A solid bar of length $L$ has a mass $m_1.$ The bar is fastened by a pivot at one end to a wall which is at an angle $$ with respect to the horizontal. The bar is held horizontally: by a vertical cord that is fastened to the bar at a distance $x$ from the wall. A mass $m_2$ is suspended from the free end of the bar.

Find the tension in the cord.

$T=(m_1+m_2)(\frac{L}{x})(\frac{g}{2})$

$T=(m_1+m_2)gsin$

$T=(\frac{1}{2}{m}_1+m_2)(\frac{L}{x})gcos$

$T=(\frac{1}{2}{m}_1+m_2)(\frac{L}{x})g$

$T_5=(m_1+m_2)gcos$

$T=(\frac{1}{2}{m}_1+m_2)(\frac{L}{x})gsin$

$T=(m_1+\frac{1}{2}{m}_2)(\frac{L}{x})gcos$

$T=(m_1+\frac{1}{2}{m}_2)(\frac{L}{x})g$

$T=0$

$T=(m_1+\frac{1}{2}{m}_2)(\frac{L}{x})gsin$