Lesson Explainer: Pascal’s Principle Physics • 9th Grade

In this explainer, we will learn how to use Pascal’s principle to analyze the magnitude and direction of fluid pressure on an object.

Pascal’s principle states that at a point within a fluid, the pressure of the fluid at the point is equal in all directions.

To understand what this means, it is necessary to understand what is meant by pressure at a point.

We recall that there is a relationship between the pressure on an area and the force acting perpendicular to the area.

Relationship: The Pressure on an Area and the Force Acting on the Area

The pressure 𝑝 on an area 𝐴 is given by 𝑝=𝐹𝐴, where 𝐹 is the component of the force acting perpendicularly to the area.

We can calculate the pressure on an area, but it is not obvious how we can calculate the pressure at a point, as the area of a point is zero.

Let us consider the pressure on a surface of an object that is within a fluid.

The following figure shows a set of objects within a fluid.

We see that the tops of these objects are all at the same depth below the surface of the fluid.

The weight of the fluid directly above the object is the force on the top side of each object.

The volume of the fluid directly above an object is proportional to the area of the top side of the object, as shown by the following figure.

We can use the formula 𝑝=𝐹𝐴, where 𝐹 is the weight of the fluid directly above an object and 𝐴 is the area of the top side of the object.

We see that the areas of the top sides of the smaller objects have proportionally less weight of fluid acting on them. The pressure must then be equal on the top sides of all these objects.

This means that we can take an area of an object in a fluid and consider an otherwise identical object with a smaller area. The pressure on the area would not change as the area decreases. This would even apply at a single point.

We see then that the pressure at the point would be the same as that on an area.

We can now consider an object suspended in a fluid as shown by the following figure.

The top side and base of the object have an area 𝐴. The weight of the object is 𝑊. As the object is suspended, it is not accelerating. The resultant force on the object is zero.

The weight of the fluid directly above the object is 𝐹. 𝐹 acts vertically downward. A force, 𝐹, must act vertically upward for there to be no resultant force on the object, as is required by Newton’s second law of motion.

The force 𝐹 is due to the pressure of water below the object.

The pressure on the top side of the object is 𝑝. The pressure on the base of the object is 𝑝.

We see then that 𝐹=𝑊+𝐹.

Using the formula 𝑝=𝐹𝐴, we see that 𝐹=𝑝×𝐴 and 𝐹=𝑝×𝐴.

We can then see that 𝑝×𝐴=𝑊+𝑝×𝐴.

This means that (𝑝×𝐴)𝑊=𝑝×𝐴, and hence, 𝑝>𝑝.

The pressure of the fluid upward on the base of the object is greater than the pressure on the top side of the object.

Let us now consider two objects that are suspended in a fluid as shown by the following figure.

The objects have unequal heights and .

The top sides of each object are at the same depth. The pressures on the top sides of each object are then both equal to 𝑝.

The pressure on the base of each object must be greater than 𝑝. We can see, though, that the pressure on the bases of the object with the height exceeds the value of 𝑝 more than for the object with the height . This means that the smaller the height of a suspended object, the less the difference between the pressure on its top side and on its base.

If we imagine an object of zero height, the pressure on the top side and on the base of the object are equal.

We saw earlier that the pressure at a point at some depth below the surface of a fluid is the same as the pressure on an area at that depth. This means that we can now consider a point that has zero area and see that the pressure on that point from above and from below are equal.

If the pressures at the point from above and below are equal, the force acting downward and the force acting upward on the point have equal magnitudes, as shown in the following figure.

Recalling that these forces are acting on a point and that a point has zero height, we see that 𝐹down acts on whatever is directly below the point and 𝐹up acts on whatever is directly above the point.

The following figure shows the forces acting from the point in the vertical direction.

We see then two important facts about pressure in the vertical direction.

Firstly, we see that the pressure exerted by a fluid on a point is the same both upward and downward.

Secondly, we see that the pressure exerted by a fluid from a point is the same both upward and downward. At a point, the same pressure is exerted upward and downward on whatever is in contact with the point.

What has been shown for the vertical forces at a point and so for pressure upward and downward at a point also applies to horizontal forces and so to pressure on the left and right of a point.

