A steel cable with a radius 2 cm supports a chairlift at a ski area. If the maximum stress is not to exceed `10^8 N m^(-2)`, the maximam load the cable can support

Options

A. `4pi xx 10^(5)N`
B. `4pi xx 10^(4) N`
C. `2pi xx 10^(5) N`
D. `2pi xx 10^(4)`

Explanation

To solve the problem of determining the maximum load that a steel cable can support without exceeding a maximum stress of \(10^8 \, \text{N/m}^2\), we can follow these steps:

### Step 1: Understand the relationship between stress, force, and area
Stress (\(\sigma\)) is defined as the force (F) applied per unit area (A):
\[ \sigma = \frac{F}{A} \]

### Step 2: Rearrange the formula to find the force
From the formula, we can rearrange it to find the force:
\[ F = \sigma \cdot A \]

### Step 3: Calculate the area of the cable
The area (A) of a circular cross-section can be calculated using the formula:
\[ A = \pi r^2 \]
where \(r\) is the radius of the cable. Given that the radius is 2 cm, we need to convert it to meters:
\[ r = 2 \, \text{cm} = 0.02 \, \text{m} \]
Now, we can calculate the area:
\[ A = \pi (0.02)^2 = \pi (0.0004) = 0.0004\pi \, \text{m}^2 \]

### Step 4: Substitute the values into the force equation
Now we can substitute the maximum stress and the area into the force equation:
\[ F = (10^8 \, \text{N/m}^2) \cdot (0.0004\pi \, \text{m}^2) \]
\[ F = 10^8 \cdot 0.0004\pi = 4 \times 10^4 \pi \, \text{N} \]

### Step 5: Calculate the numerical value
Using the approximate value of \(\pi \approx 3.14\):
\[ F \approx 4 \times 10^4 \times 3.14 \approx 125600 \, \text{N} \]

### Step 6: Final result
Thus, the maximum load the cable can support is approximately:
\[ F \approx 125600 \, \text{N} \]