In the following figure we have an homogeneous L-shaped bar that has weight 120N. It's articulated without friction at point A. It's vertical part has length 1m and its horizontal part has length 3m. Find the intensity of F that maintains the bar in equilibrium.
2 . Relevant equations
ΣTorque = 0
The Attempt at a Solution
The book solves the problem considering that the weight of the vertical part (30N) doesn't exert Torque, so the CM that exerts torque gets dislocated to the middle of the horizontal part of the bar. From there it's trivial.
The thing is that I didn't think of doing this way initially. I first calculated the CM of the Bar, which is at (1.5, 0.5) considering the origin at the "bottom of the L". Then I defined a vector from A to CM and found it's length is 2.5. From there I had to find the component of the weight which is perpendicular to the vector. ( This is drawn below).
From all of this we get F (1) = P cos(β)(2.5) ⇒ F = 120 (3/5)(5/2) ⇔ F = 180N . This, however, is wrong.
I thought doing it this way would do the trick, but if, instead of 120, I had put 90N the above equation would give me the correct solution. Is there any way, other than the one the book did, that could solve this problem?
I appreciate, in advance, any assistance.