A flat plate of contact area 0.80 m² slides at 3.20 m/s over a stationary surface separated by an oil film 4.50 mm thick. The oil has dynamic viscosity 0.0288 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 3.20/4.50e-3 = 711.1 s^{-1}$$

3. Substituting

   $$\tau = 0.0288\left(711.1\right) = 20.5\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 20.5(0.80) = 16.4\ \text{N}$$

Why the other options are there

• 0.000415 Pa (multiplied by film thickness)
• 25.6 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 0.90 m² slides at 3.60 m/s over a stationary surface separated by an oil film 2.50 mm thick. The oil has dynamic viscosity 0.0298 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 3.60/2.50e-3 = 1,440 s^{-1}$$

3. Substituting

   $$\tau = 0.0298\left(1,440\right) = 42.9\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 42.9(0.90) = 38.6\ \text{N}$$

Why the other options are there

• 0.000268 Pa (multiplied by film thickness)
• 47.7 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 1.15 m² slides at 3.50 m/s over a stationary surface separated by an oil film 1.50 mm thick. The oil has dynamic viscosity 0.0846 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 3.50/1.50e-3 = 2,333 s^{-1}$$

3. Substituting

   $$\tau = 0.0846\left(2,333\right) = 197.4\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 197.4(1.15) = 227.0\ \text{N}$$

Why the other options are there

• 0.000444 Pa (multiplied by film thickness)
• 171.7 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 0.55 m² slides at 5.60 m/s over a stationary surface separated by an oil film 4.00 mm thick. The oil has dynamic viscosity 0.0066 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 5.60/4.00e-3 = 1,400 s^{-1}$$

3. Substituting

   $$\tau = 0.0066\left(1,400\right) = 9.2\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 9.2(0.55) = 5.1\ \text{N}$$

Why the other options are there

• 0.000148 Pa (multiplied by film thickness)
• 16.8 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 0.10 m² slides at 0.80 m/s over a stationary surface separated by an oil film 1.50 mm thick. The oil has dynamic viscosity 0.0868 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 0.80/1.50e-3 = 533.3 s^{-1}$$

3. Substituting

   $$\tau = 0.0868\left(533.3\right) = 46.3\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 46.3(0.10) = 4.6\ \text{N}$$

Why the other options are there

• 0.000104 Pa (multiplied by film thickness)
• 462.9 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 0.80 m² slides at 5.70 m/s over a stationary surface separated by an oil film 4.00 mm thick. The oil has dynamic viscosity 0.0620 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 5.70/4.00e-3 = 1,425 s^{-1}$$

3. Substituting

   $$\tau = 0.0620\left(1,425\right) = 88.4\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 88.4(0.80) = 70.7\ \text{N}$$

Why the other options are there

• 0.001414 Pa (multiplied by film thickness)
• 110.4 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 1.15 m² slides at 6.00 m/s over a stationary surface separated by an oil film 5.50 mm thick. The oil has dynamic viscosity 0.0318 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 6.00/5.50e-3 = 1,091 s^{-1}$$

3. Substituting

   $$\tau = 0.0318\left(1,091\right) = 34.7\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 34.7(1.15) = 39.9\ \text{N}$$

Why the other options are there

• 0.001049 Pa (multiplied by film thickness)
• 30.2 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 0.15 m² slides at 1.20 m/s over a stationary surface separated by an oil film 3.00 mm thick. The oil has dynamic viscosity 0.0724 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 1.20/3.00e-3 = 400.0 s^{-1}$$

3. Substituting

   $$\tau = 0.0724\left(400.0\right) = 29.0\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 29.0(0.15) = 4.3\ \text{N}$$

Why the other options are there

• 0.000261 Pa (multiplied by film thickness)
• 193.1 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 0.25 m² slides at 5.30 m/s over a stationary surface separated by an oil film 3.00 mm thick. The oil has dynamic viscosity 0.0468 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 5.30/3.00e-3 = 1,767 s^{-1}$$

3. Substituting

   $$\tau = 0.0468\left(1,767\right) = 82.7\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 82.7(0.25) = 20.7\ \text{N}$$

Why the other options are there

• 0.000744 Pa (multiplied by film thickness)
• 330.7 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as

A flat plate of contact area 1.10 m² slides at 5.50 m/s over a stationary surface separated by an oil film 1.50 mm thick. The oil has dynamic viscosity 0.0652 Pa·s and the velocity profile across the film is linear. Compute the shear stress on the plate and the force needed to keep it moving.

Start with the thinking

• Fluid stress is defined as force per unit area acting tangentially; for a Newtonian fluid it is proportional to the velocity gradient.
• A linear profile makes du/dy = U/h, so no calculus is needed.

Step-by-step solution

1. Formula

   $$\tau = \mu \dfrac{du}{dy}$$

2. Velocity gradient

   $$du/dy = U/h = 5.50/1.50e-3 = 3,667 s^{-1}$$

3. Substituting

   $$\tau = 0.0652\left(3,667\right) = 239.1\ \text{Pa}$$

4. Formula

   $$F = \tau A$$

5. Substituting

   $$F = 239.1(1.10) = 263.0\ \text{N}$$

Why the other options are there

• 0.000538 Pa (multiplied by film thickness)
• 217.3 N (divided by area)

Reference: FE Reference Handbook — Fluid Mechanics → Stress is defined as