9th Forces And Law of Motion

Force 

A force is a push or pull upon an object resulting from the object's interaction with another object. Whenever there is an interaction between two objects, there is a force upon each of the objects. When the interaction ceases, the two objects no longer experience the force. Forces only exist as a result of an interaction.

Balanced and Unbalanced Forces
Balanced forces do not cause a change in motion. They are equal in size and opposite in direction.

Example:- If any one man compete against another one who is just about as strong as second is , there will probably be a time when both men are pushing as hard as they can, but both men’s arms stay in the same place. This is an example of balanced forces. The force exerted by each person is equal, but they are pushing in opposite directions, in this case together.


Because the force that each man is exerting is equal, the two forces cancel each other out and the resulting force is zero. Therefore, there is no change in motion.

Unbalanced forces always cause a change in motion. They are not equal and opposite.

When two unbalanced forces are exerted in opposite directions, their combined force is equal to the difference between the two forces and is exerted in the direction of the larger force.

LAW OF MOTION
First Law of Motion:- Every object in a state of uniform motion tends to remain in that state of motion unless an external force is applied to it.


Second Law of Motion- The change of momentum of a body is proportional to the impulse impressed on the body, and happens along the straight line on which that impulse is impressed.

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.

MATHEMATICAL FORMULATION OF SECOND LAW OF MOTION

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.

which is the rate of change of velocity. 

The quantity, k is a constant of proportionality. 

The SI units of mass and acceleration are kg and ms-2 respectively. 

The unit of force is so chosen that the value of the constant, k becomes one. 

For this, one unit of force is defined as the amount that produces an acceleration of 1 m s-2 in an object of 1 kg mass.

That is,  1 unit of force = k × (1 kg) × (1 m s-2).

Thus, the value of k becomes 1.
F = ma

The unit of force is kg m s-2 or newton, which has the symbol N.

The second law of motion gives us a method to measure the force acting on an object as a product of its mass and acceleration.

The first law of motion can be mathematically stated from the mathematical expression for the second law of motion
F = ma 

F = m(v-u)/t

or 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, u throughout the time, t.

If u is zero then v will also be zero. That is, the object will remain at rest.

Example:-

1) A constant force acts on an object of mass 5 kg for a duration of 2 s. It increases the object’s velocity from 3 m s–1to 7 m s-1. Find the magnitude of the applied force. Now, if the force was applied for duration of 5 s, what would be the final velocity of the object?

Solution:
We have been given that u = 3 m s–1

and v = 7 m s-1, t = 2 s and m = 5 kg.

we have,

F = m(v-u)/t = 5 (7-3)/2

Now, if this force is applied for a duration of 5 s (t = 5 s), then the final velocity can be calculated by

v = u + (Ft/m)
On substituting the values of u, F, m and t, we get the final velocity, v = 13 m s-1.

2) Which would require a greater force –– accelerating a 2 kg mass at 5 m s-2 or a 4 kg mass at 2 m s-2?


Solution: we have F = ma.
Here we have m1 = 2 kg; a1 = 5 m s-2
and m2 = 4 kg; a2 = 2 m s-2.

Thus, F1 = m1a1 = 2 kg × 5 m s-2 = 10 N;
    and F2 = m2a2 = 4 kg × 2 m s-2= 8 N.

     Or, F1 > F2.

Thus, accelerating a 2 kg mass at 5 m s-2 would require a greater force.

3) A motorcar is moving with a velocity of 108 km/h and it takes 4 s to stop after the brakes are applied. Calculate the force exerted by the brakes on the motorcar if its mass along with the passengers is 1000 kg.

Solution:
The initial velocity of the motorcar    u = 108 km/h = 108 × 1000 m/(60 × 60 s) = 30 m s-1

and the final velocity of the motorcar    v = 0 m s-1.

The total mass of the motorcar along with its passengers = 1000 kg

and the time taken to stop the motorcar, t = 4 s.

we have the magnitude of the force applied by the brakes F as   m(v – u)/t.

On substituting the values, we get

F = 1000 kg × (0 – 30) m s-1/4 s
   = – 7500 kg m s-2 or – 7500 N.

The negative sign tells us that the force exerted by the brakes is opposite to the direction of motion of the motorcar.

4) A force of 5 N gives a mass m1, an acceleration of 10 m s–2 and a mass m2, an acceleration of 20 m s-2. What acceleration would it give if both the masses were tied together?
Solution:  we have m1 = F/a1; and    m2 = F/a2. Here, a1 = 10 m s-2;

a2 = 20 m s-2 and F = 5 N.

Thus, m1 = 5 N/10 m s-2 = 0.50 kg; and m2 = 5 N/20 m s-2 = 0.25 kg.

If the two masses were tied together, the total mass, m would be m = 0.50 kg + 0.25 kg = 0.75 kg.

The acceleration, a produced in the combined mass by the 5 N force would be,
a = F/m = 5 N/0.75 kg = 6.67 m s-2.

Third Law of Motion: - For a force there is always an equal and opposite reaction: or the forces of two bodies on each other are always equal and are directed in opposite directions.

For more clear cut explanation: jsuniltutorial