Question

A load of 5KN to be raised with the help of a steel wire. Find the minimum diameter of the steel wire, if the stress is not exceed 100Mpa

Answer 1

Answer:

Diameter will be $2.523\times 10^{-3}m$

Explanation:

We have given that load $=5KN=5000N$

Maximum stress $=100Mpa=100\times 10^6Pa$

We know that force is given by $Force=sress\times area$

So area will be equal to $Area=\frac{force}{pressure}=\frac{5000}{100\times 10^6}=5\times 10^{-6}m^2$

Now we know that area is given by

$A=\pi r^2$

$5\times 10^{-6}=3.14\times r^2$

$r=1.261\times 10^{-3}m$

So diameter will be $d=2r=2\times 1.261\times 10^{-3}=2.523\times 10^{-3}m$

Answer 2

Correct Question :

A load of 5KN to be raised with the help of a steel wire. Find the minimum diameter of the steel wire, if the stress is not exceed 100Mpa

Explanation :

Load = 5KN = 5000 N

Stress = 100 Mpa = 100 × 10⁶ Pa

WE KNOW THE FORMULA TO CALCULATE THE FORCE

⇒ Force = Stress × Area

→ Area = Force/Pressure

→ πr² = 5000/100 × 10⁶

→ 3.14 × r² = 5 × 10⁻⁶

→ r = 5 × 10⁻⁶/3.14

→ r = 1.26 × 10⁻³ m

∴ D = 2r = 2(1.26 × 10⁻³)

⇒ D = 2.52  × 10⁻³ m