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.
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
To solve this problem, you are indeed correct in using shear deformation and the equation x = (F / (SA)) * L, where x is the deformation, F is the force, S is the shear modulus of aluminum, A is the cross-sectional area, and L is the height of the pole.
However, there seems to be an error in the calculation of the cross-sectional area. To find the area of a solid cylinder, you correctly used A = π * r^2. Given that the diameter is 3.50 cm, the radius would be 1.75 cm or 0.0175 m.
Therefore, the cross-sectional area should be A = π * (0.0175 m)^2 = 9.621 m^2 (rounded to three decimal places).
Now, let's substitute the values into the equation:
x = (900 N / (25 × 10^9 N/m^2 * 9.621 m^2)) * 25.0 m
= (900 N / (240.525 × 10^9 N/m)) * 25.0 m
≈ 0.0015 m
So, the top of the flagpole flexes approximately 0.0015 meters or 1.5 millimeters to the side due to the force exerted by the wind.
To solve the problem, you correctly identified that it is a shear deformation problem. However, the formula you mentioned is not applicable in this case because it calculates the shear strain, not the lateral displacement.
To find the lateral displacement, you can use the concept of flexural rigidity. The formula for the lateral displacement is given by:
x = (F * L^3) / (3 * E * I)
where x is the lateral displacement, F is the force applied at the top, L is the height of the pole, E is the Young's modulus of aluminum (which is 70 GPa or 70 x 10^9 N/m^2), and I is the moment of inertia of the pole's cross-sectional shape.
To find the moment of inertia for a hollow cylinder, you need to subtract the moment of inertia of the inner cylinder from the moment of inertia of the outer cylinder. The formula for the moment of inertia (I) of a hollow cylinder is:
I = (π/64) * (D^4 - d^4)
where D is the outer diameter and d is the inner diameter.
Given that the diameter of the solid cylinder is 3.50 cm, the outer diameter of the hollow cylinder is 3.50 cm, and the height (L) of the pole is 25.0 m, we can calculate the moment of inertia (I) as follows:
D = 3.50 cm = 0.035 m
d = 0 (as it is a solid cylinder)
I = (π/64) * (0.035^4 - 0^4)
Now you can plug these values into the formula for the lateral displacement:
x = (900 N * (25.0 m)^3) / (3 * (70 x 10^9 N/m^2) * I)
Solving this equation will give you the lateral displacement (x) of the top of the pole.