The elastic limit of a steel cable is `3.0xx10^(8) N//m^(2)` and the cross-section area is `4 cm^(2)`. Find the maximum upward acceleration that can be given to a `900 kg` elevator supported by the cable if the stress is not to exceed one - third of the elastic limit.

Explanation

To solve the problem, we need to determine the maximum upward acceleration that can be given to a 900 kg elevator supported by a steel cable, ensuring that the stress does not exceed one-third of the elastic limit. Here’s how to approach the problem step by step:

### Step 1: Understand the Given Data
- Elastic limit (σ) of the steel cable = \(3.0 \times 10^8 \, \text{N/m}^2\)
- Cross-sectional area (A) of the cable = \(4 \, \text{cm}^2 = 4 \times 10^{-4} \, \text{m}^2\) (conversion from cm² to m²)
- Mass (m) of the elevator = \(900 \, \text{kg}\)
- Gravitational acceleration (g) = \(9.8 \, \text{m/s}^2\)

### Step 2: Calculate the Maximum Allowable Stress
The maximum stress that the cable can withstand without exceeding one-third of the elastic limit is:
\[ \sigma_{\text{max}} = \frac{1}{3} \sigma = \frac{1}{3} \times 3.0 \times 10^8 \, \text{N/m}^2 = 1.0 \times 10^8 \, \text{N/m}^2 \]

### Step 3: Relate Stress to Force
Stress (σ) is defined as force (F) per unit area (A):
\[ \sigma = \frac{F}{A} \]
Thus, the maximum force that the cable can support without exceeding the maximum stress is:
\[ F_{\text{max}} = \sigma_{\text{max}} \times A = (1.0 \times 10^8 \, \text{N/m}^2) \times (4 \times 10^{-4} \, \text{m}^2) = 4.0 \times 10^4 \, \text{N} \]

### Step 4: Set Up the Equation of Motion
When the elevator is accelerating upward, the net force acting on it can be expressed using Newton's second law:
\[ F_{\text{net}} = m(g + a) \]
Where \(a\) is the upward acceleration. The tension in the cable (T) must balance the weight of the elevator and provide the necessary upward acceleration:
\[ T = mg + ma \]
Since the maximum tension (T) is equal to \(F_{\text{max}}\):
\[ F_{\text{max}} = mg + ma \]

### Step 5: Substitute Known Values
Substituting the known values into the equation:
\[ 4.0 \times 10^4 = 900 \times 9.8 + 900a \]
Calculating \(mg\):
\[ mg = 900 \times 9.8 = 8820 \, \text{N} \]
So the equation becomes:
\[ 4.0 \times 10^4 = 8820 + 900a \]

### Step 6: Solve for Upward Acceleration (a)
Rearranging the equation to solve for \(a\):
\[ 900a = 4.0 \times 10^4 - 8820 \]
Calculating the right side:
\[ 900a = 40000 - 8820 = 31180 \]
Now, divide by 900:
\[ a = \frac{31180}{900} \approx 34.64 \, \text{m/s}^2 \]

### Final Answer
The maximum upward acceleration that can be given to the elevator is approximately:
\[ a \approx 34.64 \, \text{m/s}^2 \quad \text{(rounded to 35 m/s²)} \]

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