https://file.aiquickdraw.com/mathbot/file/1728398689481p5yfvcnx.jpg

Let's break down the problem:

We are asked to find the minimum length of a 16 mm diameter cable that supports a 10 kN weight at point B, with the cable tension not exceeding 40 MPa.

### Given Data:
- Force \( W = 10 \, \text{kN} = 10,000 \, \text{N} \)
- Maximum allowable tension \( \sigma = 40 \, \text{MPa} = 40 \times 10^6 \, \text{Pa} \)
- Diameter of the cable \( d = 16 \, \text{mm} = 0.016 \, \text{m} \)
- Geometry: 
  - Distance between points A and B = 4 m,
  - Distance from B to C vertically = 1 m,
  - Total span of the cable = 2 m horizontally on both sides.

### Step 1: Calculate the cross-sectional area of the cable
The cable has a circular cross-section, so its area is given by:

\[
A = \frac{\pi d^2}{4}
\]

Substitute \( d = 0.016 \, \text{m} \):

\[
A = \frac{\pi (0.016)^2}{4} = 2.0106 \times 10^{-4} \, \text{m}^2
\]

### Step 2: Find the maximum tension force
We can use the maximum stress to find the maximum allowable tension force \( T \) using:

\[
T = \sigma \times A
\]

Substituting the values for \( \sigma \) and \( A \):

\[
T = 40 \times 10^6 \times 2.0106 \times 10^{-4} = 8042.4 \, \text{N}
\]

So, the maximum tension that the cable can handle is approximately 8042.4 N.

### Step 3: Analyze the forces in the cable
The forces in the cable can be broken into two components:
- Horizontal component: \( T_x = T \cos(\theta) \)
- Vertical component: \( T_y = T \sin(\theta) \)

At equilibrium, the sum of the vertical forces must equal the weight \( W \):

\[
2T_y = W \quad \Rightarrow \quad 2T \sin(\theta) = W
\]

Thus,

\[
T \sin(\theta) = \frac{W}{2} = \frac{10,000}{2} = 5000 \, \text{N}
\]

Now, we know that the vertical height from B to C is 1 m and the horizontal distance between A and B is 2 m. Therefore, \( \tan(\theta) \) can be written as:

\[
\tan(\theta) = \frac{1}{2}
\]

Using trigonometry:

\[
\sin(\theta) = \frac{1}{\sqrt{1^2 + 2^2}} = \frac{1}{\sqrt{5}} = 0.4472
\]

### Step 4: Solve for the tension in the cable
Now substitute \( \sin(\theta) \) into the equation:

\[
T \times 0.4472 = 5000
\]

\[
T = \frac{5000}{0.4472} = 11,184.7 \, \text{N}
\]

This exceeds the maximum tension of 8042.4 N, meaning that the current configuration would exceed the allowable tension. Therefore, we need to adjust the geometry to reduce the angle \( \theta \), effectively increasing the length of the cable to reduce the tension.

### Step 5: Calculate the minimum length of the cable
We need to find the total length of the cable from A to C. The length \( L \) of the cable is given by:

\[
L = \frac{h}{\sin(\theta)}
\]

Would you like more details on this or have any specific questions?

### Here are five follow-up questions to explore the problem further:

1. How does changing the height from B to C affect the tension in the cable?
2. What would happen if the diameter of the cable were increased?
3. Can you calculate the required cable diameter for this current setup to handle the 10 kN load without exceeding the 40 MPa tension limit?
4. What if the allowable tension was reduced to 30 MPa—how would that affect the required cable length?
5. How does this setup relate to real-world applications like suspension bridges or construction cranes?

### Tip:
Always check the units and convert all measurements to consistent SI units to avoid calculation errors when working with stress, tension, and forces.

The cable ABC supports a 10 kN weight hanging on a pulley at B. What must be the minimum length of this 16 mm diameter cable if its tension must not exceed 40 MPa?

Find the minimum length of a 16 mm diameter cable supporting a 10 kN weight on a pulley at B, ensuring tension does not exceed 40 MPa. Learn about key engineering concepts, stress analysis, and trigonometry.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation