Question

The elastic limit of an alloy is 5.0×108 N/m2. What is the minimum radius rmin of a 4.0 m long wire made from the alloy if a single strand is designed to support a commercial sign that has a weight of 8000 N and hangs from a fixed point? To stay within safety codes, the wire cannot stretch more than 5.0 cm.

Answer 1

4.5x 10^ -9m

Step-by-step explanation:

See attached file

Answer 2

The radius is
`r_(min) = 0.00226 \ m`

Step-by-step explanation:

From the question we are told that

The elastic limit(stress) is
`\sigma = 5.0*10^(8) \ N /m^2`

The length is
`L = 4.0 \ m`

The weight of the commercial sign is
`F_s = 8000 \ N`

The maximum extension of the wire is
`\Delta L = 5.0 \ cm = 0.05 \ m`

Generally the elastic limit of an alloy (stress) is is mathematically represented as

`\sigma = ( F_s )/( A )`

Where A is the cross-sectional area of the wire which is mathematically represented as

`A = \pi r^2`

here
`r = r_(min)` which is the minimum radius of the wire that support the commercial sign

So

`\sigma = ( F_s )/( \pi r_(min)^2 )`

=>
`r_(min) = \sqrt{(F_s)/(\sigma * \pi) }`

substituting values

`r_(min) = \sqrt{(8000)/( 5.0* 10^8 * 3.142) }`

`r_(min) = 0.00226 \ m`