# Why is this force completely vertical?

I have the following system, which is in equilibrium:

It's a bar that can pivot around A, and is held in place by a string at B

I'm supposed to find the force applied to the bar at the pivot. I assumed that said force would have a component on the x axis and a component on the y axis (where the x axis is parallel to the ground)

The solution in my book, however, has the following diagram:

The way it's solved in my book, along with the diagram, imply that the force at the pivot ONLY has a vertical (on the y axis) component.

Why is this true? I would have assumed that, in order for the system to be in equilibrium, this force would have a component on the x axis with the same value and opposite direction than that of the string tension projected on the x axis.

Is my reasoning incorrect? If so, why?

• There is something odd in the first figure. What keeps the "L" shaped bar from tipping over? I see nothing supporting it. – Bob D Jan 28 at 19:17
• the string at B. The bar isn't L shaped, it's straight – Francisco José Letterio Jan 28 at 19:18
• Is there any friction between the bar and the ground? – Clara Diaz Sanchez Jan 28 at 19:20
• no, there is no friction – Francisco José Letterio Jan 28 at 19:22
• Seems to me that in order for the whole thing not to tip over at the end of the foot of the L shaped bar, the sum of the moments about the end of the bar due to $Q$ and $T_B$ needs to be zero. – Bob D Jan 28 at 20:44

If $$T_B \neq 0$$ then that cannot be true because there would be a NET horizontal force acting on the bar, i.e. the horizontal component of $$T_B$$.