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 the shear modulus of 1 × 10⁹ , {N/m}². The disk is equivalent to a solid cylinder 0.700 , {cm} high and 4.00 , {cm} in diameter.

A. 0.067 , {cm}
B. 0.092 , {cm}
C. 0.115 , {cm}
D. 0.138 , {cm}

Answer 1

The shear deformation of the disk between vertebrae in the spine is approximately 0.092 cm.

Step-by-step explanation:

To find the shear deformation of the disk between the vertebrae in the spine, we can use the formula:

Shear deformation = (shear force * height) / (shear modulus * area)

First, we need to calculate the area of the disk. The diameter of the disk is given as 4.00 cm, so the radius is half of that, which is 2.00 cm. The area can be calculated using the formula for the area of a circle: Area = π * radius^2.

After calculating the area, we can substitute the values into the shear deformation formula and solve for the shear deformation.

Substituting the given values, the shear deformation is found to be approximately 0.092 cm.