Week 9: Fluids and Pressure

[1] Equations⚓︎

I wrote a little cheatsheet for this unit, I'll bring printed copies for everyone in my recitation sections.

[2] Helpful Resources⚓︎

Yau Jong Twu's videos on Fluid Mechanics (see next section for an idea of what topics to review).

[3] Recitation Prep⚓︎

I recommend reviewing the following concepts, they will probably come up in recitation:

• Floating/submerged objects (buoyant force + force balance).
• Pressure differences in fluids at rest (especially U-tubes).
• How fluid speed changes when water flows through pipes or out of a container.

[4] UPDATE (April 8): Practice Problems⚓︎

If you're comfortable with the "★★★" problems, you should do great in recitation.

These are meant to target the same skills as the worksheet, without being near-copies of the actual recitation questions.

If you want to print these questions, simply press Ctrl + P while on this page, and it should come out formatted nicely.

Difficulty key:

• ★☆☆ = beginner
• ★★☆ = standard
• ★★★ = challenging / multi-step

[4.1] Buoyancy and Apparent Weight⚓︎

1. ★☆☆
   A solid plastic cube with side length \(8.0 \ \mathrm{cm}\) is held completely submerged in water. What buoyant force acts on the cube?

   Solution

   Use Archimedes' principle:

   \[ F_B = \rho_{\mathrm{water}} V g. \]

   The cube's volume is

   \[ V = (0.080)^3 = 5.12 \times 10^{-4} \ \mathrm{m^3}. \]

   So

   \[ F_B = (1000)(5.12 \times 10^{-4})(9.80) = 5.02 \ \mathrm{N}. \]

   
   Answer: \(5.02 \ \mathrm{N}\) upward.

2. ★★☆
   A \(2.40 \ \mathrm{kg}\) rock of density \(2600 \ \mathrm{kg/m^3}\) hangs from a spring scale while fully submerged in fresh water. What does the scale read?

   Solution

   The scale reads the tension, which is the true weight minus the buoyant force:

   \[ T = mg - F_B. \]

   First find the rock's volume:

   \[ V = \frac{m}{\rho_{\mathrm{rock}}} = \frac{2.40}{2600} = 9.23 \times 10^{-4} \ \mathrm{m^3}. \]

   Then the buoyant force is

   \[ F_B = \rho_{\mathrm{water}} V g = (1000)(9.23 \times 10^{-4})(9.80) = 9.05 \ \mathrm{N}. \]

   The rock's weight is

   \[ mg = (2.40)(9.80) = 23.5 \ \mathrm{N}. \]

   Therefore,

   \[ T = 23.5 - 9.05 = 14.5 \ \mathrm{N}. \]

   
   Answer: \(14.5 \ \mathrm{N}\).

3. ★★★
   A hollow metal sphere has outer volume \(3.20 \times 10^{-3} \ \mathrm{m^3}\) and mass \(1.80 \ \mathrm{kg}\). It is held completely submerged in water by a light cord attached to the bottom of a tank. Find the tension in the cord, and state whether the cord pulls upward or downward on the sphere.

   Solution

   Since the sphere is fully submerged, the buoyant force is

   \[ F_B = \rho_{\mathrm{water}} V g = (1000)(3.20 \times 10^{-3})(9.80) = 31.4 \ \mathrm{N}. \]

   Its weight is

   \[ mg = (1.80)(9.80) = 17.6 \ \mathrm{N}. \]

   Because \(F_B > mg\), the sphere would rise on its own, so the cord must pull downward to keep it at rest.

   For equilibrium,

   \[ F_B = mg + T \]

   so

   \[ T = F_B - mg = 31.4 - 17.6 = 13.7 \ \mathrm{N}. \]

   
   Answer: \(13.7 \ \mathrm{N}\), and the cord pulls downward on the sphere.

[4.2] Fluid Pressure, Connected Fluids, and Pascal's Principle⚓︎

1. ★☆☆
   A diver is \(6.50 \ \mathrm{m}\) below the surface of a freshwater lake. Find the gauge pressure and the absolute pressure at that depth.

   Solution

   Gauge pressure is

   \[ P_g = \rho g h = (1000)(9.80)(6.50) = 6.37 \times 10^4 \ \mathrm{Pa}. \]

   Absolute pressure is

   \[ P_{\mathrm{abs}} = P_{\mathrm{atm}} + P_g = 1.013 \times 10^5 + 6.37 \times 10^4 = 1.65 \times 10^5 \ \mathrm{Pa}. \]

   
   Answer: Gauge pressure \(= 6.37 \times 10^4 \ \mathrm{Pa}\); absolute pressure \(= 1.65 \times 10^5 \ \mathrm{Pa}\).

