Lesson B · 5.5–5.7
By the end you should choose $p=F/A$ or $p=h\rho g$, convert $\text{cm}^2$, and know when to add atmospheric pressure.
How I start this lesson
Pressure is force spread over an area. Same weight, smaller area, larger pressure — that is a drawing pin. Same weight, larger area, smaller pressure — that is a snowshoe. The first equation is $p=F/A$. The second, $p=h\rho g$, is for a column of fluid. They are not interchangeable. If the question gives a force and a contact area, it is $F/A$. If it gives a depth and a density, it is $h\rho g$.
5.5
$F$ is the force normal (perpendicular) to the surface. For an object sitting still on the floor, that force is usually its weight, $W=mg$. $A$ is the contact area in $\text{m}^2$. Rearrange: $F=pA$ and $A=F/p$.
Want a large pressure
Make $A$ small. Knife edge, needle, stiletto heel, drawing-pin point. The force is unchanged; it is concentrated.
Want a small pressure
Make $A$ large. Snowshoes, skis, camel feet, a wide book on soft sand. Same weight, less chance of sinking.
Board example — convert $\text{cm}^2$
A force of $400\,\text{N}$ acts on an area of $20\,\text{cm}^2$. Calculate the pressure.
$A=20\div 10^4=0.0020\,\text{m}^2$.
$$p = \frac{400}{0.0020} = 2.0\times 10^5\,\text{Pa}$$Sense-check: twice atmospheric pressure (about $1.0\times 10^5\,\text{Pa}$). If you left $A$ as $20$, you would get $20\,\text{Pa}$ — far too small for $400\,\text{N}$ on a thumbnail-sized patch.
$\text{cm}^2$ is $\div 10^4$, not $\div 100$
$1\,\text{m}=100\,\text{cm}$, so $1\,\text{m}^2=10\,000\,\text{cm}^2$. I see $A=0.20\,\text{m}^2$ written from $20\,\text{cm}^2$ every year. That error makes $p$ one hundred times too small.
2-mark “explain why snowshoes stop you sinking”
Snowshoes increase the area in contact with the snow [1]. Pressure $p=F/A$ decreases (weight unchanged) so you are less likely to sink [1].
5.6
A fluid is a liquid or a gas — anything that can flow. The particles move and collide with every surface they meet. At a given point in a fluid that is not flowing, the pressure is the same up, down and sideways.
That is why a dam must be strong at the bottom in every direction the water can push, why a punctured tyre leaks sideways as well as “out”, and why a diver feels pressure on the ears, chest and mask, not only on the top of the head.
The arrows are equal at one depth. Deeper down, every arrow would be longer — that is 5.7.
5.7
$h$ is the vertical depth below the free surface, in metres. $\rho$ is the density of the fluid, in $\text{kg/m}^3$. $g=10\,\text{N/kg}$ in this course unless a paper prints another value. The product is a pressure difference — the extra pressure caused by that column of fluid.
Board example — difference only
A point is $4.0\,\text{m}$ below the surface of fresh water, $\rho=1000\,\text{kg/m}^3$. Calculate the pressure due to the water.
$$p = h\rho g = 4.0 \times 1000 \times 10 = 4.0\times 10^4\,\text{Pa}$$The question said “due to the water” / “pressure difference”. Do not add the atmosphere.
Board example — total pressure
Seawater $\rho=1020\,\text{kg/m}^3$, depth $2.5\,\text{m}$. Atmospheric pressure $=1.00\times 10^5\,\text{Pa}$. Find the pressure difference due to the seawater, then the total pressure at that depth.
$$p_{\text{water}} = 2.5 \times 1020 \times 10 = 2.55\times 10^4\,\text{Pa}$$ $$p_{\text{total}} = 2.55\times 10^4 + 1.00\times 10^5 = 1.255\times 10^5\,\text{Pa}$$When do I add the atmosphere?
Add it only if the question asks for the total pressure, the pressure “acting on the object”, or gives you atmospheric pressure and clearly wants it used. If it says “pressure difference”, “pressure due to the liquid”, or “increase in pressure”, leave $h\rho g$ alone.
Do not mix the two pressure equations
$p=F/A$ needs a force and an area. $p=h\rho g$ needs a depth and a density. Using $F/A$ on a swimming-pool depth, or $h\rho g$ on a person standing on snow, is a zero for the physics mark.
Watch
1.
A person of weight $800\,\text{N}$ stands on snowshoes of total area $0.040\,\text{m}^2$. Calculate the pressure. What happens to the pressure if they stand on boots of area $0.010\,\text{m}^2$ instead?
2.
State the direction in which the pressure acts at a point in a liquid at rest.
3.
Oil of density $900\,\text{kg/m}^3$ is $3.0\,\text{m}$ deep. Atmospheric pressure is $1.00\times 10^5\,\text{Pa}$. Calculate the pressure due to the oil and the total pressure at the bottom of the tank.