LAWS OF MOTION
Created by C. Mani, Assistant Commissioner, KVS RO Silchar
Aristotle’s Fallacy
Aristotelian law of motion states that “an external force is required to keep a body in motion”.
Galileo’s observations
It was proved wrong by Galileo Galilei and Isaac Newton.
A ball released from rest on one plane rolls down and climbs up the other.
In the absence of friction, the heights of the ball on both the planes are the same (ideal condition).
In practical situation, the height on the second plane is always a little less but never greater.
A ball released from rest in case-1 reaches the same height on the other side (in the absence of friction).
Case-1
Case-2
Case-3
In case-2, the ball reaches the same height on the other side (in the absence of friction) but travels longer distance.
In case-3, which is the limiting case, the ball rolls down and continues to move in the horizontal direction through infinite distance (in the absence of friction).
Galileo arrived at a new insight that “the state of rest and the state of uniform linear motion are equivalent”.
In both the cases, there is no net force acting on the body.
Therefore, there is no need to apply an external force to keep a body in uniform motion.
We need to apply external force only to encounter the frictional or any other force, so that the forces sum up to zero net external force.
To summarise, if the net external force is zero, a body continues to remain at rest and a body in uniform motion continues to move with uniform velocity.
This property of the body is called ‘inertia’.
Inertia means ‘resistance to change’.
Galileo’s Law of Inertia:
A body does not change its state of rest or of uniform motion, unless an external force compels to change its 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 and until compelled by an external force to change its state of rest or of uniform motion.
In simple words, if the net external force on a body is zero, its acceleration is zero. Or, acceleration can be non-zero only if there is a net external force on the body.
Reaction (R) (Exerted by the table)
Weight (W) (Force of gravity)
Since the block is at rest, the normal reaction R must be equal and opposite to its weight W.
The car continues to be in uniform motion……
in the absence of an external force.
F
The block remains at rest…..
unless and until it is acted upon by an external force.
F
The ball continues to be in uniform motion……
unless and until it is acted upon by an external force.
Inertia
Inertia is the inherent property of a body due to which it resists a change in its state of rest or of uniform motion.
Inertia can be understood in parts, viz. inertia of rest, inertia of motion and inertia of direction.
Mass is a measure of the inertia of a body.
Heavier objects have more inertia than lighter objects.
Eg. 1. A stone of size of a football has more inertia than football.
2. A cricket ball has more inertia than a rubber ball of the same size.
Inertia of rest
Examples of Inertia of rest:
Examples of inertia of motion:
MOMENTUM
Momentum is the quantity of motion in a body and it depends on its mass and velocity.
Momentum of a body is defined as the product of its mass and velocity.
i.e. Momentum = mass x velocity
If a cricket ball and a tennis ball move with same velocity, momentum of
cricket ball is more because its mass is larger than that of the tennis ball.
If two cricket balls move with different velocities, then the momentum of
the ball with greater velocity possesses more momentum.
or p = m x v
Importance of Momentum for considering the effect of force on motion:
Eg.
The same force for the same time causes the same change in momentum in different bodies.
The force not only depends on the change in momentum, but also on how fast the change is brought. The greater the rate of change of momentum, the greater is the force.
The above examples do not indicate that momentum is a vector.
The following example shows that momentum is a vector.
Suppose a stone is whirled with uniform speed in a horizontal circle by means of a string.
The magnitude of momentum is constant but its direction keeps changing continuously.
A force is needed to change this momentum vector.
This force is provided by hand through the string and it is also felt by the hand.
NEWTON’S SECOND LAW OF MOTION
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.
Force α
Time taken
Change in momentum
or
where k is a constant of proportionality.
or
Δp
Δt
lim
Δt→0
F = k
or
If force F changes the velocity of a body of mass m from v to v + Δv, then its initial momentum p = mv will change by Δp = m Δv.
F α
Δp
Δt
F = k
Δp
Δt
dp
dt
F = k
According to Newton’s Second Law,
dt
(mv)
F = k
d
dt
F = k m
dv
F = k m a
If force of 1 N applied on a body of mass 1 kg produces an acceleration of 1 m/s2 on the body, then
1 = k x 1 x 1
or k = 1
Force = mass x acceleration
F =
m a
The acceleration produced in a body is directly proportional to the force acting on it and inversely proportional to the mass of the body.
Acceleration a =
F
m
Eg.: Accelerator of a car accelerates it and brakes decelerate it.
mass 1 kg produces an acceleration of 1 m/s2 in it.
100 g
Place an 100 g on your outstretched palm. The force you feel is nearly 1 newton !
