Answer:
Same.
Explanation:
I made a diagram with the description that is given. So things to be able to perform. In this diagram I consider the recessed bar.
For the solution to this problem it is necessary to take into account the concepts of static force and torque.
For case 1,
We perform sum of moments in the embedded end,
[tex]\sum T = 0[/tex]
[tex]F(L)(sin\theta) -mg (\frac{L}{2})=0[/tex]
clearing F,
[tex]F=\frac{mg}{2sin\theta}[/tex]
For case 2,
We perform sum of moments in the embedded end
[tex]F(\frac{L}{2})sin\theta - mg \frac{L/2}{2}=0[/tex]
[tex]F\frac{L}{2}sin\theta = mg \frac{L}{4}[/tex]
[tex]F = \frac{mg}{2sin\theta}[/tex]
We can verify that the tension for both cases is same.