Before studying Newton’s laws of motion, we shall understand a very important concept of reference frames in this article.

Motion is a Relative Concept:

Motion is always with respect to the observer. Consider a person A is standing in a train moving in the north direction with a uniform velocity of 5 m/s in a straight line. Let us consider another person B standing on the ground.

Now B as an observer would say that he himself is stationary and the person A is moving in the north direction with a uniform velocity of 5 m/s. Now A as an observer would say that he himself is stationary and the person B is moving in the south direction with a uniform velocity of 5 m/s.

Thus there is a difference in the interpretation of motion of each other it is because their states of motion are different. It is to be noted that both are in the state of uniform motion with respect to earth. Thus we can say that the motion is a relative concept and not the absolute one.

Reference Frame:

It is a system of co-ordinate axes with reference to which position or motion of an object in space is described. The rectangular Cartesian coordinate system of axes is considered as the simplest reference system.

If (x, y, z) are the coordinates of the point in this system of reference, then the position vector of the particle w.r.t. the origin is given by

Let us consider two persons A and B traveling by train with uniform velocity 5 m/s towards right as shown.

There is a third observer C standing on the ground near the track. observer A and B are moving with uniform velocity w.r.t. earth similarly observer C is at rest w.r.t. earth. Hence A, B, C are the inertial frame of reference. We shall ask an opinion of motion of each observer. A will say that he is stationary, B is stationary and C is moving backward with uniform velocity 5 m/s and no force is acting on him. The opinion of B will be similar. But C will say he is stationary and B and C are moving towards the right with uniform velocity 5 m/s. No force acts on them. Then who is correct. Everybody is correct because the motion is always with respect to the state of the observer. Thus Newton’s laws are not violated.

Concept of Inertial Reference Frame:

The reference frame which is at rest or moving with uniform velocity with respect to earth is called inertial reference frame.

Characteristics of Inertial Reference System:

• The reference frames which are at rest or moving with uniform motion with respect to earth are called as inertial reference frames.
• Newton’s laws of motion are applicable in inertial reference frames
• Concept of pseudo force is not required in inertial reference frames.
• A person sitting in a car moving with uniform motion in a straight line is the inertial frame of reference.

Concept of Non-Inertial Reference Frame:

The reference frame which is moving with uniform acceleration w.r.t. earth is called non-inertial reference frame

Characteristics of Non-inertial Reference System:

• The reference frames which are moving with acceleration with respect to earth are called non-inertial reference frames.
• Newton’s laws of motion are not applicable in non-inertial reference frames.
• To apply Newton’s laws of motion in non-inertial reference frames we have to consider the existence of an imaginary force called pseudo force.
• A person sitting in a car negotiating a curve is non-inertial frame of reference.

Distinguish between inertial reference frame and non inertial reference frame.

Concept of Pseudo Force:

Pseudo force is imaginary or hypothetical force whose existence is to be granted in non inertial reference frames so that Newton’s laws of motion are applicable in non inertial reference frame.

Let us consider two persons A and B traveling by train with uniform acceleration 5 ms^-2 towards the right as shown. There is a third observer C standing on the ground near the track. observer A and B are moving with uniform acceleration w.r.t. earth while observer C is at rest w.r.t. earth. Hence A and B are the non-inertial frame of reference, while C is still inertial frame of reference.

We shall ask the opinion of motion of each observer. C will say he is stationary and B and C are moving towards the right with uniform acceleration 5 m/s². Necessary force for acceleration is provided by the engine of the train. Thus he is not violating Newton’s laws of motion. 

A or B will say that he is stationary and C is moving backward with uniform acceleration 5 m/s² and no force is acting on C. It is a contradiction to Newton’s laws of motion because he is saying acceleration is there but no force present to cause it. Thus Newton’s laws of motion are violated in the non inertial reference frame.

To correct A and B the observer in non-inertial reference frame, imaginary or hypothetical force acts on C. This force is called pseudo force.

Examples of Pseudo Forces:

An object revolving in a circle experiences a centrifugal force directed radially outward. This force tends to move the object away from the centre. This centrifugal force is a pseudo force because the circular motion accelerated motion. Thus the object is in acceleration w.r.t. earth. The reference frames which are moving with acceleration with respect to earth are called non-inertial reference frames.

When bus negotiates a curve on the right side, the passengers are pushed on the left side and if the bus negotiates a curve on left side passengers are pushed on the right side. As the bus is negotiating a curve the passengers are in accelerated motion. Thus they constitute the non-inertial reference frame.

