https://r2.erweima.ai/mathbot/image/1724487252191xnhu4oto.jpg

Find the tension in the cable, the wheel reaction R, and the distance “e” from the center of gravity to the line of action of the cable

To solve this problem, we need to analyze the forces acting on the airplane as it lands on the carrier, specifically focusing on the tension in the cable, the reaction at the wheel (denoted by \( R \)), and the distance \( e \) from the center of gravity to the line of action of the cable.

Let's break this down step by step.

### Given:
1. **Weight of the airplane** (\( W \)): 10,000 lb
2. **Acceleration** (\( a \)): 3 times the acceleration due to gravity, so \( a = 3g \), where \( g = 32.2 \, \text{ft/s}^2 \).
3. **Cable tension** (\( T \)): Unknown
4. **Reaction at the wheel** (\( R \)): Unknown
5. **Distance \( e \)**: Distance from the center of gravity to the line of action of the cable.

### 1. Analyzing Forces:
- The forces acting on the airplane include:
  - The weight \( W \) acting vertically downward.
  - The tension \( T \) in the cable acting horizontally to the left.
  - The reaction force \( R \) at the wheel acting vertically upward.

### 2. Setting up Equations of Equilibrium:

**Horizontal Equilibrium:**
\[
\sum F_x = 0 \quad \Rightarrow \quad T = ma
\]
Where \( m \) is the mass of the airplane.

Since \( W = mg \):
\[
m = \frac{W}{g} = \frac{10,000 \, \text{lb}}{32.2 \, \text{ft/s}^2}
\]

So, the tension \( T \) becomes:
\[
T = \frac{W}{g} \times 3g = 3W = 3 \times 10,000 \, \text{lb} = 30,000 \, \text{lb}
\]

**Vertical Equilibrium:**
\[
\sum F_y = 0 \quad \Rightarrow \quad R = W = 10,000 \, \text{lb}
\]

**Moment about point \( A \):**

Taking moments about the wheel at point \( A \) (assuming \( e \) is the distance from the center of gravity to the line of action of the tension \( T \)):
\[
T \cdot h = W \cdot e
\]
Where \( h \) is the vertical distance from the point \( A \) to the line of action of \( T \). Solving for \( e \):
\[
e = \frac{T \cdot h}{W}
\]

### 3. Final Expressions:
- **Tension \( T \) in the cable**: \( T = 30,000 \, \text{lb} \)
- **Reaction \( R \) at the wheel**: \( R = 10,000 \, \text{lb} \)
- **Distance \( e \)**: \( e = \frac{30,000 \times h}{10,000} = 3h \)

The exact value of \( e \) depends on the value of \( h \), which isn't provided, but the relationship \( e = 3h \) is key.

Would you like any further details or have any questions?

### Related Questions:
1. How does the location of the center of gravity affect the tension in the cable?
2. What would happen to the wheel reaction \( R \) if the acceleration was doubled?
3. How would the tension \( T \) change if the weight of the airplane was reduced by half?
4. If the cable angle was not horizontal, how would that affect the tension calculation?
5. What would be the impact on the distance \( e \) if the plane’s acceleration was less than 3g?

### Tip:
Always ensure the force vectors are properly aligned when analyzing equilibrium problems to avoid errors in moment calculations.

Learn how to calculate tension in a cable, wheel reaction R, and distance 'e' from the center of gravity to the line of action of the cable in a statics problem. This detailed solution covers the equilibrium of forces and moments, applicable to college-level mechanics.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation