An elevator cable is to have a maximum stress of `7xx10^(7) Nm^(-2)` to allow for appropriate safety factors. Its maximum upward acceleration is `1.5 ms^(-2)` . If the cable has to support the total weight of 2000 kg of a loaded elevator , the area of cross-section of the cable should be

Options

A. `3.22 cm^(2)`
B. `2.38 cm^(2)`
C. `0.32 cm^(2)`
D. `8.23 cm^(2)`

Explanation

To find the area of cross-section of the elevator cable, we can follow these steps:

### Step 1: Identify the forces acting on the elevator
The total force (F) acting on the cable when the elevator is accelerating upwards is the sum of the gravitational force and the force due to acceleration. This can be expressed as:
\[ F = mg + ma \]
where:
- \( m = 2000 \, \text{kg} \) (mass of the elevator)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( a = 1.5 \, \text{m/s}^2 \) (upward acceleration)

### Step 2: Calculate the total force
Substituting the values into the equation:
\[ F = 2000 \times 9.8 + 2000 \times 1.5 \]
\[ F = 2000 \times (9.8 + 1.5) \]
\[ F = 2000 \times 11.3 \]
\[ F = 22600 \, \text{N} \]

### Step 3: Use the maximum stress to find the area of cross-section
The maximum stress (\( \sigma \)) allowed in the cable is given as:
\[ \sigma = 7 \times 10^7 \, \text{N/m}^2 \]

The relationship between stress, force, and area is given by:
\[ \sigma = \frac{F}{A} \]
Rearranging this gives us:
\[ A = \frac{F}{\sigma} \]

### Step 4: Substitute the values to find the area
Substituting the values we calculated:
\[ A = \frac{22600}{7 \times 10^7} \]
\[ A = \frac{22600}{70000000} \]
\[ A = 3.22857 \times 10^{-4} \, \text{m}^2 \]

### Step 5: Convert the area from square meters to square centimeters
Since \( 1 \, \text{m}^2 = 10^4 \, \text{cm}^2 \):
\[ A = 3.22857 \times 10^{-4} \, \text{m}^2 \times 10^4 \, \text{cm}^2/\text{m}^2 \]
\[ A = 3.22857 \, \text{cm}^2 \]

### Final Answer
The area of cross-section of the cable should be approximately:
\[ A \approx 3.23 \, \text{cm}^2 \]