2.4 Pressure and pressure differences in fluids — Practise

Work through these on your own. Keep your working neat: the method matters as much as the answer.

Both the questions and the answers below come from a set that is checked by computer every time the book is built.

A block pushes down on the ground with a force of 200 N. The area in contact with the ground is 0.080 m $^2$ . Calculate the pressure on the ground.
A tyre pushes on the road with a force of 540 N. The contact patch between the tyre and the road has an area of 0.0012 m $^2$ . Calculate the pressure in the tyre.
A pressure of 4500 Pa acts on an area of 0.60 m $^2$ . Calculate the force normal to the surface.
A force of 600 N produces a pressure of 15000 Pa. Calculate the area that the force acts on.
The area of a small tile is 25 cm $^2$ . Convert this area to square metres.
A force of 600 N acts through the point of a stiletto heel. The area of the point is 0.00020 m $^2$ . Calculate the pressure. Explain why a wide heel is kinder to a wooden floor.
Explain why the pressure in a fluid causes a force normal to a surface, and give one example of the effect.
Explain why atmospheric pressure decreases as the height above the ground increases.
H Calculate the pressure due to a column of water 10 m high. The density of water is 1000 kg/m $^3$ and $g = 9.8\,\text{N/kg}$ . Use $p = h\rho g$ .
H A tank holds a layer of oil 0.50 m deep. The density of the oil is 800 kg/m $^3$ and $g = 9.8\,\text{N/kg}$ . Calculate the pressure due to the oil at the bottom of the layer.
H A diver moves from a depth of 1.0 m to a depth of 3.0 m in water of density 1000 kg/m $^3$ . Take $g = 9.8\,\text{N/kg}$ . Calculate the increase in pressure.
H Explain why a partly submerged object feels an upward force, and describe one factor that decides whether the object floats or sinks.

Answers and concise solutions

1. Pressure on the ground

Answer: $p = \dfrac{F}{A} = \dfrac{200}{0.080} = 2500\,\text{Pa} = 2.5 \times 10^3\,\text{Pa}$ .

2. Pressure in the tyre

Answer: $p = \dfrac{F}{A} = \dfrac{540}{0.0012} = 450000\,\text{Pa} = 4.5 \times 10^5\,\text{Pa}$ .

3. Force normal to the surface

Answer: $F = pA = 4500 \times 0.60 = 2700\,\text{N} = 2.7 \times 10^3\,\text{N}$ .

4. Area the force acts on

Answer: $A = \dfrac{F}{p} = \dfrac{600}{15000} = 0.040\,\text{m}^2$ .

5. Area in square metres

Answer: There are $10000\,\text{cm}^2$ in $1\,\text{m}^2$ , so $25\,\text{cm}^2 = 25 \div 10000 = 0.0025\,\text{m}^2$ .

6. Pressure from the heel

Answer: $p = \dfrac{F}{A} = \dfrac{600}{0.00020} = 3000000\,\text{Pa} = 3.0 \times 10^6\,\text{Pa}$ . A wide heel spreads the same force over a larger area, so the pressure is smaller and the floor is less likely to be damaged.

7. The force from fluid pressure

The particles of the fluid collide with the surface. Each collision pushes on the surface, and the total force is at right angles to the surface.
The air inside a bicycle tyre pushes outwards on the tyre, at right angles to the rubber at every point.

8. Why atmospheric pressure falls with height

The atmosphere is a thin layer of air around the Earth, and there are fewer air molecules above a high surface.
So the weight of the air above the surface is smaller.
Fewer collisions from the air molecules give a smaller pressure, so atmospheric pressure decreases with height.

9. Pressure at a depth of 10 m

Answer: $p = h\rho g = 10 \times 1000 \times 9.8 = 98000\,\text{Pa} = 9.8 \times 10^4\,\text{Pa}$ .

10. Pressure due to the oil

Answer: $p = h\rho g = 0.50 \times 800 \times 9.8 = 3920\,\text{Pa} = 3.9 \times 10^3\,\text{Pa}$ .

11. Increase in pressure on the diver

Answer: The extra depth is $3.0 - 1.0 = 2.0\,\text{m}$ . Then $\Delta p = h\rho g = 2.0 \times 1000 \times 9.8 = 19600\,\text{Pa} = 2.0 \times 10^4\,\text{Pa}$ .

12. Upthrust and floating

The bottom surface of the object is deeper than the top surface, so the pressure on the bottom is greater.
The greater upward force on the bottom and the smaller downward force on the top give a resultant upward force called upthrust.
An object floats if its average density is less than the density of the fluid, and sinks if its average density is greater.