The equation to be used is x=((F)/(EA))(L)
x is the deformation, F is the force, E is the Young's modulus of aluminum (70 GPa), A is the cross sectional area, and L is the height of the pole.
A = pi*r^2 = pi*(1.75e-2)^2 = 9.62e-4 m^2
x = ((900 N)/(70 GPa * 9.62e-4 m^2))(25 m) = 0.0042 m = 4.2 mm
A 25.0 m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder of aluminum 3.50 cm in diameter. A strong wind bends the pole much as a horizontal force of 900 N exerted at the top would. How far to the side does the top of the pole flex?
I am almost 100% sure that this is a shear deformation problem. The equation to be used is x=((1F)/(SA))(L)
x is the deformation, F is the force, S is the shear modulus of aluminum (25x10^9), A is the cross sectional area, and L is the height of the pole. I calculated A to be pi*r^2 and got 9.62e-4 m^2. This may have been where I went wrong. Any help is much appreciated.
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