# Normal force

## Normal Force

1. Normal Force:

What is a Normal force?
• .The normal force comes into play any time two bodies are in direct contact with one another
• It always acts perpendicular to the body that applies the force.
Let’s take the example of a block resting on a horizontal surface.

Clearly a gravitational force acts on the block, pulling it down, perpendicular to the surface. Since the block is not accelerating, another force must act to counteract the gravitational force. This force is applied by the surface and is called the normal force, and is referred to as FN .Let’s now draw a free body diagram to understand the case. Free body diagram (FBD) : While studying mechanics, when we examine the forces acting on an object there are five "classic" types that are usually considered:

weight normal friction tensions applied forces

We use freebody diagrams to illustrate the magnitude and direction of all of the forces acting directly on a single object (usually represented by a rectangle). Consider a scenario in which a mass is being pulled across a table by a cord.

The weight vector begins at the object's center of mass and points towards the center of the earth.

A normal vector begins at the point of contact between the mass and its supporting surface. It is directed perpendicularly away from the surface and passes through the object's center of gravity.

Tensions are forces conducted along strings, ropes, and wires. They begin at the point of contact and point in the direction in which they are pulling.

Friction forces begin at the same point as the normal and act parallel to the sliding surface. They always oppose motion.

Applied forces is a catch-all, generic category encompassing any other interactions. In our current example, there are no generic applied forces.

If a force acts at an angle, then we usually work with its x- and y-components.

If an object is in static (at rest) or dynamic (constant velocity) equilibrium, then all of the forces acting on it are balanced.

The magnitude of the forces acting to the left equals the magnitude of the forces acting to the right. The magnitude of the forces acting upwards equals the magnitude of the forces acting downwards.

In this case:

x: f = T cos θ y: + T sin θ = mg

If the forces were not balanced, then the object would be accelerated in the direction of the unbalanced force. For example, using the same forces as in our previous example, if T cos θ were greater than f, then Newton's Second Law will allow us the ability to calculate the object's acceleration towards the right as it starts gaining speed.

net F = ma T cos θ - f = ma (a > 0)

However, if T cos θ were less than f, then the object would still move towards the right but it would be losing speed.

net F = ma T cos θ - f = ma (a < 0)

Now let’s apply the concept of free body diagram.

There are only two forces in the system in figure above. The forcemg which is the weight of the block, and is acting on the surface. R1 = mg through its centre (since the body is symmetric). From FBD, we can see that net external force on the block is zero. That is why; it is stationary on the surface.

• The normal force can also be seen as a direct consequence of Newton's Third Law. • Since the normal force is a reactive force, its magnitude is independent of the nature of the force causing it. • The most common normal force is caused by gravity, as seen on the block resting on a plane surface. However, there can be additional forces that also cause a normal force.

1. The surface experiences a total force of 25 N, and reacts with a normal force of 25 N, keeping the block in equilibrium. 2.

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2. The Normal Force on an Inclined Plane

Consider the case of a block resting on an inclined plane. In this instance, the gravitational force on the block is not perpendicular to the plane.

Free Body Diagram of an Inclined Plane

In order to calculate the normal force for this situation we must find the component of the gravitational force that is perpendicular to the plane. What does our free body diagram predict in this case? To find out we analyze all forces acting upon the object. • The perpendicular gravitational force ( F cosθ ) cancels exactly with the normal force ( F N ) • The left out parallel gravitational force ( F sinθ ), which points down the plane. Thus the block will accelerate down the incline. • The normal force applies in any situation in which a force is exerted on an object by direct contact from another object.

1.We solve the problem by drawing a free body diagram, and resolving all force vectors into components parallel and perpendicular to the plane:

The component of the gravitational force perpendicular to the plane is given by: F GP = F Gsin 45 o = 10 sin 45 o = 7.07N

Similarly, the component of the applied force perpendicular to the plane is:
F HP = F sin 45 o = 10 sin 45 o = 7.07N

Thus the normal force on the block is simply the sum of the two perpendicular forces, or 14.14N .

Question 1 What is the normal force acting on a 8-kg mass that is at rest on a horizontal surface?

0.8 N 0 N 1.2 N 78.4 N

Question 2 What is the normal force acting on a 8-kg mass that is sitting on the floor of an elevator which is accelerating upwards at a rate of 10 m/sec2?

89.8 N 78.4 N 158.4 N 1.60 N

Question 3 What is the normal force acting on a 8-kg mass that is sitting on the floor of an elevator which is accelerating downwards at a rate of 5 m/sec2?

3.02 x 101 N 3.84 x 101 N 7.84 x 101 N 4.98 x 101 N

Question 4 What is the normal force acting on a 8-kg mass that is being pulled up a 40º frictionless inclined plane at a constant velocity of 10 m/sec?

7.84 x 101 N 6.01 x 101 N 5.04 x 101 N 5.04 x 102 N

Question 5 What is the normal force acting on a 8-kg mass which is being pulled at a constant velocity of 10 m/sec by a rope having a tension of 28 newtons at an angle of 40º to the horizontal?

78.4 N 96.4 N 60.4 N 57.0 N

Question 6 What is the normal force acting on a 8-kg mass which is being pushed at a constant velocity of 10 m/sec by a force of 28 newtons along a rigid handle that makes an angle of 40º to the horizontal?

96.4 N 60.4 N 99.8 N 57.0 N

References:
1)http://dev.physicslab.org/PracticeProblems/Worksheets/APB/normals/assortment.aspx
2) http://dev.physicslab.org/Document.aspx?doctype=3&filename=Dynamics_FreebodyDiagrams.xml
3) http://www.askiitians.com/iit-jee-physics/mechanics/free-body-diagram.aspx
4)http://www.sparknotes.com/physics/dynamics/newtonapplications/section3.rhtml