Question

A disk between vertebrae in the spine is subjected to a shearing force of 425 N. Find its shear deformation, taking it to have a shear modulus of 1.70×10^9 N/m^2. The disk is equivalent to a solid cylinder 0.700 cm high and 6.50 cm in diameter. a) 5.27 x 10^-7 m
b) 1.54 x 10^-6 m
c) 3.08 x 10^-6 m
d) 6.16 x 10^-6 m

Answer 1

Answer:

option A

Explanation:

given,

shear force = 425 N

Shear modulus = 1.7 × 10⁹ N/m²

disk equivalent to solid cylinder

height = 0.7 cm  and diameter = 6.50 cm

$\delta = \dfrac{VL}{AG}\\\\\delta = \dfrac{425\times 0.007}{0.25 \times \pi d^2\times 1.7\times 10^9}\\\\\delta = \dfrac{425\times 0.007}{0.25 \times \pi\times 0.065^2\times 1.7\times 10^9}\\\\\delta = 5.27 \times 10^{-7} m$

hence, the correct answer is option A