91Ó°ÊÓ

The inclination angle of${\text{T}}_{\text{1}}$is${\mathrm{θ}}_1=60°$.

The inclination angle of${\mathrm{T}}_2$is${\mathrm{θ}}_2=20°$ .

The mass of the ball, $\text{M}=2.0\text{ k²µ}$.

We draw the free body diagram. The system is at equilibrium. For such a system, the vector sum of the forces acting on it is zero. We can apply this concept along the x and y axes separately and find out${\mathbf{\text{T}}}_{\mathbf{\text{1}}}$and${\mathbf{T}}_{\mathbf{2}}$,

Formulae:

$\mathbf{Î\pounds }{\stackrel{\mathbf{→}}{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}$

According to the free body diagram, you can apply static equilibrium conditions along the x-axis as

$\begin{array}{c}\mathrm{Î\pounds }{\stackrel{→}{\mathrm{F}}}_{\mathrm{x},\mathrm{net}}=0\\ {\mathrm{T}}_1\cos {\mathrm{θ}}_1−{\mathrm{T}}_2\cos {\mathrm{θ}}_2=0\\ {\mathrm{T}}_1\cos {\mathrm{θ}}_1={\mathrm{T}}_2\cos {\mathrm{θ}}_2\\ {\mathrm{T}}_1={\mathrm{T}}_2\frac{\cos {\mathrm{θ}}_2}{\cos {\mathrm{θ}}_1}\text{ â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰â¶Ä‰}\left(1\right)\end{array}$

You can apply static equilibrium conditions along the y-axis as

$\begin{array}{c}\mathrm{Î\pounds }{\stackrel{→}{\mathrm{F}}}_{\mathrm{y},\mathrm{net}}=0\\ {\mathrm{T}}_1\sin {\mathrm{θ}}_1−{\mathrm{T}}_2\sin {\mathrm{θ}}_2−\mathrm{Mg}=0\\ \left({\mathrm{T}}_2\frac{\cos {\mathrm{θ}}_2}{\cos {\mathrm{θ}}_1}\right)\sin {\mathrm{θ}}_1−{\mathrm{T}}_2\sin {\mathrm{θ}}_2−\mathrm{Mg}=0\\ {\mathrm{T}}_2(\cos {\mathrm{θ}}_2\tan {\mathrm{θ}}_1−\sin {\mathrm{θ}}_2)−\mathrm{Mg}=0\\ {\mathrm{T}}_2(\cos {\mathrm{θ}}_2\tan {\mathrm{θ}}_1−\sin {\mathrm{θ}}_2)=\mathrm{Mg}\\ \left({\mathrm{T}}_2\frac{\cos {\mathrm{θ}}_2}{\cos {\mathrm{θ}}_1}\right)\sin {\mathrm{θ}}_1−{\mathrm{T}}_2\sin {\mathrm{θ}}_2−\mathrm{Mg}=0\\ {\mathrm{T}}_2=\frac{\mathrm{Mg}}{(\cos {\mathrm{θ}}_2\tan {\mathrm{θ}}_1−\sin {\mathrm{θ}}_2)}\\ {\mathrm{T}}_2=\frac{2.0\text{ k²µ}×9.8{\text{″¾/²õ}}^2}{(\cos 20°\tan 60°−\sin 20°)}\\ {\mathrm{T}}_2=15\text{ N}\end{array}$

$\begin{array}{c}{\mathrm{T}}_1={\mathrm{T}}_2\frac{\cos {\mathrm{θ}}_2}{\cos {\mathrm{θ}}_1}\\ =15.25\text{ }\mathrm{N}\left(\frac{\cos 20°}{\cos 60°}\right)\\ =29\text{ N}\end{array}$