Question

During heavy lifting, a disk between spinal vertebrae is subjected to a 5000-N compressional force. (a) What pressure is created, assuming that the disk has a uniform circular cross section 2.00 cm in radius? (b) What deformation is produced if the disk is 0.800 cm thick and has a Young's modulus of 1.5×109 N/m2?

Answer 1

Answer:

$3978873.58\ \text{Pa}$

$0.00002122\ \text{m}$

Explanation:

F = Force = 5000 N

r = Radius of circular cross section = 2 cm

l = Length of disk = 0.8 cm

A = Area = $\pi r^2$

Y = Young's modulus = $1.5\times 10^9\ \text{N/m}^2$

Pressure is given by

$P=\dfrac{F}{A}\\\Rightarrow P=\dfrac{5000}{\pi (2\times 10^{-2})^2}\\\Rightarrow P=3978873.58\ \text{Pa}$

The pressure on the cross section is $3978873.58\ \text{Pa}$

The change in length of the cross section is given by

$\Delta L=\dfrac{PL}{Y}\\\Rightarrow \Delta L=\dfrac{3978873.58\times 0.8\times 10^{-2}}{1.5\times 10^9}\\\Rightarrow \Delta L=0.00002122\ \text{m}$

The deformation produced is $0.00002122\ \text{m}$