We can consider a volume of a fluid in a container that has a gap in one of its vertical sides, from which a stream of fluid flows. This is shown in the following figure.

We see that there are forces 𝑊 and 𝐹 acting on the point in a fluid that is at the same height as the gap in the container’s vertical side.

The force 𝑊 is the weight of the fluid above the height of the gap. The weight acts vertically downward.

The force 𝐹 acts horizontally. This force produces the horizontal component of the motion of the fluid leaking through the gap.

Now, let us consider two such gaps in a container that are at different heights. This is shown in the following figure.

We see that the stream of fluid from each gap travels the same horizontal distance, but the stream from the lower gap travels a lesser vertical distance.

This means that the fluid from the lower gap has a greater horizontal velocity than the fluid from the higher gap.

Had no gaps existed, the fluid would have no horizontal velocity. At a point in the fluid, the forces in each direction would be equal, as shown in the following figure.

In the case where there is a gap in the container, the horizontal acceleration of the fluid through the gap that is observed shows that the fluid must exert a net horizontal force at the position of the gap. The side of the point that is not within the fluid does not exert a force to the right to balance the force to the left that the fluid exerts.

We see that the lower the gap, the greater the weight acting on the point in the fluid next to the gap and the greater the horizontal force exerted by the fluid at that position of the gap.

According to Pascal’s principle, the weight and the horizontal force at the point have equal magnitudes, as the force in every direction at a point in a fluid is equal.

Principle: Pascal’s Principle

Pascal’s principle states that at a point within a fluid, the pressure at that point is equal in all directions.

This has two implications for forces acting at the point:

  • The force exerted on the point by the fluid is the same in every direction.
  • The force exerted on anything that is in contact with the point is the same in every direction.

Let us now look at an example involving Pascal’s principle.

Example 1: Determining Changes in the Pressure Exerted by a Fluid on Its Container

A cubic container holds water. The lid of the container can be pushed downward, exerting pressure on the water in the container. Which of the following statements most correctly describes how the pressure exerted by the water is changed when the lid is pushed downward?

  1. The pressure exerted on the base of the cube, its vertical sides, and its lid is increased.
  2. The pressure exerted on the base of the cube is increased.
  3. The pressure exerted on the base of the cube and its lid is increased.
  4. The pressure exerted on the base of the cube and its vertical sides is increased.
  5. The pressure exerted on the cube’s vertical sides is increased.

Answer

The question states that a cubic container holds some water. The lid of the container is pushed down, applying pressure to the top surface of the water.

Pascal’s principle states that at a point within a fluid, the pressure at that point is equal in all directions.

We can consider a force applied at the center of the top surface of the water as shown in the following figure.

We can clearly see that the top surface of the water at the point where the force acts will transmit the force vertically downward, as shown in the following figure.

According to Pascal’s principle, the force on the top surface of the water will also result in a force in all directions in the water from the point where the force acts. The following figure shows some of the directions in which the force is transmitted horizontally.

Forces would though act in every direction from the point both vertically and horizontally. The following figure shows some of the directions in which the force is transmitted both vertically and horizontally.

We can see from this that there must be a force acting on each side of the container. This tells us that there will be an increase in pressure on each side of the container. This includes the lid, as the top surface of the water will exert a force upward on the lid.

The correct option is then that the pressure exerted on the base of the cube, its vertical sides, and its lid is increased.

Pascal’s principle means that a force can be transmitted through a fluid and change direction as it is transmitted.

A vertically downward force can act on a horizontal surface of a fluid at one end of a pipe. The force applies pressure throughout the fluid. As a result of the applied pressure, a vertically upward force acts on the surface of the fluid at the other end of the pipe.

The following figure shows this process for two similar pipes.

It is important to notice that the ends of each pipe are at the same height.

For each pipe, the areas of the ends of the pipe are equal.

For the curved pipe, the area of the pipe is constant along the length of the pipe. For the straight pipe, the cross-sectional area of the pipe is greater where the pipe turns. For both pipes, only the areas of the ends of the pipes need to be considered.

For each pipe, the magnitudes of the forces at the ends of the pipe can be related by the formula 𝑝=𝐹𝐴,𝑝=𝐹𝐴.