Some important points about Second Law
dpx
dt
Fx =
= max
dpy
dt
Fy =
= may
dpz
dt
Fz =
= maz
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 e.g. in the motion of a projectile under the vertical gravitational force, the horizontal component of velocity remains unchanged.
3. The equation
dp
dt
F =
is applicable to a single point particle.
= m a
However, it is same for a rigid body or, even more generally, to a system of particles. In this case, m refers to the mass of rigid body or total mass of the system of the particles and a refers to the acceleration of the centre of mass of the system. Any internal forces in the system are not to be included in F.
IMPULSE
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 on the ball.
In these situations, it is difficult to ascertain the force and time duration separately.
However, the product of force and time, which is the change in momentum is a measurable quantity. This product is called ‘impulse’.
Impulse = Force x time duration
= Change in momentum
j = F t
A large force acting for a short time to produce change in momentum is called ‘impulsive force’.
The cushion or sand helps to increase the time in which the momentum comes to zero. This decreases the rate of change of momentum and hence the force. So, the athlete does not get hurt.
These materials help to increase the time in which the momentum comes to zero when jolting and jerking take place. This decreases the rate of change of momentum and hence the force. So, the articles do not get broken.
Importance of Impulse in Practice
A fast moving cricket ball has a large momentum. In stopping the cricket ball, its momentum has to be reduced to zero. When a player moves his hands back, the time taken to stop the ball increases and hence the rate of change of momentum decreases. i.e. the force exerted by the ball on the hands decreases. So, the hands of the player do not get hurt.
NEWTON’S THIRD LAW OF MOTION
To every action, there is always an equal and opposite reaction.
Note:
The statement means that in every interaction, there is a pair of forces acting on the two interacting objects. The size of the forces on the first object equals the size of the force on the second object. The direction of the force on the first object is opposite to the direction of the force on the second object. Forces always come in pairs - equal and opposite (action-reaction) force pairs.
FAB = - FBA
F
F
0
20
40
80
60
100
0
20
40
80
60
100
Reaction = 58 gwt
Action = 58 gwt
0
20
40
80
60
100
0
20
40
80
60
100
Examples / Applications of Newton’s Third Law of motion
Reaction
Action
Weight
Vertical component of reaction
Horizontal component of reaction
Reaction
Weight
Forward Motion
Horse and Cart Problem
A horse is urged to pull a cart. The horse refuses to try, citing Newton’s third law as his defence. “The pull of me on the cart is equal but opposite to the pull of the cart on me. If I can never exert a greater force (action and reaction are always equal) on the cart than it exerts on me, how can I ever set the cart moving?”, asks the horse.
How would you reply?
WC
RC
TCH
THC
Action
Reaction R
H
V
f
Why don’t you educate the Horse properly?
Reaction ‘RC’ on the cart offered by the ground must be equal and opposite to weight of the cart ‘WC’.
Reaction pull of the cart on the horse ‘TCH’ must be equal and opposite to forward pull of the horse on the cart ‘THC’.
If the horse pushes the ground in a slanting manner (Action), the Reaction offered by the ground is resolved into Vertical and Horizontal components.
The Vertical component ‘V’ must be equal and opposite to weight of the horse ‘WH’.
If the Horizontal component ‘H’ is greater than the Friction ‘f’, then the horse-cart system will move forward with acceleration.
WH
LAW OF CONSERVATION OF MOMENTUM
The total momentum of an isolated system of interacting particles is conserved.
OR
When two or more bodies act upon one another, their total momentum remains constant provided no external forces are acting on them.
Suppose a big and a small car move in the same direction with different velocities.
Let the mass of the bigger car be ‘m1’ and its initial velocity is ‘u1’.
Let the mass of the smaller car be ‘m2’ and its initial velocity is ‘u2’ such that u2 < u1.
Suppose both the cars collide for a short time ‘t’.
Due to the collision, the velocities will change.
Let the velocities after the collision be v1 and v2 respectively.
Scalar Treatment
Suppose that during collision, the bigger car exerts a force F21 on the smaller car and in turn, the smaller car exerts a force F12 on the bigger one.
When the force F21 acts on the smaller car, its velocity changes from u2 to v2.
According to Newton’s third law, F21 = – F12
F21 = m2a2
When the force F12 acts on the bigger car, its velocity changes from u1 to v1.
