How To Find Normal Force Between Two Blocks: Several Approaches and Problem Examples

The push or pull exerted on body externally which tends to change its position at rest or its uniform motion is know as force.

Forces can be classified in two groups:- Contact force : this force comes into play only when two bodies come in contact. Example, frictional force. Non contact forces: these forces come into play without any contact between two bodies. Example: magnetic force, gravitational force, electrostatic force.

In order to find, “how to find normal force between two blocks’ lets go through following.

The normal force is always perpendicular to the  to the the spot where two bodies are in proximity.

The resultant force can be calculated between two bodies, by the given formula:-

fssR

Where, we know that fs is the static friction, and the latter is know as coefficient of the  limiting friction. And R is the normal reaction which is normal force exerted on body.

What is the force between two blocks

The force between two blocks can be explained by the help of newtons third law of motion.

According to newtons third law of motion, for every action there is an equal and opposite reaction, or we can say that the mutual force exerted by two bodies on one another are always equal in magnitude and always in opposite direction.

We can write the force exerted on them as follow:-

Fab+Fba=0

Fab=-Fba

Here, the action force and reaction never exerts on the same body.

Now, to calculate the force between two blocks ,lets go through the following:-

Let there be two objects having mass m and 2m respectively.

Now to calculate the force between these two blocks having mass of m and 2m, we will derive the equation:-

Let force F acts on the block having mass m:-

We know that F=ma so , a=F/m

Now let the force be f for the other block having mass 2m, we will calculate its acceleration  as follow:-

Now on equation acceleration served upon both the blocks can be equated as follow:-

F/m=f/2m

F/1=f/2

f=2F

So now we can say that a block having mass ‘m’ exerts a force 2F on block having having mass 2m.

How do you calculate the force between two forces

When an object faces several forces, then the resultant force on the object is the force that will produce the equal acceleration collective to all forces.

To calculate the force between two forces, lets understand it through the example:-

Let us consider a body on which two forces act . One force is of 10N from the right direction and the force is of 6N from the left direction.

In the given figure the resultant force will be in right direction and its computation will be of 4N . As the force with larger magnitude is in right direction. Here the resultant force will 4N, as these forces are in opposite direction so they get subtracted.

Let us consider a body exerting forces in the same direction:-

Suppose a force of 3N and 4N acts in right direction only. Here the resultant force on the body will be in right direction itself and the force will be 7N. The forces gets add up as they were in same direction.

Let us consider a body exerting force at an angle:-

Here two forces exerts  at an angle 45 degree. Having force of magnitude 30 N and 40 N respectively.

To calculate the force between these two forces we will break the force in its component and use the formula, as shown below:-

R=F1+F2

How do you find the net force between two objects

Lets go through following examples to understand, how to find net force between two blocks,

Q. When a bullet strikes a heavy wooden block and travels inside it where its mass is 0.08 kg and its is moving at a speed of 180 ms2. The bullet  after striking is stopped at a distance of 60 cm. Now , Find the net resistive force exerted on the bullet by the block.

Let us take a as the retardation by the bullet. Then from Newton’s law we know that:-

v2=u2-2as

Now, on putting the values we will get,

Now the resistive force will be:-

F=ma=0.08kg*27000ms-2=2160N

Q. Consider a wooden block having mass 4 kg which is resting on a floor. When a cylinder of mass 50 kg is put on the block, the floor yields and the block and the cylinder together go under an acceleration of 0.2ms2. Now Find the exertion of the block on the floor  in the given condition (a) before and (b) after the floor yields. where g= 10ms2.

Here two forces act upon the block. So the gravitational attraction by earth will be 40N. And the normal force will be R. As the block is at rest the net force will be 0.

R=mg=40N ( By first law)

Now by Newton’s third law the action of the block on the floor will be equal and opposite to R. Which will be 40N directed vertically downward.

Now for block and cylinder, the force of gravity will due to earth will be 500N. And R will be the normal force.

Now we know the system is accelerating downward, then the net force will be:-

From the above calculation we got to know that, the action of block and cylinder on the floor is 490N, which is directed vertically downwards.

Q. There is a toy train which is at rest position, where the frictional force is 15 N , now when a force of 40 N is applied on it. Then find the  the net force between them.

We know that the exerted force on toy train is 40 N

And the Frictional force acting is , -15N

For the calculation of net force , we will use the below formula :-

Net force =force exerted + frictional force

Net force = 40 – 15

Net force= 25 N

Therefore, the net force will be given as 25 N.

Q. A door exerts a pull on both of its faces. Where on side it faces a pull of 78 N, whereas on the other side it faces a pull force of 39 N. now calculate the net forces exerted on the door.

We know that the exerted force on the door is 78 N and 39 N respectively.

For the calculation of net force , we will use the below formula :-

Net force =force exerted on one side + force exerted on the second side

Net force =78-39

Net force=39N

Riya Pandey

I am Riya Pandey. I have completed Post Graduation in physics in 2021. Currently I am working as a Subject Matter Expert in Physics for Lambdageeks. I try to explain Physics subject easily understandable in simple way.