- The Pressure Field in a Static Liquid
- Manometers
- Forces on Submerged Surfaces and the Center of Pressure
- Archimedes Principle and Bouyancy

**1)**
The atmospheric pressure on an Earth-like planet (far far away) is
\(80 \, kPa\). If humans were to build a habitat on this planet what
should the gage pressure on the inside be

- \(17 \, kPa\)
- \(21 \, kPa\)
- \(26 \, kPa\)
- \(31 \, kPa\)

**The correct answer is (B)**

**Solution:**
Based on the local atmospheric pressure of \(80 \, kPa \) and assuming
that the desired habitat pressure is \(101 \, kPa\) abs. the gage
pressure inside the hab should be
$$ \text{gage pressure} = \text{abs pressure in hab} - \text{local atm pressure} $$
$$ \begin{align}
\text{gage pressure} &= 101 kPa - 80 kPa \\
&= 21 kPa
\end{align} $$
**The correct answer is (B)**

**2)**
The absolute pressure inside a container is \(25 \, kPa\). What is the
vacuum pressure. Assume the local atmospheric pressure to be
\(101 \, kPa\)

- \(55 \, kPa\)
- \(68 \, kPa\)
- \(76 \, kPa\)
- \(91 \, kPa\)

**The correct answer is (C)**

**Solution:**
Vacuum pressure is the difference between atmospheric pressure and abs
pressure inside the container
$$ \text{vacuum pressure} = 101 \, kPa - 25 \, kPa = 76 \, kPa $$
**The correct answer is (C)**

**3)**
What pressure is equivalent to \(600 \, mm\) of mercury
(SG \(13.6\)).

- \(43 \, kPa\)
- \(58 \, kPa\)
- \(72 \, kPa\)
- \(80 \, kPa\)

**The correct answer is (D)**

**Solution:**
Given the height of mercury in the column
$$ \begin{align}
P_{gage} &= \rho g h \\
&= 13{,}600 \, kg/m^3 \times 9.81 \, m/s^2 \times 0.6 \, m \\
&= 80{,}050 \, Pa \\
&= 80 \, kPa
\end{align} $$
**The correct answer is (D)**

**4)**
A closed cylindrical tank, shown in the figure has a
pressure \( 60 \, kPa\) inside the dome. Determine the pressure in pipe
B.

- \(89 \, kPa\)
- \(103 \, kPa\)
- \(124 \, kPa\)
- \(135 \, kPa\)

**The correct answer is (B)**

**Solution:**
Note that \( P_1 = P_2 = P_3\), so
$$ \begin{align}
P_B &= P_A + (\rho g h)_{oil} + (\rho g h)_{water} \\\\
P_B &= 60{,}000 Pa \\
&+ 800 \, kg/m^3 \times 9.81 \, m/s^2 \times 3 m \\
&+ 1000 \, kg/m^3 \times 9.81 \, m/s^2 \times 2 m\\\\
P_B &= 103 \, kPa
\end{align} $$
**The correct answer is (B)**

**5)**
Pipe \(A\) and \(B\) both contain water. The gage fluid in the manometer
shown has a specific gravity of \(3.46\). Calculate the differential
pressure

- \(15.5 \, kPa\)
- \(17.9 \, kPa\)
- \(21.5 \, kPa\)
- \(30.1 \, kPa\)

**The correct answer is (A)**

**Solution:**
Note that the pressure at points marked 2 and 3 are going to be the same.
$$ P_A + \gamma_w \cdot (0.6 m + 0.4 m) + \gamma_{gf} \cdot 0.4 m = P_B
+ \gamma_w \cdot (0.4 m + 0.4 m) $$
Rearrange and cancel terms to get
$$ \begin{align}
P_A - P_B &= \gamma_w \cdot 0.2 \, m + \gamma_{gf} \cdot 0.4 \, m \\
&= 9810 \, N/m^3 \times 0.2 \, m + 3.46 \times 9810 \, N/m^3 \times 0.4 \, m \\
&= 15{,}539 \, Pa \\
&= 15.5 \, kPa
\end{align}$$
**The correct answer is (A)**

**6)**
A tank containing water has a gate hinged at the bottom as shown in the
figure below. The gate is \(5 \, m\) wide. Calculate the force on the
gate due to water pressure.

