# FHSST Physics/Newtonian Gravitation/Normal Forces

Newtonian Gravitation The Free High School Science Texts: A Textbook for High School Students Studying Physics Main Page - << Previous Chapter (Collisions and Explosions) - Next Chapter (Pressure) >> Properties - Mass and Weight - Normal Forces - Comparative Problems - Falling Bodies - Terminal Velocity - Drag Force - Important Equations and Quantities

# Normal Forces

If you put a book on a table, it does not accelerate; it just lies on the table. We know that gravity is acting on it with a force

 $\begin{matrix}F= G\frac{m_Em_{book}}{r^2}\end{matrix}$ (9.12)

but if there is a net force there MUST be an acceleration and there isn't. This means that the gravitational force is being balanced by another force [Newton's third law!].

This force we call the normal force. It is the reaction force between the book and the table. It is equal to the force of gravity on the book. This is also the force we measure when we measure the weight of something.

The most interesting and illustrative normal force question, that is often asked, has to do with a scale in a lift. Using Newton's third law we can solve these problems quite easily.

When you stand on a scale to measure your weight you are pulled down by gravity. There is no acceleration downwards because there is a reaction force we call the normal force acting upwards on you. This is the force that the scale would measure. If the gravitational force were less than the reading on the scale would be less.

## Worked Example 53 Normal Forces 1

Question: A man weighing 100kg stands on a scale (measuring newtons). What is the reading on the scale?

Step 1 :

We are given the mass of the man. We know the gravitational acceleration that acts on him.

Step 2 :

The scale measures the normal force on the man. This is the force that balances gravity.

Step 3 :

Firstly we determine the net force acting downwards on the man due to gravity.

$\begin{matrix}F_g&=&mg \\&=& 100kg\times 9.8\frac{m}{s^2} \\&=& 980\frac{kgm}{s^2} \\&=& 980N\ downwards \\\end{matrix}$

Step 4 :

We now know the gravitational force downwards. We also know that there must be a reactionary force FN upwards. To determine the magnitude of this we use newton's laws again. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is 0. if we write out the equation:

$\begin{matrix}F_r&=&F_g+F_N \\0 &=& -980N+F_N\\F_N = 980N\ upwards\\\end{matrix}$

The normal force is then 980N upwards. It exactly balances the gravitational force downwards so there is no net force and no acceleration on the man.

Now we are going to add things to exactly the same problem to show how things change slightly. We will now move to a lift moving at constant velocity. Remember if velocity is constant then acceleration is zero.

## Worked Example 55 Normal Forces 2

Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating downwards at $2\frac{m}{s^2}$. What is the reading on the scale?

Step 1 : We are given the mass of the man. We know the gravitational acceleration that acts on him.

Step 2 :

The scale measures the normal force on the man.

Step 3 :

Firstly we determine the net force acting downwards on the man due to gravity.

$\begin{matrix}F_g&=&mg \\&=& 100kg\times 9.8\frac{m}{s^2} \\&=& 980\frac{kgm}{s^2} \\&=& 980N\ downwards \\\end{matrix}$

Step 4 :

We now know the gravitational force downwards. We also know that there must be a reactionary force FN upwards. To determine the magnitude of this we use newton's laws again. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is $2\frac{m}{s^2}$. if we write out the equation:

$\begin{matrix}F_r&=&F_g+F_N \\100kg\times (-2)\frac{m}{s^2} &=& -980N+F_N\\-200\frac{kgm}{s^2} &=& -980N+F_N\\-200N &=& -980N+F_N\\F_N = 780N\ upwards\\\end{matrix}$

The normal force is then 780N upwards. It balances the gravitational force downwards just enough so that the man only accelerates downwards at $2\frac{m}{s^2}$.

## Worked Example 56 Normal Forces 3

Question: A man weighing 100kg stands on a scale (measuring newtons) inside a lift accelerating upwards at $4\frac{m}{s^2}$>. What is the reading on the scale?

Step 1 : We are given the mass of the man. We know the gravitational acceleration that acts on him.

Step 2 :

The scale measures the normal force on the man.

Step 3 :

Firstly we determine the net force acting downwards on the man due to gravity.

$\begin{matrix}F_g&=&mg \\&=& 100kg\times 9.8\frac{m}{s^2} \\&=& 980\frac{kgm}{s^2} \\&=& 980N\ downwards \\\end{matrix}$

Step 4 :

We now know the gravitational force downwards. We also know that there must be a reactionary force FN upwards. To determine the magnitude of this we use newton's laws again. We know that the sum of all the forces must equal the resultant acceleration times the mass. The overall resultant acceleration of the man on the scale is $4\frac{m}{s^2}$. if we write out the equation:

$\begin{matrix}F_r&=&F_g+F_N \\100kg\times (4)\frac{m}{s^2} &=& -980N+F_N\\400\frac{\mbox{kg} \cdot \mbox{m}}{\mbox{s}^2} &=& -980 \mbox{ N}+F_N\\400\mbox{ N} &=& -980 \mbox{ N}+F_N\\F_N = 1380 \mbox{ N}\ upwards\\\end{matrix}$

The normal force is then 1380 N upwards. It balances the gravitational force and then in addition applies sufficient force to accelerate the man upwards at $4\frac{m}{s^2}$.