Answer
(a) $0.134N$
(b) $60^{\circ}$
Work Step by Step
(a) We can find the required tension as follows:
$\Sigma F_y=Tcos\theta-mg=0$
This can be rearranged as:
$T=\frac{mg}{cos\theta}$
We plug in the known values to obtain:
$T=\frac{(0.0136Kg)(9.81m/s^2)}{cos6.44^{\circ}}$
$T=0.134N$
(b) The required angle can be determined as follows:
As given that $T=2mg$
We know that
$Tcos\theta =mg$
$\implies 2mgcos\theta=mg$
$\implies cos\theta =\frac{1}{2}$
$\theta=cos^{-1}({\frac{1}{2}})=60^{\circ}$
