# Lesson 10: Normal Forces

**Notice**: Trying to get property 'post_author' of non-object in

**/home/forge/cilearning.ca/web/app/themes/buddyboss-theme/learndash/ld30/lesson.php**on line

**168**

Download here: Ontario Curriculum Expectations

A firm faith in the universal providence of God is the solution of all earthly troubles.B.B. Warfield

Chances are that even as a child you knew that the force of gravity causes things to fall towards the ground.

The downward force of gravity causes the ball to fall faster and faster towards the ground. Yet, what happens when the ball reaches the ground? You know from experience that it stops falling and comes to rest on the ground. Why doesn’t it get pulled straight through the ground? Has the gravity stopped acting on it? Of course not. The force of gravity is still pulling down on the ball. However, because it is not moving, there must be another force acting upwards on it that balances out the downward force of gravity. We call this force the normal force.

#### The Normal Force

In physics, the word ānormalā means perpendicular, or at a 90Ā° angle. The normal force is the force that prevents an object falling through whatever it is that it is resting on. In the case of the ball in the above example, it is the force of the ground that acts upwards on the ball in order to prevent it from being pulled through the ground.

It is important to remember that although the force of gravity will always act down, the normal force does not necessarily act up. It acts in the direction perpendicular to the surface that the object is resting on. The diagram below shows how this may be the case.

Watch this video to find out more about the normal force:

#### Calculating the Normal Force

The normal force will not cause an object to start moving. It only provides enough force to prevent an object from falling through the surface it is resting on.

#### Example 1

The box below is resting on the ground and has a mass of 10 kg. Find the force of gravity and the normal force acting on the box.

**Given:**

m = 10 kg

g = 9.8 m/s^{2}

**Required:**

F_{g} = ?

F_{N} = ?

**Solution:**

F_{g} = mg

F_{g} = (10 kg)( 9.8 m/s^{2})

F_{g} = 98 N

Since the box is at rest, the amount of force acting up must be equal to the amount of force acting down. F_{N} is the only force acting up and F_{g} is the only force acting down. Therefore, they must be equal.

F_{N} = F_{g}

F_{N} = 98 N [up]

Consequently, the force of gravity is 98 N [down] and the normal force is 98 N [up].

In many situations, the force of gravity will equal the normal force. However, this is not always true:

F_{g} = F_{N} will only occur when:

- The object is on a horizontal surface.
- F
_{g}= F_{N}are the only vertical forces acting on the object.

#### Example 2

If a boy with a mass of 50 kg sits on the box from example one, what is the new normal force acting on the box?

The force of gravity acting on the boy must be added to the force of gravity already acting on the box. As a result, there will be a much greater downward force acting on the box. Consequently, there must be a greater upward normal force acting on the box.

**Given:**

F_{g} = 98 N (From example 1)

m_{boy} = 50 kg

g = 9.8 m/s^{2}

**Required:**

F_{g (of boy)} = ?

F_{N} = ?

**Solution:**

F_{g (of boy)} = (m_{boy})g

F_{g (of boy)} = (50 kg)(9.8 m/s^{2})

F_{g (of boy)} = 490 N

Total force acting down on the box = F_{g(box)} + F_{g(of boy)} = 98 + 490 = 588 N

Since the box is at rest, the total force acting up on the box must be the same as the total force acting down on the box. Therefore, F_{N} = 588 N [up].

#### Practice Questions

- A car has a mass of 1 200 kg. A very strong weightlifter attempts, unsuccessfully, to lift the car by applying an upward force of 2 500 N. What is the normal force acting on the car as the weightlifter is applying this force? Be sure to include a force diagram in your answer.

**Answer**

**Given:**

m_{car} = 1 200 kg

F_{weightlifter} = F_{w} = 2 500 N [up]

g = 9.8 m/s^{2}

**Required:**

F_{g(of car)} = ?

F_{N} = ?

**Solution:**

F_{g(of car)} = (m_{car})(g)

F_{g(of car)} = (1200 kg)(9.8 m/s^{2})

F_{g(of car)} = 11 760 N

In order for the car to be at rest the forces up must equal the forces down.

F_{up} = F_{down}

F_{N} + F_{weightlifter} = F_{g}

F_{N} + 2500 N = 11 760 N

F_{N} = 9260 N [up]

The normal force acting on the car is 9 260 N [up].