Normal Force Formula: Concept, Derivation, Sample Questions

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The normal force, is also known as the normal reaction force or a contact force, t. On two surfaces that are not connected to one another, a normal force cannot be applied. Consider a table and a container that cannot exert normal force on one another while they are not in contact. When two objects collide, they exert normal force on each other, which is vertical to the contacting surface. The term "normal" here refers to a perpendicular. The normal force during deceleration is equal to the body's weight. When a body is about to fall, its location on the ground determines when it is about to fall.

Table of Content

Concept of normal force
Normal force formula
Things to remember
Sample questions

Key Takeaways: Normal force, Normal force formula, Force, Gravitational force, Mass, Angle

Concept of Normal Force

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The component of force that is vertical to any contact surface is known as the normal force. It also controls how much force is exerted on the ground by the body. Only if the item is not accelerating, i.e. decelerating, is the normal force equal to its weight. The position in which an object falls on the ground impacts the outcome when it is about to fall. It is denoted by the symbol Fn and is measured in newtons (N).

Also Read: Centripetal force formula

Normal Force Formula

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The normal force Fn on an object at rest on a flat surface equals

Fn= mg

Where,

g = Gravitational Force

m = mass

When a force acting on a falling object causes it to fall at an angle of θ, the Fn is more than the calculated weight.

Fn= mg + Fsinθ

Where,

Fn= normal force

g = Gravitational Force

m = mass

θ = angle of object falling

When a force pulls an object upward, Fnis less than its weight.

Fn = mg – Fsinθ

Where,

Fn= normal force

g = Gravitational Force

m = mass

θ = angle with which body upward

When an object is put on an inclined plane, normal force (Fn) is applied:

Fn= mg cos θ

Where,

Fn= normal force

g = Gravitational Force

m = mass

θ = angle of the inclined surface

Also Read: Force and Laws of motion formula

Things to Remember

The normal force is the force exerted by any surface on any other object.
The net force acting on the object is equal to zero when it is at rest.
It is a proven fact that the descending force, or weight, must equal the upward force, or normal force.
When an object is about to fall, the position in which it falls on the ground is important.

Also Read: Newton’s second law of motion

Sample Questions

Ques: A text-book of mass 1.7 kg is kept on the table. Calculate the normal force being applied on the book. (2 marks)

Ans: Given,

m = 1.7 kg

g = 9.8 m/s

The normal force is

Fn= mg

FN = 1.7 × 9.8

FN = 16.66 N

Thus, the normal force applied is 16.66 N.

Ques: A text-book of mass 0.3 kg is kept on the table. Calculate the normal force being applied on the book. (2 marks)

Ans: Given,

m = 0.3 kg

g = 9.8 m/s

The normal force is

Fn= mg

FN = 0.3 × 9.8

FN = 2.94 N

Thus, the normal force applied is 2.94 N.

Ques: An iron ball of 5 kg is kept on the table. Calculate the normal force being applied on the iron ball. (2 marks)

Ans: Given,

m = 5 kg

g = 9.8 m/s

The normal force is

Fn= mg

Fn = 5 × 9.8

Fn = 49 N

Thus, the normal force applied is 49 N.

Ques: With a force of 300 N, the body falls. If the object has a mass of 20 kg and is inclined at a 30 degree angle. Then calculate the normal force acting on the body. (3 marks)

Ans: Given,

m = 20 kg

F = 300 N

g = 9.8m/s

θ=30°

Sin 30°= ½

The normal force formula is calculated as,

Fn=mg+Fsinθ

Fn=20×9.8+300×sin30

Fn=196+150

Fn=246N.

Thus, the normal force is 246 N.

Ques: With a force of 160 N, the body falls. If the object has a mass of 10 kg and is inclined at a 30 degree angle. Then calculate the normal force acting on the body. (3 marks)

Ans: Given,

m = 10 kg

g = 9.8m/s

F = 160 N

θ=30°

Sin 30°= ½

The normal force formula is calculated as,

Fn=mg+Fsinθ

Fn=10×9.8+160×sin30

Fn= 98+80

Fn= 178N.

