Answer:40.603 MPa
Explanation:
Given
Car weighs (m)4000 lbs [tex]\approx 1814.37 kg[/tex]
diameter of cable(d)[tex]=0.75 in.\approx 19.05 mm[/tex]
Hill angle [tex]\theta =15^{\circ}[/tex]
Now
tension in cable will bear the weight of car acting parallel to rope which is [tex]mgsin\theta [/tex]
Thus
[tex]T=mgsin\theta [/tex]
[tex]T=1814.37\times 9.81\times sin(15)=11,574.45 N[/tex]
T=11.57 kN
thus stress([tex]\sigma [/tex])=[tex]\frac{T}{A}[/tex]
where A=cross section of wire[tex]=285.059 mm^2[/tex]
[tex]\sigma =\frac{11574.45}{285.059}=40.603 MPa[/tex]
