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

Answer:

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

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

$G=\frac{\tau}{\gamma}$

Where

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

$G=\frac{F/A}{\triangle x/L}$

$G=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900/\pi r^2}{\triangle x/1.8}$

$4.0*10^1^0=\frac{900}{\pi r^2} *\frac{1.8}{\triangle x}$

${\triangle x} =\frac{1620}{\pi r^2* 4.0*10^1^0}$

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