A wind with speed $40 ext{ m/s}$ blows parallel to the roof of a house. The area of the roof is $250 ext{

Question

A wind with speed $40 	ext{ m/s}$ blows parallel to the roof of a house. The area of the roof is $250 	ext{ m}^2$. Assuming that the pressure inside the house is atmospheric pressure, the force exerted by the wind on the roof and the direction of the force will be: $(ho_{air}=1.2 	ext{ kg/m}^3)$

Accepted Answer

1. **Bernoulli's Principle:** Apply Bernoulli's equation for the air inside and outside the roof. Neglecting the difference in height (potential energy), the equation is:
   $$ P_{in} + \frac{1}{2}ho v_{in}^2 = P_{out} + \frac{1}{2}ho v_{out}^2 $$
   
2. **Conditions:** 
   - Inside the house, the air is static, so $v_{in} = 0$. The pressure is atmospheric, $P_{in} = P_{atm}$.
   - Outside the house, the wind speed is $v_{out} = 40 	ext{ m/s}$. The pressure is $P_{out}$.
3. **Pressure Difference:** Rearranging the equation:
   $$ P_{in} - P_{out} = \frac{1}{2}ho (v_{out}^2 - v_{in}^2) $$
   $$ \Delta P = \frac{1}{2} 	imes 1.2 	imes (40^2 - 0) = 0.6 	imes 1600 = 960 	ext{ Pa (N/m}^2) $$
4. **Calculate Force:** The aerodynamic lift (force) is the pressure difference multiplied by the area ($A = 250 	ext{ m}^2$).
   $$ F = \Delta P 	imes A = 960 	imes 250 = 240,000 	ext{ N} = 2.4 	imes 10^5 	ext{ N} $$
5. **Determine Direction:** Since the velocity outside ($40 	ext{ m/s}$) is greater than inside ($0 	ext{ m/s}$), the pressure outside is lower than the pressure inside ($P_{out} < P_{in}$). Therefore, the net force acts from high pressure to low pressure, which is **upwards** .

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