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In this article, we shall study to solve numerical problems to calculate the magnitude of force, momentum, and change in momentum.

Example – 01:

The speed of a tortoise and hare are 1 m/s and 3 m/s respectively. The mass of the hare is 5 kg while that of tortoise is 20 kg. Which of the two has greater momentum?

Solution:

The momentum of hare = Mass of hare x Speed of hare= 5 x 3 = 15 kg m/s

The momentum of tortoise = Mass of tortoise x Speed of tortoise= 20 x 1 = 20 kg m/s

Thus the momentum of tortoise is more than that of the hare.

Example – 02:

What is the magnitude of force exerted by a horse in pulling a cart of mass 600 kg and accelerating at the rate of 1.2 m/s²?

Given: mass of cart = m = 600 kg, acceleration = a = 1.2 m/s².

To Find: F =?

Solution:

By Newton’s second law of motion, magnitude of force

F = ma = 600  x 1.2

∴ F = 720 N

Ans: The required force is 720 N.

Example – 03:

An object of mass 10 kg is moving with the initial velocity of 10 m/s. A constant force acts on it for 4 s giving it a speed of 2 m/s in opposite direction. Find the acceleration and the force.

Given: mass of object = m = 10 kg, initial velocity = u = 10 m/s, Final velocity = v = – 2m/s (negative since it is in opposite direction), time for which force is acting = t = 4 s

To Find: acceleration = a =? Force acting = F =?

Solution:

a = (v – u)/t = (-2 – 10)/4 = -12/4 = -3 m/s²

The negative sign indicated retardation.

By newton’s second law of motion we have

F = ma = 10 x (-3) = -30 N

Negative sign indicated retarding force.

Ans: The acceleration = 3 m/s² and Force = -30 N

Example – 04:

A constant force of 2 N acts on a body for 5 seconds to change its velocity. Calculate the change in the momentum.

Given: Force acting = F = 2 N, time for which the force is acting = t = 5s.

To Find:  Change in momentum =?

Solution:

Change in momentum = F x t = 2  x 5  = 10 kg m/s.

Ans: the change in momentum is 10 kg m/s.

Example – 05:

An empty truck of mass 1000 kg is moving at a speed of 36 km/hr. It is loaded with 500 kg of material on its way and again moves at the same speed. Will the momentum of the truck remain the same after loading? if not, find the momentum of the truck after loading.

Given: Mass of a truck = 1000 kg, Mass of Load = 500 kg, Speed of the vehicle = v = 36 km/hr = 36 x 5/18 = 10 m/s.

To Find: Momentum of truck = p =?

Solution:

The momentum of a body depends on its mass. In this case, the truck is loaded on the way, hence its momentum should change.

Total Mass = m = 1000 + 500 = 1500 kg

New momentum = Total Mass x Velocity = 1500 X 10 = 15000 kg m/s

Ans: The momentum of the truck after loading is 15000 kg m/s.

Example – 06:

A railway wagon of mass 1000 kg is pulled with a force of 10000 N. What is the acceleration?

Given : Force applied = F = 10000 N, mass of wagon = m = 1000 kg.

To Find:  acceleration = a =?,

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴ 10000 = 1000 x a

∴ a = 10000/1000 = 10 m/s².

Ans: The acceleration is 10 m/s².

Example – 07:

A car of mass 1000 kg is moving at a certain speed when a constant braking force 1000 N acts on it for 5 s and speed of the car reduced to half the original speed. Find the further time required to stop the car, if the same constant force acts on it.

Given: mass of car = m = 1000 kg, Force acting = F = 1000 N,time taken =  t = 5 s, Final speed (v) = 1/2 Initial speed (u) = u/2

To find: t =? when v = 0

Solution:

By Newton’s second law of motion, magnitude of force

F = ma

∴1000 = 1000 x a

∴ a = 1000/1000 = 1 m/s².

By the first equation of motion

∴ a = (v – u)/t  = (u/2 – u)/5 = (-u/2)/5 = (-u/10)

∴ a = -u/10 = – 1

∴ u = 10 m/s

We have v = u + at

∴ 0 = 10 + (-1)t

∴ -10 = – t

∴ t = 10 s

Ans: The further time required to stop the car is 10 s

Example – 08:

Find the magnitude of the force applied to a block of mass 5 kg at rest, if it moves 36 m, in first 3 seconds. Neglect the force of friction.

Given: mass of block = m = 5 kg, Initial velocity = u = 0, Distance traveled = s = 36 m, time taken = t = 3 s,

To find: F =?