F12 = m1a1
Cancelling t on both sides, we get
m2(v2 – u2) = m1(v1 – u1)
m2v2 – m2u2 = m1v1 – m1u1
Total momentum after collision = Total momentum before collision
Initial momentum of the bigger car = m1u1
Initial momentum of the smaller car = m2u2
Final momentum of the bigger car = m1v1
Final momentum of the smaller car = m2v2
m1v1 + m2v2 = m1u1 + m2u2
F21 = m2
v2 – u2
t
F12 = m1
v1 – u1
t
m2
v2 – u2
t
m1
v1 – u1
t
= -
Consider two bodies A and B of mass mA and mB moving with velocities vA and vB respectively. Their initial momenta are pA and pB. Let the bodies collide, get apart and move with final momenta pA’ and pB’ respectively.
By Newton’s Second Law,
and
where Δt is the common interval of time for which the bodies are in contact.
By Newton’s Third Law,
FAB = - FBA
or
FAB Δt = - FBA Δt
or
pA’ - pA
pB’ - pB)
= - (
i.e.
pA’ + pB’
pA + pB
=
Click to see the collision…
The above equation shows that the total final momentum of the isolated system equals its total initial momentum.
vA
vB
vA’
vB’
FBA
FAB
mA
mB
mA
mB
mA
mB
Vector Treatment
This is true for elastic as well as inelastic collision.
FAB Δt = pA’ - pA
FBA Δt = pB’ - pB
EQUILIBRIUM OF A PARTICLE
Equilibrium of a particle refers to the situation when the net external force on the particle is zero.
Equilibrium of a particle requires not only translational equilibrium (zero net external force) but also rotational equilibrium (zero net external torque).
According to Newton’s law, this means that, the particle is either at rest or in uniform motion.
If two forces F1 and F2 act on a particle, equilibrium requires
F1 = - F2
i.e. the two forces must be equal and opposite.
or
F1 + F2 = 0
Equilibrium under three concurrent forces F1, F2 and F3 requires that the vector sum of them is zero.
F1
F2
F1
F2
F3
The resultant of two forces F1 and F2 obtained by parallelogram law of forces must be equal and opposite to the third force F3.
F1
F2
F3
F1 + F2 + F3 = 0
By triangle law of vectors, the three sides of the triangle with arrows taken in the same order represent the equilibrium of forces.
-F3
A particle is in equilibrium under the action of forces F1, F2, …. Fn if they can be represented by the sides of a closed n-sided polygon with arrows directed in the same sense.
F1 + F2 + F3 = 0
implies that
F1x + F2x + F3x = 0
F1y + F2y + F3y = 0
F1z + F2z + F3z = 0
where F1x + F1y + F1z are the components of F1 along x, y and z directions,
F2x + F2y + F2z are the components of F2 along x, y and z directions
and F3x + F3y + F3z are the components of F3 along x, y and z directions
respectively.
O
Q
P
R
F1
F2
F3
R = 0
The gravitational force is the force of mutual attraction between any two objects by virtue of their masses.
Every object experiences this force due to every other object in the universe.
All objects on the earth, experience the force of gravity due to the earth.
In particular, gravity governs the motion of the moon and artificial satellites around the earth, motion of the earth and planets around the sun, and, of course, the motion of bodies falling to the earth.
It plays a key role in the large-scale phenomena of the universe, such
as formation and evolution of stars, galaxies and galactic clusters.
The gravitational force of attraction between two objects of masses m1 and m2 held at a distance r apart is given by
COMMON FORCES IN MECHANICS
where G = Universal constant of gravitation.
F = G
m1 m2
r2
Gravitational Force:
Some of the important properties of Gravitational forces:
force).
(iv) It obeys inverse square law.
(v) It is always attractive in nature.
(vi) It is a long range force.
(vii) It is the weakest force operating in nature.
(viii) It is a conservative force.
All other forces in mechanics are contact forces.
Contact force arises due to contact of one body with another (solid or fluid).
Contact forces are mutual forces (for each pair of bodies) satisfying Newton’s third law.
Other Forces:
The component of contact force normal to the surfaces in contact is called ‘normal reaction’.
The component of contact force parallel to the surfaces in contact is called ‘friction’.
For a solid immersed in a liquid, there is an upward buoyant force equal to the weight of the liquid displaced.
The viscous force, air resistance, etc. are also examples of contact forces.
Two other common forces are tension in a string and the force due to spring.
The restoring force in a string is called tension. It is customary to use the same tension T throughout the string (of negligible mass).
Buoyant Force
Weight
When a spring is compressed or extended by an external force, a restoring force is generated which is proportional to the compression or elongation (for small displacements).
The spring force is written as F = - kx where x is the displacement and k is a constant.
The different contact forces mentioned above arise fundamentally from electrical forces.
