Find the tension in each cord in (Figure 1) if the weight of the suspended object is w.
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The figure is divided into two parts. In part (a), an object of weight w is suspended on vertical cord C, which is held by two cords, 'A' and B. The two cords are attached to the ceiling.
Cord 'A' extends to the left from C at an angle of 30 degrees to the horizontal. Cord B extends to the right from C at an angle of 45 degrees to the horizontal. In part (b), the object of weight w is suspended on vertical cord C, which is held by the two cords, 'A' and B. Cord 'A' is attached to a vertical wall on the left at an angle of 60 degrees to the vertical.
Cord B is attached to a horizontal ceiling on the right at an angle of 45 degrees to the horizontal. - Two cases of an object suspended from cords.
Find the tension of the cord A in the figure (b).
Express your answer in terms of  w .

To find the tension in cord \( A \) for **Figure (b)**, we will apply the principles of static equilibrium. The object is in equilibrium, so the forces must balance both horizontally and vertically.

### Step 1: Analyzing the forces

In Figure (b), the weight \( w \) acts downward, and the tension forces in cords \( A \) and \( B \) have components both horizontally and vertically. The angles for the cords are:

- Cord \( A \): Angle with the vertical is \( 60^\circ \).
- Cord \( B \): Angle with the horizontal is \( 45^\circ \).

Let:
- \( T_A \) be the tension in cord \( A \),
- \( T_B \) be the tension in cord \( B \),
- \( w \) be the weight of the suspended object.

Since the object is in equilibrium, the sum of forces in both the horizontal and vertical directions must be zero.

### Step 2: Equilibrium conditions

#### Vertical equilibrium:
The total vertical forces must add up to zero:
\[
T_A \cos(60^\circ) + T_B \sin(45^\circ) = w
\]
This equation accounts for the fact that the upward components of the tensions in cords \( A \) and \( B \) must balance the downward force of weight \( w \).

#### Horizontal equilibrium:
The total horizontal forces must add up to zero:
\[
T_A \sin(60^\circ) = T_B \cos(45^\circ)
\]
This equation shows that the horizontal components of the tensions in cords \( A \) and \( B \) must cancel each other out.

### Step 3: Solving the system of equations

From the horizontal equilibrium equation, we can express \( T_B \) in terms of \( T_A \):
\[
T_B = \frac{T_A \sin(60^\circ)}{\cos(45^\circ)}
\]
Substitute \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \) and \( \cos(45^\circ) = \frac{\sqrt{2}}{2} \):
\[
T_B = \frac{T_A \frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{2}} = T_A \frac{\sqrt{3}}{\sqrt{2}}
\]

Now substitute this expression for \( T_B \) into the vertical equilibrium equation:
\[
T_A \cos(60^\circ) + \left(T_A \frac{\sqrt{3}}{\sqrt{2}}\right) \sin(45^\circ) = w
\]
Substitute \( \cos(60^\circ) = \frac{1}{2} \) and \( \sin(45^\circ) = \frac{\sqrt{2}}{2} \):
\[
T_A \frac{1}{2} + T_A \frac{\sqrt{3}}{\sqrt{2}} \cdot \frac{\sqrt{2}}{2} = w
\]
Simplifying:
\[
T_A \frac{1}{2} + T_A \frac{\sqrt{3}}{2} = w
\]
\[
T_A \left( \frac{1}{2} + \frac{\sqrt{3}}{2} \right) = w
\]
\[
T_A \left( \frac{1 + \sqrt{3}}{2} \right) = w
\]

### Step 4: Solve for \( T_A \)
\[
T_A = \frac{2w}{1 + \sqrt{3}}
\]

Thus, the tension in cord \( A \) is:
\[
T_A = \frac{2w}{1 + \sqrt{3}}
\]

### Summary:
The tension in cord \( A \) is \( \frac{2w}{1 + \sqrt{3}} \), expressed in terms of the weight \( w \).

Would you like more details on this, or have any questions?

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**Related Questions:**
1. What is the tension in cord \( B \) in Figure (b)?
2. How would the tension change if the angle of cord \( A \) were different?
3. How can we calculate the horizontal force components in equilibrium problems?
4. What if the angles were reversed between cords \( A \) and \( B \)?
5. How does the equilibrium principle apply to three-dimensional systems?

**Tip**: In problems involving angles and equilibrium, always break forces into horizontal and vertical components for easier calculations.

Find the tension in each cord in (Figure 1) if the weight of the suspended object is w. The figure is divided into two parts. In part (a), an object of weight w is suspended on vertical cord C, which is held by two cords, 'A' and B. The two cords are attached to the ceiling. Cord 'A' extends to the left from C at an angle of 30 degrees to the horizontal. Cord B extends to the right from C at an angle of 45 degrees to the horizontal. In part (b), the object of weight w is suspended on vertical cord C, which is held by the two cords, 'A' and B. Cord 'A' is attached to a vertical wall on the left at an angle of 60 degrees to the vertical. Cord B is attached to a horizontal ceiling on the right at an angle of 45 degrees to the horizontal. Find the tension of the cord A in the figure (b). Express your answer in terms of w.

Solve for the tension in cord A when an object of weight w is suspended and held by cords at angles of 60° to the vertical and 45° to the horizontal. This problem uses principles of static equilibrium and trigonometry to find the tension forces in a two-dimensional system. Suitable for Grades 11-12.

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