Question

Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 1100 N exerted at the top would. How to the side does the top of the pole flex?

Answer 1

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the  area

Using formula of area

$A=\pi\times r^2$

$A=\pi\times(2.00\times10^{-2})^2$

$A=0.00125\ m^2$

$A=1.25\times10^{-3}\ m^2$

We need to calculate the deformation

Using formula of deformation

$\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)$

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

$\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)$

$\Delta x=0.000704\ m$

$\Delta x=7.04\times10^{-4}\ m$

$\DElta x=0.70\ mm$

Hence, The deformation in the pole due to force is 0.70 mm.