1st WEEK:HYDROSTATICS

QUESTION 1

1. By how much does the pressure under the seawater for every 10 m increase in depth?

ANSWER 1:
1. 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.

QUESTION 2


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 10 ˉ² 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 10 ˉ⁵ = 2690 kg/m ³
specific gravity = 2690kg/m ³ ÷ 1000kg/m ³

QUESTION 3

3. An 80kg metal cyclinder 2 m long and with each end of area 25 cm ² 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/s ² ) / 25 x 10 ˉ ⁴ m ²

= 3.14 x 10 ⁵N/m ²

= 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 10 ³kg/m ³ and 13.6 x 10 ³kg/m ³.


ANSWER 4;

4. p = patm + pgy = phghgh + pgy
= (13.6 x 10 ³)(9.8)(0.760) + (10 ³)(9.8)(10)
= 1.99 x 10 ⁵ 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