Q10P - Fundamentals Of Physics

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 B weighs  50 N , and angle $\mathit{\varphi }$ is 35⁰ . Find (a) tension T₁ , (b) tension  T₂ , (c) tension T₃ , and (d) angle $\mathit{\theta }$ .

Step-by-Step Solution

Answer

Answer:

 

1. The tension T₁   in the string is 49 N 
2. The tension  T₂  in the string is  28 N
3. The tension  T₃  in the string is  57 N
4. The angle $\theta$ made by  T₃  with vertical is 29⁰ .

Step 1: 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,  $\varphi ={35}^0$

Step 2: Concept and formula used in the given question

 

Using the free body diagram and 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

Step 3: (a) Calculation for the tension T 1

 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

 

Step 3: (b) Calculation for the tension T 2

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 is  28 N

Step 3: (c) Calculation for the tension T 3

 

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 tension T₃    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

Step 3: (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 .