Chapter 12, Problem 12.3DQ

Comparing Example 12.1 (Section 12.1) and Example 12.2 (Section 12.2), it seems that 700 N of air is exerting a downward force of 2.0 × 10⁶ N on the floor. How is this possible?

EXAMPLE 12.1 THE WEIGHT OF A ROOMFUL OF AIR

Find the mass and weight of the air at 20°C in a living room with a 4.0 m × 5.0 m floor and a ceiling 3.0 m high, and the mass and weight of an equal volume of water.

SOLUTION

IDENTIFY and SET UP: We assume that the air density is the same throughout the room. (Air is less dense at high elevations than near sea level, but the density varies negligibly over the room’s 3.0-m height; see Section 12.2.) We use Eq. (12. 1) to relate the mass m_air to the room’s volume V (which we’ll calculate) and the air density ρ_air (given in Table 12.1).

EXECUTE: We have V = (4.0 m) (5.0 m) (3.0 m) = 60m³, so from Eq. (12.1),

$\begin{array}{l}{m}_{air}={\rho }_{air}V=\left(1.20kg/m^3\right)\left(60m^3\right)=72kg\hfill \\ W_{air}=m_{air}g=\left(72kg\right)\left(9.8m/s^2\right)=700N=160lb\hfill \end{array}$

The mass and weight of an equal volume of water are

$\begin{array}{l}{m}_{water}={\rho }_{water}V=\left(1000kg/m^3\right)\left(60m^3\right)=6.0\times {10}^{4}kg\\ W_{water}=m_{water}g=\left(6.0\times {10}^{4}kg\right)\left(9.8m/s^2\right)\\ =5.9\times {10}^{5}N=1.3\times {10}^5\text{ lb }=66\text{ tons}\end{array}$

EVALUATE: A roomful of air weighs about the same as an average adult. Water is nearly a thousand times denser than air, so its mass and weight are larger by the same factor. The weight of a roomful of water would collapse the floor of an ordinary house.

EXAMPLE 12.2 THE FORCE OF AIR

In the room described in Example 12.1, what is the total downward force on the floor due to an air pressure of 1.00 atm?

SOLUTION

IDENTIFY and SET UP: This example uses the relationship among the pressure p of a fluid (air), the area A subjected to that pressure, and the resulting normal force F the fluid exerts. The pressure is uniform, so we use Eq. (12.3), F_⊥ = pA, to determine F_⊥. The floor is horizontal, so F_⊥ is vertical (downward).

EXECUTE: We have A = (4.0 m) (5.0 m) = 20 m², so from Eq. (12.3),

$\begin{array}{l}{F}_{\perp }=pA=(1.013\times {10}^5{\text{N/m}}^2)(20{\text{ m}}^{\text{2}}\text{)}\\ \text{ =2}\text{.0 }\times {\text{ 10}}^{\text{6}}\text{ N = 4}\text{.6 }\times {\text{ 10}}^{\text{5}}\text{ lb = 230 tons }\end{array}$

EVALUATE: Unlike the water in Example 12.1, F_⊥ will not collapse the floor here, because there is an upward force of equal magnitude on the floor’s underside. If the house has a basement, this upward force is exerted by the air underneath the floor. In this case, if we ignore the thickness of the floor, the net force due to air pressure is zero.