontal force component at C, which is pointing to the right.
Then the corresponding torque is:
$$-H_C \cdot (300 \text{ mm} + 300 \text{ mm}) = -H_C \cdot 600\text{ mm} \tag 2$$
This is negative, since it would make the structure turn counter clockwise.
So the total sum of torques is (1) and (2) together:
$${}_+^\curvearrowright\sum T_A = +600\text{ N} \cdot 200\text{ mm} - H_C \cdot 600\text{ mm}$$
Btw, I don't know why the symbol $R_{AH}$ was chosen for $H_C$.