2. ★★☆
   A U-tube is open to the atmosphere on both sides. Water stands \(18.0 \ \mathrm{cm}\) above the common interface on the left side, and an unknown liquid stands \(24.0 \ \mathrm{cm}\) above the same interface on the right side. What is the density of the unknown liquid?

   Solution

   At the same horizontal level in the connected fluid, pressures must match:

   \[ P_{\mathrm{left}} = P_{\mathrm{right}}. \]

   Since both sides are open to air, atmospheric pressure cancels, giving

   \[ \rho_{\mathrm{water}} g h_{\mathrm{water}} = \rho_u g h_u. \]

   So

   \[ \rho_u = \rho_{\mathrm{water}} \frac{h_{\mathrm{water}}}{h_u} = (1000)\frac{0.180}{0.240} = 750 \ \mathrm{kg/m^3}. \]

   
   Answer: \(\rho_u = 750 \ \mathrm{kg/m^3}\).

3. ★★★
   A hydraulic lift has a small piston of radius \(2.50 \ \mathrm{cm}\) and a large piston of radius \(18.0 \ \mathrm{cm}\). A \(1200 \ \mathrm{kg}\) car rests on the large piston.

   (a) What force must be applied to the small piston to support the car? (b) If the large piston rises by \(6.0 \ \mathrm{cm}\), how far must the small piston move?

   Solution

   (a) By Pascal's principle,

   \[ \frac{F_1}{A_1} = \frac{F_2}{A_2}. \]

   The large piston supports the car, so

   \[ F_2 = mg = (1200)(9.80) = 1.176 \times 10^4 \ \mathrm{N}. \]

   Since area is proportional to \(r^2\),

   \[ F_1 = F_2 \frac{r_1^2}{r_2^2} = (1.176 \times 10^4)\frac{(0.0250)^2}{(0.180)^2} = 2.27 \times 10^2 \ \mathrm{N}. \]

   (b) The fluid volume pushed on the small side equals the volume raised on the large side:

   \[ A_1 \Delta x_1 = A_2 \Delta x_2. \]

   Thus,

   \[ \Delta x_1 = \frac{A_2}{A_1}\Delta x_2 = \frac{r_2^2}{r_1^2}(0.0600). \]

   So

   \[ \Delta x_1 = \frac{(0.180)^2}{(0.0250)^2}(0.0600) = 3.11 \ \mathrm{m}. \]

   
   Answer: (a) \(2.27 \times 10^2 \ \mathrm{N}\) (b) \(3.11 \ \mathrm{m}\)

[4.3] Continuity and Bernoulli in Pipes⚓︎

1. ★☆☆
   Water flows through a hose of radius \(1.20 \ \mathrm{cm}\) at a speed of \(1.80 \ \mathrm{m/s}\). It exits a nozzle of radius \(0.40 \ \mathrm{cm}\). What is the exit speed?

   Solution

   Use continuity:

   \[ A_1 v_1 = A_2 v_2. \]

   Since area is proportional to \(r^2\),

   \[ v_2 = \frac{r_1^2}{r_2^2}v_1 = \frac{(1.20)^2}{(0.40)^2}(1.80). \]

   This gives

   \[ v_2 = 9(1.80) = 16.2 \ \mathrm{m/s}. \]

   
   Answer: \(16.2 \ \mathrm{m/s}\).

2. ★★☆
   Water flows horizontally through a pipe that narrows from diameter \(4.0 \ \mathrm{cm}\) to diameter \(2.0 \ \mathrm{cm}\). The speed in the wide section is \(1.50 \ \mathrm{m/s}\), and the pressure there is \(2.20 \times 10^5 \ \mathrm{Pa}\). Find the pressure in the narrow section.

   Solution

   First use continuity:

   \[ A_1 v_1 = A_2 v_2 \quad \Rightarrow \quad v_2 = \frac{d_1^2}{d_2^2}v_1 = \frac{(4.0)^2}{(2.0)^2}(1.50) = 6.00 \ \mathrm{m/s}. \]

   Since the pipe is horizontal, Bernoulli becomes

   \[ P_1 + \frac12 \rho v_1^2 = P_2 + \frac12 \rho v_2^2. \]

   Solve for \(P_2\):

   \[ P_2 = P_1 + \frac12 \rho (v_1^2 - v_2^2). \]

   Substituting,

   \[ P_2 = 2.20 \times 10^5 + \frac12 (1000)\left(1.50^2 - 6.00^2\right) = 2.03 \times 10^5 \ \mathrm{Pa}. \]

   
   Answer: \(2.03 \times 10^5 \ \mathrm{Pa}\).

3. ★★☆
   Water flows upward through a vertical pipe from a basement to a faucet \(3.20 \ \mathrm{m}\) higher. The pipe diameter is the same everywhere, and the water speed is \(2.40 \ \mathrm{m/s}\) throughout. If the pressure in the basement pipe is \(2.65 \times 10^5 \ \mathrm{Pa}\), what is the pressure at the faucet?

   Solution

   Because the diameter is constant, the speed is the same at both points, so the kinetic-energy terms cancel in Bernoulli:

   \[ P_1 + \rho g y_1 = P_2 + \rho g y_2. \]

   Let \(y_2 - y_1 = 3.20 \ \mathrm{m}\). Then

   \[ P_2 = P_1 - \rho g (y_2-y_1). \]

   So

   \[ P_2 = 2.65 \times 10^5 - (1000)(9.80)(3.20) = 2.34 \times 10^5 \ \mathrm{Pa}. \]

   
   Answer: \(2.34 \times 10^5 \ \mathrm{Pa}\).

4. ★★★
   On the ground floor, water flows through a pipe of diameter \(5.0 \ \mathrm{cm}\) at speed \(1.20 \ \mathrm{m/s}\) and pressure \(3.00 \times 10^5 \ \mathrm{Pa}\). Upstairs, \(4.0 \ \mathrm{m}\) higher, the pipe narrows to diameter \(2.0 \ \mathrm{cm}\). Find the water pressure upstairs.

   Solution

   First use continuity to find the upstairs speed:

   \[ A_1 v_1 = A_2 v_2 \quad \Rightarrow \quad v_2 = \frac{d_1^2}{d_2^2}v_1 = \frac{(5.0)^2}{(2.0)^2}(1.20) = 7.50 \ \mathrm{m/s}. \]

   Now use Bernoulli:

   \[ P_1 + \frac12 \rho v_1^2 + \rho g y_1 = P_2 + \frac12 \rho v_2^2 + \rho g y_2. \]

   With \(y_2-y_1=4.0 \ \mathrm{m}\),

   \[ P_2 = P_1 + \frac12 \rho (v_1^2 - v_2^2) - \rho g (4.0). \]

   Substituting,

   \[ P_2 = 3.00 \times 10^5 + \frac12 (1000)(1.20^2 - 7.50^2) - (1000)(9.80)(4.0). \]

   This gives

   \[ P_2 = 2.33 \times 10^5 \ \mathrm{Pa}. \]

   
   Answer: \(2.33 \times 10^5 \ \mathrm{Pa}\).

[4.4] Mixed Practice / Synthesis⚓︎

1. ★★☆
   A large open tank of water has a small hole in its side. The water surface is \(1.25 \ \mathrm{m}\) above the hole, and the hole is \(0.80 \ \mathrm{m}\) above the ground. How far horizontally from the tank will the water land?

   Solution

   First use Torricelli's result for the exit speed:

   \[ v = \sqrt{2gh} = \sqrt{2(9.80)(1.25)} = 4.95 \ \mathrm{m/s}. \]

   The water then falls like a projectile from height \(0.80 \ \mathrm{m}\), so the fall time is

   \[ t = \sqrt{\frac{2y}{g}} = \sqrt{\frac{2(0.80)}{9.80}} = 0.404 \ \mathrm{s}. \]

   Horizontal distance:

   \[ x = vt = (4.95)(0.404) = 2.00 \ \mathrm{m}. \]

   
   Answer: \(2.00 \ \mathrm{m}\).

2. ★★★
   A block of volume \(0.0200 \ \mathrm{m^3}\) and mass \(12.0 \ \mathrm{kg}\) floats in water. A student gently pushes straight downward on the block until it is just completely submerged and held at rest. What downward force is required?

   Solution

   When the block is just fully submerged, the buoyant force is

   \[ F_B = \rho_{\mathrm{water}} V g = (1000)(0.0200)(9.80) = 196 \ \mathrm{N}. \]

   The block's weight is

   \[ mg = (12.0)(9.80) = 117.6 \ \mathrm{N}. \]

   Since the block is held at rest while fully submerged, the downward push must balance the excess buoyant force:

   \[ F_{\mathrm{push}} + mg = F_B. \]

   Therefore,

   \[ F_{\mathrm{push}} = F_B - mg = 196 - 117.6 = 78.4 \ \mathrm{N}. \]

   
   Answer: \(78.4 \ \mathrm{N}\) downward.