The maximum stress that can be applied to the material of a wire used to suspend an elevator is `1.3 xx 10^(8) Nm^(-2)`. If the mass of the elevator is 900 kg and it moves up with an acceleration of `2.2 ms^(-2) ` , what is the minimum diameter of the wire?

Explanation

To find the minimum diameter of the wire used to suspend an elevator, we will follow these steps:

### Step 1: Calculate the total force acting on the wire
The total force (F) acting on the wire when the elevator is moving upwards is the sum of the gravitational force (mg) and the force due to the upward acceleration (ma).

\[ F = mg + ma \]

Where:
- \( m = 900 \, \text{kg} \) (mass of the elevator)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
- \( a = 2.2 \, \text{m/s}^2 \) (upward acceleration)

### Step 2: Substitute the values into the equation
Substituting the values into the equation for force:

\[ F = 900 \times 9.8 + 900 \times 2.2 \]

Calculating each term:

\[ F = 8820 + 1980 = 10800 \, \text{N} \]

### Step 3: Use the formula for stress
Stress (σ) is defined as the force (F) divided by the cross-sectional area (A) of the wire:

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

Rearranging for area gives:

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

### Step 4: Substitute the maximum stress value
The maximum stress that can be applied to the material of the wire is given as \( \sigma = 1.3 \times 10^8 \, \text{N/m}^2 \).

Substituting the values:

\[ A = \frac{10800}{1.3 \times 10^8} \]

Calculating the area:

\[ A \approx 8.3077 \times 10^{-5} \, \text{m}^2 \]

### Step 5: Relate area to diameter
The area of a circular cross-section is given by:

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

Rearranging for diameter \( d \):

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

### Step 6: Substitute the area value into the diameter equation
Substituting the area we calculated:

\[ d^2 = \frac{4 \times 8.3077 \times 10^{-5}}{\pi} \]

Calculating \( d^2 \):

\[ d^2 \approx \frac{3.3231 \times 10^{-4}}{3.1416} \approx 1.059 \times 10^{-4} \]

### Step 7: Calculate the diameter
Taking the square root to find \( d \):

\[ d \approx \sqrt{1.059 \times 10^{-4}} \approx 0.0103 \, \text{m} = 1.03 \times 10^{-2} \, \text{m} \]

### Final Answer
The minimum diameter of the wire is approximately \( 1.03 \times 10^{-2} \, \text{m} \) or \( 10.3 \, \text{mm} \).

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