A 40 N block is supported by two ropes. One rope is horizontal and other makes an angle of `30^(@)` with the ceiling. The tension in the rope attached to the ceiling is approximately

Options

A. 80 N
B. 40 N
C. 34.6 N
D. 46.2 N

Explanation

To solve the problem of finding the tension in the rope attached to the ceiling, we can follow these steps:

### Step 1: Identify the Forces Acting on the Block
The block has a weight acting downward, which is given as 40 N. There are two tension forces acting on the block:
1. The tension \( T_1 \) in the horizontal rope.
2. The tension \( T_2 \) in the rope that makes an angle of 30° with the ceiling.

### Step 2: Resolve the Tension in the Rope at an Angle
The tension \( T_2 \) can be resolved into two components:
- A vertical component \( T_{2y} = T_2 \sin(30°) \)
- A horizontal component \( T_{2x} = T_2 \cos(30°) \)

### Step 3: Apply the Equilibrium Condition in the Vertical Direction
Since the block is in equilibrium, the sum of the vertical forces must equal zero. Therefore, we can write the equation:
\[ T_{2y} + T_1 = 40 \, \text{N} \]
Substituting the vertical component:
\[ T_2 \sin(30°) + T_1 = 40 \]

### Step 4: Calculate the Value of \( \sin(30°) \)
From trigonometric values, we know:
\[ \sin(30°) = \frac{1}{2} \]
Thus, we can rewrite the equation:
\[ T_2 \cdot \frac{1}{2} + T_1 = 40 \]

### Step 5: Apply the Equilibrium Condition in the Horizontal Direction
In the horizontal direction, the only force acting is the horizontal component of the tension \( T_2 \):
\[ T_{2x} = T_1 \]
Substituting the horizontal component:
\[ T_2 \cos(30°) = T_1 \]

### Step 6: Calculate the Value of \( \cos(30°) \)
From trigonometric values, we know:
\[ \cos(30°) = \frac{\sqrt{3}}{2} \]
Thus, we can rewrite the equation:
\[ T_2 \cdot \frac{\sqrt{3}}{2} = T_1 \]

### Step 7: Substitute \( T_1 \) into the Vertical Force Equation
Now we can substitute \( T_1 \) from the horizontal equation into the vertical equation:
\[ T_2 \cdot \frac{1}{2} + T_2 \cdot \frac{\sqrt{3}}{2} = 40 \]

### Step 8: Factor Out \( T_2 \)
Factoring out \( T_2 \):
\[ T_2 \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) = 40 \]

### Step 9: Simplify the Equation
Combining the terms inside the parentheses:
\[ T_2 \left( \frac{1 + \sqrt{3}}{2} \right) = 40 \]

### Step 10: Solve for \( T_2 \)
Now, we can solve for \( T_2 \):
\[ T_2 = \frac{40 \cdot 2}{1 + \sqrt{3}} \]

### Step 11: Calculate the Numerical Value
Calculating the value:
1. \( 1 + \sqrt{3} \approx 1 + 1.732 = 2.732 \)
2. \( T_2 \approx \frac{80}{2.732} \approx 29.3 \, \text{N} \)

Thus, the tension in the rope attached to the ceiling \( T_2 \) is approximately **29.3 N**.