Calculating Maximum Lift Acceleration

The problem requires finding the maximum acceleration a lift can have, given the wire's properties and the lift's mass. The wire's strength limits the maximum tension it can endure.

Determining Wire Properties

• The radius of the iron wire is given as $r = 4 \text{ mm}$. Convert this to meters: $r = 4 \times 10^{-3} \text{ m}$
• Calculate the cross-sectional area ($A$) of the wire using the formula for the area of a circle, $A = \pi r^2$: $A = 3.14 \times (4 \times 10^{-3} \text{ m})^2$ $A = 3.14 \times (16 \times 10^{-6} \text{ m}^2)$ $A = 50.24 \times 10^{-6} \text{ m}^2$

Calculating Maximum Tension

The maximum stress ($\sigma_{max}$) the wire can withstand is $4 \times 10^8 \text{ N/m}^2$. The maximum tension ($T_{max}$) is the product of maximum stress and the cross-sectional area:

$T_{max} = \sigma_{max} \times A$

$T_{max} = (4 \times 10^8 \text{ N/m}^2) \times (50.24 \times 10^{-6} \text{ m}^2)$

$T_{max} = 200.96 \times 10^2 \text{ N}$

$T_{max} = 20096 \text{ N}$

Applying Newton's Second Law

Consider the lift accelerating upwards with maximum acceleration $a$. The forces acting on the lift are the tension ($T$) upwards and the weight ($mg$) downwards. According to Newton's second law ($F_{net} = ma$):

$T - mg = ma$

For the maximum possible acceleration ($a_{max}$), the tension must be at its maximum value ($T_{max}$):

$T_{max} - mg = m a_{max}$

Rearrange the formula to solve for $a_{max}$:

$a_{max} = \frac{T_{max} - mg}{m}$

$a_{max} = \frac{T_{max}}{m} - g$

Finding the Maximum Acceleration

Substitute the known values:

• Mass of the lift, $m = 1600 \text{ kg}$
• Maximum tension, $T_{max} = 20096 \text{ N}$
• Acceleration due to gravity, $g = 10 \text{ m/s}^2$

Calculate $a_{max}$:

$a_{max} = \frac{20096 \text{ N}}{1600 \text{ kg}} - 10 \text{ m/s}^2$

$a_{max} = 12.56 \text{ m/s}^2 - 10 \text{ m/s}^2$

$a_{max} = 2.56 \text{ m/s}^2$

Therefore, the maximum acceleration the lift can take is $2.56 \text{ m/s}^2$.