Showing posts with label NEWTON LAWS. Show all posts
Showing posts with label NEWTON LAWS. Show all posts

Newton Laws Free body Diagram

To handle a typical problem in mechanics systematically, one should use the following steps :

(1) Draw a diagram showing schematically the various parts of the assembly of bodies, the links, supports, etc.

(2) Choose a convenient part of the assembly as one system.

(3) Draw a separate diagram which shows this system and all the forces on the system by the remaining part of the assembly. Include also the forces on the system by other agencies. Do not include the forces on the environment by the system. A diagram of this type is known as ‘a free-body diagram’.

(4) In a free-body diagram, include informationabout forces (their magnitudes and directions) that are either given or you are sure of (e.g., the direction of tension in a string along its length). The rest should be treated as unknowns to be determined using laws of motion.

(5) If necessary, follow the same procedure for another choice of the system. In doing so, employ Newton’s third law. That is, if in the free-body diagram of A, the force on A due to B is shown as F, then in the free-body diagram of B, the force on B due to A should be shown as –F.

Example

A wooden block of mass 2 kg rests on a soft horizontal floor. When an iron cylinder of mass 25 kg is placed on top of the block, the floor yields steadily and the block and the cylinder together go down with an acceleration of 0.1 m s–2.

What is the action of the block on the floor (a) before and (b) after the floor yields ? Take g = 10 m s–2. Identify the action-reaction pairs in the problem.

Answer

(a) The block is at rest on the floor. Its free-body diagram shows two forces on the block, the force of gravitational attraction by the earth equal to 2 × 10 = 20 N; and the normal force R of the floor on the block. By the First Law,the net force on the block must be zero i.e., R = 20 N. Using third law the action of the block (i.e. the force exerted on the floor by the block) is equal to 20 N and directed vertically downwards.

(b) The system (block + cylinder) accelerates downwards with 0.1 m s-2. The free-body diagram of the system shows two forces on the system : the force of gravity due to the earth (270 N); and the normal force R′ by the floor. Note, the free-body diagram of the system does not show the internal forces between the block and the cylinder. Applying the second law to the system.

270 – R′ = 27 × 0.1N
ie. R′ = 267.3 N

By the third law, the action of the system on the floor is equal to 267.3 N vertically downward.

Action-reaction pairs

For (a): (i) the force of gravity (20 N) on the block by the earth (say, action); the force of gravity on the earth by the block (reaction) equal to 20 N directed upwards .

(ii) the force on the floor by the block (action); the force on the block by the floor (reaction).

For (b): (i) the force of gravity (270 N) on the system by the earth (say, action); the force of gravity on the earth by the system (reaction), equal to 270 N,directed upwards .

(ii) the force on the floor by the system (action); the force on the system by the floor (reaction). In addition, for (b), the force on the block by the cylinder and the force on the cylinder by the block also constitute an action-reaction pair.

An action-reaction pair consists of mutual forces which are always equal and opposite between two bodies. Two forces on the same body which happen to be equal and opposite can never constitute an action-reaction pair. The force of gravity on the mass in (a) or (b) and the normal force on the mass by the floor are not actionreaction pairs. These forces happen to be equal and opposite for (a) since the mass is at rest. They are not so for case (b), as seen already. The weight of the system is 270 N, while the normal force R′ is 267.3 N.
Related posts :

Friction introduction
Rolling Friction
Newton's First law of motion




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Conservation of Linear Momentum

When no external forces are acting on a system , the total momentum of the system remains constant.

Example

A bullet is fired from a gun. If the force on the bullet by the gun is F, the force on the gun by the bullet is – F, according to the third law. The two forces act for a common interval of time Δt. According to the second law, F Δt is the change in momentum of the bullet and – F Δt is the change in momentum of the gun. Since initially, both are at rest, the change in momentum equals the final momentum for each.

Thus if pb is the momentum of the bullet after firing and pg is the recoil momentum of the gun,
pg = – pb i.e. pb + pg = 0. That is, the total momentum of the (bullet + gun) system is conserved.

Thus in an isolated system (i.e. a system with no external force), mutual forces between pairs of particles in the system can cause momentum change in individual particles, but since the mutual forces for each pair are equal and opposite, the momentum changes cancel in pairs and the total momentum remains unchanged.

The total momentum of an isolated system of interacting particles is conserved.

This law can be applied for two bodies in collision.Consider two bodies A and B, with initial momenta pA and pB. The bodies collide, get apart, with final momenta p′ A and p′ B respectively. By the Second Law
which shows that the total final momentum of the isolated system equals its initial momentum. Notice that this is true whether the collision is elastic or inelastic.

Related posts :

Newton's First law of motion
Newton's second law of motion
Newton's third law of motion
Concept of momentum
Uniform circular motion
Projectile motion





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Newton Third Law of motion-Mechanics

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

Forces always occur in pairs. Force on a body A by B is equal and opposite to the force on the body B by A.

