91Ó°ÊÓ

$\mathrm{ϕ}=40°$

From the Free Body Diagram and condition for static equilibrium of the system, we can find the angle and the tension in cord 2.

Equations:

${\mathbf{∑}}_{}^{}{\mathit{F}}_{\mathbf{x}}\mathbf{=}\mathbf{0}$

${\mathbf{∑}}_{}^{}{\mathit{F}}_{\mathbf{y}}\mathbf{=}\mathbf{0}$

$\mathbf{\text{Weight}}\mathbf{=}\mathbf{\text{mg}}$

$\begin{array}{l}{∑}_{}^{}{\text{F}}_{\text{x}}=0\\ {\text{T}}_{\text{1}}\text{cos40}°-{\text{T}}_{\text{2}}\text{³¦´Ç²õθ}=0\\ {\text{T}}_{\text{1}}\text{cos40}°={\text{T}}_{\text{2}}\text{³¦´Ç²õθ}\\ {\text{T}}_{\text{1}}=\frac{{\text{T}}_{\text{2}}\text{³¦´Ç²õθ}}{\text{cos40}°}\end{array}$

$\begin{array}{l}{∑}_{}^{}{\text{F}}_{\text{y}}=0\\ {\text{T}}_{\text{1}}{\text{²õ¾±²Ô40°+°Õ}}_{\text{2}}\text{²õ¾±²Ôθ}−\text{mg}=0\\ \frac{{\text{T}}_{\text{2}}\text{³¦´Ç²õθ}}{\text{cos40}°}×{\text{²õ¾±²Ô40°+°Õ}}_{\text{2}}\text{²õ¾±²Ôθ}=\text{mg}\\ {\text{T}}_{\text{2}}(\text{²õ¾±²Ôθ}+\text{³¦´Ç²õθtan40}°)=\text{mg}\\ {\text{T}}_{\text{2}}=\frac{\text{mg}}{(\text{²õ¾±²Ôθ}+\text{³¦´Ç²õθtan40}°)}\end{array}$

As the value in the ${\text{T}}_{\text{2}}$should be minimum, we take derivative at both sides, and we get,

$\begin{array}{l}\frac{d}{d\mathrm{θ}}\left(T_2\right)=\frac{d}{d\mathrm{θ}}\left[\frac{\text{mg}}{(\text{²õ¾±²Ôθ}+\text{³¦´Ç²õθtan40}°)}\right]\\ 0=\frac{-\text{mg}(\text{tan40}°×-\text{²õ¾±²Ôθ}+\text{³¦´Ç²õθ})}{{\left((\text{tan40}°\text{׳¦´Ç²õθ})\text{+²õ¾±²Ôθ}\right)}^{\text{2}}}\\ (\text{tan40}°×-\text{²õ¾±²Ôθ}+\text{³¦´Ç²õθ})=0\\ \text{tan40}°=\cot \mathrm{θ}\\ \mathrm{θ}={\cot }^{−1}(\text{tan40}°)\\ \mathrm{θ}=50°\end{array}$

As we know,

$\begin{array}{c}{\text{T}}_{\text{2}}\text{=}\frac{\text{mg}}{(\text{tan40}°\text{׳¦´Ç²õθ})\text{+²õ¾±²Ôθ}}\\ \text{=}\frac{\text{mg}}{(\text{tan40}°\text{׳¦´Ç²õ50}°)\text{+sin50}°}\\ \text{=0.77mg}\end{array}$

The minimum tension in cord 2 in terms of mg is the$\text{0.77mg}$.