A brief note on Pressure and Fluids at Rest . Read it before a test. Try each question at the end before you open its answer.
The big ideas Pressure : p = F A p = \dfrac{F}{A} p = A F , in pascal. A fluid pushes at right angles to every surface. Depth : p = p 0 + ρ g h p = p_0 + \rho gh p = p 0 + ρ g h , the same at every point at one depth, whatever the vessel's shape. Pascal's law : a change of pressure reaches every part of an enclosed fluid, so F 2 = F 1 A 2 A 1 F_2 = F_1\dfrac{A_2}{A_1} F 2 = F 1 A 1 A 2 . The air : p 0 = 1.013 × 10 5 Pa p_0 = 1.013 \times 10^{5}\ \text{Pa} p 0 = 1.013 × 1 0 5 Pa , 76 cm 76\ \text{cm} 76 cm of mercury. Gauge pressure is the part above it. Archimedes : the buoyant force is the weight of fluid pushed aside; a floating body has ρ b ρ f \dfrac{\rho_b}{\rho_f} ρ f ρ b of its volume under. The facts at a glance Situation Formula Below a surface p 0 + ρ g h p_0 + \rho gh p 0 + ρ g h Hydraulic lift F 2 = F 1 A 2 A 1 F_2 = F_1\dfrac{A_2}{A_1} F 2 = F 1 A 1 A 2 Barometer p 0 = ρ g h p_0 = \rho gh p 0 = ρ g h Manometer p − p 0 = ρ g h p - p_0 = \rho gh p − p 0 = ρ g h Buoyant force ρ f V under g \rho_f V_{\text{under}}\,g ρ f V under g Relative density W W − W ′ \dfrac{W}{W - W'} W − W ′ W
A worked example A wooden buoy of volume 0.080 m 3 0.080\ \text{m}^3 0.080 m 3 and density 450 kg/m 3 450\ \text{kg/m}^3 450 kg/m 3 floats in the sea ( 1030 kg/m 3 1030\ \text{kg/m}^3 1030 kg/m 3 ).
Mass. 450 kg/m 3 × 0.080 m 3 = 36 kg 450\ \text{kg/m}^3 \times 0.080\ \text{m}^3 = 36\ \text{kg} 450 kg/m 3 × 0.080 m 3 = 36 kg . Fraction under. 450 kg/m 3 1030 kg/m 3 = 0.437 \dfrac{450\ \text{kg/m}^3}{1030\ \text{kg/m}^3} = 0.437 1030 kg/m 3 450 kg/m 3 = 0.437 , so 43.7 % 43.7\% 43.7% is under. Extra load. Fully under it pushes aside 1030 kg/m 3 × 0.080 m 3 = 82.4 kg 1030\ \text{kg/m}^3 \times 0.080\ \text{m}^3 = 82.4\ \text{kg} 1030 kg/m 3 × 0.080 m 3 = 82.4 kg of sea, so it can carry 82.4 kg − 36 kg = 46.4 kg 82.4\ \text{kg} - 36\ \text{kg} = 46.4\ \text{kg} 82.4 kg − 36 kg = 46.4 kg before it goes under. Practice with new numbers
A program to try The program works out the buoy of the worked example.
Read it, then run it.
Float it in fresh water instead. How much less can it carry?
Mistakes to avoid Gauge for absolute pressure, or the other way round. Read the question for which one it wants. Diameters for areas in a hydraulic lift. Square the ratio. The vessel's shape changing the pressure. Only the depth matters. The body's density in the buoyant force. Use the fluid's. Forgetting the air behind a window or a mask. It cancels the air outside. Check yourself What is the gauge pressure 7.0 m 7.0\ \text{m} 7.0 m down in fresh water? Stuck? Show a hint Show the answerA lift's pistons are 3 cm 3\ \text{cm} 3 cm and 30 cm 30\ \text{cm} 30 cm across. What force on the small one holds up 5000 N 5000\ \text{N} 5000 N ? Stuck? Show a hint Show the answerA barometer reads 72 cm 72\ \text{cm} 72 cm of mercury. Is the place above or below sea level? Stuck? Show a hint Show the answerA block floats in water with 1 3 \tfrac13 3 1 of its volume above the surface. What is its density? Stuck? Show a hint Show the answerWhy does a helium balloon rise? Stuck? Show a hint Show the answer