3.1 INTRODUCTION
There is an interesting story that Newton was sitting under an apple tree, then an apple fell on him.
He knew from the first law of motion that a net force should have acted on the apple to make it fall from the tree. This event made Newton to imagine that perhaps all bodies in the universe are attracted towards each other in the same way the apple was attracted towards the earth, and this force of attraction is called gravitational force or force of gravitation. Now in this chapter we will study gravitation in more detail.
3.2 GRAVITATION
When a body is thrown up, it reaches a certain height and then falls down. The downward pull of the earth on the body decreases its velocity in the upward direction to zero at some height. The body cannot rise further. This is therefore, the maximum height. The same downward pull of earth on the body makes it fall downwards from the maximum height. Based on these findings, Newton postulated his famous Universal Law of Gravitation in
1687. This law is also known as Newton’s law of Gravitation.
He proposed that it was the gravitational force due to the earth which was responsible for keeping the moon in its orbit around the earth. He argued that at each point of its orbit, the moon falls towards the earth, instead of going off in a straight line. So it must be attracted by
the earth. But we do not really see the moon falling towards the earth. Why? To understand the motion of the moon around the earth, let us perform the following activity.
Tie a small piece of stone at one end of a thread. Hold the other end of the thread in your hand and whirl it round. Observe the motion of the stone. Now release the thread and again observe the motion of the stone. (Try this activity in an open space).
You can notice that after releasing, the stone will follow a path whose direction is tangential to the point from which it is released and finally it falls on the ground.
The force that causes acceleration and keeps the stone moving along the circular path is acting towards the centre, i.e. towards our hands. This force is called centripetal force. The force which is needed to make an object travel in a circular path is called centripetal force.
3.3 UNIVERSAL LAW OF GRAVITATION
Statement
Newton’s law of gravitation states that every body in this universe attracts every other body with a force, which is directly proportional to the product of the masses and inversely proportional to the square of the distance between their centres and acts along their line joining their centres.
Mathematical Expression
Consider two bodies of masses ${m}_{1}and{m}_{{2}_{}}$ lying at a distance r apart. Then, according to Newton’s law of gravitation, the force of attraction between the two bodies i.e.,
$\mathrm{F}\propto {\mathrm{m}}_{1}{\mathrm{m}}_{2}\u2013\u2013\u2013\left(1\right)\phantom{\rule{0ex}{0ex}}\mathrm{F}\propto \frac{1}{{\mathrm{r}}^{2}}\u2013\u2013\u2013\u2013\u2013\left(2\right)$
Combining equations (1) and (2), we get $\mathrm{F}\propto \frac{{\mathrm{m}}_{1}{\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}\text{or}\mathrm{F}=\mathrm{G}\frac{{\mathrm{m}}_{1}{\mathrm{m}}_{2}}{{\mathrm{r}}^{2}}\u2013\u2013\u2013\u2013\left(3\right)$
where the constant of proportionality G is known as universal gravitational constant.
Definition of G
$\text{Suppose that}{m}_{1}={m}_{2}=1\text{unit and}r=1\text{unit.}\phantom{\rule{0ex}{0ex}}\text{Then,}\phantom{\rule{1em}{0ex}}\mathrm{F}=\mathrm{G}\frac{1\times 1}{{1}^{2}}\text{or}\mathrm{G}=\mathrm{F}$
Therefore, universal gravitational constant is numerically equal to the force of attraction between two unit masses placed at unit distance apart.
Value of G
The value of G was experimentally found by the scientist, Henry Cavendish. He found that the value of G is $6.67\times {10}^{-8}dynec{m}^{2}{\mathrm{g}}^{\u20132}$in CGS system and $6.67\times {10}^{-11}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{\u20132}$in S.I.
Note: G is a scalar quantity.
Importance of Universal Law of Gravitation
The gravitational force is one of the fundamental forces in nature. The gravitational force is responsible, for the following:
• the existence of the solar system (motion of planets around the sun).
• holding the atmosphere near the surface of the earth.
• the flow of water in rivers.
• for rainfall and snowfall.
• motion of the moon around the earth.
• occurrence of tides.
• prediction about solar and lunar eclipses made on the basis of this law always come out to be true.
Characteristics of Gravitational Force
1. Gravitational force does not depend on the medium which lies between the two bodies.
2. Gravitational force obeys Newton’s third law of motion i.e., the gravitational force of attraction between two bodies form an action and reaction pair.
3. Gravitational force is always attractive.
4. Gravitational force is the weakest force among all the forces.
3.4. ESTIMATION OF GRAVITATIONAL FORCE BETWEEN DIFFERENT OBJECTS
The formula applied for calculating gravitational force between light objects and heavy objects is the same, $\text{i.e.,}\mathrm{F}=\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{{\mathrm{R}}^{2}}$
Let us take three different cases:
The gravitational force between a body of 1 kg and the earth
$\text{[mass of earth}=6\times {10}^{24}\mathrm{kg},\text{radius of earth is}\left.6.4\times {10}^{6}\mathrm{m}.\right]$
Let a body of mass 1 kg be placed on the surface of the earth as shown in figure. The distance between the centre of the earth and body is equal to the radius of the earth
$\text{i.e.,}R=6.4\times {10}^{6}\mathrm{m}$
The magnitude of gravitational force between the earth and the body is given by $\mathrm{F}=\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{{\mathrm{R}}^{2}}$
Where ${m}_{l}=1\mathrm{kg}$
$\begin{array}{l}{\mathrm{m}}_{2}=6\times {10}^{24}\mathrm{kg}\text{(mass of the earth)}\\ R=6.4\times {10}^{6}\mathrm{m}\\ \mathrm{G}=6.67\times {10}^{-11}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{-2}\end{array}$
$\therefore \frac{6.67\times {10}^{-7}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{-2}\times 1\mathrm{kg}\times 6\times {10}^{24}\mathrm{kg}}{{\left(6.4\times {10}^{6}\mathrm{m}\right)}^{2}}=9.8\mathrm{N}$
Thus, a body of mass 1 kg is attracted by the earth with a force of 9.8 N.
Gravitational force between the sun and the earth
$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{earth},{\mathrm{m}}_{\mathrm{l}}=6\times {10}^{24}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\text{Mass of the sun,}{\mathrm{m}}_{2}=2\times {10}^{30}\mathrm{kg}$
Distance between the sun and the earth, $R=1.5\times {10}^{11}\mathrm{m}$
Gravitational force between the sun and the earth,$\mathrm{F}=\frac{{\mathrm{Gm}}_{1}{\mathrm{m}}_{2}}{{\mathrm{R}}^{2}}$
$\begin{array}{l}=\frac{6.67\times {10}^{-7}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{-2}\times 6\times {10}^{24}\mathrm{kg}\times 2\times {10}^{30}\mathrm{kg}}{{\left(1.5\times {10}^{11}\mathrm{m}\right)}^{2}}\\ =3.6\times {10}^{22}\mathrm{N}\end{array}$
The gravitational force between the sun and the earth is very large ($\text{i.e.,}3.6\times {10}^{22}\mathrm{N}$). This force keeps the earth bond to the sun.
Gravitational force between the moon and the earth.
$\mathrm{Mass}\mathrm{of}\mathrm{the}\mathrm{earth},{\mathrm{m}}_{\mathrm{l}}=6\times {10}^{24}\mathrm{kg}\phantom{\rule{0ex}{0ex}}\mathrm{Massofthemoon},{\mathrm{m}}_{2}=7.4\times {10}^{22}\mathrm{kg}$
$\therefore $Gravitational force between the earth and the moon.$=\frac{6.67\times {10}^{-7}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{-2}\times 6\times {10}^{24}\mathrm{kg}\times 7.4\times {10}^{22}\mathrm{kg}}{{\left(3.8\times {10}^{8}\mathrm{m}\right)}^{2}}=2.05\times {10}^{20}\mathrm{N}$
This large gravitational force keeps the moon to move around the earth. We find that,
(i) the gravitational force between two small bodies is very small.
(ii) the gravitational force between a small body and a larger body (e.g. the earth) is large.
(iii) the gravitational force between two large bodies (e.g., the sun and the earth) is very large.
3.5 GRAVITATION AND NEWTON’S THIRD LAW OF MOTION
According to Newton’s third law, to every action, there is always an equal and opposite reaction. It means if an object A exerts some force on another object B, then the object B exerts an equal and opposite force on the object A at the same instant. This law applies to the force of gravitation also.
(a) Gravitational force follows Newton’s third law of motion
(b) Force of gravitation between a stone and the earth According to Newton’s second law of motion,
force = mass × acceleration
i.e., for a given force, acceleration produced, varies inversely as the mass.
