A 21.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder of aluminum 6.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 940 N exerted at the top would. How far to the side does the top of the pole flex? The shear modulus for aluminum is 25 ⨯ 109 N/m2.

Question

A 21.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder of aluminum 6.00 cm in diameter. A strong wind bends the pole much as a horizontal force of 940 N exerted at the top would. How far to the side does the top of the pole flex? The shear modulus for aluminum is 25 ⨯ 109 N/m2.

mm

Answer 1

To calculate the displacement or flexion of the top of the pole, we need to consider the bending of the pole due to the horizontal force. This can be found using the formula for the flexion of a beam:

Δx = (F * L^3) / (48 * E * I)

Where:
Δx is the displacement of the top of the pole (flexion)
F is the horizontal force exerted on the top of the pole (940 N in this case)
L is the length of the pole (21.0 m)
E is the shear modulus of aluminum (25 ⨯ 10^9 N/m^2)
I is the moment of inertia of the pole

The moment of inertia can be calculated for the hollow cylinder using the formula:

I = (π / 64) * (D_outer^4 - D_inner^4)

Where:
D_outer is the outer diameter of the pole (6.00 cm or 0.06 m)
D_inner is the inner diameter of the pole

However, the inner diameter of the pole is not given. So we need to find it using the information that the hollow pole is equivalent in strength to a solid cylinder of aluminum 6.00 cm in diameter.

The moment of inertia for a solid cylinder is given by:

I_solid = (π / 64) * D^4

Equating I to I_solid, we can solve for D_inner:

(π / 64) * (D_outer^4 - D_inner^4) = (π / 64) * D^4

Simplifying and rearranging, we get:

D_inner^4 = D_outer^4 - D^4

Since D_outer is given as 6.00 cm (or 0.06 m) and D is given as 6.00 cm, we can substitute these values:

D_inner^4 = (0.06^4) - (0.06^4)
D_inner^4 = 0
D_inner = 0

This means that the inner diameter of the pole is zero, and hence it is a solid cylinder.

Now we can calculate the displacement using the formula mentioned earlier:

Δx = (F * L^3) / (48 * E * I)

Plugging in the values:

Δx = (940 N * (21.0 m)^3) / (48 * (25 ⨯ 10^9 N/m^2) * ((π / 64) * (0.06 m)^4))

Simplifying, we get:

Δx ≈ 0.115 mm

Therefore, the top of the pole flexes by approximately 0.115 mm to the side due to the wind.