Respuesta :
Answer:
The shear deformation is [tex]\Delta x=3.34\times10^{-6}\ m[/tex].
Explanation:
Given that,
Shearing force F = 600 N
Shear modulus [tex]S = 1\times10^{9}\ N/m^2[/tex]
length = 0.700 cm
diameter = 4.00 cm
We need to find the shear deformation
Using formula of shear modulus
[tex]S=\dfrac{Fl_{0}}{A\Delta x}[/tex]
[tex]\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}[/tex]
[tex]\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}[/tex]
Put the value into the formula
[tex]\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}[/tex]
[tex]\Delta x=3.34\times10^{-6}\ m[/tex]
Hence, The shear deformation is [tex]\Delta x=3.34\times10^{-6}\ m[/tex].