We know that acceleration produced in a body due to gravitational pull of earth on it is 9.8 $m{s}^{\u20132}$ As this acceleration is very large, we can see the body falling towards earth. We shall show that when gravitational pull of same magnitude acts on earth (where mass is $6\times {10}^{24}$ kg), the acceleration produced in earth is $1.6\times {10}^{\u201324}m{s}^{\u20132}$. As this value of acceleration is too small, we cannot see the earth moving towards the falling body
3.6 CENTRE OF MASS AND CENTRE OF GRAVITY
The point in a body where its whole mass is assumed to be concentrated is called its centre of mass. The centre of mass of a homogeneous sphere or cube must lie at its geometric centre. A rigid extended body is a continuous distribution of mass. Each particle or portion of the body experiences the force of gravity. The net effect of all these forces is equivalent to the effect of a single force, mg acting through a point called centre of gravity of the body, or we can say that a point in any body at which the of gravity of force of gravity on the whole of
the body can be assumed to act is called its centre of gravity.
On the surface of the earth, or near it, where the force of gravity is constant, the centre of mass also becomes the centre of gravity.
If we assume the earth to be a sphere of uniform density, then its centre of mass lies at its centre. The force of attraction of the earth on any body is, therefore, towards its centre.
3.7 APPLICATION OF NEWTON’S LAW OF GRAVITATION
Determination of the masses of planets and stars Knowing precise values of g, R and G, it is possible to determine accurately the mass of any planet or star by using the relationship,
$\mathrm{M}=\frac{{\mathrm{gR}}^{2}}{\mathrm{G}}$
Estimating the masses of double stars – A double star is a system consisting of two stars orbiting round their common centre of mass. From the extent of irregularity in the motion of a star due to the gravitational pull by some other star bound to it, can be used for estimating their masses. Such a small irregularity in motion is called wobble.
Many planets outside our solar system have been detected in recent years by measuring the irregularity (called wobble) in the motion of stars.
3.8 FREE FALL
The motion of a body towards the earth when no other force except the force of gravity acts on it is called a free fall. Thus all the freely falling bodies, lighter or heavier, fall towards the earth with the same acceleration.
According to Galileo Galilei, if there were no air, all the bodies having different masses when dropped simultaneously from the same height would hit the ground at the same time.
Later Robert Boyle proved this experimentally. He placed a coin and a feather in a long glass tube and removed its air with the help of vacuum pump. When the tube was inverted both the coin and the feather fell to the bottom of the tube at the same time. Thus the acceleration produced in all the freely falling bodies is the same, and does not depend upon the mass of the falling body.
3.9 ACCELERATION DUE TO GRAVITY
The earth attracts every other body lying near or on its surface towards its centre. The origin of this force between the earth and a body is due to interaction between their masses.
The force of attraction exerted by the earth on a body is called gravitational pull or gravity.
We know that when force acts on a body, it produces acceleration. Therefore, a body under the effect of gravitational pull must accelerate.
The acceleration produced in the motion of a body under the effect of gravity is called acceleration due to gravity. It is denoted by g.
The value of acceleration due to gravity near the surface of the earth is $9.8m{s}^{\u20132}$
The value of acceleration due to gravity does not depend upon, whether the body falling under gravity is a light one or a heavy one i.e., the value of g is independent of the mass of the body.
If ‘g’ is acceleration due to gravity, ‘m’ is the mass of a body; then the gravitational force on the body is ‘mg’ and is called the weight of the body.
weight = mass × acceleration due to gravity
i.e. W = mg.
Expression of Acceleration due to Gravity or (Relation between g and G)
Consider that the earth is a sphere of mass M and radius R and a body of mass m is lying on the surface of the earth. If the size of the body is very small as compared to that of the earth, then the distance between the centre of the body and that of the earth will be approximately equal to the radius of the earth. Therefore, the force of attraction on the body due to the earth is given by
The force F on the body of mass m due to the earth produces acceleration due to gravity (g) in the motion of the body. Therefore, force on the body due to earth, $\overline{)\mathrm{F}=\mathrm{mg}}$
F = mg ——-(2)
From the equation (1) and (2),
$\text{We have}\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^{2}}\u2013\u2013\u2013\u2013\u2013\left(3\right)$
This is the relation between g and G and shows that g depends on reference body i.e., $M/{R}^{2}$
Gravity and gravitation
According to Newton, the gravitational force of attraction between the moon and earth is responsible for providing the necessary centripetal force for moon to revolve around earth.
The terms gravity and gravitation are not the same. The force of attraction between any two objects by virtue of their masses is called gravitation (or gravitational force).
For example, force of attraction between any two objects such as books, tables, chairs, and between any two heavenly bodies are the examples of gravitation.
The force of gravitation exerted by a huge heavenly body such as, the earth, the moon, or the sun etc., on a smaller object near its surface is called its gravity (or force of gravity). For example, earth pulls an object of mass 1 kg towards it with a force of 9.8 N. So, force of 9.8 N is the gravity (or, force of gravity) of the earth. Similarly, on the surface of moon, we can talk of the gravity of the moon. Thus, we see that the gravity is a particular case of gravitation.
Acceleration due to Gravity in terms of Density
We know $g=\frac{GM}{{R}^{2}}=\frac{G\times V\times \rho}{{R}^{2}}$ where V is the volume of the planet, R is the radius of the planet and $\rho $ is the density of the planet. Assuming that planet is spherical in shape we can write the above equation as $g=\frac{G\times \frac{4}{3}\pi {R}^{3}\times \rho}{{R}^{2}}$
since Volume of sphere (planet) $=\frac{4}{3}\pi {R}^{3}$
$g=G\times \frac{4}{3}\pi R\times \rho \text{or}g=\frac{4}{3}\pi G\rho R$
i.e., the greater the value of ($M/{R}^{2}$) or $\rho $R , the greater will be g and as with change of reference body ($M/{R}^{2}$ ) or $\rho $R will change, so will g.
$\overline{)\mathrm{g}\propto \frac{1}{{\mathrm{R}}^{2}}\text{if Misconstant}}\phantom{\rule{0ex}{0ex}}\overline{)\mathrm{g}\propto \mathrm{R}\text{if}\rho \text{is constant}}$
3.10 FACTORS AFFECTING G
i) Shape of Earth
The earth is not perfectly spherical in shape but is an oblate sphere: it is bulged at the equator and flattened at the poles (its equatorial radius is about 21km more than polar radius).
We know, $\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^{2}}\Rightarrow \mathrm{g}\propto \frac{1}{{\mathrm{R}}^{2}}$ (as G and M of a given planet is constant).
i.e., R decreases, g increases. From the figure ${\mathrm{R}}_{\mathrm{e}}>{\mathrm{R}}_{\mathrm{p}}$$\Rightarrow {\mathrm{g}}_{\mathrm{e}}<{\mathrm{g}}_{\mathrm{p}}$
The value of g at the poles is greater than the value of g at the equator.
${\mathrm{g}}_{\text{equator}}=9.78{\mathrm{ms}}^{-2}\text{and}{\mathrm{g}}_{\text{poles}}=9.83{\mathrm{ms}}^{-2}$
ii) Altitude (height)
Let M and R be the mass and radius of Earth respectively and g be the acceleration due to gravity on the surface of the Earth. Let us find the value of acceleration due to gravity (${g}_{h}$
) at a height h from the surface of the Earth.
We know that acceleration due to gravity on the surface of Earth.
${\mathrm{g}}_{\mathrm{s}}=\mathrm{g}=\frac{\mathrm{GM}}{{\mathrm{R}}^{2}}$ —–(1)
At a height h from the Earth’s surface, the distance from the centre of Earth becomes (R + h) and acceleration due to gravity becomes
${g}_{h}=\frac{GM}{(R+h{)}^{2}}$ ——(2)
Now dividing (2) by (1), we have, $\frac{{g}_{b}}{g}=\frac{\frac{GM}{(R+h{)}^{2}}}{\frac{GM}{{R}^{2}}}=\frac{GM}{(R+h{)}^{2}}\times \frac{{R}^{2}}{GM}$
$\Rightarrow \frac{{g}_{h}}{g}={\left(\frac{R}{R+h}\right)}^{2}$
$\Rightarrow {\mathrm{g}}_{\mathrm{h}}=\mathrm{g}{\left(\frac{\mathrm{R}}{\mathrm{R}+\mathrm{h}}\right)}^{2}$
The above formula is used for any value of h
$\Rightarrow {g}_{\mathrm{h}}=\mathrm{g}\frac{{\mathrm{R}}^{2}}{{\mathrm{R}}^{2}+2\mathrm{Rh}+{\mathrm{h}}^{2}}$
Dividing numerator and denominator on R.H.S by ${R}^{2}$, we get
$\begin{array}{l}{g}_{h}=g\frac{{R}^{2}/{R}^{2}}{\frac{{R}^{2}}{{R}^{2}}+\frac{2Rh}{{R}^{2}}+\frac{{h}^{2}}{{R}^{2}}}\\ \Rightarrow {g}_{h}=g\frac{1}{1+\frac{2h}{R}+\frac{{h}^{2}}{{R}^{2}}}\end{array}$
If h < < < R, then $\frac{{h}^{2}}{{R}^{2}}$can be neglected.
