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

To solve this, we first find the maximum tension the wire can handle before breaking. This is determined by the maximum stress and the cross-sectional area of the wire. The concept involved is the definition of stress: $\text{Stress} = \frac{\text{Force}}{\text{Area}}$.

Step 1: Calculate the cross-sectional area of the wire.
The radius $r = 4\text{ mm} = 4 \times 10^{-3}\text{ m}$.
Area $A = \pi r^2 = 3.14 \times (4 \times 10^{-3})^2 = 3.14 \times 16 \times 10^{-6} = 50.24 \times 10^{-6}\text{ m}^2$.

Step 2: Calculate the maximum tension $T_{max}$.
$$T_{max} = \text{Maximum Stress} \times A$$
$$T_{max} = (4 \times 10^8) \times (50.24 \times 10^{-6}) = 4 \times 50.24 \times 10^2 = 200.96 \times 10^2 = 20096\text{ N}$$

Step 3: Apply Newton's Second Law to the lift.
When the lift is accelerating upwards with acceleration $a$, the equation of motion is:
$$T - mg = ma$$
To find the maximum acceleration $a_{max}$, we use the maximum tension $T_{max}$:
$$20096 - (1600 \times 10) = 1600 \times a_{max}$$
$$20096 - 16000 = 1600 \times a_{max}$$
$$4096 = 1600 \times a_{max}$$
$$a_{max} = \frac{4096}{1600} = 2.56\text{ m/s}^2$$
Thus, the maximum acceleration is $2.56\text{ m/s}^2$.