In previous lesson, we have discussed measuring gas pressure. In this lesson, we are going to discuss Pascal’s Principle, including its formula and applications.

In SPM, there are a few thing s that we need to know about the Pascal’s Principle. First of all, of course we need to know what is stated in Pascal’s Principle. Second, we need to know the experiment that proves the Pascal Principle. Third, we need to know the hydraulic system and the equations related to the hydraulic system. Last but not least, we need to know the applications of hydraulic system such as hydraulic jack, hydraulic brake and hydraulic cylinder.

Let’s begin with the statement of Pascal’s Principle.

Pascal’s principle states that any change in pressure applied to a confined liquid will be **transmitted** to every point of the fluid without any change in magnitude. In other words, when there is an increase in pressure at any point in the confined fluid, there is an equal increase at every other point in the container.

For example, this is closed container. If a pressure of 10 Pa is exerted on the liquid, every point in the fluid will experience an increase of pressure of 10 Pa.

As we mentioned before, in SPM, we need to know the experiment that prove Pascal’s Principle.

The figure shows that when the plunger is pushed in, the water **squirts equally from all the holes**. This shows that the pressure applied to the plunger has been **transmitted uniformly** throughout the water. This is one of the experiments that support Pascal’s Principle.

Hydraulic system is the most important part in Pascal’s Principle. In the coming slide, we are going to discuss how a hydraulic system function and the equations related to hydraulic system.

The figure shows the illustration of the structure of a hydraulic system. In a hydraulic system, a force applied at one point can be transmitted to another point by a confined fluid. In this example, a force is applied on surface X. The force is then transmitted to surface Y by the fluid.

In SPM, There are a few things that you need to know about a hydraulic system. First, we need to know that the working principle of a hydraulic system is based on the Pascal’s Principle. We can use Pascal’s Principle to explain how a hydraulic system works. For example, if a force is exerted on surface X, the increase of pressure at surface X will be equal the increase of pressure at surface Y.

Second, we need to know that a hydraulic system can act as a force multiplier. The transmitted force can be magnified if the area of the second surface is larger than the first surface. For example, the force F_{2} is much greater than the force F_{1} because the area A_{2} is much bigger than A_{1}.

Third, we need to know the equation that we can use to calculate the magnification of the force. In a hydraulic system, the magnification of force can be calculated by the equation F_{1}/A_{1} = F_{2}/A_{2}. You need to memorise this equation because usually it is not given in the formula list in exam.

Click on the example button for a sample question or the continue button to skip the example and proceed to next section. The diagram shows a hydraulic system where the large piston has cross-sectional area A_{2} = 5000 cm^{2} and the small piston has cross-sectional area A_{1} = 20 cm^{2}. A 200kg load is placed on the large piston. A force is exerted on the small piston to support the load. Find the magnitude of force F.

The cross sectional area of the large piston, A_{1} = 5000cm^{2} and the cross sectional area of the small piston A_{2} = 20cm^{2}. On the large piston, there is a 200kg mass. The weight of the mass is 2000N. This is the force exerted on the large piston. Therefore, F_{1} = 2000N. We are asked to find the force exerted on the small piston, F_{2}.

The pressure exerted on the large piston is F_{1}/A_{1} and the pressure exerted on the small piston is F_{2}/A_{2}. According to Pascal’s Principle, the pressure at both pistons must be the same. Therefore, F_{1}/A_{1} = F_{2}/A_{2}. Plug in F_{1}, A_{1} and A_{2}. Solve the equation. We get F_{2} = 8N. The magnitude of the force exerted on the small piston is 8N.

In this case we can see that a hydraulic system act as a force multiplier. We can use a force as small as 8N to support a mass of weight 2000N.

Other than the equation for force magnification in a hydraulic system, F_{1}/A_{1} = F_{2}/A_{2} there is one more equation which is also important in the calculation related to a hydraulic system, namely the equation to calculate the change of moving distance of the pistons in a hydraulic system. This is given by the equation h_{1}A_{1} = h_{2}A_{2}. We are going to discuss this in the very next slide.

When a force is applied on surface X, surface X will move down and force the liquid in the small cylinder flow to the large cylinder.

Let’s say the small piston moves down a distance h_{1}, causes the large piston raises a distance h_{2}. The decrease of the liquid volume in the small cylinder will be equal to the increase of the liquid volume in the large cylinder. Therefore, V_{1} = V_{2}. For a cylinder, its volume is equal to the height of the cylinder x the surface area, or V = hA. Therefore, we get h_{1}A_{1} = h_{2}A_{2}. This is the formula to find the change of the level of the liquid in a hydraulic cylinder.

You need to memorise this equation because usually it is not given in the formula list in exam.

Click on the example button for a sample question or the continue button to skip the example and proceed to next section. The figure shows a hydraulic system. The area of surface X is 10cm^{2} and the area of surface Y is 50 cm^{2}. Surface X has been pushed down 20cm. what is the change of liquid level, h, at surface Y?

As what we have discussed in previous slide, this question can be solved by using the formula h_{1}A_{1} = h_{2}A_{2}. In this case, the first piston has been pushed down 20cm. Therefore h_{1} = 20cm. A_{1} = 10cm^{2}; h_{2} = h and A_{2} = 50cm^{2}.Substitute this number into the formula. h = 4 cm. Piston Y increases by 4 cm.

This is how we find the change of distance of the pistons in a hydraulic system. We have just concluded our discussion on hydraulic system where we have learned 2 new equations, F_{1}/A_{1} = F_{2}/A_{2} and h_{1}A_{1 }= h_{2}A_{2}. In coming slides, we are going to discuss the applications of hydraulic system, including hydraulic jack, hydraulic brake and hydraulic cylinder.

Pascal’s Principle is applied in hydraulic jack, hydraulic brake and hydraulic cylinder. Click on any of it to see the explanation.

This is the illustration of the structure of a hydraulic jack.

When the handle is pushed forward, valve A will be closed whereas valve B will opened. The hydraulic fluid is forced into the large cylinder and hence pushes the piston moving upward. This can be done easily because the hydraulic system acts as a force multiplier to increase the force exerts on the large cylinder.

When the handle is pulled backward, valve B will be closed while valve A will open. The close of valve B can prevent the large cylinder fall back to its original position. The open of valve A enable the hydraulic fluid from the buffer tank sucked into the small cylinder.

This process is repeated until the load is sufficiently lifted up. The large piston can be lowered down by releasing the hydraulic fluid back to the buffer tank through the release vale.

The diagram shows the illustration of the structure of a hydraulic brake system. Normally, the front wheels of a car use the disc brakes and the rear wheels use the drum brakes.

In a hydraulic brake system, when the brake pedal is pressed, the piston of the master cylinder applies a **pressure on the brake fluid**. This pressure is **transmitted uniformly** to each cylinder at the wheels, causes the pistons at the wheels to **push the brake shoes** to press against the surface of the brake. The friction between the brakes and brake shoes causes the vehicle to slow down and stop.

Hydraulic cylinder is used in excavators, bulldozers and also in hydraulic garage lift. This is all you need to know in SPM. How the hydraulic cylinder work is not important in SPM.