Question

You push on the top edge of a 1.8 m tall solid circular cylinder of iron which is 4.00 cm in diameter. You push with a horizontal force of 900 N to make the top of the pole flex to the right.

Y = 10.0 x 1010 N/m2 B = 9.0 x 1010 N/m2 S = 4.0 x 1010 N/m2
How far does the top of the pole flex to the right?

Answer 1

`\triangle x=3.2*10^-^5 m`

Step-by-step explanation:

From the question we are told that

Height of circular cylinder is
`h= 1.8m`

Diameter of cylinder
`D=4cm=>0.04m`

Horizontal Force
`F=900N`

`Y = 10.0 *10^1^0 N/m^2 \\B = 9.0 * 10^1^0 N/m^2\\S = 4.0 * 10^1^0 N/m^2\\`

Generally the formula for shear modulus is mathematically represented by

`G=(\tau)/(\gamma)`

Where

`G=Shear modulus\\\tau= shear\ stress\\\gamma=shear strain`

`G=(F/A)/(\triangle x/L)`

`G=(900/\pi r^2)/(\triangle x/1.8)`

`4.0*10^1^0=(900/\pi r^2)/(\triangle x/1.8)`

`4.0*10^1^0=(900)/(\pi r^2) *(1.8)/(\triangle x)`

`{\triangle x} =(1620)/(\pi r^2* 4.0*10^1^0)`

`\triangle x=3.2*10^-^5 m`