$\therefore {g}_{\mathrm{h}}=\mathrm{g}{\left(1+\frac{2\mathrm{h}}{\mathrm{R}}\right)}^{-1}$
[from Binomial Theorem we know that $(1\pm x{)}^{n}=1\pm nx$ when x is small compared to1]
$\therefore {g}_{h}=g\left(1-\frac{2h}{R}\right)$ ——-(3)
The above formula should be used for h < < < R.
From (3) it is clear that, as height increases, acceleration due to gravity decreases.
Note:
The formula ${g}_{b}=g{\left(\frac{R}{R+h}\right)}^{2}$ is used for any value of ‘h’.
Whereas the formula ${g}_{h}=g\left(1-\frac{2h}{R}\right)$ is used only for h < < < R.
iii) Depth
Let d and R be the density and radius of Earth respectively and g be the acceleration due to gravity on the surface of the Earth. Let us find the value of acceleration due to gravity (${g}_{0}$) at a depth D from the surface.
We know that acceleration due to gravity on the surface of Earth.
$g=\frac{4}{3}\pi \mathrm{GdR}$ ——(1)
where G is the universal gravitational constant R is the radius of the Earth and d is the density.
When a body is at depth D from the Earth’s surface, its distance from the centre of the Earth is (R – D) and its acceleration due to gravity is
${g}_{D}=\frac{4}{3}\pi Gd(R-D)$ ——(2)
Dividing (2) by (1), we have, $\frac{{\mathrm{g}}_{\mathrm{D}}}{\mathrm{g}}=\frac{\frac{4}{3}\pi \mathrm{Gd}(\mathrm{R}-\mathrm{D})}{\frac{4}{3}\pi \mathrm{GdR}}=\frac{\mathrm{R}-\mathrm{D}}{\mathrm{R}}=\left(1-\frac{\mathrm{D}}{\mathrm{R}}\right)$
$\therefore {g}_{\mathrm{D}}=\mathrm{g}\left(1-\frac{\mathrm{D}}{\mathrm{R}}\right)$ ——(3)
From (3) it is clear that acceleration due to gravity decreases with increase of depth from the surface of the Earth.
With depth, acceleration due to gravity decreases as the effective mass pulling a body towards the centre decreases.
Zero Gravity
Consider the centre of Earth where depth (D) = Radius of Earth (R)
$\text{Substituting}D=R\text{in}{g}_{D}=g\left(1-\frac{D}{R}\right)\text{}\phantom{\rule{0ex}{0ex}}\text{We get,}{g}_{D}=g\left(1-\frac{R}{R}\right)=0$
Thus, at the centre of the Earth acceleration due to gravity is zero.
The variation of g, with both depth and height is shown in the graph drawn between g and the distance r of a place from the centre of the earth.
(fig. Variation of g with distance from centre of the earth)
We know, inside the earth, ${g}_{d}=g\left(1-\frac{d}{R}\right)=g\left(\frac{R-d}{R}\right)=g\frac{r}{R}$ $(\because \mathrm{R}-\mathrm{d}=\mathrm{r})$
$\Rightarrow {g}_{h}$versus r graph is a straight line passing through the origin as shown in figure.
We know, above the earth, ${g}_{h}=g\frac{{R}^{2}}{(R+h{)}^{2}}\Rightarrow {g}_{h}=\frac{g{R}^{2}}{{r}^{2}}\phantom{\rule{1em}{0ex}}(\because R+h=r)$
$\Rightarrow {g}_{h}$versus r graph is a curve as shown in the figure.
Difference between acceleration due to gravity and Universal Gravitational Constant
Acceleration due to Gravity (g) |
Universal Gravitational Constant (G) |
1. It is the acceleration acquired by a body due to the earth’s gravitational pull on it. 2. The value of g is different at different places on the surface of the earth. Its va lue varies from one celestial (heavenly) body to another. 3. It is a vector quantity. 4. The value of g at the centre of the earth is zero. |
1. It is equal to the force of attraction between two masses of 1 kg each separated by a distance of 1 m. 2. G is a universal constant, i.e., its value is the same everywhere in the universe $\left(G=6.67\times {10}^{-11}{\mathrm{Nm}}^{2}{\mathrm{kg}}^{-2}\right)$ 3. It is a scalar quantity. 4. The value of G is not zero at the centre of the earth or anywhere else. |
3.11. MASS AND WEIGHT
Mass: The quantity of matter in a body is called its mass. Mass is a measure of the number of atoms contained in any body. Since the number of atoms in any object remains constant, hence its mass remains constant. Mass is expressed in the mass units. The SI unit of mass is kilogram (kg). Mass is usually denoted by m.
The characteristics of mass of a body are as follows:
• Mass is a scalar quantity.
• Mass of a body does not depend on the shape, size and the state of the body.
• Mass of a body is proportional to the quantity of matter contained in it.
• Mass of a body can be measured with the help of a common balance.
Weight: The force with which a body is attracted by the earth is known as the weight of the body. It varies from place to place. The weight of a body on earth is equal to the force of gravity exerted by the earth on that body. We have, F = ma (From Newton’s second law)
W = mg…(i)
where g is the acceleration due to gravity.
The S.I. unit of weight is same as that of the force, i.e., newton (N).
From equation (i) it is clear that weight of a body depends upon the mass of the body and value of acceleration due to gravity g at a place.
The characteristics of weight of the body are as follows:
• Weight is a vector quantity.
• Weight is measured with a spring balance, or on a weighing machine.
• The weight of a body is directly proportional to its mass.
• The weight of a body changes with the value of g. So when g decreases, the weight of the body also decreases.
• The value of g at the poles is higher than that at the equator.
• The value of g on the surfaces of different planets of the solar system is different, therefore, the weight of a body is different on the different planets.
• The value of g decreases with height from the surface of the earth. Therefore, the weight of the body also decreases with height from the surface of the earth. That is why, the weight of a man is less on the peak of Mount Everest than the weight of the man at Delhi.
• The value of g decreases with depth from the surface of the earth. Therefore, the weight of a body decreases with depth from the surface of the earth.
Mass |
Weight |
1. The matter contained in a body is known as mass and is denoted by ‘m’. 2. Units: C.G.S: Gram. S.I: Kilogram. 3. Mass remains constant at any place. 4. Mass is measured by common balance (physical balance) or beam balance. 5. Mass is a scalar quantity. |
1. The gravitational force acting on a body is known as weight and is denoted by ‘W’. W = mg 2. Units: CGS: dynes or gram weight (gm.wt.). 1 gm. wt. = 980dynes. S.I: Newton / Kilogram weight (Kg. wt.) 3. Weight changes from place to place due to variation in acceleration due to gravity. 4. Weight is measured by spring balance. 5. Weight is a vector quantity. |
Weightlessness
The weight of a body is the force with which it is attracted towards the centre of the earth. When a body is stationary with respect to the earth, its weight equals that of gravity. This weight of the body is known as its static or true weight.
We become conscious of our weight only when our weight (which is gravity) is opposed by some other object. Actually, the secret of measuring the weight of a body with a weighing machine lies in the fact that as we place the body on the machine, the weighing machine opposes the weight of the body. The reaction of the weighing machine to the body gives
the measure of the weight of the body.
The state of weightlessness can be observed in the following situations:
1. When objects fall freely under gravity. For example, a lift falling freely, or an airship showing a feat in which it falls freely for a few seconds during its flight, are in a state of weightlessness.
2. When a satellite revolves in its orbit around the earth, such that the surrounding atmosphere does not offer any appreciable frictional force. Weightlessness poses many serious problems to the astronauts. It becomes quite difficult for them to control their movements. Everything in the satellite has to be keep tied down. Creation of artificial gravity is the answer to this problem
3. When bodies are at null points in outer space. On a body projected up, the pull of the earth goes on decreasing, but at the same the gravitational pull of the moon on the body goes on increasing. At one particular position, the two gravitational pulls may be equal and opposite and the net pull on the body becomes zero. This is zero gravity region or the null point and the body in question appears weightless.
3.12 EQUATION OF MOTION UNDER GRAVITY
The force with which the earth pulls the objects towards it is called the gravitational force of earth or gravity of the earth.
Acceleration due to Gravity
The uniform acceleration produced in a freely falling body due to the gravitational pull of the earth is known as acceleration due to gravity and is denoted by ‘g’.
The value of g on the surface of the earth is nearly 9.8 $m/{s}^{2}$.
It is a vector quantity.
