NEWTON’S LAWS OF MOTION – MOTION PHYSICS FORM 5
Motion occurs when a body covers distance with time. The quantity of motion is the product of the mass of a body and the speed at which it moves. Momentum represents the quantity of motion and it is the time rate of change of momentum that determines which keepsthe body moving.
Through ages, man has observed the motions of various bodies both in space and on smooth and rough planes. Out of these observations some laws that govern motion in general.
It was Isaac Newton who formulated the three laws we now call Newton’s laws of motion. By applying these laws with certain conditions motion problems can be solved.
Laws of motion
(a) First law of motion
“A body remains in a state of rest or uniform motion in a straight line unless acted upon by the external force”.
This law is sometimes referred to as the law of inertia. Inertia means reluctance of a body to be set into motion or to stop if already moving. This inertia depends very much on the mass the body possesses?
A body with less mass has small inertia and vice versa. On the other hand mass is the measure of inertia of a given body. The greater the mass of the body the less the acceleration when an external force is applied.
Acceleration
a
If a force F is applied on a body of mass M_{1}, and then on another body of mass M_{2} as shown in figure, the corresponding accelerations a_{1} and a_{2} are related by
/
=
/
OR
/
=
/ Figure 4.1
Mass is an inherent property of a body. It is independent of its surroundings and the method used in measuring it. Under the first law if there is no external force, a Stationary body is supposed to remain in one position forever; likewise a body in motion continues moving along the same direction indefinitely.
However in reality when a body is pushed along a horizontal plane, it just moves for a short time and stops. If the plane is polished and the same body pushed over it, it moves for a little longer time before coming to rest
(b) Second law of motion.
“The rate of change of momentum of a body is directly proportional to the external force applied and takes place in the direction of the force” We can use the above statement to derive an expression for the force that keeps the body accelerating or decelerating.
To do this let us consider the momentum and momentum change of the body under the influence of the force.
Momentum
Momentum is the product of mass of a moving body and the velocity at which it moves Momentum = mass x velocity
The unit of momentum is a
Momentum change
For a solid body, what can possibly change during motion is the velocity because the mass remains constant. If initially a body moves at a velocity V and after sometime t the velocity changes to v then its momentum changes from initial value P, to the final value
P_{f }as illustrated below
Figure 4.2
The change in momentum = Final momentum – Initial momentum
The time rate of change in momentum = Change in momentum/Time interval =
It is this rate of change of momentum which is proportional to the applied force F.
Since
and
The unit of force is Newton by definition:
A Newton is a force that makes a mass of 1kg to move with an acceleration of 1 ms^{2}
Which means that, when m = 1kg and , the force F = 1N
Substituting for m, a and F we have
And therefore
Thus the expression for the force is given by
(c) The third law
For every action there is an equal and opposite reaction
Fig 4.3 gives an illustration on the third law of motion. A body rest on the top reaction of the table exerts the weight W in return the table through the point of contact supports it by exerting a reaction force R equal to that of W only that the two are acting in opposite direction.
Figure 4.3
Implications of Newton’s laws of motion
The laws we have seen above can be used to explain some of the occurrences that we encounter in daily life. Things like friction, impulse, weightlessness, collisions, motion in fluids and so on. We shall consider few of them in general, in common cases and try to use the appropriate laws to explain the situation.
(a) Friction and Frictional force
Friction is the rubbing between two bodies in contact when moving relative to one another or stationery. If the surfaces in contact are rough there is opposition that takes place giving rise to the friction force.
To illustrate this consider a block resting on a plane as in fig 4.4. The molecular moles sticking out of each surface interlock when the surfaces come in contact. The extent of interaction between these protrusions depends on the weight a block exerts on the plane.
When the action force F is applied to the body with an intention of pulling it, the obstruction by the interlocking protrusions translates into a friction ‘f’ force / as in fig 4.4.
figure 4.4
To maintain the motion of the block, the applied force F must be greater than factional force “ƒ” such that the difference (F – ƒ) produces the acceleration ‘a’
This means the applied force partly goes into overcoming frictional force and partly accelerates the body. The difference (F – ƒ) is sometimes referred to as the effective force that causes acceleration.
If the molecular protrusions are crashed and leveled, the surfaces become smoothed and the opposition to motion is drastically reduced to the extent that very little force is required to keep the body moving. In addition if the surface is polished, frictional force is further minimized to the level where it can be regarded as smooth.
