Answer to Question #100430 in Mechanics | Relativity for Bairaiza Yahya Tawakim

Question

This circular railroad crossing sign has a diameter of 1.3 m. The distance from the bottom of the sign to the ground is 3m. Determine the moment at the base 'A' of the sign due to the wind drag on the front of the sign. Wind speed is 150 kilometer per hour. Air temperature 20 degree celsius. Neglect the drag on the pole.

Answer 1

At the flow rate specified in the problem $v$ , the air flow can be considered as an incompressible fluid flow. Since the wind is blowing perpendicular to the surface of the sign, the Bernoulli equation gives pressure on the front surface of sign as

$P=\rho\cdot\frac{v^2}{2}$ ,

where $\rho=1,2041kg/m^3$ density of air at temperature 20 degree celsius. https://en.wikipedia.org/wiki/Density_of_air

On the back surface of the sign, the air is at rest (the sign is a bad streamlined body) and its excess pressure on the sign is negligible. Thus fluid dynamic force acts on circular sign is

$F_w=P\cdot S$ , where $S=\pi \cdot \frac{D^2}{4}=3.14\cdot \frac{1.3^2}{4}=1.3 m^2$

This force is applied to the center of the circle. Therefore, the shoulder of force that we must use when calculating the moment is $H=h+\frac{D}{2}=3.65 m$. The moment of force equals

$M=F_w\cdot H$

In calculations, all quantities must be written in the SI system.

$v=150 km/h = 150\cdot \frac{1000m}{3600s}=41.7m/s$

$P=1.204 (kg/m^3)\frac{(41.7 m/s)^2}{2}=1047 N/m^2$

$F_w=1047\cdot 1.3=1361 N$

$M=1361\cdot 3.65=4967 N\cdot m\approx 5000 Nm$

Answer: The moment of force at the base of the sign due to the wind drag on the front of the sign is

$M\approx 5000 Nm$