Units
S.I. unit of g is $m/{s}^{2}$
C.G.S unit of g is $cm/{s}^{2}$
Under gravity g may be positive or negative. Linear distance will be height. Therefore equations of motion are
$\begin{array}{l}v=u\pm gt\\ h=ut\pm \frac{1}{2}g{t}^{2}\\ {\mathrm{v}}^{2}-{\mathrm{u}}^{2}=\pm 2\mathrm{gh}\text{and}{S}_{nth}=u\pm \frac{g}{2}(2n-1)\end{array}$
Equation of Motion for Freely Falling Body
For a freely falling body, with initial velocity, u = 0, the velocity continuously increases as it falls through a height. The direction of ‘g’ and the direction of motion of the body are same (i.e. downwards). Therefore g is taken as positive under gravity.
Here $\begin{array}{l}g=+ve\\ u=0,\phantom{\rule{1em}{0ex}}s=h\end{array}$
Thus the first equation of motion $\mathrm{v}=\mathrm{u}+\mathrm{at}\Rightarrow \mathrm{v}=0+\mathrm{gt}\Rightarrow \mathrm{v}=\mathrm{gt}$
The ${2}^{nd}$ equation of motion $s=ut+\frac{1}{2}a{t}^{2}\Rightarrow h=0+\frac{1}{2}g{t}^{2}$$\Rightarrow h=\frac{1}{2}g{t}^{2}$
The ${3}^{rd}$ equation of motion ${\mathrm{v}}^{2}-{\mathrm{u}}^{2}=2\mathrm{as}\Rightarrow {\mathrm{v}}^{2}-{0}^{2}=2\mathrm{gh}\Rightarrow {\mathrm{v}}^{2}=2\mathrm{gh}$ $\Rightarrow v=\sqrt{2gh}$
The equation for the distance covered in ${n}^{th}$second ${s}_{nbh}=u+\frac{a}{2}(2n-1)$$\Rightarrow {h}_{nth}=0+\frac{g}{2}(2n-1)\Rightarrow {h}_{nth}=\frac{g}{2}[2n-1]$
Note:
i) When a body is dropped freely from a height, its initial velocity ‘u’ becomes zero.
ii) g does not depend upon the mass of the body.
Equations of Motion for a Body Projected Upwards
Can a body move away from the earth on its own? You know that it is not possible because of the gravitational attraction.
Therefore, if you want a body to move upwards, you have to use force and throw it up with an initial velocity u.
However, the body will not go up indefinitely but will fall back after reaching a certain height. the acceleration due to gravity g acts in a direction opposite to the motion of the body. Therefore, velocity continuously decreases and finally becomes zero.
Therefore, the sign of g is negative for bodies projected vertically upwards.
The Equations of motion get modified as
$\begin{array}{r}\begin{array}{c}\mathrm{v}=\mathrm{u}-\mathrm{gt}\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\left(1\right)\\ \mathrm{h}=\mathrm{ut}-\frac{1}{2}{\mathrm{gt}}^{2}\u2013\u2013\u2013\u2013\u2013\u2013\u2013\u2013\left(2\right)\end{array}\\ {\mathrm{v}}^{2}-{\mathrm{u}}^{2}=-2\mathrm{gh}\u2013\u2013\u2013\u2013\u2013\u2013\u2013\left(3\right)\\ \mathrm{h}=\mathrm{u}-\mathrm{g}\left(n-\frac{1}{2}\right)\u2013\u2013\u2013\u2013\u2013\u2013\left(4\right)\end{array}$
3.13 DERIVATION OF DIFFERENT FORMULAE RELATED TO MOTION UNDER GRAVITY
1. Maximum Height reached by a body Thrown Up:
When a body is projected upwards, at certain height, its velocity becomes zero. This height is called Maximum height reached by the body.
Initial velocity, u = u m/s
Final velocity, v = 0
Acceleration due to gravity = $\u2013gm/{s}^{2}$
Distance travelled = h
Then, using the formula ${\mathrm{v}}^{2}-{\mathrm{u}}^{2}=2\mathrm{gh}$
We get $0-{\mathrm{u}}^{2}=-2\mathrm{gh}\Rightarrow \mathrm{h}=\frac{{\mathrm{u}}^{2}}{2\mathrm{g}}$
which is the maximum height reached by the body if it is projected vertically upwards with a velocity of ‘u’ m/s.
Hence, maximum height attained by the body is directly proportional to the square of the initial velocity u. i.e., $\mathrm{h}\propto {\mathrm{u}}^{2}\Rightarrow \frac{{h}_{1}}{{h}_{2}}=\frac{{u}_{1}^{2}}{{u}_{2}^{2}}$
2. Time of Ascent and Time of Descent
Time of ascent: The time taken by the body to reach its maximum height is called time of ascent. (${t}_{a}$)
Let the initial velocity = u m/s
Acceleration of the body = – g $m/{s}^{2}$
Final velocity of the body = v = 0
The time of ascent = ‘t’ s
Now using formula v = u – gt for the projected body, we can find the time of ascent i.e. time taken to reach the height.
$\Rightarrow 0=\mathrm{u}-\mathrm{gt}\Rightarrow \mathrm{u}=\mathrm{gt}\Rightarrow \mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}$
Time of descent: The time taken by a body to reach ground from certain height is called its time of descent.
Initial velocity, u = 0 m/s
Acceleration due to gravity = g
Final velocity = v m/s
Time of descent = ‘t’ s
Let the height reached by the body be ‘h’ m.
We know that, $h=\frac{1}{2}g{t}^{2}\Rightarrow {t}^{2}=\frac{2h}{g}\Rightarrow t\sqrt{\frac{2h}{g}}\dots ..\left(1\right)$
Now, in order to eliminate ‘h’, which is an unknown quantity, let us use the formula…(2)
${\mathrm{v}}^{2}-{\mathrm{u}}^{2}=2\mathrm{gh}\Rightarrow {\mathrm{v}}^{2}=2\mathrm{gh}\Rightarrow \mathrm{h}=\frac{{\mathrm{v}}^{2}}{2\mathrm{g}}$
Substituting (2) in (1), we get $\mathrm{t}=\sqrt{\frac{2{\mathrm{xv}}^{2}}{2\mathrm{gxg}}}=\frac{\mathrm{v}}{\mathrm{g}}$
But we know that u = v i.e. projected velocity of a body is equal to the velocity of the body on reaching the ground.
$\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}=\text{time of descent}$
$\therefore $Time of ascent = Time of descent
3. Time of Flight
Time of flight is the time for which a body remains in the air and is given by the sum of time of ascent and time of descent.
$\text{Hence,}\phantom{\rule{1em}{0ex}}\mathrm{t}={\mathrm{t}}_{1}+{\mathrm{t}}_{2}\phantom{\rule{1em}{0ex}}\text{or}\mathrm{t}=\frac{\mathrm{u}}{\mathrm{g}}+\frac{\mathrm{u}}{\mathrm{g}}\text{,}$$\text{time of flight},\mathrm{t}=\frac{2\mathrm{u}}{\mathrm{g}}$
4. Velocity on reaching the Ground
When a body is dropped from a height h, its initial velocity is zero. Let the final velocity on reaching the ground be v. For a freely falling body,
${\mathrm{v}}^{2}-{\mathrm{u}}^{2}=2\mathrm{gh}\text{, but}\mathrm{u}=0\phantom{\rule{1em}{0ex}}\text{therefore,}{\mathrm{v}}^{2}-0=2\mathrm{gh}\text{or}\mathrm{v}=\sqrt{2\mathrm{gh}}$ ——(10)
Note:
The upwards velocity at any point in its flight is the same as its downward velocity at that point.
Vertical projection of an object from a tower
Assume that an object is thrown with an initial velocity u from the top of a tower of height h. Let it reach the ground after t seconds after its projection. As shown in the fig if the initial and final positions are represented by the point ‘B’ and ‘A’, the displacement of the object
becomes s = –h. In this case, the distances in the direction of projection are taken as positive and the distances against the direction of projection, as negative. From the equation of motion
$s=ut+\frac{1}{2}a{t}^{2}$
We have $\mathrm{s}=-\mathrm{h};\mathrm{u};\mathrm{t}=\mathrm{t};\mathrm{a}=-\mathrm{g}$
then we get, $-h=ut-\frac{1}{2}g{t}^{2}$
$h=ut\text{\u0964}\frac{1}{2}g{t}^{2}\text{or}h=ut\text{\u0964}\frac{1}{2}g{t}^{2}$
3.14 KEPLER’S LAWS OF PLANETARY MOTION
First Law or Law of orbits
Each planet revolves around the sun in an elliptical orbit with the sun as one of the foci of the ellipse.
Here AB and CD are the major and minor axes of the ellipse. The planet revolves round the Sun in this elliptical orbit. The sun is at the focus ${S}_{1}$ of the ellipse.
Second Law or Law of areas
The aerial velocity of a planet round the sun is constant or the line joining the planet to the Sun sweeps equal areas in equal intervals of time.
The imaginary line joining the sun and the planet is called radius vector which sweeps out equal areas n equal intervals of time.
Let the time taken by the planet to move from A to B be equal to time taken by it to move from C to D. Thus area of SAB is equal to area SCD. When the planet is nearest to the Sun, its speed is the maximum, but when it is farthest from the Sun, its speed is the minimum.
The distance between the planet and Sun when the planet is nearest to the Sun is called the perihelion distance and when the planet is farthest from the Sun, it is called
aphelion distance.
The planet moves fastest when it is near perihelion and slowest when it is near aphelion.
Third Law or Law of periods
The square of the period of revolution of a planet around the Sun is directly proportional to the cube of the average distance between the Sun and the planet.
If R is the average distance of the planet from Sun and T is the period of its revolution, then ${\mathrm{T}}^{2}\propto {\mathrm{R}}^{3}\Rightarrow \frac{{\mathrm{T}}^{2}}{{\mathrm{R}}^{3}}$
$\Rightarrow {\mathrm{T}}^{2}={\mathrm{KR}}^{3}$ where K is constant for the planets.
If the average distance increases, its period also increases. $\Rightarrow {\mathrm{T}}^{2}\propto {\mathrm{R}}^{3}$
Kepler’s III law implies that the period for a planet to orbit the Sun increases rapidly with the increase in R of its orbit. Thus, we find that Mercury, the innermost planet, takes only 88 days to orbit the sun but the outermost planet, Pluto, requires 248 years to do the same.
Summary of Kepler’s laws
i) Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse. This law is also known as the law of elliptical orbits and obviously gives the shape of the orbits of the planets round the sun.
ii) The radius vector, drawn from the sun to a planet, sweeps out equal areas in equal intervals of time, i.e., its areal velocity (or the area swept out by it per unit time) is constant. This is referred to as the law of areas and gives the relationship between the orbital speed of the planet and its distance from the sun.
iii) The square of the planet’s time period is proportional to the cube of the average distance
between the sun and the planet. This is known as the harmonic law and gives the relationship between the size of the orbit of a planet and its time of revolution. Mathematically, ${T}^{2}\propto {R}^{3}$ where T is the time period of revolution of a planet around the
sun and R is the average distance between the sun and the planet.
THURST AND PRESSURE
3.15 INTRODUCTION
Have you ever thought of these:
i. How does a glider move in the air?
ii. Why a sudden outbreak of sand storm blew off the tops of the tents in desert.
iii. Why the paste comes out from the front end when it is pressed from the rear end?
iv. Why a camel moves faster than a horse in the desert?
v. Why a sharp knife can cut easily in contrast to a blunt knife.
vi. How could we empty a cool drink bottle using a straw. The answers for these questions can be answered from this topic.
3.16 THRUST
All bodies have a certain weight and a certain base area. When a body rests on a surface, it exerts a force equal to its weight, normal to its surface.
This force is called thrust.
Thrust is the total force or total weight acting normal to the surface.
Weight of the body is the product of mass and acceleration due to gravity i.e., W = mg (where ‘g’ is acceleration due to gravity)
$\overline{)Thrust\left(T\right)=weightofthebody\phantom{\rule{0ex}{0ex}}T=mg}$
The value of acceleration due to gravity (g) is 9.8 $m{s}^{\u20132}$ in S.I. system and 980 cm s2
in C.G.S. system. This is considered to be constant on the earth’s surface.
Units: Since thrust is defined as the total force or weight, its units are the same as force and weight.
Kilogram Weight (or) Kilogram Force
The weight of one kilogram object is called kilogram weight (or) the force exerted by one kilogram body on its resting surface is called kilogram force. It is represented by kg wt (or) kgf. $\overline{)1kgf=19.8N=9.8N}$
Gram Weight (or) Gram Force
Weight of one gram object is called gram weight (or) the force exerted by one gram object on its resting surface is called gram force. It is represented by g wt (or) gf respectively.
$\overline{)1gf=1980dyne=980dyne}$
3.17 PRESSURE
Let us try to understand the same from the following activity.
Try to fix a nail into the wall (i) with its tip pointing towards the wall (ii) with its head pointing towards the wall with same amount of force.
What do you observe?
In first case, the nail moves easily into the wall and in second case, it is almost impossible to fix the nail.
What is the reason for the above result?
In first case thrust (force) applied is transmitted to the wall over a small area of contact and in second case thrust is transmitted to the wall over a large area of contact.
What can be concluded out of this?
The effect on the wall depends on both thrust and area of contact.
By changing either thrust or area of contact, the effect on the wall can be changed. This effect is called pressure.
$\overline{)Pressureisdefinedastheratioofthethrust\left(F\right)\phantom{\rule{0ex}{0ex}}andareaofcontact\left(A\right)onwhichthethrustacting.}$
$\text{Pressure}\left(P\right)=\frac{\text{Thrust}\left(T\right)}{\text{Area}\left(A\right)}$
$\overline{)\mathbf{Units}\mathbf{:}\phantom{\rule{0ex}{0ex}}\mathrm{C}.\mathrm{G}.\mathrm{S}.\mathrm{units}:\mathrm{dyne}/{\mathrm{cm}}^{2}\text{or}\mathrm{g}.\mathrm{wt}/{\mathrm{cm}}^{2}\phantom{\rule{0ex}{0ex}}\mathrm{S}.\mathrm{I}.\mathrm{units}:\mathrm{N}/{\mathrm{m}}^{2}\text{or}\mathrm{kg}.\mathrm{wt}/{\mathrm{m}}^{2}}$
Note:
i) The S.I. unit $N/{m}^{2}$is also known as pascal. It is denoted by ‘Pa’. The other unit of pressure is ‘bar’.
ii) $1\mathrm{Pa}=1\mathrm{N}/{\mathrm{m}}^{2}\text{and}1\mathrm{bar}={10}^{5}\mathrm{Pa}\text{.}$
iii) Pressure on a body can be exerted from any direction. Hence, it has a no particular direction and is considered as a scalar quantity.
Applications of Pressure
i) Design of Wheels of different vehicles:
Depending on the requirement, wheels of different vehicles are designed accordingly. Two illustrations are cited here under:
Double wheel at rear end
In general, the weight (force) is more at the rear end of the trucks, buses etc.
This results in the development of greater pressure between the tyres and the road.
If the pressure is more, the rate of damage is more and it may lead to accidents.
Decreasing the force (weight) is practically not feasible, but the pressure can be decreased by
increasing the area of contact between the tyres and the road.
This is done by using double tyres in the case of lorries and buses.
Broad chains for tankers
Army tankers are much heavier than the ordinary vehicles.
If they are provided with tyres, they cannot withstand the weight of the tanker and may result in the bursting of tyres.
In addition to this, the tankers have to move through deserts, forests and hill places.
To satisfy all these conditions and make the usage of tankers effective, wheels are designed with broad chains.
These broad chains occupy more area than the tyres, thereby exerting less pressure on the ground. This enables an easy movement.
This is done by using double tyres in the case of lorries and buses.
ii) Easy walking of camels in deserts
Animals like camels walk easily in deserts a s compared to horses or donkeys because they have broad feet. Broad feet exert less pressure on sandy ground.
iii) Usage of flat and long skies to slide: The skiers use flat and long skies to slide on the snow because the larger the area of cross-section, the less is the pressure on the snow.
Hence, a skier can easily slide over snow without sinking his feet in the ground.
It is for the same reason that sledges are never provided with wheels because the wheels increase pressure on snow.
iv) Sharpening of cutting edge of knives, blades and axes:
The cutting edge of knives, blades and axes are sharpened. As the cutting edge is sharpened, the area of contact decreases and hence pressure exerted by it increases. Thus, they can easily penetrate a given surface.
v) Wide foundations for high-rise buildings:
Foundations of high-rise buildings are kept wide so that they exert less pressure on the ground and do not sink in due to the extremely high pressure of the building.
3. 18 FLUID PRESSURE
A substance which can flow from one point to another is called fluid. Since liquids and gases can flow, so they are known as fluids.
A fluid exerts three types of pressure
1. Downward Pressure
2. Upward Pressure
3. Lateral Pressure
1. Downward Pressure of Water: Consider a vessel (V) containing some water of weight (W), say 500 gm. wt. Let the area of its base (A) be 100 $c{m}^{2}$. Then the weight of water (W) is the thrust acting on the base of area 100 cm2$c{m}^{2}$. This shows that liquids exert downward pressure
2. Lateral Pressure of Water: Take a long cylindrical vessel containing water and puch a hole (H) on its wall as shown in the above figure. The water comes out with a speed and falls at distance. This proves that liquid has lateral pressure.
3. Upward Pressure of Water: Take a football and immerse it in water in a vessel (V) and leave it. The ball immediately comes up and floats on water. This shows that water (or fluid) exerts pressure in the upward direction.
A direct-reading instrument used for measuring the density or relative density of a liquid is called hydrometer
(i) Hydrometer is based on the principle of floatation, i.e., the weight of the liquid displaced by the immersed portion of the hydrometer is equal to the weight of the hydrometer.
