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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

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

Respuesta :

cjmejiab

We need to consider for this exercise the concept Drag Force and Torque. The equation of Drag force is

[tex]F_D = c_D A \frac{\rho V^2}{2}[/tex]

Where,

F_D = Drag Force

[tex]c_D[/tex] = Drag coefficient

A = Area

[tex]\rho[/tex]= Density

V = Velocity

Our values are given by,

[tex]c_D = 0.5[/tex] (That is proper of a cone-shape)

[tex]A = 9m^2[/tex]

[tex]\rho = 1.2Kg/m^3[/tex]

[tex]V = 6.5m/s[/tex]

Part A ) Replacing our values,

[tex]F_D = 0.5*9*\frac{1.2*6.5^2}{2}[/tex]

[tex]F_D = 114.075N[/tex]

Part B ) To find the torque we apply the equation as follow,

[tex]\tau = F*d[/tex]

[tex]\tau = (114.075N)(7)[/tex]

[tex]\tau = 798.525N.m[/tex]

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