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

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

Step-by-step 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* r^2`

`A=\pi*(2.00*10^(-2))^2`

`A=0.00125\ m^2`

`A=1.25*10^(-3)\ m^2`

We need to calculate the deformation

Using formula of deformation

`\Delta x=(1)/(s)((F)/(A)* L)`

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

`\Delta x=(1)/(2.5*10^(10))*((1100)/(1.25*10^(-3))* 20.0)`

`\Delta x=0.000704\ m`

`\Delta x=7.04*10^(-4)\ m`

`\DElta x=0.70\ mm`

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