(ii) Hydrometer is a glass instrument. It consists of a cylindrical bulb having a spherical bulb at its lower end and a long stem (narrow tube) at its upper end. The lower spherical bulb is
partially filled with lead shots. This helps hydrometer to float (or stand) vertical in liquids.
The stem has markings so that relative density of a liquid can be read off directly. The hydrometer is placed in the liquid taken in a glass jar. The reading against the level of liquid gives the relative density of the liquid.
Mathematical Expression for Pressure in Fluids
Consider a beaker filled with a liquid of density d. Now imagine a cylindrical column of the liquid of height h and cross-sectional area A.
The force acting on the base of this imaginary cylindrical column of liquid is equal to the weight of the liquid contained in this column of liquid.
i.e, F = weight of liquid in cylindrical column of liquid F = mg —————- (1)
Now, mass of liquid in the cylinder column of liquid.
m = volume of cylindrical column × density of liquid = Area of cross section × length of cylindrical column × density of liquid
m = A h d
$\therefore $weight of liquid in cylindrical column of liquid = mg = Ah d g
Putting this value in equation (1) we get F = Ah d g
$\therefore $ pressure exerted by liquid at depth ‘h’ $P=\frac{F}{A}=\frac{Ahdg}{A}\Rightarrow P=hdg$
Does Fluid Pressure depend on Size, Shape and Area?
You might be surprised to know that fluid pressure does not depend on size shape and area.
You have learnt earlier that pressure depends on area. This is true only in the case of solids.
But, when it comes to fluids (liquids and gases), the pressure exerted is independent of area. Let us understand it from the following activity.
Consider five containers A, B, C, D and E containing the same liquid of the same height from the bottom.
The shape, size and area of the containers are different from each other.
Do they exert the same (or) different pressure.
Let us check:
We know that the pressure exerted by the liquid depends on the height of the liquid, the density of the liquid and the acceleration due to gravity.
Pressure = height (depth) × density of liquid × acceleration due to gravity
Pressure = hdg
‘h’ is same in all the cases ($\therefore $ fluid level is same), ‘d’ is same in all the cases ($\therefore $ liquid is same) and ‘g’ is constant at a given place.
Thus, fluid pressure is the same in all the cases and it is independent of the shape, size and area of the container.
Liquids seek their own level
The apparatus in the above figure , shows five tubes of various shapes which are connected to each other by a horizontal pipe.
Fix the apparatus and pour water through any one of the vertical tubes.
It is observed that the level of the water is same in all the tubes though they have different shapes and areas of cross section.
Note: Fluids (Liquids and gases) move from a region of high pressure to that of low pressure.
Use: As we know water seeks its own level. When we open the tap, the water in the tap tries to maintain the same level as the water in the tank.
But taps are kept below the level of the tank, so water comes out through the tap. Similarly, large water tanks are constructed at a higher level and supplied to the locality.
3.19 PASCAL’S LAW – TRANSMISSION OF FLUID PRESSURE
This law is also known as ‘the principle of transmission of fluid-pressure’.
This law states that ‘The pressure exerted at any point in an enclosed and incompressible liquid is transmitted equally in all 4 directions’.
Example: Take a round bottom flask with a number of narrow openings on its sides, provided with an air-tight piston in its mouth. Fill it to the brim with water. Now push the piston down. Water will come out in jets with equal force from all the openings. This shows that the pressure exerted at one point is transmitted equally in all directions.
Explanation: Consider a vessel full of water and filled with air tight piston in different positions as shown in the diagram.
Let the piston at A be pushed down with a force ${F}_{1}$. Pressure P on the piston is $\mathrm{P}=\frac{{\mathrm{F}}_{1}}{{\mathrm{a}}_{1}}$.
Where ‘${a}_{1}$’ is the area of cross-section of piston at A. It will be observed that to hold pistons at B, C, D and E we have to apply forces ${\mathrm{F}}_{2},{\mathrm{F}}_{3},{\mathrm{F}}_{4}\text{and}{\mathrm{F}}_{5}$ on them such that $\frac{{\mathrm{F}}_{1}}{{\mathrm{a}}_{1}}=\frac{{\mathrm{F}}_{2}}{{\mathrm{a}}_{2}}=\frac{{\mathrm{F}}_{3}}{{\mathrm{a}}_{3}}=\frac{{\mathrm{F}}_{4}}{{\mathrm{a}}_{4}}=\frac{{\mathrm{F}}_{5}}{{\mathrm{a}}_{5}}$
where ${a}_{2},{a}_{3},{a}_{4}\text{and}{a}_{5}$ are the areas of cross-section of pistons at B, C, D and E respectively. This indicates that pressure is transmitted equally in all directions as stated by Pascal’s law.
Application of Pascal’s Law
Pascal’s law is used to multiply force in machines such as the hydraulic press or Bramah press, hydraulic lift and hydraulic brake.
Principle of a Hydraulic Machine
( Hydraulic Press or Bramah Press)
Hydraulic press works on the principle of Pascal’s law.
Construction: It consists of two cylindrical vessels X and Y of different cross sections. Let the cross sectional area of X be $\u2018{A}_{1}\u2018$ and that Y be $\u2018{A}_{2}\u2018$
The vessels contain a liquid. The vessel X is provided with water tight piston M. The vessel X has smaller diameter than vessel Y. The vessel Y is provided with watertight piston N. r and R are the radii of vessel X and vessel Y respectively. Let the area of cross-section of vessel X with a smaller diameter be ${A}_{1}$ and that of vessel Y be ${A}_{2}$. So ${A}_{2}>{A}_{1}$
A weight is placed on piston M of the vessel X. It exerts a force ${F}_{1}$ on the piston M. Therefore, the pressure applied on piston M in the downward direction $\frac{{\mathrm{F}}_{1}}{{\mathrm{A}}_{1}}$.
According to Pascal’s law, the pressure exerted on the piston M is transmitted by the liquid to the piston N.
$\therefore $ The upward pressure exerted on the piston N is $\frac{{\mathrm{F}}_{1}}{{\mathrm{A}}_{1}}$.
Hence, the upward force (${F}_{2}$) exerted on piston N is ${F}_{2}$ = Pressure on the piston N × Area [Since force = pressure × area ]
${\mathrm{F}}_{2}=\frac{{\mathrm{F}}_{1}}{{\mathrm{A}}_{1}}\times {\mathrm{A}}_{2}\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}\frac{{\mathrm{F}}_{2}}{{\mathrm{F}}_{1}}=\frac{{\mathrm{A}}_{2}}{{\mathrm{A}}_{1}}$
$\text{Since}{\mathrm{A}}_{2}{\mathrm{A}}_{1}\text{therefore}{\mathrm{F}}_{2}{\mathrm{F}}_{1}$
Thus a small force ${F}_{1}$ applied on the smaller piston M can be used to exert a large force ${F}_{2}$
on the bigger piston N.
This is the principle of a hydraulic machine.
Note: Applying the principle of hydraulic machines, we get:
$\text{Mechanical advantage}=\mathrm{MA}=\frac{\text{Load}}{\text{Effort}}$$=\frac{\text{Area of larger piston}}{\text{Area of smaller piston}}=\frac{\pi {\mathrm{R}}^{2}}{\pi {\mathrm{r}}^{2}}=\frac{{\mathrm{R}}^{2}}{{\mathrm{r}}^{2}}$
$\mathrm{MA}=\frac{(\text{Radius of larger piston}{)}^{2}}{(\text{Radius of smaller piston}{)}^{2}}$
Note:
i) Larger piston is generally referred as press plunger
ii) Smaller piston is referred as pump plunger
Uses of Hydraulic Press:
1. It is used for compressing the cotton bales and straw.
2. It is used for extracting oil from oil seeds.
3. It is used for punching holes in metals.
4. It is used for giving specific shapes to metal sheets.
5. It is used for servicing automobiles in service stations.
3.20 ATMOSPHERIC PRESSURE
The atmosphere is a mixture of gases. Above the ground, the Earth’s atmosphere extends to a few hundred kilometers in height.
Very close to the Earth’s surface, we find millions of gas molecules in every cubic centimeter.
It is calculated that roughly around molecules per cubic centimeter are present.
These molecules are all zipping around in different directions, colliding into each other and rebounding.
They make an impact on each and every object by bombarding their surface resulting in pressure.
The weight of the atmosphere also exerts force resulting in pressure. This pressure is known as Atmospheric pressure.
The weight of thrust of air exerted on unit area of earth’s surface is called Atmospheric pressure.
It is to be noted that the atmospheric pressure decrease as the altitude increases.
The reason being that the number of molecules present at a higher altitude is less than the number of molecules present at the sea level.
So, as the number of air molecules decreases with increases in altitude, the air pressure also decrease.
