- Define The Newton.
- Use free body diagrams to analyse the forces acting on an body.
- Carry out calculations involving the relationship between the unbalanced force, mass and accelerations where the resolution of forces is not required.
- Carry out calculations involving work done, potential energy, kinetic energy and power.

Labelling Force Vectors

Work Done = kinetic Energy and Potential Energy

Aids to Understanding

International System of Units

The second, kilogram and meter are called SI units (International System or Système International d'unités)
There are seven units in total. The units are often prefixed by a term that denotes multiples of thousandth or thousands.
For example a millisecond is one thousandth or a second and a microsecond is one thousand thousandth (millionth) of a second
These terms are precisely measured. For example the actual mass of a **kilogram** is almost exactly equal to the mass of 1 litre of water and its
prototype lies in a vault in **International Bureau of Weights and Measures** in France. The **metre** is defined as the distance
light travels through
a vacuum 1/299,792,458 of a second. (We clearly know the speed of light very acurately). The **second** is defined to be
the duration of 9,192,631,770 periods of the radiation corresponding to the radiation produced during an electronic transition
between the two hyperfine levels of the ground state of the caesium 133 atom. We will appreciate this meaning when we have studied
Unit 3.

Note: A point about Labelling Force Vectors

If we have a a cart being pulled by a horse, then the
the force vector is labelled F_{CH}, however
in the free body diagram of the cart it is acceptable
to write F_{H}; the body diagram indicates the
body experiencing the force.
Example:
Draw a free body diagram for ..

i) a log floating on water

ii) a helicopter beginning to rise

iii) a ball freely falling

iv) a ball freely rising

(Note: iii) and iv) have identical body diagrams)

Analysing the Lift

The motion of a lift is controlled by the cable attached to it.
This cable can exert a force on the lift (F_{LC})that is ..

- equal to to weight of the lift
- greater than the weight of the lift
- less than he weight of the lift

However the lift can have the following motions

- the lift can be at rest
- the lift can pick up speed going up
- the lift can travel upwards at a uniform (steady) speed
- the lift can slow down going up
- the lift can pick up speed going down
- the lift can travel downwards at a uniform speed
- the lift can slow down going down

So we have three different cable tensions enabling seven different stages of a lift's motion.

Problem:

i) Draw a free body diagram for a lift of mass 3000 kg for each of the seven
stages. Assume the (F_{LC}) (aka Tension) can be no more
than 35000 N or no less than 25 000 N and use these accordingly.

ii) Find the accelerations of the lift and find the reading on Newton scales if a woman of
50 kg is standing on them at each of the seven stages.

Forces Down Slopes

A block on a slope will accelerate down the slope but clearly the acceleration is less than -9.8 ms^{-2}.
To find the force on the block down the slope due to the Earth, we draw a block of mass m on a slope of angle q.
We draw in the Force on the block due to the Earth, resolve this force into two rectangular components, one of which is parallel
to the slope (blue). Put these vectors head to toe, use simple trigonometry and you will find the force on a block down a slope = mgSinq.

Unbalanced Force (N) | Mass (kg) | Acceleration (ms^{-2}) |
---|---|---|

2 | 1 | ? |

4 | ? | 4 |

? | 4 | 2 |

Look at starred * rows for the car and ** rows for the parachutist, notice we have identical free body diagrams but the motion is different in each case.

Direction of velocity | Nature of motion | Free body diagram | free body diagram showing Resultant Force only(N) | Accel = F_{r}/m (ms^{-2}) |
---|---|---|---|---|

Velocity to the Right | A car is travelling at a steady speed of 3ms^{-1}. |
0 | ||

Velocity to the Right | A car is getting faster | 0.5 | ||

*Velocity to the right | A car is slowing down | -0.5 | ||

*Velocity to the left | The car is speeding up | -0.5 | ||

Velocity of skydiver is always downwards (negative) |
A skydiver is travelling at -20ms^{-1} and getting faster |
5 | ||

**Velocity of skydiver is always downwards (negative) |
The skydiver is falling at a steady speed of 60ms^{-1} |
0 | ||

Velocity of skydiver is always downwards (negative) |
The skydiver has pulled the ripcord | 40 | ||

**Velocity of skydiver is always downwards (negative) |
The skydiver is falling with a gentle constant speed of 5 ms^{-1>} |
0 |

Situation | Force Notation | How it is read | Action or Reaction? |
---|---|---|---|

Lift pulled vertically by cable | F_{LC} |
Force on Lift due to Cable | Action |

Man pulled by Earth | F_{ME} |
Force on Man due to Earth | Action |

Man pushed up by scales | F_{MS} |
Force on Man due to Scales | Reaction |

Apple supported by balance | F_{AB} |
Force on Apple due to Balance | Reaction |

Tug boat pulling an oil Rig | F_{RT} |
Force on Rig due to Tugs | Action |

Sprinter pushed by ground | F_{SG} |
Force on Sprinter due to Ground | Reaction |

- The acceleration of the lift.
- The reading on the Newton balance when a 160 g apple is hanging from it.
- The reading on the scales when an 80 kg man is standing on them.

The Work Done by a force on a body is the energy transferred to the body. In mechanics the energy the body acquires can manifest itself as kinetic energy, potential energy or heat energy. The following argument shows that Work Done on a body equals its change in Kinetic Energy. (provided there are no other forces present).

As you read this argument draw a diagram to support it. Suppose a block is at rest and u = 0. Now suppose a uniform force pulls (Works) the block until its velocity is v. If this change in velocity occurred in a time t, then the average velocity of the block is v/2 and its acceleration is v/t. If the block was displaced s metres during this time then the average velocity of the block is s/t. Thus s/t = v/2 or s = vt/2. Now E

A similar argument can be used to show E