Respuesta :
Answer:
The wind force on the screen is approximately 5,341.5936 [tex]lb_f[/tex]
Explanation:
The speed of the wind, v = 36 mph
The width of the outdoor movie, w = 70 ft. wide
The height of the outdoor movie, h = 20 ft. tall
The drag coefficient, Cd = 1.15
We have;
[tex]C_d = \dfrac{F}{\dfrac{\rho \cdot v^2 \cdot A}{2} }[/tex]
From which we have;
The wind force, F = 0.00256·[tex]C_d[/tex]·v²·A
Where;
A = The cross sectional area of the rectangular outdoor movie screen, A = w × h
∴ A = 70 ft. × 20 ft. = 1,400 ft.²
The wind force, F = 0.00256 × 1.15 × (36 mph)² × 1,400 ft.² = 5,341.5936 [tex]lb_f[/tex]
The wind force on the screen, F = 5,341.5936 [tex]lb_f[/tex].
The wind force on the outdoor movie screen is;
F = 23437.26 N
We are given;
Wind speed; v = 36 mph = 16.0934 m/s
Width of screen; w = 70 ft = 21.336 m
Height of screen; h = 20 ft = 6.096 m
Drag coefficient; Cd = 1.15
Now formula for the drag force is;
F = ½•C_d•ρ•A•v²
Where;
F is drag force
C_d is drag coefficient
ρ is density of object
v is speed of object
A is area
Let us use density as 1.21 kg/m³
Area; A = wh = (21.336 × 6.096)
Thus;
F = ½ × 1.15 × 1.21 × (21.336 × 6.096) × 16.0934²
F = 23437.26 N
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