Physics A Level
Chapter 7: Matter and materials 7.3 Archimedes’ principle
Physics A Level
Chapter 7: Matter and materials 7.3 Archimedes’ principle
The variation of pressure with depth can be used to explain Archimedes’ principle.
Archimedes’ principle states that the upthrust acting on a body is equal to the weight of the liquid or gas that it displaces.
When the object is placed in a liquid, it displaces some of the liquid. In other words, it takes up some of the space of the liquid. The volume of the liquid displaced is equal to the volume of the liquid taken up by the object. If the object floats, the volume displaced is equal to the volume of the part of the object that is under the surface of the liquid.
Consider a rectangular shaped object immersed in a liquid (Figure 7.4). There is a larger pressure on the bottom surface than there is on the top surface because the bottom surface is deeper in the liquid.
The pressure on the top surface produces a force downwards on the top. It may seem surprising, but the pressure on the bottom surface actually produces a force upwards on the object. This is because pressure can act in any direction and always acts at right angles to a surface in a liquid. You may also be surprised to know that pressure is a scalar quantity even though it is defined in terms of force (which is a vector).
Since pressure acts in all directions at a point it is not possible to define a single direction for it!
Because the pressure is larger on the bottom surface, the force acting upwards on the bottom surface is larger than the force acting downwards on the top surface. This is the cause of the upthrust, which you experience when you swim. Because your density is less than that of water, when you are underwater, the weight of water you displace is greater than your own weight. The upthrust is, therefore, greater than your own weight and there is a resultant force upwards to bring you to the surface.
To calculate this upthrust:
The force due to water on the top surface ${F_1} = \rho \times g \times {h_1} \times A$
Similarly, the force due to the water on the bottom surface is ${F_2} = \rho \times g \times {h_2} \times A$
$\begin{array}{l} upthurst = {F_2} - {F_1} = \rho \times g \times ({h_2} - {h_1}) \times A = \rho \times g \times h \times A\\ = \rho \times g \times V \end{array}$
where the volume of the object $V = h \times A$
= the weight of the liquid displaced
Questions
7) a: Why is it difficult to hold an inflated plastic ball underwater?
b: A submarine floats at rest under the water. To rise to the surface compressed air is used to push water out of its ‘ballast’ tanks into the sea. Why does this cause the submarine to rise?
8) A boat has a uniform cross-sectional area at the water line of $750{m^2}$. Fifteen cars of average mass $1200 kg$ are driven on board. Calculate the extra depth that the boat sinks in water of density
$1000\,kg\,m\,{s^{ - 3}}$
9) Describe how to use a newton-meter, a micrometer screw gauge, a metal cube of side approximately $1.0 cm$ and a beaker of water to show experimentally that Archimedes’ principle is correct. The density of water is known to be $1000\,kg\,m\,{s^{ - 3}}$.
10) A balloon of volume $3000\,{m^{ - 3}}$ is filled with hydrogen of density $0.090\,kg\,m\,{s^{ - 3}}$. The mass of the fabric of the balloon is $100 kg$. Calculate the greatest mass that the balloon can lift in air of density $102\,kg\,m\,{s^{ - 3}}$