Force and Laws of Motion

Force and Laws of Motion

Force is defined as something that causes a change in the motion of an object. Forces may be defined as :Applied force.

Gravitational force.

Normal force.

Frictional force.

Air resistance force.

Tension force.

Spring force.

Balanced and Unbalanced forces

Balanced forces applied on an object do not affect the object in terms of speed, or acceleration. An object moves with a uniform velocity when the forces (pushing force and frictional force) acting on the object are balanced and there is no net external force on it.

Whereas when unbalanced forces are applied, the object will change its state . If an unbalanced force is applied on the object, there will be a change either in its speed or in the direction of its motion. Thus, to accelerate the motion of an object, an unbalanced force is required. And the change in its speed (or in the direction of motion) would continue as long as this unbalanced force is applied.

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.

Inertia is a measure of mass of a body. Greater the mass of a body greater will be its inertia or vice-versa.

Inertia is of three types:

(i) Inertia of Rest When a bus or train starts to move suddenly, the passengers sitting in it falls backward due to inertia of rest.

(ii) Inertia of Motion When a moving bus or train stops suddenly, the passengers sitting in it jerks in forward direction due to inertia of motion.

(iii) Inertia of Direction We have seen that the rotating wheels of a vehicle throughout mud in a tangential direction. This is due to inertia of direction. This is the reason that the mud guards are provided in vehicles over the wheels to stop this mud so that to protect the clothes.

Law of motion

First law of motion

Also called the law of inertia.

An object remains in a state of rest or of uniform motion in a straight line unless compelled to change that state by an applied force. In other words, all objects resist a change in their state of motion. In a qualitative way, the tendency of undisturbed objects to stay at rest or to keep moving with the same velocity.

It can also be stated 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

Second Law of Motion

The second law of motion states that the rate of change of momentum of an object is proportional to the applied unbalanced force  in the direction of force.

The momentum, p of an object is defined as  the product of its mass, m and velocity, v. That is,
p = mv

Suppose an object of mass, m is moving along a straight line with an initial velocity, u. It is uniformly accelerated to velocity, v in time, t by the application of a constant force, F  throughout the time, t.

The initial and final momentum of the object will be, p1 = mu and p2 = mv respectively.

The change in momentum p2 – p1
                                                 ∝ mv – mu
                                                 ∝ m × (v – u)

The rate of change of momentum m( v -u )/t

Or, the applied force,  F m( v- u )/t

F =k m( v- u )/t

Thus F = k m a,

k is a constant of proportionality

Mathematical representation   F = k m a, which shows that force is proportional to the  product of mass m and acceleration and where k is a constant of proportionality.

NOTE: The first law of motion can be mathematically stated from the mathematical expression for the second law of motion .

F   m( v- u )/t

Ft = mv – mu

That is, when F = 0, v = u for whatever time, t is taken. This means that the object will continue moving with uniform velocity, throughout the time, t. If u is zero then v will also be zero. That is, the object will remain
at rest.

Third  Law of Motion

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

The terms action and reaction in the third law mean nothing else but ‘force.

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.

The force on A by B and the force on B by A act at the same instant. By the same reasoning, any one of them may be called action and the other reaction.

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)

Momentum

Momentum can be defined as “mass in motion.” All objects have mass; so if an object is moving, then it has momentum – it has its mass in motion.

Momentum = mass • velocity

In physics, the symbol for the quantity momentum is the lower case p. Thus, the above equation can be rewritten as

p = m • v       where m is the mass and v is the velocity.

Momentum is a vector quantity i.e  a vector quantity is a quantity that is fully described by both magnitude and direction.

Conservation of Momentum

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

We can also say that the sum of momenta  of the two objects before collision is equal to the sum of momenta after the collision provided there is no external unbalanced force acting on them(called an isolated system). This is known as the law  of conservation of momentum.

This statement can alternatively be given as the total momentum of the two objects is unchanged or conserved by the collision.

Let us consider two bodies A and B, with initial momenta pA=mAuA  and  pB=mBuB
The bodies collide, get apart, with final momenta p′A =mAvA and p′B=mBvB respectively

By the Second Law
FAB=pA-p′A=mA( vA- uA)/t ,  and 

FBA=pB-p′B=mB( vB- uB)/t

By the Third Law

FAB = – FBA

pA-p′A=pB-p′B

pA+pB=p′A+p′B

mAuA+mBuB=mAvA+mBvB

which shows that the total final momentum of the isolated system equals its initial momentum.

