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Consider the 663 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 810 N
and makes an angle of 26◦ 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 need to break down the forces acting on the system into horizontal and vertical components. Here's the approach step-by-step:

### Step 1: Identify the forces
- The weight \( W = 663 \, \text{N} \) is acting vertically downward.
- The right-hand cable has a tension of \( 810 \, \text{N} \) and makes an angle of \( 26^\circ \) with the ceiling.
- The left-hand cable has a tension \( T \), which makes an angle \( \theta \) with the wall (we will solve for this later).

### Step 2: Break forces into components
We need to break the forces into their horizontal and vertical components and use the fact that the system is in equilibrium. In equilibrium:
- The sum of horizontal forces is zero: \( \sum F_x = 0 \)
- The sum of vertical forces is zero: \( \sum F_y = 0 \)

#### Vertical components:
- The vertical component of the tension in the right cable is: 
  \[
  T_{\text{right}, y} = 810 \, \text{N} \cdot \sin(26^\circ)
  \]
- The vertical component of the tension in the left cable is:
  \[
  T_{\text{left}, y} = T \cdot \sin(\theta)
  \]
- The total vertical force must balance the weight:
  \[
  T_{\text{left}, y} + T_{\text{right}, y} = W = 663 \, \text{N}
  \]
  So,
  \[
  T \cdot \sin(\theta) + 810 \, \text{N} \cdot \sin(26^\circ) = 663 \, \text{N}
  \]

#### Horizontal components:
- The horizontal component of the tension in the right cable is:
  \[
  T_{\text{right}, x} = 810 \, \text{N} \cdot \cos(26^\circ)
  \]
- The horizontal component of the tension in the left cable is:
  \[
  T_{\text{left}, x} = T \cdot \cos(\theta)
  \]
  Since there is no horizontal movement:
  \[
  T_{\text{left}, x} = T_{\text{right}, x}
  \]
  Therefore:
  \[
  T \cdot \cos(\theta) = 810 \, \text{N} \cdot \cos(26^\circ)
  \]

### Step 3: Solve the system of equations
We now have two equations:
1. \( T \cdot \sin(\theta) + 810 \cdot \sin(26^\circ) = 663 \)
2. \( T \cdot \cos(\theta) = 810 \cdot \cos(26^\circ) \)

Let me calculate these values and solve for \( T \).The tension \( T \) in the left-hand cable is \( 810 \, \text{N} \).

Would you like further details on how this result was obtained or have any additional questions?

Here are five follow-up questions to expand your understanding:
1. How do horizontal and vertical force components ensure equilibrium in this scenario?
2. What happens to the tension \( T \) if the angle \( \theta \) increases?
3. How would you calculate the angle \( \theta \) if the problem provided it explicitly?
4. Could you apply this method to a similar setup but with different angles or tensions?
5. How would the equilibrium change if the weight of the object increased?

**Tip:** Always decompose forces into horizontal and vertical components when solving equilibrium problems involving angled forces.

Consider the 663 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 810 N and makes an angle of 26° 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.

Solve for the tension in a cable supporting a 663 N weight, with one cable making a 26° angle with the ceiling and another at an unknown angle θ. Apply trigonometric functions to decompose forces and solve for equilibrium.

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