## Essential University Physics: Volume 1 (3rd Edition)

$0.87$
As we know that $N=Tsin(\theta)$ and the force of friction is given as $f=\mu N$ $\implies f=\mu Tsin(\theta)$ Now we can find the torque as $fR-Tcos(\theta)(\frac{1}{2}R)=0$ This simplifies to: $\mu sin(\theta)-cos(\theta)(\frac{1}{2})=0$ $\implies \mu=\frac{1}{2tan(\theta)}=\frac{1}{2tan(30)}=0.87$