PHSX 220 Physics I (w/ Calculus) · Unit 2: Applying Newton, statics, fluids, energy · on Exam 2
Ch 14.1–14.4 Static fluids
Study notes for Static fluids in PHSX 220 Physics I (w/ Calculus) at Montana State University: the big ideas, key formulas, a worked example, common mistakes and an exam tip. The same topic has endless practice questions and flashcards in StudyCat.
Big ideas
- Density $\rho = m/V$ (water: $1000\ \text{kg/m}^3$). Pressure $P = F/A$ (Pa); atmospheric pressure $P_0 = 1.013\times10^5$ Pa.
- Pressure increases with depth: $P = P_0 + \rho g h$. Gauge pressure is $\rho g h$ (what a tire gauge reads); absolute pressure adds $P_0$. Pressure depends only on depth, not on the container's shape.
- Pascal's principle: a pressure change is transmitted undiminished through an enclosed fluid, so a hydraulic lift multiplies force: $F_2 = F_1 A_2/A_1$ (but the small piston moves farther: work is conserved).
- Archimedes: the buoyant force equals the weight of the displaced fluid, $F_B = \rho_{\text{fluid}} V_{\text{sub}}\, g$. A floating object has $F_B = mg$, so the submerged fraction is $\rho_{\text{obj}}/\rho_{\text{fluid}}$. A fully submerged object sinks if $\rho_{\text{obj}} > \rho_{\text{fluid}}$.
Key formulas
Pressure with depth
$$P = P_0 + \rho g h,\qquad P_{\text{gauge}} = \rho g h$$
Pascal
$$\frac{F_1}{A_1} = \frac{F_2}{A_2}$$
Buoyancy
$$F_B = \rho_{\text{fluid}}\, V_{\text{sub}}\, g,\qquad \frac{V_{\text{sub}}}{V} = \frac{\rho_{\text{obj}}}{\rho_{\text{fluid}}}\ \text{(floating)}$$
Worked example
A 2.0 kg block of density $800\ \text{kg/m}^3$ floats in water. What fraction is submerged, and what is the buoyant force?
Show the solution
Fraction $= 800/1000 = 0.80$. Floating means $F_B = mg = 19.6$ N. (Check: $V = 2/800 = 2.5\times10^{-3}\ \text{m}^3$; $0.8V\rho_w g = 2.0\times10^{-3}(1000)(9.8) = 19.6$ N.)
Common mistakes
- Using the object’s density instead of the fluid’s in the buoyant force.
- Forgetting that gauge and absolute pressure differ by one atmosphere.
- Thinking a wider tank means more pressure at the bottom.
Exam tipApparent weight under water = true weight − buoyant force. That one line solves most "scale reading in water" problems.
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