Question

A disk between vertebrae in the spine is subjected to a shearing force of 600 N. Find its shear deformation, taking it to have a shear modulus of 1x10^9 N/m^2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter. a)3 μm
b)3 mm
c)3 cm
d)3 km

Answer 1

Answer:

The shear deformation is $\Delta x=3.34\times10^{-6}\ m$.

Explanation:

Given that,

Shearing force F = 600 N

Shear modulus $S = 1\times10^{9}\ N/m^2$

length = 0.700 cm

diameter = 4.00 cm

We need to find the shear deformation

Using formula of shear modulus

$S=\dfrac{Fl_{0}}{A\Delta x}$

$\Delta x=\dfrac{Fl_{0}}{(\dfrac{\pi d^2}{4})S}$

$\Delta x=\dfrac{4Fl_{0}}{\pi d^2 S}$

Put the value into the formula

$\Delta x=\dfrac{4\times600\times0.700\times10^{-2}}{3.14\times1\times10^{9}\times(4.00\times10^{-2})^2}$

$\Delta x=3.34\times10^{-6}\ m$

Hence, The shear deformation is $\Delta x=3.34\times10^{-6}\ m$.