Solution:

s = ut + 1/2 at²

∴   36 = (0)(3) + 1/2 a(3)²

∴   36 = 1/2 a(9)

∴   72 = a (9)

∴ a = 72/9 = 8 m/s².

By Newton’s second law of motion, the magnitude of force

F = ma = 5 x 8 = 40 N

Ans: The force applied = 40 N

Example – 09:

Two spheres of mass 10 g and 100 g each fall on the two pans of a table balance from a height of 40 cm and 10 cm respectively. If both are brought to rest in 0.1 seconds. Determine the force exerted by each sphere on the pans.

Solution:

For the first sphere: 

Given : m₁ = 10 g = 0.01 kg, h = – 40 cm  = – 0.4 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴  v² = (0)² + 2(-9.8)(-0.4)

∴  v² =  7.84

∴  v = 2.8 m/s

F₁ = m₁a

F₁ = m₁ (v – u)/t  = 0.01 x (2.8 – 0)/0.1 = 0.28 N

For the second sphere: 

Given : m₁ = 100 g = 0.1 kg, h = – 10 cm  = – 0.1 m,  t = 0.1 s, g = – 9.8 m/s²

v² = u² + 2gh

∴ v² = (0)² + 2(-9.8)(-0.1)

∴ v² =  1.96

∴ v = 1.4 m/s

F₂ = m₂a

F₂ = m₂ (v – u)/t  = 0.1 x (1.4 – 0)/0.1 = 1.4 N

Ans: The force exerted by the first sphere is 0.28 N and that by second sphere is 1.4 N

Example – 10:

Calculate the density of a cubical ice block of side 50 cm. If a force of 1125 N applied to it produces an acceleration of 10 m/s² in it. Neglect the force of friction. Assume the ice block remains in solid state without melting.

Given: Force applied = F = 1125 N, acceleration = a = 10 ms-1, Side of a block = 50 cm = 0.5 m

To Find:  Density = ρ = ?,

Solution:

By Newton’s second law of motion

F = m.a

∴ 1125 = m x 10

∴ m = 1125/10 = 112.5 kg

Ans: The density of ice block is 900 kg/m³.

Example – 11:

A ball of mass 50 g at rest is hit by a bat and the ball covers a distance of 400 m, in 2 seconds. If the ball was in contact with it for 0.1 s, find the magnitude of the force acting on it. Assuming no other force acts on a ball after it is hit by the bat.

Given: mass of ball = m = 50 g = 0.05 kg,  time of contact = 0.1 s, distance covered  s= 400 m, time taken to cover the distance = t = 2 s.

To find: Force acting =F =?

Solution:

No other force acts on a ball after it is hit by the bat. Thus it is in uniform motion after hit.

v = s/t = 400/2 = 200 m/s

Now Force, F = ma

F = m (v – u)/t  = 0.05 x (200 – 0)/0.1 = 100 N

Ans: The force acting = 100 N

Example – 12:

A force of 500 N acts on a body of mass 1000 kg and the body is brought to rest within a distance of 64 m. Find the initial velocity and time taken by the body to come to rest.

Given: mass of body = m = 1000 kg, Force acting = F = 500 N, Final velocity = v = 0 ms-1, distance traveled = s = 64m.

To find: initial velocity =u =? time taken = t = ?

Solution:

We have     F = ma

∴   500 = 1000 x a

∴ a = 500/1000 = 0.5 m/s²

As the body is brought to rest a = – 0.5 m/s²

v² = u² + 2as

∴ (0)² = u² + 2(-0.5)(64)

∴ (0)² = u² – (64)

∴ u² = 64

∴ u = 8 m/s

By first equation of motion

v = u + at

∴ 0 = 8 + (- 0.5) x t

∴ 8 = – 0.5 x t

∴ t = 8/0.5 = 16 s

Ans: Initial velocity = 8 m/s, time taken to come to rest = 16 s

Example – 13:

A car of mass 1000 kg is moving uniformly with 10 m/s. If the engine of the car develops an extra linear momentum of 1000 kg m/s. Calculate the new velocity with which the car runs.

Solution:

Given : mass of car = m = 100 kg, initial velocity = u = 10 m/s, Extra momentum = 1000 kg m/s,

To find: Final velocity = v = ?

Initial momentum = p₁ = mu = 1000 x 10 = 10000 kg m/s

Final momentum = p₂ = 10000 + 1000 = 11000 kg ms-1.