Thus, the normal force is 178N

Ques: With a force of 210 N, the body falls. If the object has a mass of 17 kg and is inclined at a 30 degree angle. Then calculate the normal force acting on the body. (3 marks)

Ans: Given,

m = 17 kg

g = 9.8m/s

F = 210 N

θ=30°

Sin 30°= ½

The normal force formula is calculated as,

Fn=mg+Fsinθ

Fn=17×9.8+210×sin30

Fn= 166.6+105

Fn= 271.6 N

Thus, the normal force is 271.6 N.

Ques: A 10kg block of iron is placed on a 30 degree slope above horizontal. What is the normal force acting on the iron block? (2 marks)

Ans: Given,

g = 9.8m/s

m = 10kg

θ = 30 degree

cos 30°=√3/2

Fn= mg cos θ

Fn= 10×9.8×√3/2

Fn= 77.9N

Thus, the normal force is 77.9N.

Ques: A 20kg block of iron is placed on a 30 degree slope above horizontal. What is the normal force acting on the iron block?(2 marks)

Ans: Given,

g = 9.8m/s

m = 20kg

θ = 30 degree

Cos 30°=√3/2

Fn= mg cos θ

Fn= 20×9.8×√3/2

Fn= 169.7 N

Thus, the normal force is 169.7 N.

Ques: A person tries to lift a very massive 70kg rock by applying a 500N upward force, but is unable to do it. Determine how much more force was required to lift the rock off the ground. (2 marks)

Ans: Given,

m= 70kg

g= 9.8m/s

Fg= mg = 9.8×70 = 686N

Additional force required = 686-500 = 186N.

Ques: What is the magnitude of the normal force exerted on a 5kg object that is inclined at 60 degrees to the horizontal? (2 marks)

Ans: Given,

m= 5kg

g= 9.8m/s

θ = 60°

cos 30°= ½

Fn= mg cos θ

Fn= 5×9.8×1/2= 24.5N

Thus, the normal force is 24.5N.

Ques: What is the magnitude of the normal force exerted on a 2kg object that is inclined at 60 degrees to the horizontal? (2 marks)

Ans: Given,

m= 2kg

g= 9.8m/s

θ = 60°

cos 30°= ½

Fn= mg cos θ

Fn= 2×9.8×1/2= 9.8N

Thus, the normal force is 9.8N.

Also Read:

Acceleration	Elastic Collision	Pulley
Kinetic energy	Unit of Moment of Inertia	Torque
Angular Momentum	Angular Speed	Rotational kinetic energy
Radial acceleration	Unit of Power	Unit of work

Normal Force Formula: Concept, Derivation, Sample Questions

Concept of Normal Force

Normal Force Formula

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

3.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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Normal Force Formula: Concept, Derivation, Sample Questions

Concept of Normal Force

Normal Force Formula

Things to Remember

Sample Questions

CBSE CLASS XII Related Questions

1.Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

3.Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4. Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

6.The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be:

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SUBSCRIBE TO OUR NEWS LETTER

1.
Draw a circuit diagram of a full-wave rectifier using p-n junction diodes. Explain its working and show the input-output waveforms.

2.
Assertion (A) : The mass of a nucleus is less than the sum of the masses of the constituent nucleons. Reason (R) : Energy is absorbed when the nucleons are bound together to form a nucleus.

3.
Four independent waves are expressed as \[ (i)\; y_1=A_1\sin\omega t, \] \[ (ii)\; y_2=A_2\sin 2\omega t, \] \[ (iii)\; y_3=A_3\cos\omega t, \] \[ (iv)\; y_4=A_4\sin\left(\omega t+\frac{\pi}{3}\right) \] The interference between two of these waves is possible in

4.
Draw the number of scattered particles versus the scattering angle graph for scattering of alpha particles by a thin foil. Write two important conclusions that can be drawn from this plot.

6.
The figure shows three point charges kept at the vertices of triangle ABC. The net electric field, due to this system of charges, at the midpoint M of base BC will be: