Since the pulley does not have friction, the two segments of the cable are subject to the same tension force T. Based on the free-body diagram shown above, we have
[tex]EM _{A} = 0 {\to T x 5 + \frac{2}{ \sqrt{5} } T x 10 - W x 13 = 0[/tex]
Thus, we get
[tex]T = \frac{13W}{(5 + 20/ \sqrt{5)} } = \frac{13 x 80}{(5 + 20/ \sqrt{5)}} = 74.583 (Ib)[/tex]
To find the horizontal support force, we have
[tex]EF_{X} = 0 \to A_{X} - \frac{1}{ \sqrt{5} } T = 0[/tex]
Thus,
[tex]A_{X} = \frac{1}{ \sqrt{5} } T = \frac{1}{ \sqrt{5} } x 74.583 = 33.354 (Ib)[/tex]
