Page 207 - Schaum's Outline of Theory and Problems of Applied Physics

192 FLUIDS AT REST [CHAP. 16

A tire whose gauge pressure is 2 bar contains air at an absolute pressure of about 3 bar, since sea-level atmospheric
pressure is about 1 bar.

SOLVED PROBLEM 16.5
A 65-kg woman balances on the heel of her right shoe, which has a circular base 1 cm in radius. How
much pressure does she exert on the ground?
2
2
2
The area of the heel is A = πr = 3.14 cm = 3.14 × 10 −4 m . Hence the pressure is
2
F mg (65 kg)(9.8 m/s )
6
p = = = = 2.03 × 10 Pa = 20.3 bar
A A 3.14 × 10 −4 m 2
Since atmospheric pressure is 1.013 bar, this pressure is 20 times atmospheric pressure.
SOLVED PROBLEM 16.6
The weight of a car is equally supported by its four tires. The gauge pressure of the air in the tires is 2.0
2
bar and each tire has an area of 140 cm in contact with the ground. What is the mass of the car?
The load on each tire consists of one-quarter of the car’s weight plus the weight of the column of air directly
above the area of the tire in contact with the ground, since this part of the tire has no air under it to provide an
equal upward force. Therefore, only the gauge pressure of the air in the tires, which is the excess over atmospheric
5
pressure, acts to support the car’s weight. Since p gauge = 2.0 bar = 2.0 × 10 Pa, each tire supports a weight of
2
2
2
5
w = mg = p gauge A = 2.0 bar = (2.0 × 10 Pa)(140 cm )(10 −4 m /cm ) = 2800 N
The weight of the entire car is 4w and its mass is
4w (4)(2800 N)
M = = 2 = 1143 kg
g 9.8 m/s
SOLVED PROBLEM 16.7
The flat roof of a house is 30 ft long and 25 ft wide and weighs 15,000 lb. Before a severe storm the
doors and windows of the house are closed so tightly that the air pressure inside remains at a normal
2
2
14.7 lb/in. even when the outside pressure falls to 14.3 lb/in. . Compare the upward force on the roof
with its weight.
2
The area of the roof is A = (30 ft)(25 ft) = 750 ft . The difference between the pressure on the inside and
2
2
outside of the roof is p = (14.7 − 14.3) ft/in. = 0.4 lb/in. . Because the pressure on the inside of the roof is
greater, the net force on it is upward with the magnitude
lb 144 in. 2
2
F = A p = (750 ft ) 0.4 = 43,200 lb
in. 2 ft 2
This is nearly three times the roof’s weight. If the roof is not securely attached to the walls of the house and if the
windows do not break first, the roof will be lifted off during the storm. Evidently a building should not be sealed
when a large drop in pressure due to a storm is forecast.

PRESSURE IN A FLUID
Pressure is a useful quantity where fluids (gases and liquids) are concerned because of the following properties
of fluids:
1. The forces that a fluid exerts on the walls of its container, and those that the walls exert on the fluid,
always act perpendicular to the walls.
2. The force exerted by the pressure in a fluid is the same in all directions at a given depth.