Answer:
(a) Shear stress will be 4780487.80 [tex]N/m^2[/tex]
(b) [tex]\Delta l=7.648\times 10^{-6}m[/tex]
Explanation:
We have given length of the aluminium rod l = 4.8 cm = 0.048 m
Diameter [tex]d=5.6cm,\ so\ r=\frac{5.6}{2}=2.8cm=0.028m[/tex]
Area [tex]A=\pi r^2=3.14\times 0.028^2=0.00246m^2[/tex]
Shear modulus [tex]G=3\times 10^{10}N/m^2[/tex]
Mass m = 1200 kg
So weight [tex]w=1200\times 9.8=11760N[/tex]
(a) So hear stress P [tex]=\frac{F}{A}=\frac{11760}{0.00246}=4780487.80N/m^2[/tex]
(b) Now shear modulus is [tex]G=\frac{stress}{strain}[/tex]
So [tex]3\times 10^{10}=\frac{4780487.80}{strain}[/tex]
[tex]strain=1.593\times 10^{-4}[/tex]
Now strain is given by
[tex]strain=\frac{\Deltal}{l}[/tex]
So [tex]1.593\times 10^{-4}=\frac{\Delta l}{0.048}[/tex]
[tex]\Delta l=7.648\times 10^{-6}m[/tex]
