This problem can be solved by considering it as a statics problem with a suspended weight at the center of the cable, forming a catenary curve. However, given the relatively small sag in comparison to the horizontal distance, we can approximate the situation using principles of equilibrium and trigonometry. Here's the approach:

Step 1: Analyze the system

• Weight (W): 4.0 kN (kilonewtons), which is equivalent to 4000 N (newtons).
• Distance between the houses (L): 10.0 m.
• Sag (h): The vertical distance from the cable attachment points (at 10.0 m) to the lowest point where the weight is hanging, which is 3.0 m (since 10.0 m - 7.0 m = 3.0 m).

Step 2: Forces acting on the system

The forces in the cable can be broken down into two components:

1. Tension in the horizontal direction (T_h): The horizontal tension is constant throughout the cable.

2. Tension in the vertical direction at the center (T_v): The vertical tension must balance the weight at the center.

Step 3: Equilibrium conditions

At equilibrium, the vertical and horizontal components of the tension force at the point of attachment (on the walls) must satisfy the following conditions:

Vertical equilibrium:

At the lowest point (the center), the vertical component of the tension, $T_v$, must balance the weight of the mass.

$T_v = W = 4000 \, \text{N}$

Horizontal equilibrium:

Let $T$ represent the total tension in the cable at the attachment point on the wall. The angle $\theta$ is the angle the cable makes with the horizontal at the attachment points. The horizontal component of the tension $T_h$ is related to the total tension by:

$T_h = T \cos(\theta)$

The vertical component at the ends is:

$T_v = T \sin(\theta)$

From trigonometry, the tangent of the angle $\theta$ is given by:

$\tan(\theta) = \frac{T_v}{T_h} = \frac{4000}{T_h}$

Geometry of the problem:

From the geometry of the situation, the sag of the cable (3.0 m) and half the horizontal distance (5.0 m) gives:

$\tan(\theta) = \frac{3.0 \, \text{m}}{5.0 \, \text{m}} = 0.6$

So,

$\frac{4000}{T_h} = 0.6 \quad \Rightarrow \quad T_h = \frac{4000}{0.6} = 6666.67 \, \text{N}$

Step 4: Calculate the total tension

Now that we know $T_h$, we can calculate the total tension $T$ using the Pythagorean theorem:

$T = \sqrt{T_h^2 + T_v^2}$

Substitute the values:

$T = \sqrt{(6666.67)^2 + (4000)^2}$

Let me calculate that:The total tension $T$ in the cable is approximately $7774.61 \, \text{N}$.

Summary:

• The horizontal tension component is $T_h = 6666.67 \, \text{N}$.
• The vertical tension component is $T_v = 4000 \, \text{N}$.
• The total tension in the cable is $T \approx 7774.61 \, \text{N}$.

Would you like more details or have any further questions? Here are a few related questions:

1. How does the sag of the cable affect the tension forces?
2. What if the sag were smaller (e.g., only 2 meters instead of 3)?
3. How would the tension change if the weight were doubled?
4. What assumptions are involved in approximating the cable as a straight-line tension system?
5. How could this calculation change if the cable were much longer or had significant stretch?

Tip: In problems involving suspended weights and cables, approximating using small angles or sags often simplifies the math significantly, but it's important to check if the assumption is reasonable.