A lift of mass 1600 kg is supported by thick iron wire. If the maximum stress which the wire can withstand is $$4 \times 10^8$$ N/m$$^2$$ and its radius is 4 mm, then maximum acceleration the lift can take is _______ m/s$$^2$$.
(take g = 10 m/s$$^2$$ and $$\pi$$ = 3.14)

The maximum stress is the maximum force per unit area. Therefore, the maximum tension ($$ T_{max} $$) is:

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

Where the cross-sectional area $$ A = \pi r^2 $$.

Substitute the values:

$$ T_{max} = (4 \times 10^8) \times \left(3.14 \times (4 \times 10^{-3})^2\right) $$

$$ T_{max} = (4 \times 10^8) \times (3.14 \times 16 \times 10^{-6}) $$

$$ T_{max} = 64 \times 3.14 \times 10^2 $$

$$ T_{max} = 200.96 \times 100 $$

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

When the lift is accelerating upwards with a maximum acceleration $$ a $$, the net upward force is $$ T_{max} - mg $$.

According to the equation of motion:

$$ T_{max} - mg = ma $$

$$ T_{max} = m(g + a) $$

Substitute the known values into the equation:

$$ 20096 = 1600(10 + a) $$

Solve for $$ a $$:

$$ \frac{20096}{1600} = 10 + a $$

$$ 12.56 = 10 + a $$

$$ a = 12.56 - 10 $$

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