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Fluid Mechanics and Properties of Matter formulas

5 formulas and 9 common traps for SPI Physics Fluid Mechanics and Properties of Matter, grouped by subtopic.

Full notes with worked examples

Thrust and Liquid Pressure

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Thrust, pressure and the pressure of a liquid column

Pressure

P=FAPliquid=hρg1 g/cm3=1000 kg/m3P = \frac{F}{A} \qquad P_{\text{liquid}} = h\rho g \qquad 1\ \text{g/cm}^3 = 1000\ \text{kg/m}^3
  • FFthrust (force at right angles), in newtons
  • AAarea, in square metres
  • hhdepth of liquid, in metres
  • ρρdensity of the liquid, in kg/m³

Pressure in a liquid depends only on depth

Pressure at a depth

P=P0+hρgP = P_0 + h\rho g
  • P₀air pressure on the open surface
  • hhdepth below the free surface

Common traps

Using g/cm³ in h ρ g

With h in metres and g in m/s², the density must be in kg/m³. An oil of 0.9 g/cm³ is 900 kg/m³; leaving it as 0.9 makes the answer a thousand times too small.

Thinking a heavier-looking face presses harder

The weight is the same whichever face is down. Only the area changes, so the smallest face gives the largest pressure, and opposite faces give the same pressure.

Thinking the wall changes the pressure

Two points at the same depth have the same pressure, whether one is near the wall and the other in the middle. Only depth and the liquid's density matter.

Thinking a wider vessel gives more pressure

A wide tank holds more water than a narrow one of the same height, but the pressure at the bottom is the same. The extra water is spread over a larger base.

Density, Buoyancy and Flotation

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Density and relative density

ρ=mVrelative density=ρsubstanceρwater\rho = \frac{m}{V} \qquad \text{relative density} = \frac{\rho_{\text{substance}}}{\rho_{\text{water}}}

Buoyant force and the loss of weight in a liquid

Archimedes' principle

FB=weight of liquid displaced=Vρliquidgapparent weight=W−FBF_B = \text{weight of liquid displaced} = V \rho_{\text{liquid}} g \qquad \text{apparent weight} = W - F_B

Floating and the extra load a floating body can carry

Flotation

V=mρgreatest load (in kg)=V×1000−mvolume under watertotal volume=ρbodyρwaterV = \frac{m}{\rho} \qquad \text{greatest load (in kg)} = V \times 1000 - m \qquad \frac{\text{volume under water}}{\text{total volume}} = \frac{\rho_{\text{body}}}{\rho_{\text{water}}}

Common traps

Giving relative density a unit

Relative density divides one density by another, so the units cancel. Writing kg/m³ after it is wrong.

Forgetting the push back on the water

When a hanging body is dipped in water, the water's balance reading goes UP by the weight the body loses. The total held by the two balances does not change.

Thinking the buoyant force depends on the body's weight

The buoyant force depends on the volume under the liquid and the liquid's density, not on how heavy the body is. Two bodies of the same volume, fully under water, get the same push.

Stopping at the total load

The greatest load a raft can hold includes the people on it. If a person is already aboard, take his mass away too before giving the luggage he can bring.

Using the body's density for the water

The water displaced is the body's volume times the density of WATER, 1000 kg/m³. Multiplying the volume by the body's own density only gives its own mass back.

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