Physics: A20.0-m-tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. a strong wind bends the pole as much as a horizontal 900.0-n force on the top would do. how far to the side does the top of the pole flex?

Rearranging it, we get;Where a is the acceleration, f is the force, and m is the mass...

A20.0-m-tall hollow aluminum flagpole is equivalent - id2003900779, sgillespie006p3ute8

Daking23

10.01.2022

To solve the problem it is necessary to apply the concepts related to deflection in a metal.

By definition the deflection is given by the equation

$\Delta x = \frac{1}{s}(\frac{F_w}{A}L_o)$

Where

F_w = Force

A =Area

s = shear modulus

L_o = Original length

To calculate the area we must obtain the radius generated by the deflection, that is to say

r = \frac{D}{2}

Where,

D = Diameter

r = Radius of the aluminum flagpole

$r = \frac{4*10^{-2}}{2}$

$r = 2*10^{-2}m$

Therefore the cross sectional area of the pole is given as

$A=\pi r^2$

$A = \pi * 2*10^{-2}$

$A=1.257*10^{-3}m^2$

Replacing our values at the previous equation we have that

$\Delta x = \frac{1}{s}(\frac{F_w}{A}L_o)$

$\Delta x = \frac{1}{25*10^9}(\frac{900}{1.257*10^{-3}}20)$

$\Delta x = 0.57mm$

Therefore the deflection in the pole is 0.57mm

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