Answer:
A,)FD= 114.1N
B)Torque=798.5Nm
Explanation:
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 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)
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/m^3
B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?
A)The equation of Drag force equation can be expressed below,
FD =[ CD × A × ρ × (v^2/ 2)]
Where CD= Drag coefficient for cone-shape = 0.5
ρ = Density
Area of of the tree canopy = 9.0 m^2
density of air of = 1.2 kg/m^3
V= wind velocity= 6.5 m/s,
If we substitute those values to the equation, we have;
FD =[ CD × A × ρ × (v^2/ 2)]
F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]
FD= 114.1N
B) the torque can be calculated using below formula below
Torque= (Force × distance)
= 114.1 × 7
= 798.5Nm
