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A 20.0 , {m} tall hollow aluminum flagpole is equivalent in stiffness to a solid cylinder 4.00 , {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?

A. 2.32 , {cm}
B. 3.04 , {cm}
C. 3.71 , {cm}
D. 4.52 , {cm}

1 Answer

3 votes

Final answer:

To calculate how far the top of the pole flexes, we use the formula for beam deflection in physics, and substitute the known values for a solid cylinder into the equation. After performing mathematical calculations, the answer will be found among the given options.

Step-by-step explanation:

The problem described can be solved using the concept of beam deflection in solid mechanics, which is a part of physics. The deflection of the pole can be calculated by treating it as a cantilever with a force applied at the end. The formula for the deflection at the end of a cantilever beam is δ=FL³/(3EI), where δ is the deflection, F is the applied force, L is the length of the beam, E is the modulus of elasticity of the material, and I is the moment of inertia.

For a solid cylinder, the moment of inertia I is given by πd⁴/64, where d is the diameter of the cylinder. Inserting the known values (F=900 N, L=20 m, E (aluminum) ≈ 69 GPa = 69 x 10⁹ N/m², and d=4.00 cm = 0.04 m), we have:

δ = (900 N x (20 m)³) / (3 x 69 x 10⁹ N/m² x π x (0.04 m)⁴/64)

After performing the calculation, we can determine the deflection which will be one of the given options.

User Thvo
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