Question

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 $\mathit{\theta }$ .

Short answer

Answer:

1. The tension T₁ in the string is 49 N
2. The tension T₂ in the string is28 N
3. The tension T₃ in the string is 57 N
4. The angle$\theta$ made by T₃ with vertical is 29⁰ .

Step-by-step solution

Understanding the given information

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, $\varphi ={35}^0$

Concept and formula used in the given question

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

(a) Calculation for the tension  T1

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

(b) Calculation for the tension  T2

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

(c) Calculation for the tension T3

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

(d) Calculation for the angle  θ

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

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