Measurement of Atmospheric Pressure
Atmospheric pressure is measured by using a barometer. It was invented by Torricelli in 1643. “The Mercury Barometer”
Measurement of Atmospheric Pressure using Barometer
Atmospheric pressure (P) can be obtained by a barometer, using the formula.
$\mathrm{P}=\mathrm{h}\times \mathrm{d}\times \mathrm{g}$
Where, ‘h’ is the height of liquid column, ‘d’ is the density of the barometeric liquid and ‘g’ is the acceleration due to gravity.
If ‘h’ is expressed in meters, ‘d’ in $kg/{m}^{3}$ and ‘g’ in $m/{s}^{2}$ , the pressure obtained by this formula is in terms of ‘Newton/${m}^{2}$ or ‘dyne/$c{m}^{2}$ .
Pressure measured in this way is called Absolute pressure. In the formula ‘P = hdg’, the terms ‘d’ and ‘g’ for a given barometer are constant and the atmospheric pressure is proportional to the height of the mercury column.
1 atm is the pressure exerted by a vertical column of mercury of 76 cm (or 760 mm) height.
$\therefore 1\mathrm{atm}=76\mathrm{cm}\times 13.6\mathrm{g}{\mathrm{cm}}^{-3}\times 9.8{\mathrm{ms}}^{-2}$(using hdg for pressure exerted by a liquid)
$\begin{array}{l}=0.76\mathrm{m}\times 13.6\times {10}^{3}{\mathrm{kgm}}^{-3}\times 9.8{\mathrm{ms}}^{-2}\\ =101292.8{\mathrm{Nm}}^{-2}=1.013\times {10}^{5}\mathrm{Pa}\end{array}$
Other units usually used in measurement of pressure are torr and bar
$\text{1torr}=1\mathrm{mm}\text{of}\mathrm{Hg},1\mathrm{bar}={10}^{5}\mathrm{Pa}$
3.21 BUOYANCE
A four – year child was questioned. ‘How do you wash your hands’?
He promptly answered, ‘With the water form the tap’.
A twelve – year boy was questioned: How do you get water in your tap?
He answered, ‘We have a storage tank the upstairs, which gets filled up when we switch on the motor’.
Okay! This is how, in this modern age, we get water effortlessly. But in most of the villages the situation is entirely different.
Even agricultural fields are watered by manual drawing, from wells, leave alone the domestic purpose.
For their daily requirements, village people draw water from wells with the help of a bucket tied to a rope.
When the bucket is released into the well, it gets filled up with water and then sinks to a certain depth in water.
Now, try to pull out the rope and observe.
While drawing water from a depth to the surface, we feel that the bucket of water is very light or almost weightless, but when we draw the water from the surface to the required height, we feel some heaviness or weight of water.
From this act, it is clear that an object can be raised or lifted easily inside water.
Have you ever thought, what makes this raising or lifting easy?
This is due to the invisible force that is exerted by the liquid on the body in an upward direction.
This upward force is called Buoyant force.
From the above activity, it is very clear that some invisible force acts in upward direction when the body is immersed in water.
This force makes the weight of the body (bucket) less than its actual weight.
The loss in weight is caused due to the upward force acting on the body. It is called buoyant force and the property is called buoyancy.
The upward force which is acting on a body when it is completely or partially immersed in a fluid is called up thrust or the buoyant force, and the phenomenon is known as buoyancy.
Examples of Buoyant Force
1. When we lift a stone in air, we feel it heavier but inside the liquid we feel it to be lighter because water exerts and invisible force (up thrust) on the body, which decreases the weight of the body.
That is why we feel it lighter inside the water than in the air.
2. When a swimmer jumps into water from a height, he finds himself being pushed towards the surface of the water without any effort on his part.
3. A stone which is dropped, falls down with an acceleration of 9.8 $m{s}^{\u20132}$ in the air. The same stone (density = 5 g $c{m}^{\u20133}$ ) moves down in water with an initial acceleration of 7.8 $m{s}^{\u20132}$ due to upward buoyant force.
Cause of Buoyant Force
We know that a liquid contained in a vessel exerts pressure at all points and in all directions.
According to Pascal’s law, the direction of pressure is normal (perpendicular) to the surface of the body.
The pressure at a given point depends on the depth (h) of the point inside the liquid. (P = hdg) Now, consider a body ABCD inside the liquid, as shown in the figure.
i) Side – wise Pressure:
Consider two points ‘x’ and ‘y’ on the sides AD and BC respectively at the same depth as shown in the figure.
As the depth of these two points is same from the surface, equal thrust acts on them in an opposite direction.
Since the forces are equal in magnitude and acting in the opposite direction they get neutralized.
So, the resultant sidewise pressure is zero on the body. If any resultant force acting on the body is the resultant of the upward and downward pressure.
ii) Upward and Downward Pressure:
The downward pressure acts on the upper surface of the body and an upward pressure acts on its lower surface.
The lower face is at a greater depth than the upper face.
Hence, the upward pressure on the lower face is more than the downward pressure on the upper face.
So, the resultant pressure is in the upward direction. Therefore, the difference in the upward and downward pressure results in a buoyant force acting upwards.
Result of Buoyant Force
We know that the feeling of easiness to lift an object in water is due to buoyant force.
This is because, the buoyant force neutralizes some downward force acting on the body.
That is, there is a loss in the weight (downward force) of a body due to buoyancy.
3.22 ARCHIMEDE’S PRINCIPLE
Two thousand years ago there lived a Greek philosopher, Archimedes’. He was a scientist in the court of a king. The king had a doubt about the purity of gold with which his crown was made.
So he asked Archimedes to invent a method of testing the purity of gold without spoiling the crown. One day, Archimedes noticed that when he had entered a fully filled bath tub, some water overflowed. Further more, he noticed that he felt considerably lighter in bath the tub. He reasoned that the loss in weight of his body in water must have some connection with the
amount of water displaced by his body.
He performed a simple experiment and came out with the relation between the apparent loss of weight and the weight of water displaced. This relation is called Archimedes’ principle.
Let us understand it form the following
Take a spring balance and suspend 200 g stone to it.
The weight of the stone in air is found to be 200 gf.
Now, take an empty jar and fill it with water, till the water is one the verge of overflowing.
Gently lower the stone, suspended from spring balance, into water, in the jar, till the whole solid is completely immersed. It is observed that the spring balance records
less value of weight equal to 150 gf.
This loss of weight is due to buoyant force.
The metal block experiences buoyant force, due to which it loses its weight.
The loss of weight = weight in air – weight in water.
$={\mathrm{W}}_{\text{air}}-{\mathrm{W}}_{\text{liquid}}=200\mathrm{gf}-150\mathrm{gf}=50\mathrm{gf}$
The liquid displaced by the metal block is collected in a beaker and its weight is obtained. Let it be W gf.
It is observed that the weight of the liquid displaced is equal to the loss of weight of the object.
Weight of liquid displaced (W) = loss of weight $\left({\mathrm{W}}_{\text{air}}-{\mathrm{W}}_{\text{liquid}}\right)$
We also know,
Loss of weight = Buoyant force ________ (2)
$\therefore \left(1\right)=\left(2\right)$
Hence, we can say that buoyant force is equal to the weight of the liquid displaced.
Thus, we can conclude that
“Whenever a body is partly or wholly immersed in a fluid, it experiences an upward buoyant force which is equal to weight of the liquid displaced by the body”.
This is known as Archimedes’ principle, which, in short, is
Buoyant force = weight of liquid displaced
Calculation of Buoyant Force
We know that buoyant force is equal to the weight of the liquid displaced.
Let the weight of the liquid displaced be ‘mg’.
Buoyant force (mg) = V × d × g
$(\because \text{mass}=\text{volume}\times \text{density})$
${\mathrm{V}}_{\text{liquid displaced}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}$ ——(1)
If the body is inside the liquid, then the volume of the liquid displaced is equal to the volume of the body, Now the equation changes as
$\text{Buoyant force}={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}$ —–(2)
$\left({\mathrm{V}}_{\text{liquid displaced}}={\mathrm{V}}_{\text{liquid}}\right)$
i) Buoyant Force and Volume of Body:
Dip an iron block and a thermocol sheet of the same mass into the water.
in which case do you feel the Buoyant force more?
We know, buoyant force $={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}$
As they are dipped in the same liquid and at the same place, ${d}_{liquid}$ and g are constants.
$\therefore \text{Buoyant force}\propto {\mathrm{V}}_{\text{body}}$
Let us check their volumes.
$\text{Volume of a body}\left(\mathrm{V}\right)=\frac{\mathrm{mass}\left(\mathrm{m}\right)}{\mathrm{density}\left(\mathrm{d}\right)}$
As mass is same for both the bodies, $\mathrm{V}\propto \frac{1}{\mathrm{d}}$
As the density of thermocol is less than that of iron, volume of thermocoal is more and therefore, it displaces more volume of water.
