Answer:
d = 2.69 mm
Explanation:
Assuming the cable is rated with a factor of safety of 1.
The stress on the cable is:
σ = P/A
Where
σ = normal stress
P: load
A: cross section
The section area of a circle is:
A = π/4 * d^2
Then:
σ = 4*P / (π*d^2)
Rearranging:
d^2 = 4*P / (π*σ)
[tex]d = \sqrt{4*P / (\pi*\sigma)}[/tex]
Replacing:
[tex]d = \sqrt{4*800 / (\pi*\90000)} = 0.106 inches[/tex]
0.106 inches = 2.69 mm
