91Ó°ÊÓ

${\text{m}}_{\text{A}}=10kg$

${\text{m}}_{\text{B}}=5\text{ }kg$

$\mathrm{θ}=30°$

By drawing the free body diagram of the situation, we can find the tension in each wire employing which they are connected. Using the conditions for equilibrium, we can find the coefficient of static friction.

Formula:

${∑}_{}^{}{\text{F}}_{\text{y}}=0$

${∑}_{}^{}{\text{F}}_{\text{x}}=0$

$F_{staticfriction}={\text{μ}}_{\text{s}}×NormalForce$

From this, we can say that,

${\text{T}}_{\text{1}}={\text{F}}_{\text{s}}$

${\text{T}}_{\text{1}}={\text{μ}}_{\text{s}}×{\text{F}}_{\text{N}}$

${\text{T}}_{\text{1}}={\text{μ}}_{\text{s}}×({\text{m}}_{\text{A}}×\text{g})$

${\text{T}}_{\text{1}}={\text{μ}}_{\text{s}}×(10\text{ k²µ}×9.8\text{″¾}/{\text{s}}^{\text{2}})$

${\text{T}}_{\text{1}}=98\text{ N}×{\text{μ}}_{\text{s}}$

From this, we can say that,

${\text{T}}_{\text{2}}={\text{m}}_{\text{B}}×\text{g}$

$\begin{array}{l}{T}_2=5\text{ }kg×9.8\text{ }m/s^2\\ T_2=49\text{ }N\end{array}$

From this, we can say that,

$\begin{array}{l}{∑}_{}^{}{\text{F}}_{\text{y}}=0\\ \left({\text{T}}_{\text{3}}\text{׳¦´Ç²õθ}\right){\text{-T}}_{\text{2}}=0\\ \left({\text{T}}_{\text{3}}\text{׳¦´Ç²õθ}\right)={\text{T}}_{\text{2}}\\ \left({\text{T}}_{\text{3}}\text{׳¦´Ç²õθ}\right)=49\text{ }N\end{array}$

$\begin{array}{l}{∑}_{}^{}{\text{F}}_{\text{x}}=0\\ ({\text{T}}_{\text{3}}×\text{²õ¾±²Ôθ}){\text{-T}}_{\text{1}}=0\\ ({\text{T}}_{\text{3}}×\text{²õ¾±²Ôθ})={\text{T}}_{\text{1}}\\ ({\text{T}}_{\text{3}}×\text{²õ¾±²Ôθ})=98\text{ N}×{\mathrm{μ}}_{\text{s}}\end{array}$

$\begin{array}{l}\frac{({\text{T}}_{\text{3}}×\text{²õ¾±²Ôθ})}{({\text{T}}_{\text{3}}×\text{³¦´Ç²õθ})}=\frac{98\text{ }N×{\text{μ}}_{\text{s}}}{49\text{ }N}\\ {\text{μ}}_{\text{s}}=\tan \text{ }\mathrm{θ}×0.5\\ {\text{μ}}_{\text{s}}=0.5×\tan (30°)\\ {\text{μ}}_{\text{s}}=0.5×0.577\\ {\text{μ}}_{\text{s}}=0.288\end{array}$

The coefficient of static friction between block A and the surface is$0.288$