ICSE Class 8 Physics Chapter 07 Pressure

Pressure

Concept of Pressure

Whenever a force is applied on a body, it generally acts on a certain area of the body. To measure the effect of force, we need to take into account both the magnitude of force and the area on which force is applied.

Consider the following figures:

Imagine a rectangular block placed on a table top. We may place it in such a away that its smallest face ABCD is in contact with the table (Fig. 7.1 (a)) or its largest face EFGH is in contact with it (Fig. 7.1 (b)).

When placed on face ABCD, the weight of the block is acting vertically downwards on the smallest face and when placed on face EFGH, the same weight of the block is acting vertically downwards on the largest face. In both the cases, the force (weight of the block) is same but the areas on which the force is acting are different.

The magnitude of force acting normally on unit area is called pressure.

If force (F) acts normally on an area (A), the pressure (P) is given by:

\[Pressure = \frac{Force}{Area} \text{ i.e., } P = \frac{F}{A}\]

In figure 7.1(a), the force (weight of the block) is acting on a smaller area, therefore pressure (P) is maximum, as \(P = \frac{F}{A}\). Whereas in Fig. 7.1(b) the same force (weight of the block) is acting on a larger area, thus pressure (P) is less.

When the force is acting vertically downwards or normal to the surface under consideration, it is called thrust. Therefore, pressure can also be defined as thrust per unit area.

\[Pressure = \frac{Thrust}{Area}\]

Unit of Pressure

In the SI system, the unit of force is Newton and that of area is m². Therefore, the unit of pressure is N/m² (Nm-2) or Pascal (Pa).

Teacher's Note

When you wear high heels versus flat shoes, the pressure on the ground differs because the same weight is distributed over a much smaller area with heels, which is why they can damage flooring.

Conclusions from Pressure Formula

From the relation, \(Pressure = \frac{Thrust}{Area}\) we conclude that:

1. Pressure is directly proportional to the thrust applied i.e., an increase in thrust causes an increase in pressure in such a way that if the thrust is doubled, the pressure also doubles and vice-versa.

2. Pressure is inversely proportional to the area on which force (thrust) is applied i.e., an increase in area of contact causes a decrease in pressure in such a way that if area of contact is doubled, the pressure becomes half and vice-versa.

Let us take the following examples to understand the fact that a decrease in area increases the pressure or conversely, an increase in area decreases the pressure.

1. A nail or a board pin has one end pointed and sharp while the other end is blunt or flat. When we apply force with our thumb on the flat surface of the pin, keeping the pointed end in contact with the board, the pin will easily penetrate into the board. When we apply the same force keeping the blunt end in contact with the board the pin will not penetrate at all. Thus, it is proved that pressure and area are inversely proportional.

2. The cutting instruments like the blade, knife, the axe, etc., have the cutting edge very sharp. This helps in decreasing the area of contact thereby increasing the pressure. Hence, they can easily penetrate through the given surface.

3. Heavy trucks have six to eight tyres instead of the conventional four. This is instead of the conventional four. This is done to increase the area of contact between the tyres and the road because of which the pressure on the ground is reduced.

4. Animals like camels can walk easily in a desert as compared to horses because camels have broader feet. The broader feet exert less pressure on the sandy ground.

5. Skiers use long flat skies to slide over the snow. The larger the area of contact, the lesser is the pressure on the snow. This helps the skier to slide comfortably without sinking in the snow.

6. The foundations of high-rise buildings are kept wide so that they may exert less pressure on the ground and do not sink in due to extremely high pressure.

Example 1

A solid weighs 80 N. When placed on a surface, the area of contact is found to be 1.6 m². Find the pressure exerted by the solid on the surface.

Solution:

Given, Force = Weight = 80 N

Area = 1.6 m²

P = ?

\[Pressure = \frac{Force}{Area} = \frac{80 \text{ N}}{1.6 \text{ m}^2} = 50 \text{ N/m}^2 = 50 \text{ Pa}\]

Example 2

A rectangular wooden piece weighs 25 newton. If it is placed in such a way that its 25 cm length and 10 cm width are in contact with a surface, what will be the pressure exerted by it on the surface?

Solution:

Given, Force = Weight = 25 newton

Length = 25 cm = 0.25 m.

Width = 10 cm = 0.1 m.

Area = Length x width = 0.25 m x 0.1 m. = 0.025 m²

\[Pressure = \frac{Force}{Area} = \frac{25 \text{ N}}{0.025 \text{ m}^2} = 1000 \text{ N/m}^2 = 1000 \text{ Pa}\]

Teacher's Note

When you press a thumbtack into a wall, the tiny pointed end creates high pressure on a small area, allowing it to pierce through easily-demonstrating how pressure depends on both force and contact area.

Pressure Due to Liquids

We know that solids have a definite shape but liquids do not have a definite shape. They acquire the shape of the vessel in which they are kept. Accordingly, their area of contact changes with the shape of the vessel.

A liquid exerts pressure on the base of the container (vessel) due to its weight and also exerts pressure on the walls of the container due to collisions of its molecules with the wall. Since the molecules of a liquid are regularly in motion, so they strike (collide) continuously with the walls of the container. To every collision, these molecules exert a thrust on the walls of the container, thus exerting a pressure.

Example:

To understand the pressure exerted by the liquid, let us take the following example.

Take an open narrow glass tube of length about 8-10 cm. On lower end of the tube, tie a thin stretched rubber membrane. Now hold the tube in a vertical position and pour water in the tube. You will see that, as the water is poured in the tube, the rubber membrane bulges outwards. This bulging increases in size as you pour more and more water into the tube. The bulge is caused by the pressure exerted by the liquid due to its own weight.

We measure liquid pressure using a manometer. It is a simple device used to measure the liquid pressure.

We may construct a manometer and show its working. For this, we need a glass made U-tube fixed on a vertical board consisting of a scale which can measure on either side. We pour some mercury or some coloured liquid into the U-tube so that it is partially filled. The level of liquid on both sides of the U-tube will be the same. Now connect one end of the U-tube with a rubber tube and attach a thistle funnel on the open end of the rubber tube. Fix a rubber membrane to the thistle funnel as shown in Fig. 7.3.

Teacher's Note

When you fill a water balloon, the rubber stretches because the water pressure pushes outward equally in all directions against the balloon's surface.

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