# How to Define Force in Physics and Some Basic Problems About Force Calculation

Updated on January 20, 2020 ## What is the Force

Force is one of the primary physical quantities being existent in the world of Physics. It is referred to in the study of almost all the physical phenomenons involving motion of objects. Physics deals with formula, mathematical calculations and properties of Force in a detailed manner. Here we will learn about all of them.

What is the definition of Force in physics?

"A push or pull that changes the momentum of a physical object is called Force"

The momentum is the product of mass and velocity of a moving object. The mass will be constant, so we can say that the application of force will change the velocity only. Hence there arise another definition of force which is as follows:

"The force may also be defined as the push or push in a specific direction which tends to change the velocity of a moving object."

Here are some examples of the force:

• When you apply brakes on your moving car, the force of friction between the tyres and the road tends to decrease the speed of the car and help in stopping it.
• A cyclist applies the pull on the pedals of his Bicycle to move fastly.
• A football player changes the speed and direction of the ball by applying the force with the kick.

These were a few examples of the application of the force in our world. Now you can find how significant is the study of this basic term in the world of physics.

## Properties of Force

All physical quantities have some basic properties which are used in the study of the science behind their effects and actions in the universe. Force has the following basic properties which the Physics considers in the study of this physical quantity:

1. Force is a vector quantity i.e it has both the direction and magnitude.
2. The value of force cannot be zero.
3. The addition and subtraction of force cannot be done algebraically.

## Derivation of Formula to Calculate the Force

According to physics, the force is the push or pull that changes the momentum of an object. Let us consider the motion of an object on which the force is applied for time "t" seconds.

Here, the mass of object = m kg

Initial velocity of the object = v1 metre/second

So the initial momentum of the object will be = mass x velocity = m x v1

Let us suppose that after applying the force for time t, the velocity becomes v2 meter per second.

The final momentum of the object will be= m x v2

As per the definition, the force is the push or pull to change the momentum of the object. Therefore

Force = Change in momentum

Time taken for the change

= (m x v2 − m x v1)÷ t

= m( v2 − v1 ) ÷ t

now v2 − v1 = change in velocity and t = time taken for this change

so ( v2 − v1 ) ÷ t = rate of change of velocity = acceleration

So our equation of force becomes:-

Force = mass x acceleration

F = m x a

where a stands for the acceleration that object recieved in the given time t.

## Units of Measurement

In the SI system, the unit of measurement of force is Newton. Here is the definition,

The force applied on an object of 1 kilogram mass will be 1 Newton if it changes its velocity by 1metre per second in one second i.e acceleration of 1 m/s2

So we can say that:-

1 Newton = 1 Kilogram x 1 metre/second2

In the CGS system, the unit of force is Dyne. Here is the definition:-

One Dyne may be defined as the force required to accelerate the object of mass 1 gram by 1 centimetre per second2

1 Dyne = 1 gram x 1 centimetre/second2

Conversion of units from one system to another:-

Here is the conversion of units from the SI system to the CGS system:

We know that

1 kilogram = 1000 gram

1 metre = 100 centimetre

And

1 Newton = 1 Kilogram x 1 Metre/ Second2

= 1000 gram x 100 centimetre/second2

= 100000 x 1 gram x 1 centimetre/second2

As we learned that 1 gram x 1 centimere/second2 = 1 Dyne

Substituting the value in above equation we get:

1 Newton = 100000 Dyne

Similarly, the unit to unit conversion can be done for systems(like FPS, MKS etc.) by changing the values of mass and acceleration for the same.

## Example of Calculation of Force

Problem: The car T is moving at a speed of 50 km/hour. It is hit by another car J moving in the same direction from the rear side. The speed of the T becomes 80 km/hour in 4 seconds. Calculate the force exerted(In Newton) by the J on T if the mass of T including the passengers is 1500 kilogram.

Solution:

Here the mass of the object = 1500 Kg

Initial Velocity = 50 Kilometre/Hour

We know that 1 kilometre = 1000 metre

1 hour = 3600 seconds

Converting the units in the SI system = 50 x 1000 metre/3600 seconds

Initial velocity = 13.8 metre/second

Final velocity = 80 km/h = 80 x 1000 metre/ 3600 seconds

= 22.2 metre/second

From the formula given above we have:

Force = Mass x (V2 − V1)/t

where V2 is the final velocity, V1 initial velocity and t stands for the time taken for the change

We have Initial velocity = 13.8 metres/second

Final velocity = 22.2 metre/second

time = 4 seconds

Mass = 1500 kg

Substituting the values in above equation we get

Force = 1500 kg x (22.2 metre/second − 13.8 metre/second)/4 seconds

Force = 3150 Newton

So we can say that the car J exerted a force of 3150 Newton on car T.

## Scope of Study of Force in Physics

The study of forces plays an important role in all systems including the motion of objects. Here are some examples:

• The working of the engine of an automobile depends upon the internal forces exerted by the combustion of fuel.
• Farming equipment is designed keeping in view the kind of forces applicable to its working.
• The design of an aeroplane significantly depends upon the buoyant forces applied by the air on its wings and body.
• The calculation of forces existent in an elevator system is very important for the smooth and safe working of the lift.

## What will be the force applied if it changed the speed of a bicycle from 3 m/s to 4 m/s in one second. The weight is 70 kg.

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This content is accurate and true to the best of the author’s knowledge and is not meant to substitute for formal and individualized advice from a qualified professional.