In solving mechanics problems the terms smooth or frictionless are used to mean that the frictional force is zero. When there is no frictional force the applied force wholly goes into accelerating the body.
(ii) Relationship between friction force and normal reaction
There is relationship between the friction “f ” and the normal reaction R. The more is the reaction force , the greater the force of friction in another words friction force is direct proportion to the reaction. The ratio of the friction force to the normal reaction is called the coefficient of friction.
Coefficient of friction (μ) =
From which,
(iii) Coefficient of static friction
A body resting on the plane does not begin to move until the external force is applied to overcome friction force. The maximum friction force that is to be overcome by applied force is sometimes referred to as limiting force
=
R
One of the methods for determining the value of
is shown in fig 4.5.
figure 4.5
In this experimental a block of known mass is placed on the horizontal plane whose coefficient of friction is sought. The plane consists of a frictionless pulley at one end. A string with a pan at one end is tied to the block and made to pass over the pulley.
Before adding anything on the pan, the only forces on the plane are the weight “W” of the block and the reaction R from the plane. As soon as the mass is added to the pan, the tension “T” created in the string tends to pull the block and at the same time the frictional force takes charge.
Initially the weight “W” arising from the mass on the pan may not be strong enough to move the block but as more and more masses are placed on the pan the frictional force grow bigger and bigger.
The time comes, when the frictional force can no longer hold the block and therefore block just begins to slide. It is at this moment we say that the tension “T” is just equal to the frictional force ”
“. This is the maximum frictional force the plane can offer in preventing the block to slide over it, By recording the total mass on the pan, the magnitude of the weight W’ can be found and from the equation
Where T = W’ and W’ =
g the numerical value for
is evaluated.
Definition
The coefficient of static friction is the ratio of the frictional force when the body is on the average slide.
Once the block has gain momentum, the coefficient of friction is now referred to as the coefficient of kinetic friction
and its value is less than that of
Table 4.1 shows the value of
and
for the surface of different materials.
Table 4.1: coefficients of friction for some materials
Surface in contact 


Steel on the steel  0.74  0.57 
Copper on steel  0.61  0.47 
Aluminium on steel  0.53  0.36 
Rubber on concrete  1.00  0.80 
Wood on wood  0.250.50  0.20 
Glass on glass  0.94  0.40 
Ice on ice  0.1  0.03 
(b)Motion on the horizontal plane
Here we shall consider a body sliding along rough plane as a general case. Being rough, the plane offers frictional force to the body on sliding it.
If the body is just projected with initial velocity u , its velocity keep on decreasing until it come to the halt due to the frictional force that opposes motion persistently as in fig 4.6
figure 4.6
The question arising from this example would be;
How far does the body go?
What is the deceleration?
What is the frictional force?
How far does the body go?
To find how far the body reaches we can start with the third equation of linear motion in which the acceleration
such that;
=
= Where R = mg and therefore S = mu^{2} / µmg or
(b) Motion on a rough inclined plane
Consider a block projected up the inclined plane with an initial velocity as it climbs, there are two forces which opposes the motion, one being frictional force and other is the component of weight parallel to the plane as shown in fig 4.7
figure 4.7
The two forces opposing the motion simultaneously bring the block to a standstill somewhere along the plane after traveling a distance S.
To find the magnitude of the distance covered we can use the same method as in the case of the horizontal plane only that the component of the block has to be considered. The total force opposing the motion is
Where F = ma,
and
a = gsin
a =
From the third equation of linear motion,
Note: Motion in viscous fluids.
1.Where v = 0 (given),
a = (
From which
If the inclined plane is smooth i.e.
there is only one opposing force and that is
and the distance becomes
(C)Motion of connected bodies
(i) On smooth horizontal plane
Fig 4.8(a) shows two bodies A and B of masses
and
respectively connected by the light in extensible string that passes over a frictionless pulley such that the body A rest on the smooth surface and B hang freely.
figure 4.8
As soon as the system is let free and given that m_{2}> m_{1, }body B falls under its own weight W_{2 }pulling body A via a string and both move with the common acceleration as shown in fig 4.8(b).
