Newtons laws of motion

Sir Isaac Newton (1642 – 1727) played a significant role in developing our idea of Force. He explained the link between force and acceleration.

Section 8A free body diagram is an outline of the object to be considered and shows the forces that act in the direction they act and the point of action.

Consider the aeroplane below:

Draw an outline

Add and label the forces

Weight mg

Lift

Drag

Engine force

Draw a free body diagram for the following:

A hovering helicopter

A down hill skier

Section 9First law:

http://teachertech.rice.edu/Participants/louviere/Newton/law1.html

A mass at rest will remain at rest or move with constant velocity unless acted on by an unbalanced force.

All stationary masses obey the first law

Any mass that moves with a constant velocity (moving in a straight line with constant speed) obeys the first law. Possibly a asteroid in deep space or a rock sliding on ice, might be considered examples.

Second law:

The rate of change of momentum is equal to the magnitude of the unbalanced force and acts in the same direction.

Any mass that accelerates obeys the second law.Example: A kicked football

Any mass that changes direction obeys the second law.Example: A car turning a corner with constant speed

Newton’s second law is written:

Or

If mass is constant: If velocity is constant:

Example 1

A car of mass 800 kg travelling at 24 m s-1, sees a red traffic light and stops in a time of 8 seconds. Calculate the braking force.

To solve this problem what do we need to assume.

•Constant acceleration

•Constant mass

Example 1

A car of mass 800 kg travelling at 24 m s-1, sees a red traffic light and stops in a time of 8 seconds. Calculate the braking force.

Write down what you know:

m = 800 kg, U = 24 m s-1, t = 8 s

The equations to use are:

F = ma (2nd law)

V = U + at

Example 1

A car of mass 800 kg travelling at 24 m s-1, sees a red traffic light and stops in a time of 8 seconds. Calculate the braking force.

Find a using V = U + at

0 = 24 + a8

Giving a = - 3 m s-2

Find F using F = ma

F = 800 x (-3)

Giving F = -2400 N

Example 2

300 kg of Coal falls vertically onto a conveyor belt in one minute. If the conveyor belt moves at a speed of 1.4 m/s, calculate the force needed to move the belt.

Example 2

300 kg of Coal falls vertically onto a conveyor belt in one minute. If the conveyor belt moves at a speed of 1.4 m/s, calculate the force needed to move the belt.

To solve this problem what do we need to assume.

No friction (No other forces)

Belt travels at constant speed

Coal falls at a fixed rate on to belt

Example 2

300 kg of Coal falls vertically onto a conveyor belt in one minute. If the conveyor belt moves at a speed of 1.4 m/s, calculate the force needed to move the belt.

Write down what you know

The equation to use is:

Mass of 300 kg per minute Speed = 1.4 m/s

Example 2

300 kg of Coal falls vertically onto a conveyor belt in one minute. If the conveyor belt moves at a speed of 1.4 m/s, calculate the force needed to move the belt.

Convert rate to per second:Mass/time = 300/60 = 5 kg s-1 Hence m / t = 5 kg s-1 Sub into

Gives F = 1.4 x 5 = 7 Newton

Extra: If this force is not applied what would happen to the belt as the coal fell on it?

The belt would slow down.

Third law:Every action has an equal but opposite reaction.

All forces occur in pairs

When a person pushes (action force) on a wall the wall pushes back (reaction force)

The forces have the same magnitude but act in different directions.

Action Reaction

Draw free body diagrams and identify the pair forces in the following situations

Air escaping from a balloon

A golf club hitting a golf ball

A hovercraft moving forward

SummaryWhat have you learned?