A box with a frontal area of 2.5 ft2 and with a drag coefficient of CD =0.9 is fastened on the roof of your car. What is the aerodynamic drag force in pounds due to the box alone at a velocity of 88 ft/sec?

Respuesta :

To solve this problem it is necessary to apply the concepts related to the drag force.

By definition we know that drag force can be expressed as

[tex]F_D = \frac{1}{2} \rho C_D AV^2[/tex]

Where,

\rho = Density

[tex]C_D =[/tex]Drag Coefficient

A = Area

V = Velocity

Our values are given as

[tex]A = 2.5ft^2[/tex]

[tex]V = 88ft/s[/tex]

[tex]C_D = 0.9[/tex]

[tex]\rho = 0.0765lb/ft^3 \rightarrow[/tex] Air at normal temperature

Replacing at the equation we have,

[tex]F_D = \frac{1}{2} \rho C_D AV^2[/tex]

[tex]F_D = \frac{1}{2} (0.0765lb/ft^3) (0.9) (2.5ft^2) (88ft/s)^2[/tex]

[tex]F_D = 666.468lbf[/tex]

The aerodynamic drag force is 666.468Lbf