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
[tex]\Delta x = \frac{1}{s}(\frac{F_w}{A}L_o)[/tex]
Where
[tex]F_w =[/tex] Force
A =Area
s = shear modulus
[tex]L_o[/tex]= 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
[tex]r = \frac{4*10^{-2}}{2}[/tex]
[tex]r = 2*10^{-2}m[/tex]
Therefore the cross sectional area of the pole is given as
[tex]A=\pi r^2[/tex]
[tex]A = \pi * 2*10^{-2}[/tex]
[tex]A=1.257*10^{-3}m^2[/tex]
Replacing our values at the previous equation we have that
[tex]\Delta x = \frac{1}{s}(\frac{F_w}{A}L_o)[/tex]
[tex]\Delta x = \frac{1}{25*10^9}(\frac{900}{1.257*10^{-3}}20)[/tex]
[tex]\Delta x = 0.57mm[/tex]
Therefore the deflection in the pole is 0.57mm
