Answer
$\phi=5.91^o$
$F_T=14.3N$
Work Step by Step
$m=0.150 kg$
$r=0.600m$
$T=0.500s$
$F_T=$Force of tension
In the horizontal direction, $\sum F_{x}=\frac{mv^2}{r}=F_T\cos(\phi)$
$F_{T}=\frac{mv^2}{r\cos(\phi)}=\frac{(0.150kg)(7.54 \frac{m}{s})^2}{(0.600m)\cos(\phi)}=\frac{14.2N}{\cos(\phi)}$
In the vertical direction, $\sum F_{y}=F_T\sin(\phi)-mg=0$
$F_T=\frac{mg}{\sin(\phi)}=\frac{(0.150kg)(9.81\frac{m}{s^2})}{\sin(\phi)}=\frac{1.47N}{\sin(\phi)}$
Equating the two equations, we get
$\frac{14.2N}{\cos(\phi)}=\frac{1.47N}{\sin(\phi)}$
$\phi=\arctan(\frac{1.47N}{14.2N})=5.91^o$
$F_T=\frac{1.47N}{\sin(5.91^o)}=14.3N$