- \(1104 \, kN\)
- \(1380 \, kN\)
- \(1655 \, kN\)
- \(2152 \, kN\)

**The correct answer is (A)**

**Solution:** The total force on the gate will be equal to
the average pressure times the area.
$$ \begin{align} F &= P_{avg} \times A \\
&= (\gamma_w \cdot h_{avg}) \times A \\
&= 9810 \, N/m^3 \times 4.5 \, m \times 25 \, m^2 \\
&= 1{,}103{,}625 \, N \\
&= 1104 \, kN \\
\end{align}$$
**The correct answer is (A)**

**7)**
A \(4 \, m\) wide curved gate is located in the bottom of a reservoir
containing water. Determine the magnitude of the resultant force of water
on the gate.

- \( 281 \, kN \)
- \( 352 \, kN \)
- \( 470 \, kN \)
- \( 587 \, kN \)

**The correct answer is (D)**

**Solution:**
For curved surfaces calculate the horizontal and vertical forces and then
combine them to find the resultant.
$$ \begin{align} F_{hor} &= P_{avg} \times A_{projected} \\
&= \gamma_w \cdot {h_1 + h_2 \over 2} \times A_{projected} \\
&= 9810 \, N/m^3 \cdot {4+6 \over 2} \, m \times (2m \cdot 4m) \\
&= 392{,}400 \, N \\\\
F_{ver} &= \gamma_w \times (V_1 + V_2) \\
&= 9810 \, N/m^3 \times \left( (4m \cdot 2m \cdot 4m) + { \pi \cdot (2m)^2 \cdot 4m \over 4} \right) \\
&= 437{,}196 \, N \\\\
F &= \sqrt{392{,}400^2 + 437{,}196^2} \\
&= 587{,}468 \, N \\
&= 587 \, kN
\end{align} $$
**The correct answer is (D)**

**8)**
A log floats with one fourth of its volume protruding above the water
surface. Determine the specific weight of the log.

- \(2345.2 \, N/m^3\)
- \(4690.4 \, N/m^3\)
- \(6253.9 \, N/m^3\)
- \(7357.5 \, N/m^3\)

**The correct answer is (D)**

**Solution:**
Since the log is in equilibrium we can equate the weight of the log
to the bouyant force.
$$ \gamma_{log} \times V_{log} = \gamma_{water} \times V_{disp} $$
$$ \begin{aligned} \gamma_{log} &= \gamma_{water} \times { V_{disp} \over V_{log} } \\
&= 9810 \, N/m^3 \times \frac 34 \\
&= 7357.5 \, N/m^3 \end{aligned} $$
**The correct answer is (D) **

**9)**
A cylindrical block of concrete is suspended by a cable and completely
immersed in seawater (density \(1025 \, kg/m^3\)). The diameter is
\(0.75 \, m\), height \(1 \, m\) and density of concrete is
\(2300 \, kg/m^3\). Calculate the tension in the cable

- \(4245 \, N \)
- \(5526 \, N \)
- \(6631 \, N \)
- \(9670 \, N \)

**The correct answer is (B)**

**Solution:**
The tension in the cable can be found using \( T = W - F_B \) Where,
$$\begin{aligned} W &= m \cdot g = \rho_C \cdot V \cdot g \text{, and} \\
F_B &= \rho_W \cdot V \cdot g \end{aligned}$$
(HINT: Use a FBD and sum forces in the vertical direction.)
$$ \begin{aligned} T &= (\rho_C - \rho_W) \cdot V \cdot g \\
&= (2300-1025) \, kg/m^3 \cdot (pi/4*0.75^2*1.0) \, m^3 \cdot 9.81 \,m/s^2 \\
&= 5526 \, N \end{aligned}$$
**The correct answer is (B)**