Therefore, buoyant force is more on the thermocol sheet.
Conclusion: If two bodies of the same mass are immersed in water, then the body with greater volume or less density experiences more buoyant force.
In other words, the loss of weight is more for the body with greater volume.
ii) Buoyant Force and Density of Liquid
Take two beakers, one filled with water and the other with oil. Try to observe the loss of weight by immersing the body completely in water and oil separately.
In which case do you feel more loss of weight?
The loss of weight can be found out, if buoyant force is known.
We know, buoyant force $={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}$
As the same body is immersed at a given place, ${V}_{body}$ and g are constant.
$\therefore \text{Buoyant force}\mu {\mathrm{d}}_{\text{liquid}}$
Let us check the densities of water and oil.
As the density of oil is less than the density of water, buoyant force is more for water.
Conclusion: If a body is immersed in water and oil, as the liquid with greater density produces more buoyant force, here, water with greater density produces more buoyant force.
iii) Buoyant Force and Gravity
Take a body of volume ‘V”, and immerse it in a liquid of density ${d}_{liquid}$ on the surface of the earth and on the surface of the moon respectively.
In which case do you feel the buoyant force more?
We know, buoyant force $={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}$
As the body and liquid are the same on both the surfaces, ${\mathrm{V}}_{\text{body}}\text{and}{\mathrm{d}}_{\text{liquid}}$ are constant
$\therefore \text{Buoyant force}\propto \mathrm{g}$
Acceleration due to gravity on the earth is six times greater than the acceleration due to gravity on the moon. So, buoyant force is more on the earth than on the moon.
Conclusion: A body in a given liquid experiences greater buoyant force at a place with greater value of ‘g’.
Planets on which a body experiences highest and least Buoyant Force
Of all the planets, the value of ‘g’ is least for Pluto and the planet with greater value of ‘g’ is Jupiter. So, buoyant force for a body in a liquid is the least on pluto and greater on Jupiter.
The value of ‘g’ on Saturn is almost equal to that on earth.
Hence, buoyant force experienced by a body in a given liquid on Saturn is the same as that on the earth.
3.23 FLOATATION
Consider the following cases:
Case – 1: Stone in water
A stone in water is acted upon by two forces as shown.
In the case of a stone, the downward weight is greater than the upward buoyant force and hence, the stone sinks down.
Case – 2: Air balloon in water
An air balloon in water is acted upon by two forces as shown.
In the case of a hydrogen balloon, the upward buoyant force is more than the downward weight and hence, the balloon moves up.
Case – 3: Wood in Water
What happens if the weight of the body is equal to the buoyant force?
If the weight of the body is equal to the buoyant force, the body neither sinks like the stone nor moves up like the balloon. But it floats.
So, floating is an equilibrium state, where the buoyant force acting upwards on a body is equal to its weight acting downwards.
Conditions
1. Sinking of a Body
A body sinks if Weight of body > Buoyant force
$\begin{array}{l}\Rightarrow {\mathrm{m}}_{\text{body}}\times \mathrm{g}{\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\mathrm{g}.\\ \Rightarrow {\mathrm{m}}_{\text{body}}{\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\cdot \\ \Rightarrow {\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{body}}{\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\\ \Rightarrow {\mathrm{d}}_{\text{body}}{\mathrm{d}}_{\text{liquid}}\text{.}\end{array}$
A body sinks if its density is more than that of the liquid.
2. Flotation of a Body
A body floats if Weight of body = Buoyant force.
$\begin{array}{l}\Rightarrow {\mathrm{m}}_{\text{body}}\times \mathrm{g}={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}.\\ \Rightarrow {\mathrm{m}}_{\text{body}}={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\cdot \\ \Rightarrow {\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{body}}={\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\\ \Rightarrow {\mathrm{d}}_{\text{body}}={\mathrm{d}}_{\text{liquid}}\end{array}$
A body floats if its density body is equal to that of the liquid.
3. Upward Motion of a Body
A body will move up in a fluid if Weight of body < Buoyant force.
$\begin{array}{l}\Rightarrow {\mathrm{m}}_{\text{body}}\times \mathrm{g}{\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\times \mathrm{g}.\\ \Rightarrow {\mathrm{m}}_{\text{body}}{\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\cdot \\ \Rightarrow {\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{body}}{\mathrm{V}}_{\text{body}}\times {\mathrm{d}}_{\text{liquid}}\\ \Rightarrow {\mathrm{d}}_{\text{body}}{\mathrm{d}}_{\text{liquid}}\end{array}$
A body floats over the surface in a fluid if density of the body is less than that of the liquid.
3.24. RELATIVE DENSITY
Density
Density of a substance is defined as its mass per unit volume,$\text{i.e., Density}=\frac{\text{mass}}{\text{volume}}$.
The SI unit of density is $kg{m}^{\u20133}$. Its CGS unit is $gc{m}^{\u20133}$.
Let us consider mercury of density 13.6 $gc{m}^{\u20133}$, alcohol of density 0.8 $gc{m}^{\u20133}$ and aluminium of density 2.7 $gc{m}^{\u20133}$.
If you compare the densities of mercury and aluminium, aluminium has less density than mercury, whereas if you compare aluminium and alcohol, aluminium is denser than alcohol.
In this way, any substance can be heavier or lighter when compared to the other substance.
To avoid this confusion, another term is devised, which is called relative density.
In relative density, the density of a substance is compared to the density of water at 4 °C.
The relative density of a substance is the ratio of the density of the substance to the density of water at 4°C.
Relative density of a substance $=\frac{\text{Density of the substance}}{\text{Density of water at}{4}^{\circ}\mathrm{C}}$ —–(1)
Note:
i) Relative density is also known as specific gravity.
ii) Relative density is the ratio of densities. So, it has no unit, but is a pure number.
Significance of Relative Density
Relative density of a substance tells us how many times a substance is denser than water.
Example: If the relative density of a substance is ‘6’, it means that the substance is 6 times denser than water.
Relative Density in terms of Weight of Body and Weight of Water displaced by the Body
We know, Relative density $=\frac{\text{Density of substance}}{\text{Density of water}}$
Relative Density = $\frac{\frac{{\mathrm{m}}_{\text{substance}}}{{\mathrm{V}}_{\text{substance}}}}{\frac{{\mathrm{m}}_{\text{water}}}{{\mathrm{V}}_{\text{water}}}}$
If the volume of a substance is equal to the volume of the water, then,
Relative density $=\frac{\text{Mass of any volume of substance}}{\text{Mass of water of an equal volume}}$
(When both the volumes are same)
Multiplying the numerator and the denominator by ‘g’, then, we get,
Relative density $=\frac{\text{Mass of any volume of substance}\times \mathrm{g}}{\text{Mass of water of an equal volume}\times \mathrm{g}}$
$\therefore $Relative density of a substance (or body) $=\frac{\text{weight of any volume body in air}}{\text{weight of equal volume of water}}$ —–(2)
We know that the weight of equal volume of water is equal to the loss of weight of the body in water.
Loss of weight in water = Weight in air – Weight in water = Weight of water displaced _________ (3)
Using (2) and (3), we can write
Relative density of a body $=\frac{\text{weight of the body}}{\text{loss of weight of the body in water}}$ ——-(4)
Relative density of a body $=\frac{\text{weight of the body in air}}{\text{weight of the body in air}-\text{weight of the body in water}}$ —–(5)
Determination of Specific Gravity or Relative Density of Liquids
Take a body which sinks in both the liquid and the water. Determine its weight in air, let it be ${W}_{1}$ .
Now, determine its weight in the given water and liquid
with a spring balance. Let it be ${W}_{2}and{W}_{3}$ respectively.
The body displaces the liquid equal to its volume (V)
Loss of weight of the body in water = Weight of the body in air – Weight of the body in water $={\mathrm{W}}_{1}-{\mathrm{W}}_{2}$
Loss of weight of the body in the given liquid = weight of the body in air – weight of the body in liquid $={\mathrm{W}}_{1}-{\mathrm{W}}_{3}$
$\text{Relative density of liquid}=\frac{{\text{weight of}}^{\mathrm{\prime}}{\mathrm{V}}^{\mathrm{\prime}}\text{volume of liquid}}{{\text{weight of}}^{\mathrm{\prime}}{\mathrm{V}}^{\mathrm{\prime}}\text{volume of water}}$
But, we know that the weight of V volume of liquid = loss of weight of the body in liquid.
Similarly, weight of ‘V’ volume of water = loss of weight of the body in water.
$\text{Relative density}=\frac{\text{Loss of weight of the body in liquid}}{\text{Loss of weight of the body in water}}$
$\text{Relative density}=\frac{{\mathrm{W}}_{1}-{\mathrm{W}}_{3}}{{\mathrm{W}}_{1}-{\mathrm{W}}_{2}}$
By substituting the measured values of ${W}_{1},{W}_{2}$ , the relative density of the liquid can be determined.