A sign is on a truck traveling 55mp/h (88 km/h). There is a crosswind on the truck of 40 mp/h (64.37 km/h). What is the torque experienced by the sign from the crosswind?
The sign is actually 2 boxes sitting side by side with identical dimensions. Following dimensions are for just one piece of the sign. Dimension of the sign perpendicular to wind is 41"x96"x8". Surface area of sign perpendicular to wind is 3936 in^2 or 2.53934976 m^2. The volume of the sign is 54,528 (in^3) or 0.893553826 m^3.
P=1/2(density air)(v^2 wind)(drag coefficient)
The Attempt at a Solution
I know I need the force of air on the sign. I have discounted the second part of the sign for this since I dont see how it is relevant (especially since they are not attached in my picture from my teacher). For this I used P=1/2(density air)(v^2 wind)(drag coefficient). That is P=1/2(1.25kg/m^3)(64.37)^2(1.05)= 2719.17 N/m^2. So, that force times surface area 2.5393 m^2 is 6904.788 Newtons.
Having that Torque=r* 6904.788 N. Can I say r is the average distance from the force on the sign to the point at which it is attached to the truck? I have no such attachment in my drawing so I'll just say center of the sign. That is 20.5" or 0.5207 m.
So, Torque= .5207 m*6904.788 N. Which is 3595.32 Nm.
For the second half of the sign: It is a mirror copy of the other. It only has only whatever fraction of wind went around the first half that will hit it, maybe 30%.. and that would likely be at a reduced velocity. So, I would just add 30% to the final torque value. If I knew exactly how much force was left in the wind and what % of wind was hitting the second half I could get a good number here. Let me know if I need to make a drawing for this since I dont know if I'm describing it very well. Thanks