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

Options

A. `4pixx10^5N`
B. `4pixx10^4N`
C. `2pixx10^5N`
D. `2pixx10^5N`

Explanation

To find the maximum load that the steel cable can support, we will use the relationship between stress, force, and area. The formula for stress (σ) is given by:

\[ \sigma = \frac{F}{A} \]

where:
- \( \sigma \) is the stress,
- \( F \) is the force (or load),
- \( A \) is the cross-sectional area of the cable.

### Step 1: Calculate the cross-sectional area of the cable

The radius of the cable is given as \( r = 2 \, \text{cm} \). We need to convert this into meters for consistency in SI units:

\[ r = 2 \, \text{cm} = 0.02 \, \text{m} \]

The cross-sectional area \( A \) of a circular cable is calculated using the formula:

\[ A = \pi r^2 \]

Substituting the value of \( r \):

\[ A = \pi (0.02)^2 = \pi (0.0004) \, \text{m}^2 \]
\[ A = 0.0004\pi \, \text{m}^2 \]

### Step 2: Use the maximum stress to find the maximum load

The maximum stress that the cable can handle is given as \( \sigma_{\text{max}} = 10^8 \, \text{N/m}^2 \). According to the stress formula, we can rearrange it to find the force:

\[ F = \sigma \cdot A \]

Substituting the values we have:

\[ F = 10^8 \, \text{N/m}^2 \cdot (0.0004\pi \, \text{m}^2) \]

Calculating the area:

\[ F = 10^8 \cdot 0.0004\pi \]
\[ F = 4 \times 10^4 \pi \, \text{N} \]

### Step 3: Calculate the numerical value

Using \( \pi \approx 3.14 \):

\[ F \approx 4 \times 10^4 \times 3.14 \approx 125600 \, \text{N} \]

Thus, the maximum load the cable can support is approximately:

\[ F \approx 125600 \, \text{N} \]

### Final Answer

The maximum load the cable can support is \( 4\pi \times 10^4 \, \text{N} \) or approximately \( 125600 \, \text{N} \).

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