Now, Final momentum = p₂ = mv

∴ 11000 = 1000 x v

∴ v = 11000/1000 = 11m/s

Ans: New Velocity = 11 m/s

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Newton’s laws of motion have everyday applications. The change in momentum of a body in a small time is a cause of large force. Most of the applications are related to the concept of change in momentum.

Cricket fielder lowers his hand while catching a ball:

If the ball is caught without lowering hands, the duration of the impact is small. As a result, the rate of change of momentum increases producing a large force. Thus the fielder will hurt his hands due to large force. When the ball is caught by moving the hand in the direction of the motion of the ball, the duration of impact increases. As a result, the rate of change of momentum decreases and thus the force exerted by the ball on hands get reduced.

An athlete taking a long jump or a high jump bends his knees before landing:

If the athlete does not bend his knees before landing, the duration of the impact is small. As a result, the rate of change of momentum increases producing a large force. Thus the athlete will hurt his legs due to large force. If the athlete bends his knees before landing, the duration of impact increases. As a result, the rate of change of momentum decreases and thus the force exerted by the ground (Reaction) on the athlete get reduced. Thus the athlete lands safely by bending knees before landing.

A Blacksmith holds the rod on the anvil with his tongs while striking with his hammer:

For reshaping of a rod, the blacksmith has to apply very large force. By holding the rod on the anvil with his tongs while striking with his hammer, the blacksmith reduces the time of impact. As a result, the rate of change of momentum increases drastically producing a very large force.

An athlete runs a certain distance before taking a large jump:

Here Newton’s First law, second law, and Third law is used. As the athlete is running a certain distance he gets additional velocity and hence additional kinetic energy, when he is actually jumping. Due to high velocity at a point where he starts his jump (Point of rising) the time of impact is reduced. As a result the rate of change of momentum increases, producing a very large force on the ground. The ground produces a large reaction and thus a long jump can be taken easily.

A clean hole is made in a glass windowpane when it is struck by a bullet fired from a gun. Whereas the same windowpane will be broken into pieces when struck by a stone of similar size:

When a bullet is fired at a glass window, a clear hole is formed in it. This is because only that part of the glass moves with the bullet, where the bullet hits the glass. The remaining part due to the inertia remains in its position. Thus bullet is able to form a hole in the glass window due to the inertia of window pane. The time for which the force is applied is very small, thus there is very less time for the reaction by the window pane.

When the same window pane is struck with a stone of a similar size, the time for which the force is applied is large compared to that in case of a fired bullet. Thus all the section around the striking point is set into motion. Due to the brittle nature of the glass, this impact causes it to break into pieces.

The engine of a moving car exerts a constant force on the car, but car movers with a constant velocity:

When a car is moving there is friction between different parts of the engine and mechanisms, besides there is friction between the road and the tyres. These frictional forces are opposed to the direction of motion. In the case of a car moving with a constant velocity the forward force exerted by the engine on the car is exactly equal and opposite to the car backward force exerted on the car due to friction and resultant force acting on the car is zero. This is the reason why the case moves with a constant velocity, even though the engine exerts a force on the car.

Apparent Weight of a Person in a Lift:

Let a person of mass ‘m’ standing on a weighing machine placed on the floor of the lift. The actual weight of the person is ‘mg’ which acts vertically down. The reaction offered by the weighing machine is ‘R’ which is also known as the apparent weight of the person.

Case – 1 (When the lift is at rest):

Since the lift is at rest, the net force acting on the person is zero.

R – mg = 0. Thus R = mg.

Thus his apparent weight is equal to his actual weight.

Case – 2 (When the lift moves upward or downward with uniform velocity):

Since lift is moving with uniform velocity, the net acceleration of the lift is zero, thus the net force acting on the person is zero.

R – mg = 0. Thus R = mg.

Thus his apparent weight is equal to his actual weight.

Case – 3 (When the lift moves upward with constant acceleration):

As the lift is moving upward with the acceleration say ‘a’, the net force acting on the person is ‘ma’ acting upward.

R – mg = ma.

Thus R = mg + ma = m(g  + a).

Thus his apparent weight is greater than his actual weight.

Case – 4 (When the lift moves downward with constant acceleration):

As the lift is moving downward with the acceleration, the net force acting on the person is ‘ma’ acting downward.

R – mg = – ma. 

Thus R = mg – ma = m(g  – a).

Thus his apparent weight is less than his actual weight.

Case – 5 (If the lift falls freely):

As the lift falls freely, its acceleration is equal to the acceleration due to gravity. Thus a = g. Substituting this value in equation of case -4 we get

R =  m(g  – g) = m(0) = 0.

