To solve the problem, we need to find the minimum diameter of the steel wire that can safely support a lift with a mass of 1000 kg while considering the maximum acceleration of the lift. Here are the steps to arrive at 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:
- \( m = 1000 \, \text{kg} \) (mass of the lift)
- \( g = 9.8 \, \text{m/s}^2 \) (acceleration due to gravity)
Calculating the weight:
\[ W = 1000 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 9800 \, \text{N} \]
### Step 2: Calculate the maximum tension in the wire.
The maximum tension (T) in the wire occurs when the lift is accelerating upwards. The formula for tension when considering acceleration is:
\[ T = W + m \cdot a \]
where:
- \( a = 1.2 \, \text{m/s}^2 \) (maximum acceleration of the lift)
Calculating the maximum tension:
\[ T = 9800 \, \text{N} + (1000 \, \text{kg} \cdot 1.2 \, \text{m/s}^2) \]
\[ T = 9800 \, \text{N} + 1200 \, \text{N} = 11000 \, \text{N} \]
### Step 3: Use the maximum safe stress to find the minimum cross-sectional area.
The maximum safe stress (σ) is given as:
\[ \sigma = 1.4 \times 10^8 \, \text{N/m}^2 \]
The formula relating stress, force, and area is:
\[ \sigma = \frac{T}{A} \]
where \( A \) is the cross-sectional area of the wire.
Rearranging the formula to find the area:
\[ A = \frac{T}{\sigma} \]
Substituting the values:
\[ A = \frac{11000 \, \text{N}}{1.4 \times 10^8 \, \text{N/m}^2} \]
\[ A = 7.857 \times 10^{-5} \, \text{m}^2 \]
### 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):
\[ r = \sqrt{\frac{A}{\pi}} \]
Substituting the area:
\[ r = \sqrt{\frac{7.857 \times 10^{-5} \, \text{m}^2}{\pi}} \]
\[ r \approx \sqrt{\frac{7.857 \times 10^{-5}}{3.14159}} \]
\[ r \approx \sqrt{2.5 \times 10^{-5}} \]
\[ r \approx 0.005 \, \text{m} \]
### Step 5: Calculate the diameter of the wire.
The diameter (d) is twice the radius:
\[ d = 2r \]
\[ d = 2 \times 0.005 \, \text{m} = 0.01 \, \text{m} \]
### Final Answer:
The minimum diameter of the wire is:
\[ \boxed{0.01 \, \text{m}} \]
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