Question

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

• A. 2.56
• B. 3.89
• C. 4.32
• D. 5.16

Hint:

Find the maximum tension using Tension = Stress × Area. Then use T - mg = ma to find the acceleration.

Answer 1

The Correct Option is A

Alternatively, we can express the total force the wire must support as the sum of the weight of the lift and the inertial force required to accelerate it. The total tension $T$ in the supporting wire is given by:
$$T = m(g + a)$$
We also know that tension is the product of the allowable stress and the area of the wire:
$$T = \sigma \times A$$
Equating these two expressions for tension:
$$\sigma \times A = m(g + a)$$
Substituting the given values:
$\sigma = 4 \times 10^8\text{ N/m}^2$
$A = \pi(4 \times 10^{-3})^2 = 3.14 \times 16 \times 10^{-6} = 5.024 \times 10^{-5}\text{ m}^2$
$m = 1600\text{ kg}$
$g = 10\text{ m/s}^2$

Putting it all into the equation:
$$(4 \times 10^8) \times (5.024 \times 10^{-5}) = 1600(10 + a)$$
$$20096 = 16000 + 1600a$$
$$20096 - 16000 = 1600a$$
$$4096 = 1600a$$
$$a = \frac{4096}{1600} = 2.56\text{ m/s}^2$$
This confirms that the maximum upward acceleration the lift can sustain without exceeding the stress limit is $2.56\text{ m/s}^2$.