Thus his apparent weight is zero and the person feels weightless. In this case, the surface on which the person is standing and the person himself have the same acceleration and the same direction of acceleration. Hence the person feels weightless.

In satellite orbiting around the earth, the astronaut and the surface of the spacecraft have the same acceleration acting towards the centre of the Earth. Hence the astronaut feels weightless.

Weightlessness is feeling and not actually weightless. The feeling of weightlessness can be experienced by a person jumping from a height.

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In this article, we shall study Newton’s third law of motion and the concept of action-reaction pair of forces.

Statement:

To every action, there is always equal, opposite and simultaneous reaction.

Explanation:

Let us consider a block of metal kept on the table. The weight of the block acts vertically downward. Thus the block tries to push the surface of the table downward by a force which is equal to its weight. This is the action. Now the surface of the table pushes the block vertically upward which is the reaction.

Here it should be noted that action and reaction act simultaneously. The action is on the table while the reaction is on the block. Thus action and reaction act on two different bodies. Hence though action and reaction are equal they can not cancel each other’s effect.

Examples to Illustrate Action Reaction Pair:

Example 1: To explain the third law of motion we can take an example of the working of a rocket. When the fuel of the rocket is ignited the combustion of fuels takes place. The hot gases produced in the chemical reaction are pushed out through the small hole at the bottom (tail) of the rocket with very high velocity. Thus the action is downward. Now, these outgoing gases produce an equal and opposite force on the rocket, which constitutes the reaction. Under the effect of upward reaction, the rocket is propelled upward. The jet engine also works on the same principle, the difference is that the rocket has to carry oxygen or an oxidizing agent with it to propel because it has to go out of the atmosphere of the Earth where oxygen is not available. While planes with jet engines fly in Earth,s atmosphere, hence they need not carry oxygen with them.

Example 2: Let us sit on a chair in front of a wall and push the wall with our legs. We will find that the chair is pushed backward. When we are pushing the wall we are applying action force, now the wall will put equal and opposite reaction on us and the chair is pushed backward.

Example 3: When we walk on the ground, as our foot pushes down (action) the ground, the ground pushes up (reaction) which is equal and opposite. This reaction is responsible for our forward movement. If the ground is not able to give a required reaction, then our foot sinks in it as in the case of sand, mud, water etc. If we push the ground harder we can jump into the air.

Example 4: Two-wheeled blocks of the same kind are at rest on the top of a smooth surface of a table. They are tied with a thread with each other. A compressed spring is placed between the two blocks. When the thread is cut the blocks move in the opposite direction.

Example 5: If we attach a spring balance to the hook on the wall and its hook is engaged with a hook of another spring balance and apply a pull on the second spring balance that both of the spring balance show the same reading but in opposite direction.

Example 6: The ball bounces back on hitting the ground. A ball strikes the ground with certain force (Action) and the ground pushes back the ball with equal force(Reaction)

Example 7: A gun recoils when a shot is fired from it. Initially both the bullet and gun are at rest, thus the total momentum of the system is constant. When a shot is fired, the bullet moves forward pushing the gun backward. This phenomenon is called the recoil of the gun. Due to recoiling of the gun, a backward jerk is experienced on the shoulders by the shooter. Hence to avoid injury he has to hold the rifle holding tight on shoulders.

Example 8: When a boy jumps from the boat to the bank. The boat moves in the direction opposite to jump. In order to jump out of the boat, he has to push the ground (surface of the boat) with greater force (action). Now the boat is in water, thus it is representing the non-rigid surface. Thus the boat moves backward under the action of pushing force.

Example 9: While swimming, the swimmer pushes the water backward with his hands. The action of pushing water backward gives rise to a reaction in the opposite direction which results in the swimmer moving forward.

Example 10: While hammering a nail. a force is experienced on the hand holding the hammer. When a nail is hammered a force is applied to it (action) due to which the nail applies equal and opposite reaction on the hammer.

Example 11: When moving train collides with a stationary one, the former experiences a reverse motion. The moving train applies a force on the stationary train and in return, the stationary train puts a reaction which is equal and opposite to the former train.

