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.