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OLEGan [10]
3 years ago
6

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 th

e 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?
Physics
1 answer:
Temka [501]3 years ago
6 0

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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