Example 12: A person falling from a certain height on a hard surface gets hurt more seriously than when he falls on a soft surface. By Newton’s third law of motion “to every action, there is always equal, opposite and simultaneous reaction. Thus when the person falls on the surface its weight (action) acts on the surface in turn the surface puts equal and opposite reaction on the person. Now the hard surface is rigid it transfers complete reaction without absorbing any impact (energy). Thus hard surface puts more reaction on the person. While the soft surface is elastic it transfers part of reaction absorbing some impact (energy). Thus soft surface put less reaction on the person. Thus hard surface is providing greater reaction than the soft surface. Hence, the person falling from a certain height on a hard surface gets hurt more seriously than when he falls on a soft surface.

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In this article, we should study the concept of momentum and Newton’s Second Law of Motion.

To apply Newton’s second law of motion, the body should be acted upon by an unbalanced force. Thus to apply this law we should consider a net force acting on the body.

Statement: The rate of change of momentum of a body is directly proportional to the impressed (applied) force and takes place in the direction of force.

Force is a vector quantity.  The direction of the net force is in the same direction as that of the acceleration. Thus, if the direction of the acceleration is known, then the direction of the net force is also known.

Concept of Momentum:

Catching a ball which is moving with a high speed is more difficult than catching a ball which is moving with lesser speed. It is easier to catch a ball of less mass than to catch a ball of more mass. The associated physical quantity is called the momentum of the body. The physical quantity which depends on the velocity of the body and its mass and which determines how much force is required to bring them to rest is known as momentum.

The product of mass and velocity of a body is called the momentum of the body.  It is denoted by p. Thus the magnitude of momentum is given by

p = mv.

Since mass is scalar and velocity is a vector, momentum is a vector quantity whose direction is the same as that of velocity.

The S.I. unit of momentum is kilogram metre per second (kg m s^-1). The c.g.s. unit is gram centimetre per second (g cm s^-1).  Dimensions of momentum are  [M¹L¹T^-1].

If the body is moving along a straight line path then it is said to possess linear momentum and if it is rotating then it is said to possess angular momentum.

Expression for Force:

Let us assume a body of mass m moving with velocity u. Let a constant force F acts on the body for t seconds and changes its velocity from u to v. then:

Where k is constant and unit of force is defined such that the value of k is 1. Therefore in scalar form mathematical expression of Newton’s second law of motion is given by 

F = ma. 

This equation is also called the fundamental equation of classical mechanics.

Note:

• When the mass of a body is constant, then its acceleration is directly proportional to the force acting on it. When the force acting on a body is constant, then its acceleration is inversely proportional to the mass of the body.
• When a rocket is launched upward in space, its acceleration increases. As the fuel in the rocket is burned and exhausted to propel the rocket, the mass of the rocket decreases continuously. But the propulsion force is almost constant. Hence acceleration increases.

Newton’s First Law as Special Case of Newton’s Second Law:

The mathematical expression of Newton’s second law of motion is given by

F = ma.

If we substitute F = 0, in the above equation, the value of ‘a’ the acceleration is found to be zero. Thus there is no change in the magnitude and direction of velocity when there is no force.  It means the body at rest is remaining at rest and the body in uniform motion in a straight line is remaining in the uniform motion in the same straight line.  Thus Newton’s first law of motion can be considered as a special case of the second law of motion.

Dimensions of force are [M¹L¹T^-2]

Units of Force:

Definition of Unit Force:

Force = Mass   x  Acceleration .

Hence   Unit force = Unit mass x  Unit acceleration.

Unit force is that force which produces a unit acceleration in a body of unit mass.

SI unit of force is newton (N) and the cgs unit is dyne. Gravitational units of force in m.k.s. system is kgf and c.g.s. system is gf

Definition of 1 newton (1N):

1 newton  = 1 kg   x  1 ms^-2

One newton is that force which produces an acceleration of 1 ms^-2  in a body of mass 1 kg.

Definition of 1 dyne:

1 dyne  = 1 gm   x 1 cm s^-2  

One dyne is that force which produces an acceleration of 1 cm s^-2 in a body of mass 1 gram.

Definition of 1 kilogram force (1 kgf):

It is a force which is required to produce an acceleration of 9.8 m s^-2 in a body of mass 1 kg.

1 kgf = 1 kg x 9.8 m s^-2 , Thus 1 kgf = 9.8 N.

Definition of 1 gram force (1 gf):

It is a force which is required to produce an acceleration of 980 cm s^-2  in a body of mass 1 g.

1 gf = 1 g X 980 cm s^-2 , Thus 1 gf = 980 dyne

The relation between newton and dyne:

1 newton  = 1 kg   x  1 m s^-2 1 , thus newton  = 1000 g   x  100 cm s^-2

Thus 1 newton  = 100000 g cm s^-2 i.e 1 newton  = 10⁵ dyne

Impulse of a Force:

When a large force acts on a body for a short interval of time then the body is said to be subjected to impulse. Impulse is denoted by letter J. e.g. Hammer struck on a nail, the ball struck by a bat.

