The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
F = P (Normal force)
Formula for moment at section
M = P(4 + 1.5/2)
= 4.75p
Solve for the cross sectional area
Area = [tex]\frac{\pi d^{2} }{4}[/tex]
d = 1.5
[tex]\frac{\pi *1.5^{2} }{4}[/tex]
= 1.767 inches²
[tex]\frac{\pi *0.75^4}{4}[/tex]
= 0.2485inches⁴
Solve for the tensile force from here
[tex]\frac{F}{A} +\frac{Mc}{I}[/tex]
30x10³ = [tex]\frac{P}{1.767} +\frac{4.75p*0.75}{0.2485} \\\\[/tex]
30000 = 14.902 p
divide through by 14.902
2013.15 = P
The largest tensile force that can be applied to the cables given a rod with diameter 1.5 is 2013.15lb
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