ρ=m/V
m = mass (kg)
V = volume (m3)
ρ = density (kg/m3)
- SI unit of density is kg/m3. But some other units are sometimes used 1 g/cm3 = 1000 kg/m3.Water at 4*C has density (ρ) of 1.0 g/cm3 = 1000 kg/m3 = 1.0 kg/L so, a litre of water has a mass of 1 kg
FORMULA FOR PRESSURE
p=F/A or F = pA
F = force (N)
p = pressure (N/m2)
A = surface area (m2)
- For small part of object where it has a cube shape,
(p2 – p1) A = ρAΔh g
- where pressure increase with depth in the fluid.
-QUESTION FOR 1 WEEK -
HYDROSTATICS
QUESTION 1:
- By how much does the pressure under the seawater for every 10 m increase in depth?
- p2 – p1 = ρsea g Δh
= (1025 kg/m3)(9.8 m/s2)(10 m)
= 100.45 N/m2
- In this question, the pressure p1 in the air above the water is already at 1 atmosphere= 1 atm = 101.3 kPa. So for the 10 m depth increase, the pressure in the water increases by very close to 1 atmosphere. Therefore, p2 = p1 + 100.45 kPa = 202 kPa.Usually, p2– p1 would be called the “gauge-pressure”(measured relative to 1 atm). While p2 by itself is called the “absolute pressure”.
pabs = patm + pgauge
- The SI unit of pressure is 1 pascal = 1 Pa = 1 N/m2 (a very small amount of pressure).
Also, 1 atm = 14.7 lbs/in2 = 760 mm-Hg. (height of mercury column supported by atmospheric pressure), so 1 mm-Hg = 133 N/m2.
2. A solid has a radius of 1.5 cm and a mass of 0.038kg. what is the specific gravity of the
sphere?
use 1.5cm= 1.5 x 10ˉ² m for the radius
ANSWER 2;
2. V= 4/3 ∏R³ = 4/3 ∏ (1.5 x 10ˉ² )³ = 1.413 x 10ˉ⁵ m³
d = m/v = 0.038/1.413 x 10ˉ⁵ = 2690 kg/m³
specific gravity = 2690kg/m³ ÷ 1000kg/m³
QUESTION 3;
3. An 80kg metal cyclinder 2 m long and with each end of area 25 cm² stands vertically
on one end. what pressure does the cylinder exert on the floor?
ANSWER 3;
3. p = normal force/ area
= (80kg)(9.81 m/s² ) / 25 x 10ˉ⁴ m²
= 3.14 x 10⁵N/m²
= 314 kPa
QUESTION 4;
4. Find the pressure at a depth of 10m in water when the atmospheric pressure is that
corresponding to a mercury column of height 760mm. The densities of water and
mercury are 10³
ANSWER 4;
4. p = patm + pgy = phghgh + pgy
= (13.6 x 10³
= 1.99 x 10⁵ Pa = 199 kPa
QUESTION 5;
5. If the blood vessels in a human being acted as simple pipes(which they do not), what
would be the difference in blood pressure between the blood in a 1.80-m- tall man's
feet and his head when he is standing? Assume the specific gravity of blood to be
1.06.
ANSWER 5.
5. Δp = pgh
= 1060(9.8)(1.8)
= 18.7 kPa
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