Page 206 - Schaum's Outline of Theory and Problems of Applied Physics

CHAP. 16] FLUIDS AT REST 191

SOLVED PROBLEM 16.2
The density of mammals is roughly the same as that of water. Find the volume of a 250-kg lion.
m 250 kg 3
V = = = 0.25 m
d 1000 kg/m 3

SOLVED PROBLEM 16.3

An oak beam 10 cm by 20 cm by 4 m has a mass of 58 kg. Find the density and specific gravity of oak.
3
The volume of the beam is V = (0.1m)(0.2m)(4m) = 0.08 m , and so its density is
m 58 kg
d = = = 725 kg/m 3
V 0.08 m 3
3
Since the density of water is 1000 kg/m , the specific gravity (sp gr) of oak is
725
d oak
sp gr = = = 0.725
d water 1000
SOLVED PROBLEM 16.4
How much does the air in a room 12 ft square and 10 ft high weigh? The weight density of air is 0.08 lb/ft 3
at sea level.
3
The volume of the room is V = (12 ft)(12 ft)(10 ft) = 1440 ft . Hence the weight of the air is
3
3
w = (dg)V = (0.08 lb/ft )(1440 ft ) = 115 lb
PRESSURE
When a force acts perpendicular to a surface, the pressure exerted is the ratio between the magnitude of the force
and the area of the surface:
F
p =
A
force
Pressure =
area
2
Pressures are properly expressed in pascals (1Pa = 1 N/m ) or in pounds per square foot, but other units
are often used:
2
1 lb/in = 144 lb/ft 2
1 atmosphere (atm) = average pressure exerted by earth’s atmosphere at sea level
5 2
= 1.013 × 10 Pa = 14.7 lb/in.
5
1 bar = 10 Pa (slightly less than 1 atm)
1 millibar (mb) = 100 Pa (widely used in meteorology)
1 torr = 133 Pa (widely used in medicine for blood pressure)

GAUGEPRESSURE

Pressure gauges measure the difference between an unknown pressure and atmospheric pressure. What they
measure is known as gauge pressure, and the true pressure is known as absolute pressure:

p = p gauge + p atm
Absolute pressure = gauge pressure + atmospheric pressure