A 900 kg elevator hangs by a steel cable for which the allowable stress is `1.15xx10^8(N)/(m^2)`. What is the minimum diameter required if the elecator accelerates upward at `1.5(m)/(s^2)`? Take `g=10(m)/(s^2)`.

Options

A. `(6xx10^-2)/(sqrt(5)pi)m`
B. `(6xx10^-2)/(sqrt(10)pi)m`
C. `(3xx10^-2)/(sqrt(10pi))m`
D. `(3xx10^-2)/(sqrt(5)pi)m`

Explanation

To find the minimum diameter required for the steel cable supporting a 900 kg elevator that accelerates upward at 1.5 m/s², we can follow these steps:

### Step 1: Calculate the total force acting on the elevator.
The total force (T) acting on the elevator can be calculated using the formula:
\[ T = m(g + a) \]
where:
- \( m = 900 \, \text{kg} \) (mass of the elevator),
- \( g = 10 \, \text{m/s}^2 \) (acceleration due to gravity),
- \( a = 1.5 \, \text{m/s}^2 \) (upward acceleration).

Substituting the values:
\[ T = 900 \, \text{kg} \times (10 \, \text{m/s}^2 + 1.5 \, \text{m/s}^2) \]
\[ T = 900 \, \text{kg} \times 11.5 \, \text{m/s}^2 \]
\[ T = 10350 \, \text{N} \]

### Step 2: Use the allowable stress to find the required cross-sectional area.
The allowable stress (\( \sigma \)) is given as:
\[ \sigma = 1.15 \times 10^8 \, \text{N/m}^2 \]

Using the formula for stress:
\[ \sigma = \frac{F}{A} \]
where \( F \) is the force (T) and \( A \) is the cross-sectional area.

Rearranging the formula to find the area:
\[ A = \frac{F}{\sigma} \]
Substituting the values:
\[ A = \frac{10350 \, \text{N}}{1.15 \times 10^8 \, \text{N/m}^2} \]
\[ A \approx 9.00 \times 10^{-5} \, \text{m}^2 \]

### Step 3: Relate the area to the diameter.
The area of a circular cross-section is given by:
\[ A = \pi r^2 \]
where \( r \) is the radius. We can express the radius in terms of the area:
\[ r = \sqrt{\frac{A}{\pi}} \]
Substituting the area:
\[ r = \sqrt{\frac{9.00 \times 10^{-5}}{\pi}} \]

### Step 4: Calculate the radius and then the diameter.
Calculating the radius:
\[ r \approx \sqrt{\frac{9.00 \times 10^{-5}}{3.14}} \]
\[ r \approx \sqrt{2.87 \times 10^{-5}} \]
\[ r \approx 5.37 \times 10^{-3} \, \text{m} \]

Now, the diameter \( d \) is:
\[ d = 2r \]
\[ d \approx 2 \times 5.37 \times 10^{-3} \]
\[ d \approx 1.074 \times 10^{-2} \, \text{m} \]
\[ d \approx 0.01074 \, \text{m} \]
\[ d \approx 10.74 \, \text{mm} \]

### Final Answer:
The minimum diameter required for the steel cable is approximately **10.74 mm**.