# Archimedes’ Principle

In this lesson, we are going to discuss Archimedes’ Principle, where we will discuss what is upthrust? Why objects float in liquid? The relationship between upthrust and density, and what is stated in Archimedes’ Principle. After that we will discuss the equation of Archimedes Principle. Under the equation, we will learn how to find the volume of the displaced liquid and how to find weight of an object in a liquid and the upthrust exerted on it by using the same equation.

After that, we will discuss forces exerted on an object immerses in water in 3 difference cases. Last but not least, we will discuss the applications of Archimedes’ Principle such as hydrometer, hot air balloon, submarine and airship.

Click on the branch “uptrust” or the “next slide” button to start.

In coming slides, we will discuss why objects float in liquid, upthrust and density and what is stated in Archimedes’ Principle and what does it mean.

Click on any of the question clouds to see the explanation.

If a block is immersed in water, the pressure exerted on the upper surface of the block will be the atmospheric pressure, while the pressure exerted on the bottom of the block will be equal to the atmospheric pressure + the liquid pressure. As we can see, the pressure exerted on lower surface is higher than the pressure exerted on the upper surface. The difference of the pressure produces a force heading upward, called the upthrust or buoyant force. This is the force that cause objects float on water.

If an object is partially or fully immerse in water, 2 forces are acting on it: its own weight and the upthrust.

If the density of the object is greater than the density of the liquid, its weight will be greater than the upthrust, and the object will sink to the bottom of the liquid.

If the density of the object is lower than the water, the object will partially immerse in the liquid and the upthrust is equal to the weight of the object.

If the density of the object is equal to the density of the liquid, the object will fully immerse in the water and the upthrust is equal to the weight of the object. The object can stays in any position inside the water.

Archimedes Principle states that when a body is wholly or partially immersed in a fluid it experiences an upthrust equal to the weight of the fluid displaced.

For example, in the figure, there is an object of weight 10N hanging on a spring balance. If the object is immerse in water, we will find that the reading on the spring balance will decrease. This is because part of the weight is supported by the upthrust produced by the water. Let’s say, the new reading is 8N. So, what is the magnitude of the upthrust? The weight reduced by 2N owing to the upthrust. Therefore, the upthrust must also be 2N.

At the same time, some water will be displaced by the object and flows into the beaker. If we measure the weight of the water displaced, we will find that it is also 2N.

According to Archimedes’ Principle, the magnitude of the upthrust must be equal to the weight of the displaced fluid. If the weight of the displaced fluid is 2N, then the upthrust must also be 2N. This is what stated in Archimedes Principle. It said that the upthrust is equal to the weight of the displaced fluid. A fluid can be liquid or gas.

According to the Archimedes’ Principle, the upthrust exerted on an object immersed in a fluid is give by the equation, F = rVg, where F is the upthrust; r is the density of the fluid; V is the volume of the displaced fluid and g is the gravitational field strength. Take notes that the rV is actually equal to the mass of the fluid. A mass of an object can be calculated by the equation m = rV. You should have learned this in your PMR science.

When you want to use this equation to solve problems related to object immersed in fluid, you need to know 2 important notes. One is related to the volume of the fluid and another is related to the density of the fluid. Let’s begin with the volume of the displaced fluid.

According to Archimedes Principle, the upthust is equal to the weight of the displaced fluid, which is given by a formula, rVg. Take notes that, the V is the volume of the displaced fluid. However, in practice, sometime, students may confuse with the volume of the displaced fluid and the volume of the object.

Let’s look at these 2 cases. If an object is fully immersed in a fluid, the volume of the displaced fluid is equal to the volume of the object. Normally, students will have no problem with this.

However, if the object is partially immerse in the fluid, , like the one in the diagram, the volume of the displaced fluid is not equal to the volume of the object, but equal to the volume of the object immerse in the fluid. For example, if the volume of the block is 100cm3, 80 cm3 is in the water, 20cm3 is above the water level, then the volume of the displaced fluid is 80 cm3, but not 100 cm3. If an object is partially or fully immerse in water, 2 forces are acting on it: its own weight and the upthrust.

The upthrust is given by the formula F = rVg. The weight of the object can be determined by the formula, W = mg. We have just learned that, m = rV. Therefore, the weight of the object is also equal to rVg. Students always confuse with these 2 formulae because it look almost the same. Is there any difference between this 2 formulae?

Well, the upthrust is equal to the weight of the displaced fluid. Therefore, for upthrust, the density must be the density of the fluid but not the density of the object and the volume V must be the volume of the displaced fluid.

For the weight of the object, the 8density must be the density of the object and the 8volume V must be the volume of the object.

In short, upthrust is equal to the weight of the displaced fluid, therefore the r must be the density of the fluid and the V must be the volume of the displaced fluid. For the weight of the object, the r must be the density of the object and the V must be the volume of the object.

Click on the “example” button” to see an example of question related to this. You may skip the example and proceed to “Forces in 3 cases” by clicking on the “continue” button. The weight of a copper block is 50N. Find the upthrust that act on the block when it is fully immerse in water. [ Density of copper =   9000 kg m-3; Density of water =  1000kg m-3 ]

Well, in this question, we are asked to find the upthrust. The upthrust is equal to rVg. g is the gravitational field strength, equal to 10N/kg. Now, we have 2 densities given in the question: the density of the copper, equal to 9000 kg/m3 and the density of the liquid, equal to 1000kg/m3.  Which density should we use for our upthrust? Well, since upthrust is equal to the weight of the displaced water, therefore, we should use the density of the water to find the upthrust, but not the density of the copper.

V is not given in the question. Instead, we are given the density and weight of the block. Since the block is fully immersed in water, hence the volume of the displaced water will be equal to the volume of the object. In previous slide, we have learned that the weight of the object is equal to rVg as well. The weight is 300N; Density is 9000kg/m3. Since this is the weight of the copper block, hence we must use the density of copper, but not the density of water. g is equal to 10N/kg. Solve the equation, we get V = 0.005m3.

Let’s come back to the upthrust. The density is 1000; the volume is 0.005; the g is 10. Therefore, the upthrust is equal to 50N. I suppose we have learned almost everything that you need to know about Archimedes principle. Here we will discuss a few cases involve a block in water. In these cases, we should pay attention on the forces involve. This is important for us to solve the calculation questions.

1. Let’s start with the first case. If an object which is less dense than water is place into water, it will partially immerse in the water. Since the object is not moving, hence the force must be in equilibrium. Therefore, the upthrust is equal to the weight of the object.

However, if the object is tied to an iron bar by a string, it may sink to the bottom. Now, in this case, there are 2 forces pulling the object down: the weight and the tension of the string. Since the object is not moving, hence all the forces must also be in equilibrium. Therefore, the upthrust is equal to the weight + the tension. In exam, normally you will be asked to find the tension of the string.

Let’s look at another case. An iron block is hung in water by a string. There are 3 forces acting on the block: the weight, the tension of the string and the upthrust. Both the tension and the upthrust are pointing upward. The block is not moving, which means all the forces are in equilibrium. Therefore we conclude that the tension + the upthrust = the weight.

Knowing the forces acting on the object may help you to solve some of the problems related to Archimedes’ principle.