The system in Fig. 12-77 is in equilibrium. The angles are ${\mathbf{\text{θ}}}_{\mathbf{1}}\mathbf{}\mathbf{=}\mathbf{60}\mathbf{\text{°}}$and ${\mathbf{\text{θ}}}_{\mathbf{2}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{20}\mathbf{\text{°}}$, and the ball has mass $\mathbf{\text{M}}\mathbf{}\mathbf{=}\mathbf{2}\mathbf{.0}\mathbf{\text{\hspace{0.17em}kg}}$. What is the tension in (a) string ab and (b) string bc?

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

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

The mass of the ball, $\text{M}=2.0\text{\hspace{0.17em}kg}$.

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:

$\mathbf{\Sigma }{\overrightarrow{\mathbf{F}}}_{\mathbf{net}}\mathbf{=}\mathbf{0}$

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

$\begin{array}{c}\mathrm{\Sigma }{\overrightarrow{\mathrm{F}}}_{\mathrm{x},\mathrm{net}}=0\\ {\mathrm{T}}_1\cos {\mathrm{\theta }}_1-{\mathrm{T}}_2\cos {\mathrm{\theta }}_2=0\\ {\mathrm{T}}_1\cos {\mathrm{\theta }}_1={\mathrm{T}}_2\cos {\mathrm{\theta }}_2\\ {\mathrm{T}}_1={\mathrm{T}}_2\frac{\cos {\mathrm{\theta }}_2}{\cos {\mathrm{\theta }}_1}\left(1\right)\end{array}$

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

$\begin{array}{c}\mathrm{\Sigma }{\overrightarrow{\mathrm{F}}}_{\mathrm{y},\mathrm{net}}=0\\ {\mathrm{T}}_1\sin {\mathrm{\theta }}_1-{\mathrm{T}}_2\sin {\mathrm{\theta }}_2-\mathrm{Mg}=0\\ \left({\mathrm{T}}_2\frac{\cos {\mathrm{\theta }}_2}{\cos {\mathrm{\theta }}_1}\right)\sin {\mathrm{\theta }}_1-{\mathrm{T}}_2\sin {\mathrm{\theta }}_2-\mathrm{Mg}=0\\ {\mathrm{T}}_2(\cos {\mathrm{\theta }}_2\tan {\mathrm{\theta }}_1-\sin {\mathrm{\theta }}_2)-\mathrm{Mg}=0\\ {\mathrm{T}}_2(\cos {\mathrm{\theta }}_2\tan {\mathrm{\theta }}_1-\sin {\mathrm{\theta }}_2)=\mathrm{Mg}\\ \left({\mathrm{T}}_2\frac{\cos {\mathrm{\theta }}_2}{\cos {\mathrm{\theta }}_1}\right)\sin {\mathrm{\theta }}_1-{\mathrm{T}}_2\sin {\mathrm{\theta }}_2-\mathrm{Mg}=0\\ {\mathrm{T}}_2=\frac{\mathrm{Mg}}{(\cos {\mathrm{\theta }}_2\tan {\mathrm{\theta }}_1-\sin {\mathrm{\theta }}_2)}\\ {\mathrm{T}}_2=\frac{2.0\text{\hspace{0.17em}kg}\times 9.8{\text{\hspace{0.17em}m/s}}^2}{(\cos 20°\tan 60°-\sin 20°)}\\ {\mathrm{T}}_2=15\text{\hspace{0.17em}N}\end{array}$

$\begin{array}{c}{\mathrm{T}}_1={\mathrm{T}}_2\frac{\cos {\mathrm{\theta }}_2}{\cos {\mathrm{\theta }}_1}\\ =15.25\mathrm{N}\left(\frac{\cos 20°}{\cos 60°}\right)\\ =29\text{\hspace{0.17em}N}\end{array}$