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Question

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determine the fluid shear stress (7) on the bottom surface of the block.
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VIPSolutions · Accepted Answer

Let's solve each part step by step:

---

## (a) **Define pressure and list three different pressure units that are frequently used.**

### **Definition of**
**Pressure** is defined as the amount of force exerted per unit area.

$$
P = \frac{F}{A}
$$

Where:  
- $ P $ = Pressure  
- $ F $ = Force  
- $ A $ = Area

### **Common Units of Pressure**
1. **Pascal (Pa)**: 1 Pa = 1 N/m² (SI unit)
2. **Bar**: 1 bar = 100,000 Pa
3. **Atmosphere (atm)**: 1 atm = 101,325 Pa

---

## (b) **Sliding Block Problem**

Given:
- Speed, $ V = .45\, \text{m/s} $
- Dynamic viscosity, $ \mu = .17\, \text{kg/(m·s)} $
- Oil layer thickness, $ t = .15\, \text{mm} = .00015\, \text{m} $

We are to find **fluid shear stress ($\tau$)** on the bottom surface.

### **Shear Stress Formula**
For a linear velocity profile:
$$
\tau = \mu \frac{V}{t}
$$

#### **Substitute the Values**
$$
\tau = .17 \times \frac{.45}{.00015}
$$
$$
\tau = .17 \times 300
$$
$$
\tau = 510\, \text{N/m}^2
$$

**Final Answer for (b):**
> The fluid shear stress on the bottom surface of the block is **510 N/m²**.

---

## (c) **Inclined-tube Manometer Problem**

Given:
- Pressure in pipe A, $ P_A = 400\, \text{Pa} $
- Fluid in A & B = water ($ \rho_{water} = 100\, \text{kg/m}^3 $)
- Manometer fluid specific gravity ($ S_g $) = 2.6
- $ h_1 = 7\, \text{cm} = .07\, \text{m} $
- $ l_2 = 20\, \text{cm} = .20\, \text{m} $
- $ \theta = 30^\circ $

#### **Step 1: Calculate $ h_2 $**
$$
h_2 = l_2 \sin \theta = .20 \times \sin 30^\circ = .20 \times .5 = .10\, \text{m}
$$

#### **Step 2: Write Pressure Equation**
Starting from left (Pipe A) to right (Pipe B):

$$
P_A + \rho_{water} g h_1 + \rho_{man} g h_2 = P_B
$$
where
- $\rho_{man} = S_g \times \rho_{water} = 2.6 \times 100 = 260\, \text{kg/m}^3$

So,
$$
P_B = P_A + \rho_{water} g h_1 + (\rho_{man} - \rho_{water}) g h_2
$$

#### **Step 3: Substitute Values**
$$
P_B = 400 + (100 \times 9.81 \times .07) + (260 - 100) \times 9.81 \times .10
$$
$$
P_B = 400 + (100 \times .687) + (160 \times .981)
$$
$$
P_B = 400 + 687 + 1569.6
$$
$$
P_B = 6256.6\, \text{Pa}
$$

---

### **Final Answers**

**(a)**  
- **Pressure** is force per unit area.  
- **Units:** Pascal (Pa), Bar, Atmosphere (atm)

**(b)**  
- Shear stress, $ \tau = \boxed{510\, \text{N/m}^2} $

**(c)**  
- Pressure in pipe B, $ P_B = \boxed{6256.6\, \text{Pa}} $

---

Let me know if you need clarification or further explanation on any part!

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