Question: The system in Fig. 12-28 is in equilibrium, with the string in the center exactly horizontal. Block A weighs 40 N , block Bweighs 50 N , and angle $\mathit{\varphi }$ is 35⁰ . Find (a) tension T₁ , (b) tension T₂ , (c) tension T₃ , and (d) angle $\mathit{\theta }$ .

The weight of the block A, W_A = 40 N

The weight of the block B, W_B =50 N

The angle made by T₁ with vertical is, $\varphi ={35}^0$

Using the free body diagramand the condition for equilibrium,you can find the tensions in the string and the angle theta. The formulas used are given below.

Newton’s second law:

F_net= ma

At equilibrium,

F_net= 0

The free body diagram:

At equilibrium,

F_ynet= 0

From the figure we can write,

$$\begin{gathered} \backslash \mathrm{begingathered}{W}_A=T_1\text{cos}\varphi \\ {T}_1=\frac{W_A}{\text{cos}\varphi } \\ =\frac{40}{\text{cos}35°} \\ =49\text{N} \\ \backslash \mathrm{endgathered} \end{gathered}$$

Hence,the tension T₁ in the string is 49 N

At equilibrium,

$$\begin{gathered} F_{xnet}=0 \\ {T}_2={\text{T}}_1\text{sin}\varphi \\ =49\text{sin}35° \\ =28\text{N} \end{gathered}$$

Hence, the tension T₂ in the string is28 N

From the free body diagram, you can write that:
$$\begin{gathered} T_{3x}=T_3\text{sin}\theta \\ =T_2 \\ =28\text{N} \end{gathered}$$

Also,
$$\begin{gathered} T_{3y}=T_3\text{cos}\theta \\ ={\text{W}}_{\text{B}} \\ =50\text{N} \end{gathered}$$

So, the tensionT₃ is,

$$\begin{gathered} T_3=\sqrt{{T_{3x}}^2+{T_{3y}}^2} \\ =\sqrt{{28}^2+{50}^2} \\ =57.3 \\ ~57\text{N} \end{gathered}$$

Hence, the tension T₃ in the string is 57 N

$$\begin{gathered} T_3\text{sin}\theta =T_2 \\ 57\text{sin}\theta =28 \\ \theta ={\sin }^{-1}\left(\frac{28}{57}\right) \\ \theta =29 \end{gathered}$$

Hence, the angle($\theta$) made by T₃ with vertical is 29⁰ .