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 x 109 N/m2. The disk is equivalent to a solid cylinder 0.700 cm high and 4.00 cm in diameter.

3.34×10^-6m

Explanation:

The shear modulus can also be regarded as the rigidity. It is the ratio of shear stress and shear strain

can be expressed as

shear stress/(shear strain)

= (F/A)/(Lo/ . Δx)

Stress=Force/Area

The sheear stress can be expressed below as

F Lo /(A *Δx)

Where A=area of the disk= πd^2/4

F=shearing force force= 600N

Δx= distance

S= shear modulus= 1 x 109 N/m2

Lo= Lenght of the cylinder= 0.700 cm=7×10^-2m

If we make Δx subject of the formula we have

Δx= FLo/(SA)

If we substitute the Area A we have

Δx= FLo/[S(πd^2/4]

Δx=4FLo/(πd^2 *S)

If we input the values we have

(4×600×0.7×10^-2)/10^9 × 3.14 ×(4×10^-2)^2

= 3.35×10^-6m

Therefore, its shear deformation is 3.35×10^-6m

A=area of the disk= πd^2/4

= [3.142×(4×10^-2)^2]/4

real, smaller and upside-down.   so the answer should be b.