The two ends of a 20-m rope are fastened at two points on the ceiling, 12 m from each other. If a load of 200 kg is placed at the center of the rope, find the tension in the rope.

We can solve this problem by using the principles of equilibrium and basic trigonometry.

### Step 1: Define Variables
Let’s denote the following:
- The distance between the two points on the ceiling as \( d = 12 \, \text{m} \).
- The length of the rope as \( L = 20 \, \text{m} \).
- The weight of the load as \( W = 200 \, \text{kg} \).
- The gravitational force \( F_g = W \cdot g \), where \( g = 9.8 \, \text{m/s}^2 \).
- The height by which the rope sags (the vertical distance from the ceiling to the load at the center) as \( h \).

### Step 2: Geometry of the Problem
Since the load is placed at the center, the rope forms two equal segments, each of length 10 m. The distance from the center to either attachment point on the ceiling is 6 m (half of 12 m).

Using the Pythagorean theorem, we can find the sag \( h \) of the rope:

\[
\left( \frac{d}{2} \right)^2 + h^2 = \left( \frac{L}{2} \right)^2
\]

Substituting the known values:

\[
6^2 + h^2 = 10^2
\]
\[
36 + h^2 = 100
\]
\[
h^2 = 64 \quad \Rightarrow \quad h = 8 \, \text{m}
\]

So, the rope sags 8 m below the ceiling at the center.

### Step 3: Calculate the Tension
The load creates a downward force equal to \( F_g = 200 \times 9.8 = 1960 \, \text{N} \).

Since the system is in equilibrium, the vertical components of the tension on both sides of the rope must balance this force.

Let \( T \) be the tension in the rope. Each side of the rope forms an angle \( \theta \) with the horizontal. Using trigonometry, we know:

\[
\sin(\theta) = \frac{h}{\frac{L}{2}} = \frac{8}{10} = 0.8
\]

The vertical component of the tension in one side of the rope is \( T \sin(\theta) \), and since there are two sides of the rope supporting the load, the total upward force is \( 2T \sin(\theta) \). This must equal the weight of the load:

\[
2T \sin(\theta) = F_g
\]
\[
2T (0.8) = 1960
\]
\[
T = \frac{1960}{2 \times 0.8} = \frac{1960}{1.6} = 1225 \, \text{N}
\]

Thus, the tension in the rope is **1225 N**.

Would you like further details on any part of this solution or have additional questions?

### Related Questions:
1. How would the tension change if the rope was 15 m long instead of 20 m?
2. What would happen to the tension if the load were placed at a point one-third along the rope instead of in the center?
3. How does the tension change if the weight of the load is doubled?
4. How could you find the tension if the ceiling points were further apart, say 14 m instead of 12 m?
5. What would happen to the sag if the rope were less stiff and more elastic?

### Tip:
In problems involving tension in ropes or cables, always consider breaking forces into their vertical and horizontal components. This simplifies calculating equilibrium conditions.

Solve a statics problem involving a 20-meter rope fastened 12 meters apart with a 200 kg load at the center. Learn how to use the Pythagorean theorem, trigonometry, and equilibrium principles to find the tension in the rope. Suitable for students in Grades 10-12.

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