At the microscopic level, all bodies are made of charged constituents (nuclei and electrons) and the various contact forces arising due to elasticity of bodies, molecular collisions and impacts, etc. can ultimately be traced to the electrical forces between the charged constituents of different bodies.
F
FRICTION
F = 0
fs = 0
F
fs
Reaction (N) (Exerted by the table)
Weight (W) (Force of gravity)
F
Reaction (N) (Exerted by the table)
Weight (W) (Force of gravity)
fk
Static Friction
The force of gravity W = mg is equal and opposite to the normal reaction force N of the table.
If the body is applied a small force F, the body, in practice, does not move.
If the applied force F were the only external force on the body, the body must have moved with acceleration a = F/m, however small.
This force fs parallel to the surface of the body in contact with the table is known as frictional force, or simply ‘friction’. The subscript s stands for ‘static friction’.
Clearly, the body remains at rest because some other force comes into play in the opposite direction to the applied force F, resulting in zero net force on the body.
Static friction does not exist by itself. It comes into play the moment there is an applied force. As F increases, fs also increases, remaining equal and opposite to the applied force (up to a certain limit), keeping the body at rest.
Since the body is at rest, the friction is called ‘static friction’.
Since the body is at rest upto the maximum limit (fs )max, it is called the limiting value of static friction.
Static friction opposes the impending motion.
The term ‘impending motion’ means motion that would take place (but does not actually take place) under the applied force, if friction were absent.
When the applied force F exceeds the limit (fs )max, the body begins to move.
2. It varies with the normal reaction force (N) approximately as
Laws of Limiting Friction:
(fs )max α N
(fs )max = μs N
(fs )max
μs =
N
μs is called coefficient of static friction.
where μs is constant of proportionality depending only on the nature of the surfaces in contact.
The law of static friction may be written as fs ≤ μs N
Sliding or Kinetic Friction
If the applied force F exceeds the limit (fs )max, the body begins to slide on the surface. When the relative motion has started, the frictional force decreases from the limiting value of static friction (fs )max.
Frictional force that opposes relative motion between the surfaces in contact is called kinetic or sliding friction.
3. It varies with the normal reaction force (N) approximately as
Laws of Kinetic Friction:
fk α N
fk = μk N
fk
μk =
N
μk is called coefficient of kinetic friction.
where μk is constant of proportionality depending only on the nature of the surfaces in contact.
μk is less than μs.
When relative motion has begun, the acceleration of the body is (F – fk) / m.
or
For a body moving with constant velocity, F = fk.
If the applied force on the body is removed, a = - fk / m and the body eventually comes to a stop.
It is denoted by fk.
The laws of friction do not have the status of fundamental laws like those for gravitational, electric and magnetic forces.
They are empirical relations that are only approximately true, but yet useful in mechanical applications.
Rolling Friction
A body like a ring or a sphere rolling without slipping over a horizontal plane will not suffer friction, in principle.
At every instant, there is just one point of contact between the body and the plane and this point has no motion relative to the plane.
In this ideal situation, kinetic or static friction is zero and the body should continue to roll with constant velocity.
However, in practice, this does not happen and the body comes to rest eventually.
To keep the body rolling, some applied force is needed. This indicates that there is a rolling friction though it is much smaller (by 2 or 3 orders of magnitude) than static or sliding friction.
Rolling friction has a complex origin.
During rolling, the surfaces in contact get momentarily deformed a little, and this results in a finite area (not a point) of the body being in contact with the surface.
The net effect is that the component of the contact force parallel to the surface opposes the motion.
Friction is a Necessary Evil
In many situations, friction has a negative role.
In many practical situations, however, friction is critically needed.
Methods of reducing friction
Lubricant
Air Jet
Bending of a cyclist
θ
O
N sinθ
θ
N cosθ
CIRCULAR MOTION
While negotiating a curve a cyclist bends towards the centre of the curve to keep a safe balance.
The angle θ through which the cyclist bends can be calculated as given below:
The weight of the cyclist plus the cycle (W = mg) acts in the vertically downward direction.
The Normal Reaction (N) acts in the direction as shown in the figure, due to bending through the angle θ.
The centripetal force Fcp = mv2/r acts towards the centre of the curve.
The Normal Reaction (N) is resolved into the components N cosθ and N sinθ as shown in the figure.
N cosθ and mg are equal and opposite and balance each other.
N sinθ provides the necessary centripetal force mv2/r to the cyclist.
N
v
W = mg
r
Fcp
N cosθ = mg ------(1)
N sinθ = Fcp
In reality, θ will be little less owing to the friction available on the road.