For each pipe, the areas of the ends of the pipe are equal, hence, it must be true that 𝐴=𝐴.

For each pipe, the pressures at the ends of the pipes are equal, as they are at the same height.

For each pipe, we see that the magnitude of the force applied to the fluid at the left-hand end of the pipe is equal to the magnitude of the force applied by the fluid at the right-hand end of the pipe. Therefore, 𝐹=𝐹.

A pipe need not have ends of equal areas.

If the areas of the ends of a pipe are not equal, then the forces acting at the ends of the pipes will not be equal.

We can apply the equations 𝑝=𝐹𝐴 and 𝑝=𝐹𝐴 to a pipe within which the pressure is the same at both ends of the pipe. From this we can see a relationship between the areas of the ends of a pipe and the forces acting on those areas.

Relationship: The Forces on the Ends of a Pipe and the Cross-Sectional Areas of the Ends of the Pipe

For a pipe that has equal pressure at its ends, the magnitudes of the forces acting at the ends are related to the cross-sectional areas of the ends by the equation 𝐹𝐴=𝐹𝐴, where 𝐹 and 𝐹 are the magnitudes of the forces acting on the ends of the pipe and 𝐴 and 𝐴 and are the cross-sectional areas of the ends of the pipe.

Let us now look at an example involving a force transmitted by the pressure of a fluid.

Example 2: Determining the Force Exerted on an Area of a Fluid due to a Force Acting on a Different Area of the Fluid

A hydraulic pump has a thin shaft with an area of 0.15 m2 and a thick shaft with an area of 1.2 m2, as shown in the diagram. At the tops of the shafts are pistons that can be pushed. A force 𝐹=85𝐹N is applied to the piston in the thin shaft and the pressure of the hydraulic fluid applies a force 𝐹 to the piston in the thick shaft. Find the magnitude of 𝐹.

Answer

In answering this question, we recall that the pressure 𝑝 on the base of the pistons due to the pressure from the enclosed fluid is equal.

This means that for the base of each piston, it is the case that 𝑝=𝐹𝐴, where 𝐹 is the magnitude of the force acting on the base of the piston and 𝐴 is the cross-sectional area of the piston.

Each piston has a different force acting on it on a different cross-sectional area. We can relate the forces on and cross-sectional areas of the small and large pistons with the equation 𝐹0.15=𝐹1.2.mm

The question asks for the magnitude of 𝐹, so we must make 𝐹 the subject of the equation as follows. 1.2×𝐹0.15=1.2×𝐹1.2,1.2×𝐹0.15=𝐹,𝐹×1.20.15=𝐹,𝐹×8=𝐹.mmmmmmmm

The question states that 𝐹=85N. We substitute this value to obtain 𝐹×85=𝐹,𝐹=680.NN

We see from the example that by applying a force of 85 N to one piston, a force of 680 N was exerted on the other piston.

Pascal’s principle means that a force can be applied to a small area in order to exert a greater force on a greater area. A machine that uses Pascal’s principle to multiply a force in this way is called a hydraulic press.

The ratio of the force exerted on the piston of a hydraulic press with the lesser area to the force exerted by the piston with the greater area is the mechanical advantage of the hydraulic lever, 𝜂. The mechanical advantage of a hydraulic press is equal to the ratio of the areas of the pistons. For the hydraulic press in the example, 𝜂=8.

Let us now summarize what has been learned in this explainer.

Key Points

  • Pascal’s principle states that at a point within a fluid, the pressure at that point is equal in all directions.
  • Two points within a fluid at the same height that have no other forces acting on them than the weight of the fluid above the points are at equal pressures.
  • For a hydraulic press that has equal pressure at its ends, the magnitudes of the forces acting at the ends are related to the cross-sectional areas of the ends by the equation 𝐹𝐴=𝐹𝐴, where 𝐹 and 𝐹 are the magnitudes of the forces acting on the ends of the hydraulic press and 𝐴 and 𝐴 and are the cross-sectional areas of the ends of the hydraulic press.
  • The ratio of the areas of the ends of a hydraulic press is the mechanical advantage 𝜂 of the hydraulic press.

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