There is no cause effect relation implied in the third law. The force on A by B and the force on B by A act at the same instant.

Action and reaction forces act on different bodies, not on the same body. Consider a pair of bodies A and B. According to the third law, FAB = – FBA.

(force on A by B) = – (force on B by A)

Thus if we are considering the motion of any one body (A or B), only one of the two forces is relevant. It is an error to add up the two forces and claim that the net force is zero.

However, if you are considering the system of two bodies as a whole, FAB and FBA are internal forces of the system (A + B). They add up to give a null force. Internal forces in a body or a system of particles thus cancel away in pairs. This is an important fact that enables the second law to be applicable to a body or a system of particles.

The great brain behind this laws Sir Isaac Newton .

Related posts :

Newton's First law of motion
Newton's second law of motion
Concept of momentum
Uniform circular motion
Projectile motion





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Newton Second Law of motion-Mechanics

The rate of change of momentum of a body is directly proportional to the applied force and takes place in the direction in which the force acts.

Thus, if under the action of a force F for time interval Δt, the velocity of a body of mass m changes from v to v + Δv i.e. its initial momentum p = m v changes by m Δ = Δ p v . According to the Second Law, F = k dp/dt where k is a constant of proportionality.

For a body of fixed mass m, F = m dv/dt = ma which shows that force is proportional to the product of mass m and acceleration a.

In SI unit force is one that causes an acceleration of 1 m s-2 to a mass of 1 kg. This unit is known as newton : 1 N = 1 kg m s-2.

1. In the second law, F = 0 implies a = 0. The second Law is obviously consistent with the first law.

2. The second law of motion is a vector law. It is equivalent to three equations, one for each component of the vectors : F along x axis is equal to m( a along x axis and so on).

This means that if a force is not parallel to the velocity of the body, but makes some angle with it, it changes only the component of velocity along the direction of force. The component of velocity normal to the force remains unchanged. For example, in the motion of a projectile under the vertical gravitational force, the horizontal component of velocity remains unchanged .

The second law of motion given by Eq. F= ma is applicable to a single point particle. The force F in the law stands for the net external force on the particle and a stands for acceleration of the particle.

3. The law in the same form applies to a rigid body or, even more generally, to a system of particles. In that case, F refers to the total external force on the system and a refers to the acceleration of the system as a whole. Here a is the acceleration of the centre of mass of the system. Any internal forces in the system are not to be included in F.

4 . The second law of motion is a local relation which means that force F at a point in space (location of the particle) at a certain instant of time is related to a at that point at that instant. Acceleration here and now is determined by the force here and now, not by any history of the motion of the particle .


Impulsewhen a large force acts for a very short duration producing a finite change in momentum of the body. For example, when a ball hits a wall and bounces back, the force on the ball by the wall acts for a very short time when the two are in contact, yet the force is large enough to reverse the momentum of the ball. Here the force and the time duration are difficult to ascertain separately. But the product of force and time, which is the change in momentum of the body remains a measurable quantity. This product is called impulse.

Impulse = Force × time duration = Change in momentum.

A large force acting for a short time to produce a finite change in momentum is called an impulsive force.Related posts :

Newton's First law of motion
Concept of momentum
Uniform circular motion
Projectile motion

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Mechanics Momentum

Momentum is the property which explains the ability of a body to transfer kinetic energy between the bodies during the collision.

Momentum, P of a body is defined to be the product of its mass m and velocity v, and is denoted by p = m v .

Momentum is clearly a vector quantity. The following common experiences indicate the importance of this quantity for considering the effect of force on motion.

1 • Suppose a light-weight vehicle (say a small car) and a heavy weight vehicle (say a loaded truck) are parked on a horizontal road. We all know that a much greater force is needed to push the truck than the car to bring them to the same speed in same time. Similarly, a greater opposing force is needed to stop a heavy body than a light body in the same time, if they are moving with the same speed.

2 • If two stones, one light and the other heavy, are dropped from the top of a building, a person on the ground will find it easier to catch the light stone than the heavy stone. The mass of a body is thus an important parameter that determines the effect of force on its motion.

3 • Speed is another important parameter to consider. A bullet fired by a gun can easily pierce human tissue before it stops, resulting in casualty. The same bullet fired with moderate speed will not cause much damage.

Thus for a given mass, the greater the speed, the greater is the opposing force needed to stop the body in a certain time. There four, the product of mass and velocity, that is momentum, is evidently a relevant variable of motion. The greater the change in the momentum in a given time, the greater is the force that needs to be applied.

4 • A seasoned cricketer catches a cricket ball coming in with great speed far more easily than a novice, who can hurt his hands in the act. One reason is that the cricketer allows a longer time for his hands to stop the ball. As you may have noticed, he draws in the hands backward in the act of catching the ball (Fig. below). The novice, on the other hand, keeps his hands fixed and tries to catch the ball almost instantly. He needs to provide a much greater force to stop the ball instantly, and this hurts.

