To solve for the tension $T$ in the left-hand cable, we can use the principles of equilibrium. The object is in static equilibrium, which means the sum of forces in both the horizontal and vertical directions must be zero.

Step 1: Break down the forces.

The 668 N weight acts directly downward. The tensions in both cables have components in the horizontal and vertical directions.

For the right-hand cable:

• Tension = 720 N
• Angle with the ceiling = 29°

Components of the tension in the right-hand cable:

• Horizontal component: $720 \, \text{N} \cdot \cos(29^\circ)$
• Vertical component: $720 \, \text{N} \cdot \sin(29^\circ)$

For the left-hand cable:

• Tension = $T$
• Angle with the wall = $\theta$

Components of the tension in the left-hand cable:

• Horizontal component: $T \cdot \cos(\theta)$
• Vertical component: $T \cdot \sin(\theta)$

Step 2: Apply equilibrium conditions.

Vertical equilibrium:

The sum of the vertical forces must equal zero: $T \cdot \sin(\theta) + 720 \cdot \sin(29^\circ) = 668$ $T \cdot \sin(\theta) + 720 \cdot 0.4848 = 668$ $T \cdot \sin(\theta) + 348.864 = 668$ $T \cdot \sin(\theta) = 668 - 348.864$ $T \cdot \sin(\theta) = 319.136$

Horizontal equilibrium:

The sum of the horizontal forces must equal zero: $T \cdot \cos(\theta) = 720 \cdot \cos(29^\circ)$ $T \cdot \cos(\theta) = 720 \cdot 0.8746$ $T \cdot \cos(\theta) = 629.712$

Step 3: Solve for $T$.

We now have two equations:

1. $T \cdot \sin(\theta) = 319.136$
2. $T \cdot \cos(\theta) = 629.712$

To solve for $T$, divide equation (1) by equation (2): $\frac{T \cdot \sin(\theta)}{T \cdot \cos(\theta)} = \frac{319.136}{629.712}$ $\tan(\theta) = \frac{319.136}{629.712}$ $\tan(\theta) = 0.5068$

Now, solve for $\theta$: $\theta = \tan^{-1}(0.5068) \approx 26.73^\circ$

Step 4: Substitute $\theta$ back into the horizontal equilibrium equation to solve for $T$:

$T \cdot \cos(26.73^\circ) = 629.712$ $T \cdot 0.8931 = 629.712$ $T = \frac{629.712}{0.8931}$ $T \approx 705 N$

Thus, the tension in the left-hand cable is approximately $705 \, \text{N}$.

Would you like further details or clarifications on any part of the solution?

Here are 5 related questions:

1. How would the solution change if the right-hand cable tension was increased?
2. What if the angle of the left-hand cable was decreased?
3. How can we compute the angle of the left-hand cable more directly?
4. What if the weight was different, say 800 N instead of 668 N?
5. Can this problem be extended to three cables holding the object?

Tip: Always resolve forces into their components to analyze equilibrium problems systematically.