Thus impulse of force = change in the momentum of the body.

The S.I. unit of impulse is newton-second (Ns). In c.g.s. system its unit is dyne second. Its dimensions are [M¹L¹T^-1] . These dimensions are the same as that of momentum.

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Newton’s laws of motion are three physical laws that, together, laid the foundation for classical mechanics.  In this article, we shall discuss Newton’s first law of motion and the concept of inertia of a body.

Statement of Newton’s First Law of Motion:

Every material body continues to remain in its state of rest or state of uniform motion in a straight line unless acted upon by an external unbalanced force to change the state of motion. This law is also called as law of inertia.

Explanation: This law has two parts, the first part is giving us the concept of inertia while the second part helps us in defining the force. The first part indicates that if a body is at rest, then it cannot start moving by itself. Similarly, if the body is in state of uniform motion in a straight line cannot increase or decrease its velocity or can not change the direction by itself. Thus each body has the inability to change the state of rest or state of uniform motion along a straight line by itself this inherent property of body is called inertia of a body. The inertia of a body depends on the mass of the body. Thus mass is a measure of inertia of a body.

The second part of the law helps us in defining the force. In the first part, we have seen that a body can not change its state of motion on its own but some external physical quantity is required to do so. This external physical quantity which is required to change the state of motion of a body is called force.

Concept of Inertia of a Body:

• The tendency of the body to oppose the change of state of rest or state of uniform motion is called inertia of the body.
• If no unbalanced force acts on a body then the body at rest remains at rest. This inertia is sometimes referred as the inertia of rest.
• If no unbalanced force acts on a body then the body in uniform motion along a straight line remains in uniform motion along the same straight line. This inertia is sometimes referred as the inertia of motion.
• The tendency of a body to continue to move with uniform motion in a linear direction is called inertia of direction.

Examples of Inertia of Rest:

Example: A coin is placed on a smooth card which serves as a lid on a glass. When the card is pulled suddenly in the horizontal direction the coin falls into the glass. (Inertia of rest)

Explanation: When the card is pulled horizontally it acquires motion due to pulling force. But as no force is acting on the coin in a horizontal direction.  The coin initially at rest on the card due to inertia remains at rest. Thus gets separated from the card. Now there is no support at the bottom the card falls in the glass under the action of gravity.

More Examples of Inertia of Rest:

• A cyclist riding along a level road does not come to rest immediately after he stops pedaling.

• On striking the coin at the bottom of a pile of carom coins with a striker, this coin only moves away, while the rest of the pile remains at the original position.
• When a hanging carpet is beaten with a stick, the dust particles start coming out of it. When a carpet is beaten by stick the carpet is set into motion. But due to inertia, the dust particles remain at rest. Thus they get separated from the carpet.
• On shaking or giving jerks to the branches of a tree, the fruits fall down. When branches are shaken in one direction, the fruits and leaves due to inertia remain at the original position due to inertia of rest. This causes the breakup of the stalk and they fall down.
• When a bullet is fired at a glass window, a hole is formed in it. This is because only that part of the glass moves with the bullet, where the bullet hits the glass. The remaining part due to the inertia remains in its position. Thus bullet is able to form a hole in the glass window due to the inertia of window pane.
• A magician snatches a table cloth from under a full set of tableware. When the table cloth is pulled it is set into motion, but the tableware due to inertia of rest remains on the table.
• When the local train starts or stops suddenly, sliding doors of some compartments may open or close.

Example of Inertia of Motion:

Example: When a stationary bus starts moving passengers in the bus get reclined back similarly when bus moving with uniform velocity stops suddenly passengers move forward. (Inertia of motion)

Explanation: When the bus is stationary the passengers are also stationary. When the bus starts moving the part of the body (lower part) in contact with bus starts moving, but due to inertia the upper part remains stationary and thus he gets reclined back. If he is standing he will fall backward. When the bus is moving with uniform motion in a straight line the passengers have the same motion. When the bus stops, the part of the body (lower part) in contact with bus stops, but due to inertia the upper part continues to move forward and thus he moves forward. If he is standing he will fall forward.