The conclusion is clear: force not only depends on the change in momentum, but also on how fast the change is brought about. The same change in momentum brought about in a shorter time needs a greater applied force. In short, the greater the rate of change of momentum, the greater is the force.
The product of mass and velocity (i.e. momentum) is basic to the effect of force on motion. Suppose a fixed force is applied for a certain interval of time on two bodies of different masses, initially at rest, the lighter body picks up a greater speed than the heavier body. However, at the end of the time interval, observations show that each body acquires the same momentum. Thus the same force for the same time causes the same change in momentum for different bodies.

5 . Suppose a stone is rotated with uniform speed in a horizontal plane by means of a string, the magnitude of momentum is fixed, but its direction changes (Fig. below). A force is needed to cause this change in momentum vector.

This force is provided by our hand through the string. Experience suggests that our hand needs to exert a greater force if the stone is rotated at greater speed or in a circle of smaller radius, or both. This corresponds to greater acceleration or equivalently a greater rate of change in momentum vector. This suggests that the greater the rate of change in momentum vector the greater is the force applied.
Related posts :

Newton's First law of motion
Uniform circular motion
Projectile motion

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Newton First Law of motion-Mechanics

Inertia is the property of any body because of which it always continue its state and always oppose its change.

It is incorrect to assume that a net force is needed to keep a body in uniform motion. To maintain a body in uniform motion, we need to apply an external force to encounter the frictional force, so that the two forces sum up to zero net external force.

If the net external force is zero, a body at rest continues to remain at rest and a body in motion continues to move with a uniform velocity. This property of the body is called inertia. Inertia means ‘resistance to change’.

A body does not change its state of rest or uniform motion, unless an external force compels it to change that state.

Newton's First Law of motion: Every body continues to be in its state of rest or of uniform motion in a straight line unless compelled by some external force to act otherwise.

The state of rest or uniform linear motion both imply zero acceleration. The first law of motion can, therefore, be simply expressed as: If the net external force on a body is zero, its acceleration is zero. Acceleration can be non zero only if there is a net external force on the body.

For example, a spaceship out in interstellar space, far from all other objects and with all its rockets turned off, has no net external force acting on it. Its acceleration, according to the First Law, must be zero. If it is in motion, it must continue to move with a uniform velocity.

For terrestrial phenomena, in particular, every object experiences gravitational force due to the earth. Also objects in motion generally experience friction, viscous drag, etc. If then, on earth, an object is at rest or in uniform linear motion, it is not because there are no forces acting on it, but because the various external forces cancel out i.e. add up to zero net external force.

Consider a book at rest on a horizontal surface Fig. a. It is subject to two external forces :

The force due to gravity (i.e. its weight W) acting downward and the upward force on the book by the table, the normal force R . R is a self-adjusting force. We observe the book to be at rest. Therefore, we conclude from the first law that the magnitude of R equals that of W. A statement often encountered is :

“Since W = R, forces cancel and, therefore, the book is at rest”. This is incorrect reasoning. The correct statement is : “Since the book is observed to be at rest, the net external force on it must be zero, according to the first law. This implies that the normal force R must be equal and opposite to the weight W”.

Consider the motion of a car starting from rest, picking up speed and then moving on a smooth straight road with uniform speed Fig. b. When the car is stationary, there is no net force acting on it. During pick-up, it accelerates. This must happen due to a net external force. Note, it has to be an external force.

The acceleration of the car cannot be accounted for by any internal force. The only conceivable external force along the road is the force of friction. It is the frictional force that accelerates the car as a whole. When the car moves with constant velocity, there is no net external force.

The property of inertia contained in the First law is evident in many situations. Suppose we are standing in a stationary bus and the driver starts the bus suddenly. We get thrown backward with a jerk.It is because our feet are in touch with the floor. If there were no friction, we would
remain where we were, while the floor of the bus would simply slip forward under our feet and the back of the bus would hit us.

However, fortunately, there is some friction between the feet and the floor. If the start is not too sudden, i.e. if the acceleration is moderate, the frictional force would be enough to accelerate our feet along with the bus. But our body is not strictly a rigid body. It is deformable, i.e. it allows some relative displacement between different parts.

What this means is that while our feet go with the bus, the rest of the body remains where it is due to inertia. Relative to the bus, therefore, we are thrown backward. As soon as that happens, however, the muscular forces on the rest of the body (by the feet) come into play to move the body along with the bus.

A similar thing happens when the bus suddenly stops. Our feet stop due to the friction which does not allow relative motion between the feet and the floor of the bus. But the rest of the body continues to move forward due to inertia. We are thrown forward.The restoring muscular forces again come into play and bring the body to rest.

Related Posts:

Uniform circular motion
Projectile motion
Vector resolution


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