# Newton’s laws & contact Forces between blocks in Mechanics

Newton’s laws of motion gives us concepts of force (first law), measurement of force with its equation (second law), and reaction of force (3rd law). Today we will discuss an interesting topic of mechanics. It will cover a few interesting applications of Newton’s Laws and analyse different force distribution on blocks in contact. We will discuss different situations like when blocks are standing side by side or when one block is resting on another block. Sometimes there is friction between common surfaces, sometimes not. Sometimes the frame of reference is not inertial. Let’s start with our case 1.

## Case 1: Application of Newton’s laws for 2 blocks in contact

Say, 2 blocks are standing side by side and a force F is applied on one surface of the left block.

The masses of the blocks are shown in the diagram.

Now what will happen?

How can I get the acceleration of the blocks m1 and m2? What is the net force acting on the block with mass m2?

### Solution:

Let’s consider the blocks as a single system and evidently the mass of that combined system is (m1 +m2). Now if the acceleration of this system is, say, a then using Newton’s 2nd law of motion we get:
F = (m1 + m2) a

=>   a = F / (m1 + m2)

This means both the blocks will move with this acceleration.

From this, we can find the fraction of F being applied on the block with mass m2 (at the common surface between the blocks).

This contact force on the block m2 is, say F1.

F1= mass of that block x Acceleration of that block = m2 .a = m2. F /(m1 +m2)

As per the diagram, the direction of F1 is left to right (same direction of F)

It clearly shows that block m2 gets a fraction of the force F.

Now following the 3rd Law of motion, block with mass m2 also exerts an equal and opposite reaction force F2 on the block of mass m1. So F2 = – F1.

So the direction of F2 as per this setup would be from right to left on the block of mass m1.

And the magnitude of F2 would be same as F1 i.e. m2. F /(m1 +m2)

Now if we consider the free body diagram of the block with mass m1 then see what we find:

So net force on the block with mass m1 = F – F2 = F – m2. F /(m1 +m2) = m1. F/(m1 +m2).

We can validate the acceleration from this expression as well.

acceleration of block m1 = net force/mass =[ m1. F/(m1 +m2)]/m1 = F/(m1 +m2).
It is same as the one we derived at the beginning.

### Case 1 Summary:

Net force on the left block with mass m1 = m1. F/(m1 +m2)

And

Net force on the right block with mass m2 = m2. F/(m1 +m2)

If you add these two net force components you will get the total force F that we started with.

So certainly a horizontal force will get proportionately distributed among the blocks resting side by side according to their masses. To be noted: In this case we didn’t consider any friction.

## Case 2: Application of Newton’s laws for 2 blocks in contact

Say a block of mass m1 is resting on another block of mass m2. Let’s call these blocks as block m1 and block m2 respectively in the rest of our discussion. Block m2 is resting on a surface.

Assumption for this case: There is no friction between the surfaces.

Action: A force F is applied on the block m1 horizontally as shown in the diagram.

Result:

Now will any component of this force F be transmitted to the block m2, in this case? Or any side effect of this force F?
Note that, there is no friction between the surfaces of m1 and m2.

• The force F is horizontal and it doesn’t have any nonzero component towards the block m2.
(component of F towards m2= F cos90 = 0)
• Without the presence of friction between m1 and m2 and between m2 and the bottommost surface, we don’t have any frictional force as well on m1 and m2.

So only m1 gets a force applied on it in this case.
Acceleration of m1 is: a = F/m1.

Now m1 will move on m2 towards right with this acceleration.

m2 doesn’t change is static condition as it doesn’t face any external force. It’s as per Newton’s first law and this is an example of Inertia of Rest.

Now, if the length of m2 block is L then the block m1 will traverse the distance L in, say, time t

L = (1/2) a t2
t = (2L/a)1/2

So after t = (2L/a)1/2     m1 block will fall from the roof of m2.

## Case 3: Application of Newton’s laws for 2 blocks in contact

Now if the force F is applied on the block m2 (the lower block) and if we consider that there is no friction between the surfaces then what will happen?

Similar to the previous case the F won’t have any nonzero component pointing towards m1 as Fcos90 equals zero. There is no friction between the surfaces of m1 and m2; so there won’t be any force component on m1 arising from frictional ground.

As a result, m2 is the only candidate facing the force F.
And as there is no friction between m2 and the bottommost surface, hence acceleration of the block m2 will be F/m2.

With this acceleration m2 block will start moving and it moves ahead without any horizontal movement of m1 block (no force on it as discussed). Eventually m1 will fall vertically downwards as the block m2 moves ahead. It’s as per Newton’s first law and this is an example of Inertia of Rest.