More Examples:

• When a passenger jumps out of a moving train he falls down. This is because as soon as the person leaves the running train, his velocity is the same as that of the train. When his feet come in contact with the ground, the lower part of his body is brought to rest, but the upper part of the body continues to travel with original velocity. This makes him fall in forward direction. To avoid this he has to run in the forward direction till his velocity is reduced to zero.
• A ball thrown vertically upward by a person in a moving train comes back to his hand. The reason is that the moment the ball was thrown, the ball was in motion along with the person and the train due to the inertia of motion. So during the time the ball remains in the air, both the person and the ball move ahead by the same distance. This makes the ball come back to his hand on its return.
• Athletes run before taking a long jump in order to increase his speed, and thereby his inertia of motion. The increased inertia of motion enables him to jump a longer distance.
• Athletes (Long jumpers / Javelin throwers /Shot putters)  often fail to stop themselves before fault line because of the inertia of motion the upper part of the athlete’s body continues to move in the forward direction while the lower part comes to halt. Thus he may not be able to stop at the fault line and crosses it.

Example of Inertia of Direction :

Example: When a vehicle takes a sudden turn towards the left, the person seated inside the vehicle is pushed towards the right. (Inertia of direction)

Explanation: When a vehicle takes a sharp left turn it changes the direction. Whereas the person seated inside tends to move in the original direction due to inertia. Thus he is pushed towards the right.

More Examples:

• When a bus makes a turn around a corner, the passengers have to hold on to some support to prevent themselves from swaying. The bus and passenger both are in a state of motion. When the bus changes the direction, the passengers continue to move in the same direction due to the inertia of direction. If passengers do not hold to some support would get thrown in that direction.

Notes:

• If a body is at rest, then the net force acting on the body is zero.
• If a body is moving in uniform motion in a straight line, then the net force acting on the body is zero.
• If a body is neither at rest nor in uniform motion, then the net force acting on the body is not zero.
• If a body is changing direction, then the net force acting on the body is not zero.
• If the net force acting on a body is zero, then the body must be at rest or uniform motion in a straight line.
• If the net force acting on a body is not zero, then the body is neither at rest nor in uniform motion in a straight line.

Stopping of a Moving Vehicle:

If the car engine is switched off or brakes are applied to stop a car, the car does not stop at once. Sometimes a driver has to apply emergency brakes. The time interval between seeing the obstacle and actually applying break is called time of response or time of thinking. The distance traveled by car during this period is called the distance of thinking. The time interval between applying brakes and the actual stopping of the car is called the time of braking. The distance traveled by car during this period is called braking distance. The sum of thinking distance and braking distance is called the stopping distance. Thus to avoid an accident the stopping distance should be less than the distance of the obstacle from the point of seeing it.

Winnowing of Grains:

By winnowing the grain get separated from the husk. Winnowing is an agricultural process in which the grain and husk are separated from each other. Grain has a larger mass than the husk. Thus the inertia of grain is more than that of husk. Thus more force is required to change their path of motion. When dropped from height in a gentle wind, due to higher inertia they just fall down vertically. The husk particles have negligible mass and very negligible inertia. Thus small force is required to change their path of motion. When dropped from height in a gentle wind, they get carried away in the direction of wind through some distance. Thus the husk and grain get separated.

Luggage on the Roof of Bus is Tied:

It is advised to tie the luggage with ropes on the roof of the bus. When a bus is moving (especially with a high speed) on the road suddenly stops or suddenly changes its direction, the luggage on the top due to inertia of motion and direction continues to remain in the motion or in the same direction of motion. As a result, the luggage may get thrown out from the bus roof if it is not tied with the rope.

Drying of Cloths By Shaking:

When a wet cloth is shaken, the water particles start coming out of it. When the cloth is shaken the cloth is set into motion. But due to inertia, the water particles remain at rest. Thus they get separated from the wet cloth. Thus cloths can be dried early.

A person sitting in a car tries to move the car by applying force to its walls. Will the car move?

Newton’s first law states that “Every material body continues to remain in its state of rest or state of uniform motion in a straight line unless acted upon by an external unbalanced force to change the state of motion.” Thus two move a body at rest some external unbalanced force is required. In this case, the force applied by the person is internal. Hence the car will not move.

Examples where low inertia is preferred and an example where high inertia is preferred (recommended):

Carpenter works with wood and nails. To drive nails in wood, less force is required. Thus low inertia of hammer is recommended. Thus carpenter’s hammer is an example of low inertia. A blacksmith works with iron, steel. To change the shape of iron or steel, large force is required. Thus high inertia of hammer is recommended. Thus blacksmith’s hammer is an example of high inertia.

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Science > Physics > Force > Newton’s First Law of Motion

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