The cyclist will bend through an angle θ from the vertical.
m v2
r
N sinθ = -----(2)
v2
rg
tanθ =
(2) / (1)
⇒
or
v2
rg
θ = tan-1
fs4
fs3
v
Motion of a car on a level circular road
N1
N3
N4
N2
mv2/r
When a car negotiates a curve it tends to slip away from the centre of the curve and hence frictional force acts towards the centre.
Static friction opposes the impending motion of the car moving away from the circle.
If the road is in horizontal level the centripetal force required is given by the static friction (self adjustable) between the road and the tyres.
Hence, it is the static friction that gives centripetal acceleration to the car.
W = mg
fs1
fs2
N
The forces acting on the car are:
v
ii) Normal reaction, N on the car (upwards)
iii) Static friction, fs
(N = N1 + N2 + N3 + N4)
(fs = fs1 + fs2 + fs3 + fs4)
For the safe turn of the car, there is a limit of magnitude of the frictional force. It can not exceed μs N.
i.e. fs ≤ μs N (where N = N1 + N2 + N3 + N4)
Since there is no acceleration in the vertical direction,
The centripetal force is provided by the friction.
m v2
r
= fs
(where fs = fs1 + fs2 + fs3 + fs4)
m v2
r
= fs ≤ μs N
m v2
r
≤ μs N
or
m v2
r
≤ μs mg
N – mg = 0
N = mg
The above equation gives the maximum velocity that can be negotiated with the given value of μs and r.
v ≤ √μsrg
or
On the other hand, μs required for negotiating a curve of radius r with velocity v is given by
vmax = √μsrg
or
μs ≥ v2 / rg
v
θ is the angle through which the road is banked.
The maximum velocity the car can negotiate the curve without skidding in the absence of friction is
Motion of a car on a banked circular road
The contribution of friction to the circular motion of the car can be reduced if the road is banked, i.e. raised on the outer edge of the circular road.
N1
θ
N sin θ
N cos θ
N2
θ
W = mg
mv2/r
N
Since there is no acceleration in the vertical direction,
In the absence of friction:
m v2
r
N sin θ = -----(2)
N cos θ = mg ------(1)
N sin θ = Fcp
v2
rg
tan θ =
(2) / (1)
⇒
or
v2
rg
θ = tan-1
vmax = √ rg tan θ
(where N = N 1 + N 2 + N 3 + N 4)
The centripetal force is provided by N sin θ.
If the friction is available:
Since there is no acceleration in the vertical direction,
N cos θ = mg + fk sin θ
mg = N cos θ - fk sin θ -----(1)
The centripetal force is provided by the horizontal components N sin θ and fk cos θ.
N sin θ + fk cos θ = Fcp
m v2
r
= N sin θ + fk cos θ -----(2)
But fs ≤ μs N
To obtain vmax, we put fs = μs N.
Then Eqns. (1) and (2) become
mg = N cos θ - μs N sin θ -----(3)
m v2max
r
= N sin θ + μs N cos θ -----(4)
From Eqn. (3), we get
cos θ - μs sin θ
mg
N =
v
N1
θ
N sin θ
N cos θ
N2
θ
W = mg
mv2/r
N
fk cos θ
fk sin θ
θ
fk
In the absence of friction, μk= 0.
Substituting value of N in Eqn. (4), we get
m v2max
r
cos θ - μs sin θ
=
mg (sin θ + μs cos θ)
Comparing the above equation with
vmax = √μsrg
or
vmax
= rg
1 - μs tan θ
μs + tan θ)
½
we see that the maximum possible speed of a car on a banked road is greater than that on a flat (level) road.
v0 = √ rg tan θ
At this speed, frictional force is not needed at all to provided the necessary centripetal force.
Driving at this speed on a banked road will cause little wear and tear of the tyres.
The above equation tell that for v < v0, frictional force will be up the slope and that a car can be parked only if tan θ ≤ μs.
SOLVING PROBLEMS IN MECHANICS
Newton’s three laws of motion are the foundation of mechanics.
More often, we may come across an assembly of different of different bodies.
Each body in the assembly experiences force of gravity.
While solving a problem, we can choose any part of the assembly and apply the laws of motion to that part provided we include all forces on the chosen part due to the remaining parts of the assembly.
The chosen part of the assembly is called ‘system’ and the remaining part of the assembly (including other agencies of forces) as the ‘environment’.
To handle a typical problem in mechanics, the following steps should be used.
Include also the forces on the system by other agencies.
Do not include the forces on the environment by the system.
Such a diagram is called ‘a free-body diagram’.