Two houses are located at a distance of 10.0 m from each other. A cable is to be stretched between
their outer walls, to which a weight of 4.0 kN (400 kg) is attached in the center. The cable is attached
to the walls of the two houses at a height of 10.0 m each. The suspension point of the weight should
not be lower than 7.0 m under load. How high are the forces that occur in the cable?


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.

Two houses are located at a distance of 10.0 m from each other. A cable is to be stretched between their outer walls, to which a weight of 4.0 kN (400 kg) is attached in the center. The cable is attached to the walls of the two houses at a height of 10.0 m each. The suspension point of the weight should not be lower than 7.0 m under load. How high are the forces that occur in the cable?

Solve a statics problem where a 4.0 kN weight is suspended between two houses, calculating the forces in the cable. Learn how to apply equilibrium conditions, trigonometry, and tension force analysis in this real-world scenario. Suitable for high school students in Grades 10-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation