Salt water is denser than fresh water because dissolved salts increase the water’s mass per unit volume. When you float, the buoyant force must balance your weight. Since salt water has higher density, you can get the same buoyant force while displacing a smaller volume of water, so less of your body is submerged and you float higher.
Mathematically proven:
| Step | Derivation/Formula | Reasoning |
|---|---|---|
| 1 | \[F_b = \rho_{\text{fluid}} g V_{\text{displaced}}\] | This is Archimedes’ principle: the buoyant force equals the weight of the displaced fluid. Here \(\rho_{\text{fluid}}\) is the fluid density and \(V_{\text{displaced}}\) is the displaced (submerged) volume. |
| 2 | \[\text{Floating equilibrium: } F_b = W = mg\] | If you are floating at rest, the upward buoyant force must balance your weight \(mg\). |
| 3 | \[mg = \rho_{\text{fluid}} g V_{\text{displaced}}\] | Combine Step 1 and Step 2: the buoyant force expression must equal the person’s weight when floating. |
| 4 | \[V_{\text{displaced}} = \frac{m}{\rho_{\text{fluid}}}\] | Solve Step 3 for the displaced volume (\(g\) cancels). For a given person (fixed \(m\)), the required displaced volume is inversely proportional to the fluid density. |
| 5 | \[\rho_{\text{salt}} > \rho_{\text{fresh}}\] | Salt water has greater density than fresh water because dissolved salts increase mass per unit volume. |
| 6 | \[V_{\text{salt}} = \frac{m}{\rho_{\text{salt}}} < V_{\text{fresh}} = \frac{m}{\rho_{\text{fresh}}}\] | Using Step 4 and Step 5: higher density means a smaller volume must be displaced to make the displaced fluid’s weight equal to your weight. |
| 7 | Less displaced volume \(\rightarrow\) less of your body needs to be submerged | Since \(V_{\text{displaced}}\) is the submerged volume, a smaller \(V_{\text{displaced}}\) means a smaller fraction of your body is underwater. |
| 8 | You float higher in salt water than in fresh water | Because you need to displace less salt water (smaller submerged volume) to balance your weight, you ride higher in the water. |
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The side of an above-ground pool is punctured, and water gushes out through the hole. If the total depth of the pool is \( 2.5 \) \( \text{m} \), and the puncture is \( 1 \) \( \text{m} \) above the ground level, what is the efflux speed of the water?
A cube of unknown material and uniform density floats in a container of water with \(60\%\) of its volume submerged. If this same cube were placed in a container of oil with density \(800\) \(\text{kg/m}^3\), what portion of the cube’s volume would be submerged while floating?
In the laboratory, you are given a cylindrical beaker containing a fluid and you are asked to determine the density \( \rho \) of the fluid. You are to use a spring of negligible mass and unknown spring constant \( k \) that is attached to a vertical stand.
Which of the following statements is an expression of the equation of continuity?
Rex, an auto mechanic, is raising a \( 1200 \) \( \text{kg} \) car on his hydraulic lift so that he can work underneath. If the area of the input piston is \( 12.0 \) \( \text{cm}^2 \), while the output piston has an area of \( 700 \) \( \text{cm}^2 \), what force must be exerted on the input piston to lift the car?
Marc’s favorite ride at Busch Gardens is the Flying Umbrella, which is lifted by a hydraulic jack. The operator activates the ride by applying a force of \( 72 \) \( \text{N} \) to a \( 30 \)\( \text{cm}^2 \) cylindrical piston, which holds the \( 20,000 \)\( \text{N} \) ride off the ground. What is the area of the piston that holds the ride?
Balsa wood with an average density of \( 130 \) \( \text{kg/m}^3 \), is floating in pure water. What percentage of the wood is submerged?
A Venturi tube has a pressure difference of \( 15\,000 \) \( \text{Pa} \). The entrance radius is \( 3 \) \( \text{cm} \), while the exit radius is \( 1 \) \( \text{cm} \). What are the entrance velocity, exit velocity, and flow rate if the fluid is gasoline \( (\rho = 700 \) \( \text{kg/m}^3 ) \)?
A pump is used to send water through a hose, the diameter of which is \( 10 \) times that of the nozzle through which the water exits. If the nozzle is \( 1 \) \(\text{m}\) higher than the pump, and the water flows through the hose at \( 0.4 \) \(\text{m/s}\), what is the difference in pressure between the pump and the atmosphere?
A pump, submerged at the bottom of a well that is \( 35 \) \( \text{m} \) deep, is used to pump water uphill to a house that is \( 50 \) \( \text{m} \) above the top of the well, as shown to the right. The density of water is \( 1000 \) \( \text{kg/m}^3 \). All pressures are gauge pressures. Neglect the effects of friction, turbulence, and viscosity.
Salt water is denser than fresh water because the dissolved salt increases its mass. This higher density creates a stronger buoyant force, allowing you to float higher in salt water.
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| Kinematics | Forces |
|---|---|
| \(\Delta x = v_i t + \frac{1}{2} at^2\) | \(F = ma\) |
| \(v = v_i + at\) | \(F_g = \frac{G m_1 m_2}{r^2}\) |
| \(v^2 = v_i^2 + 2a \Delta x\) | \(f = \mu N\) |
| \(\Delta x = \frac{v_i + v}{2} t\) | \(F_s =-kx\) |
| \(v^2 = v_f^2 \,-\, 2a \Delta x\) |
| Circular Motion | Energy |
|---|---|
| \(F_c = \frac{mv^2}{r}\) | \(KE = \frac{1}{2} mv^2\) |
| \(a_c = \frac{v^2}{r}\) | \(PE = mgh\) |
| \(T = 2\pi \sqrt{\frac{r}{g}}\) | \(KE_i + PE_i = KE_f + PE_f\) |
| \(W = Fd \cos\theta\) |
| Momentum | Torque and Rotations |
|---|---|
| \(p = mv\) | \(\tau = r \cdot F \cdot \sin(\theta)\) |
| \(J = \Delta p\) | \(I = \sum mr^2\) |
| \(p_i = p_f\) | \(L = I \cdot \omega\) |
| Simple Harmonic Motion | Fluids |
|---|---|
| \(F = -kx\) | \(P = \frac{F}{A}\) |
| \(T = 2\pi \sqrt{\frac{l}{g}}\) | \(P_{\text{total}} = P_{\text{atm}} + \rho gh\) |
| \(T = 2\pi \sqrt{\frac{m}{k}}\) | \(Q = Av\) |
| \(x(t) = A \cos(\omega t + \phi)\) | \(F_b = \rho V g\) |
| \(a = -\omega^2 x\) | \(A_1v_1 = A_2v_2\) |
| Constant | Description |
|---|---|
| [katex]g[/katex] | Acceleration due to gravity, typically [katex]9.8 , \text{m/s}^2[/katex] on Earth’s surface |
| [katex]G[/katex] | Universal Gravitational Constant, [katex]6.674 \times 10^{-11} , \text{N} \cdot \text{m}^2/\text{kg}^2[/katex] |
| [katex]\mu_k[/katex] and [katex]\mu_s[/katex] | Coefficients of kinetic ([katex]\mu_k[/katex]) and static ([katex]\mu_s[/katex]) friction, dimensionless. Static friction ([katex]\mu_s[/katex]) is usually greater than kinetic friction ([katex]\mu_k[/katex]) as it resists the start of motion. |
| [katex]k[/katex] | Spring constant, in [katex]\text{N/m}[/katex] |
| [katex] M_E = 5.972 \times 10^{24} , \text{kg} [/katex] | Mass of the Earth |
| [katex] M_M = 7.348 \times 10^{22} , \text{kg} [/katex] | Mass of the Moon |
| [katex] M_M = 1.989 \times 10^{30} , \text{kg} [/katex] | Mass of the Sun |
| Variable | SI Unit |
|---|---|
| [katex]s[/katex] (Displacement) | [katex]\text{meters (m)}[/katex] |
| [katex]v[/katex] (Velocity) | [katex]\text{meters per second (m/s)}[/katex] |
| [katex]a[/katex] (Acceleration) | [katex]\text{meters per second squared (m/s}^2\text{)}[/katex] |
| [katex]t[/katex] (Time) | [katex]\text{seconds (s)}[/katex] |
| [katex]m[/katex] (Mass) | [katex]\text{kilograms (kg)}[/katex] |
| Variable | Derived SI Unit |
|---|---|
| [katex]F[/katex] (Force) | [katex]\text{newtons (N)}[/katex] |
| [katex]E[/katex], [katex]PE[/katex], [katex]KE[/katex] (Energy, Potential Energy, Kinetic Energy) | [katex]\text{joules (J)}[/katex] |
| [katex]P[/katex] (Power) | [katex]\text{watts (W)}[/katex] |
| [katex]p[/katex] (Momentum) | [katex]\text{kilogram meters per second (kgm/s)}[/katex] |
| [katex]\omega[/katex] (Angular Velocity) | [katex]\text{radians per second (rad/s)}[/katex] |
| [katex]\tau[/katex] (Torque) | [katex]\text{newton meters (Nm)}[/katex] |
| [katex]I[/katex] (Moment of Inertia) | [katex]\text{kilogram meter squared (kgm}^2\text{)}[/katex] |
| [katex]f[/katex] (Frequency) | [katex]\text{hertz (Hz)}[/katex] |
Metric Prefixes
Example of using unit analysis: Convert 5 kilometers to millimeters.
Start with the given measurement: [katex]\text{5 km}[/katex]
Use the conversion factors for kilometers to meters and meters to millimeters: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}}[/katex]
Perform the multiplication: [katex]\text{5 km} \times \frac{10^3 \, \text{m}}{1 \, \text{km}} \times \frac{10^3 \, \text{mm}}{1 \, \text{m}} = 5 \times 10^3 \times 10^3 \, \text{mm}[/katex]
Simplify to get the final answer: [katex]\boxed{5 \times 10^6 \, \text{mm}}[/katex]
Prefix | Symbol | Power of Ten | Equivalent |
|---|---|---|---|
Pico- | p | [katex]10^{-12}[/katex] | |
Nano- | n | [katex]10^{-9}[/katex] | |
Micro- | µ | [katex]10^{-6}[/katex] | |
Milli- | m | [katex]10^{-3}[/katex] | |
Centi- | c | [katex]10^{-2}[/katex] | |
Deci- | d | [katex]10^{-1}[/katex] | |
(Base unit) | – | [katex]10^{0}[/katex] | |
Deca- or Deka- | da | [katex]10^{1}[/katex] | |
Hecto- | h | [katex]10^{2}[/katex] | |
Kilo- | k | [katex]10^{3}[/katex] | |
Mega- | M | [katex]10^{6}[/katex] | |
Giga- | G | [katex]10^{9}[/katex] | |
Tera- | T | [katex]10^{12}[/katex] |
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