Question

A wind with speed of 40m/s blows parallel to the roof of a house. The area of the roof is 250m^2. Assuming that pressure inside the house is atmospheric pressure, the force exerted by wind on the roof and the direction of the force will be?

(My doubt is how P1-P2 becomes 960 Pa)

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Answer 1

don't worry, just check every steps you will get how $(P_1-P_2)=960Pa$ :)

from Bernoulli's theorem,

$P+\frac{1}{2}\rho v^2$ = constant

or, $P_1+\frac{1}{2}\rho v_1^2=P_2+\frac{1}{2}\rho v_2^2$

where, $P_1$ and $P_2$ are pressure inside and outside the roof respectively. $v_1$ and$v_2$ are speed of wind inside and outside of the roof respectively.

now, pressure difference , $P_1-P_2=\frac{1}\rho v_2^2-\frac{1}{2}\rho v_1^2$

$P_1-P_2=\frac{1}{2}\rho(v_2^2-v_1^2)$

here, $v_2=40m/s$, $v_1=0$ and $\rho=1.2kg/m^3$

so, $P_1-P_2=\frac{1}{2}1.2(40^2-0)$

= 1/2 × 1.2 × 1600 = 960 N/m² or 960Pa

hence, $\boxed{\bf{P_1-P_2=960Pa}}$ [ I hope now you got it ]

now force acting on the roof is F = $(P_1-P_2)A$

= 960 × 250 = 2.4 × 10^5 N

as pressure inside the roof is greater than outside the roof. so, force will act upward direction.

Answer 2

Answer:

2.4*10^5N

Explanation:

Hope it helps you!!!