Relative Density

The density of a substance is defined as mass of a unit volume.

The unit of density is kilogram per metre cube (kg m–3). The density of a given substance, under specified conditions, remains the same.

For example, the density of gold is 19300 kg m-3 while that of water is 1000 kg m-3. The density of a given
sample of a substance can help us to determine its purity.

It is often convenient to express density of a substance in comparison with that of water. The relative density of a substance is  the ratio of its density to that of water:

Relative density=Density  of  a  substance/Density of water

Examples

Q1)Calculate the force required to impact to a car, a velocity of 30 m s-1 in 10 seconds. The mass of the car is 1,500 kg.
Sol. Here u = 0 m s-1; v = 30 ms-1; t = 10 s; a = ?
Using v = u+at, we have
30 = 0+ a (10)
a = 3 m s_2
Now F = ma = 1,500 × 3
or F = 4,500 N

 Q2)A cricket ball of mass 70 g moving with a velocity of 0.5 m s-1 is stopped by player in 0.5 s. What is the force applied by player to stop the ball ?
Sol. Here m = 70 g = 0.070 kg; u = 0.5 m s-1; v = 0; t = 0.5 s
F = 
or F = 
or F = _ 0.07 newton

Q3)What will be acceleration of a body of mass 5 kg if a force of 200 N is applied to it ?
Sol. Here m = 5 kg; F = 200 N
F = ma or a = F/m
a =  = 40 m s-2
a = 40 m s-2

Q4) A bullet of mass 10 g is fired from a rifle. The bullet takes 0.003 s to move through its barrel and leaves with a velocity of 300 ms-1. What is the force exerted on the bullet by the rifle?

Sol. Here m = 10 g = 0.010 kg ; u = a ; v = 300 m s-1
t = 0.003 s, F = ?

or 
or F = 1,000 N

Q5)What force would be needed to produce an acceleration of 1 ms-2 on a ball of mass 1 kg ?
Sol. Here m = 1 kg; a = 1 ms-2 ; F = ?
Now F = m a = 1 × 1
or F = 1 newton

Q6)What is the acceleration produced by a force of 5 N exerted on an object of mass 10 kg ?
Sol. Here F = 5 N; m = 10 kg; a = ?
Now F = ma or 5 = 10 (a)
a = 0.5 ms-2

Q7)How long should a force of 100 N act on a body of 20 kg so that it acquires a velocity of 100 ms-1 ?
Sol. Here v _ u = 100 m s-1, m = 20 kg; F = 100 N ;t = ?
We know F = ma = 
or t =  = 20 s.

Q8)A 1,000 kg vehicle moving with a speed of 20 m s-1 is brought to rest in a distance of 50 m,

(i) Find the acceleration;
(ii) Calculate the unbalanced force acting on the vehicle;
(iii) The actual force applied by the brakes may be slightly less than that calculated in (ii). Why? Give reason.
Sol. (i) Here u = 20 m s_1; v = 0; s = 50 m; a = ?
Using v2 _ u2 = 2as, we have
a =  = -4 ms-2
(ii) F = ma = 1,000 × (-4) = -4,000 N
(iii) Due to force of friction, the actual force applied by brakes may be slightly less than calculated one.

Q9) How much momentum will a dumb-bell of mass 10 kg transfer to the floor if it falls from a height of 80 cm? Take its downward acceleration to be 10 ms–2,

sol .  Mass of dumb-bell = 10 kg

Height, h = 80 cm = 0.8 m

a = 10 m/s2

u = 0

v2 – u2 = 2 as

v2 – (0)2 = 2 × 10 × 0.8

v2 = 16

v = 4 m/s

Momentum p = mv

= 10 × 4

= 40 kgm/s

Q10)An automobile vehicle has a mass of 1500 kg. What must be the force between the vehicle and road if the vehicle is to be stopped with a negative acceleration of 1.7 ms–2?
sol. mass = 1500 kg
      a = –1.7 m/s2
      F = ?
      F = m × a
      = 1500 × (– 1.7)
      = – 2550 N
      The force between the vehicle and road is – 2550 N.