The tension in the string is uniform throughout. Using the freebody diagram it is easy to determine both the tension and the acceleration of the system. The process is as follows:
Body B falls vertically downwards because
But
………………(1)
Body A moves horizontally in direction of T without opposing because the surface is smooth
……………………… (2)
Solving equations (1) and (2) we have
Substituting for
in equation (2) we get
(ii)On rough inclined plane
figure 4.9
For a rough plane frictional force play part by opposing the motion of body A as the system is free to move. Equation (i) above remains the same
…………………(3)
Due to the frictional force equation (2) becomes
But
……………………(4)
Add (3) and (4)
(
+
Substitute for
in equation (4) and simplify
T
(iii) Connected bodies on inclined plane
Consider a situation where two bodies connected by an extensible string that passes over a frictionless pulley rest on two planes inclined at an angle
to the horizontal as shown in fig 4.10(a)
Figure 4.10 (a) Figure 4.10 (b)
When let it free, the bodies may move as in fig 4.10(b) provided and the surface of inclined plane are alike i.e their coefficients of friction are equal. The forces involved are indicated their direction and as a results both tension T and acceleration of the system can be found using similar procedures as in the case of horizontal plane. Since body A climbs the plane it implies that
T
i.e
But,
and
………(1)
Likewise since body B moves down the plane it means
^{a}
Where
and
Or
————(2)
Add (1) and (2) to obtain
(
Substitute for
in (2) and simplify to get
(iv) Connected bodies on pulley
Again consider two unequal bodies of masses
connected by a light inextensible string that passes over a frictionless pulley as in fig 4.11
Figure 4.11 (a) Figure 4.11 (b)
In figure 4.1 1(a) the masses are mounted on a pulley with m_{2}> m_{1. } When released, the weight of the larger mass overcomes that of smaller mass and the motion takes place as in figi4.1 l(b).
Using the observation and reasoning we can determine the acceleration”a” of the system ‘and the tension “T” in the string. For the mass moving downwards W_{2}>T and therefore;
For the mass moving vertically upwards T > W_{1 }and hence
Adding (1) and (2) we have
Since and
From (1)
(v) Motion in a lift
If you have been in a lift, you may have felt a difference in your weight as a lift ascends or descends with constant acceleration. When ascending one feel heavier than normal by pressing hard on the lift floor and when descending, one becomes apparently lighter than normal by losing weight.
To get the idea of what happens consider a body of mass m on a pan of a weighing machine on the floor in a lift. Before the motion along the vertical plane begins, the weight of the body is taken to be normal, say W as in fig 4.12.
Figure 4.12(a) (b) (c)
When a lift accelerates uniformly downwards, the scale of a weighing machine shows that there is a drop in weight by indicating a lower value W_{1 }as in fig 4.12(b) and when it accelerates upwards the machine indicates a higher value in weight Was in fig 4.12(c).
To explain these observations, let us consider the body on the scale pan in each case separately.
In the body exerts the normal weight W on the pan and in return the pan exerts force of figure 4.12 (a) reaction R as shown in fig 4.13. Since the lift is at rest, meaning that
Figure 4
But
and therefore
figure 4.13
In the lift accelerates uniformly vertically downwards, this causes the apparent loss in weight meaning that the weight W_{1} indicated on the scale equals to the reaction R_{1 }such that R_{1 }
_{ }as in fig 4.14
Figure 4.14.
The apparent loss in weight as the lift goes down is
The reason for the apparent loss in weight in a body is that the surface on which the body is resting runs away or falls faster in such a way that the contact between;^{ }the pan and the body becomes lighter and hence less reaction from the surface.
In fact if the lift falls with acceleration equal to that of gravity, the reaction on the body from the surface would be zero i.e. when a = g, .R, = m(g – g) = 0 and the pointer on the scale would indicate zero weight. Under such condition the body would appear weightless.
In figure 4.12(c), the lift accelerates vertically upwards causing the scale to indicate a larger weight W_{2}. The apparent increase in weight as the lift ascends arises from the fact that the pan on which the body sits tends to rise faster the body itself
Therefore pressing harder underneath it and hence the reaction R_{2}as in fig 4.15. The effect of the pan pressing hard on the body is the apparent gain in weight W_{2}registered on the scale. The apparent gain in weight is given as
Or
Figure. 4 .15
If the lift ascends with acceleration equals to that of the gravity, the reaction R_{2 }would double. That is if
, then R_{1 }=m (