A lift is tied with thick iron and its mass is 314 kg. What should be the minimum diameter of wire if the maximum acceleration of lift is `1.2(m)/(sec^2)` and the maximum safe stress of the wire is `1xx10^7(N)/(m^2)`?

Options

A. 2 cm
B. 1 cm
C. 1.5 cm
D. none of these

Explanation

To solve the problem, we need to determine the minimum diameter of the wire that can safely support the lift under the given conditions. Here’s a step-by-step breakdown of the solution:

### Step 1: Calculate the Weight of the Lift
The weight (W) of the lift can be calculated using the formula:
\[ W = m \cdot g \]
where:
- \( m = 314 \, \text{kg} \) (mass of the lift)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)

Calculating the weight:
\[ W = 314 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 3077.2 \, \text{N} \]

### Step 2: Calculate the Maximum Tension in the Wire
When the lift accelerates upwards, the tension (T) in the wire can be calculated using the formula:
\[ T = W + F \]
where \( F \) is the force due to acceleration:
\[ F = m \cdot a \]
with \( a = 1.2 \, \text{m/s}^2 \).

Calculating the force due to acceleration:
\[ F = 314 \, \text{kg} \cdot 1.2 \, \text{m/s}^2 = 376.8 \, \text{N} \]

Now, substituting back to find the tension:
\[ T = 3077.2 \, \text{N} + 376.8 \, \text{N} = 3454 \, \text{N} \]

### Step 3: Use the Maximum Safe Stress to Find the Required Area
The maximum safe stress (\( \sigma \)) is given as:
\[ \sigma = 1 \times 10^7 \, \text{N/m}^2 \]

Stress is defined as:
\[ \sigma = \frac{T}{A} \]
where \( A \) is the cross-sectional area of the wire. Rearranging gives:
\[ A = \frac{T}{\sigma} \]

Substituting the values we have:
\[ A = \frac{3454 \, \text{N}}{1 \times 10^7 \, \text{N/m}^2} = 3.454 \times 10^{-4} \, \text{m}^2 \]

### Step 4: Calculate the Radius of the Wire
The area of a circle is given by:
\[ A = \pi r^2 \]

Rearranging to find the radius:
\[ r^2 = \frac{A}{\pi} \]
Substituting the area and using \( \pi \approx 3.14 \):
\[ r^2 = \frac{3.454 \times 10^{-4}}{3.14} \approx 1.101 \times 10^{-4} \]

Taking the square root to find the radius:
\[ r = \sqrt{1.101 \times 10^{-4}} \approx 0.0105 \, \text{m} = 1.05 \, \text{cm} \]

### Step 5: Calculate the Diameter of the Wire
The diameter (\( d \)) is twice the radius:
\[ d = 2r = 2 \times 1.05 \, \text{cm} \approx 2.1 \, \text{cm} \]

### Final Answer
The minimum diameter of the wire should be approximately **2.1 cm**.