# If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy?

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

Part A: If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m3. Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m3.

Part B:What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

### 2 Answers

- 2 years agoFavorite Answer
Part A:

A) drag F = ½ρAv²C

F = ½ * 1.2kg/m³ * 9.0m² * (6.5m/s)² * 0.5 = 114 N (or ~110N)

Part B:

The resulting triangle is made up of a side with length 7.0, a side with length 110, and unknown hypotenuse. Using tanget(angle) we get:

Tan(angle) = opposite/adjacent = 110/7

===> Angle = tan-1 (110/7) = 86.4

Torque = F * r * sin(angle) = 110N * 7.0m * sin(86.4) = 768 N*m (or ~800Nm)

- NCSLv 72 years ago
A) drag F = ½ρAv²C

F = ½ * 1.2kg/m³ * 9.0m² * (6.5m/s)² * 0.5 = 114 N ≈ 110 N

to two significant digits

B) τ = F*d = 114N * 7.0m = 798 N·m ≈ 800 N·m

Hope this helps!