Question

1. Winds of 50 mph happen frequently around high-rise buildings. What is the total lift force that such a wind exerts on a 2'x3' glass window? What is the weight of this pane, if its width is 4 inch? The density of glass is 2.5 x 10'kg/m

Answer 1

wind force over a window:

pressure P = Force(F)/Area(A)

Force = P*A

force lifting the window glass becomes $F = \Delta P*A$

Force holding on the window:

given the density of glass kg/m^3

Pu, 2.5 * 103

width of the window = 2 inch = 0.0508m, height of the window = 3 inch = 0.0762m

The weight of the window $w= mg = \rho _{w}*volume*g$

w -2.5 % 103 * 0,0508 0.0762 * 9.8

w-0.09483852 * 103N

Using Bernoulli's principle :

$P_{out}+1/2\rho v_{out}^2+\rho gh = P_{in}+1/2\rho v_{in}^2+\rho gh$

ΔΡ-Pout-Pin-1 /2ρι..nd
  $\rho\rightarrow is density of air$

AP 1/2 1.225 (0.013889)

ΔΡ 0.0001 181538966 125Pa

ForceF一0.000118 1538966 125 * 0.0508 * 0.0762

ForceFー0.000000457369007631 123N

from this, it is clear that the wind force is far less than the weight of the window glass.

(ii) when the width of the window is 1/4 inch = 0.25 inch = 0.00635m

the weight of the window will become

wt 2.5 * 103 * 0.00635 * 0.0762 * 9.8

ut 0.11854815 * 10°