Consider the 668 N weight held by two cables
shown below. The left-hand cable had tension
T and makes an angle of θ with the wall. The
right-hand cable had tension 720 N and makes
an angle of 29◦ with the ceiling. What is the tension T in the left-hand
cable slanted at an angle of θ with respect to the wall?
Answer in units of N.

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.

Consider the 668 N weight held by two cables shown below. The left-hand cable had tension T and makes an angle of θ with the wall. The right-hand cable had tension 720 N and makes an angle of 29° with the ceiling. What is the tension T in the left-hand cable slanted at an angle of θ with respect to the wall? Answer in units of N.

Learn how to solve for the tension in a cable supporting a 668 N weight using principles of force equilibrium and trigonometry. This example demonstrates how to break forces into components and apply equilibrium conditions. Suitable for students in Grades 10-12.

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