1. Topic 2.2 Forces and Dynamics
Mechanics
Topic 2.2 Forces and Dynamics

2. Forces and Free-body Diagrams
To a physicist a force is recognised by the effect or effects that it produces
A force is something that can cause an object to
Deform (i.e. change its shape)
Speed up
Slow Down
Change direction

3. The last three of these can be summarised by stating that a force produces a change in velocity
Or an acceleration

4. Free-body Diagrams
A free-body diagram is a diagram in which the forces acting on the body are represented by lines with arrows.
The length of the lines represent the relative magnitude of the forces.
The lines point in the direction of the force.
The forces act from the centre of mass of the body
The arrows should come from the centre of mass of the body

5. Example 1 Normal/Contact Force A block resting on a worktop
Weight/Force due to Gravity
A block resting on a worktop

6. Example 2 A car moving with a constant velocity Normal/Contact Force
Weight/Force due to Gravity
Motor Force
Resistance

7. Example 3 A plane accelerating horizontally Upthrust/Lift Motor Force
Weight/Force due to Gravity
Motor Force
Air Resistance

8. Resolving Forces
Q. A force of 50N is applied to a block on a worktop at an angle of 30o to the horizontal.
What are the vertical and horizontal components of this force?

9. Answer
First we need to draw a free-body diagram
30o
50N

10. We can then resolve the force into the 2 components
Vertical = 50 sin 30o
30o
50N
Horizontal = 50 cos 30o

11. Therefore Vertical = 50 sin 30o = 25N
Horizontal = 50 cos 30o = 43.3 = 43N

12. Determining the Resultant Force
Two forces act on a body P as shown in the diagram
Find the resultant force on the body.
30o
50N
30N

13. Solution
Resolve the forces into the vertical and horizontal componenets (where applicable)
30o
50N
30N
50 sin 30o
50 cos 30o

14. Add horizontal components and add vertical components.
50 sin 30o = 25N
50 cos 30o – 30N = 13.3N

15. Now combine these 2 componentsR
R2 =
R = 28.3 = 28N

16. Finally to Find the AngleR 
tan  = 25/13.3
 =
 = 62o
The answer is therefore 28N at 62o upwards from
the horizontal to the right

17. Springs
The extension of a spring which obeys Hooke´s law is directly proportional to the extending tension
A mass m attached to the end of a spring exerts a downward tension mg on it and if it is stretched by an amount x, then if k is the tension required to produce unit extension (called the spring constant and measured in Nm-1) the stretching tension is also kx and so
mg = kx

18. Spring Diagram x

19. Newton´s Laws
The First Law
Every object continues in a state of rest or uniform motion in a straight line unless acted upon by an external force

20. Examples Any stationary object!
Difficult to find examples of moving objects here on the earth due to friction
Possible example could be a puck on ice where it is a near frictionless surface

21. Equilibrium
If a body is acted upon by a number of coplanar forces and is in equilibrium ( i.e. there is rest (static equilibrium) or unaccelerated motion (dynamic equilibrium)) then the following condition must apply
The components of the forces in both of any two directions (usually taken at right angles) must balance.

22. Newton´s Laws
The Second Law
There are 2 versions of this law

23. Newton´s Second Law
1st version
The rate of change of momentum of a body is proportional to the resultant force and occurs in the direction of the force.
F = mv – mu F = t t

24. Newton´s Second Law
2nd version
The acceleration of a body is proportional to the resultant force and occurs in the direction of the force.
F = ma

25. Linear Momentum
The momentum p of a body of constant mass m moving with velocity v is, by definition mv
Momentum of a body is defined as the mass of the body multiplied by its velocity
Momentum = mass x velocity
p = mv
It is a vector quantity
Its units are kg m s-1 or Ns
It is the property of a moving body.

26. Impulse From Newtons second law F = mv – mu F = t t Ft = mv – mu
This quantity Ft is called the impulse of the force on the body and it is equal to the change in momentum of a body.
It is a vector quantity
Its units are kg m s-1or Ns

27. Law of Conservation of Linear Momentum
The law can be stated thus
When bodies in a system interact the total momentum remains constant provided no external force acts on the system.

28. Deriving This Law
To derive this law we apply Newton´s 2nd law to each body and Newton´s 3rd law to the system
i.e. Imagine 2 bodies A and B interacting
If A has a mass of mA and B has a mass mB If A has a velocity change of uA to vA and B has a velocity change of uB to vB during the time of the interaction t

29. Then the force on A given by Newton 2 is
FA = mAvA – mAuA t
And the force on B is
FB = mBvB – mBuB t
But Newton 3 says that these 2 forces are equal and opposite in direction

30. Therefore mAvA – mAuA = -(mBvB – mBuB) t t
Therefore mAvA – mAuA = mBuB – mBvB
Rearranging mAvA + mBvB = mAuA + mBuB
Total Momentum after =
Total Momentum before

31. Newton´s Laws
The Third Law
When two bodies A and B interact, the force that A exerts on B is equal and opposite to the force that B exerts on A.

32. Example of Newton´s 3rd
Q. According to Newton’s third Law what is the opposite force to your weight?
A. As your weight is the pull of the Earth on you, then the opposite is the pull of you on the Earth!

33. Newton´s 3rd Law
The law is stating that forces never occur singularly but always in pairs as a result of the interaction between two bodies.
For example, when you step forward from rest, your foot pushes backwards on the Earth and the Earth exerts an equal and opposite force forward on you.
Two bodies and two forces are involved.

34. The equal and opposite forces do not act on the same body!
Important
The equal and opposite forces do not act on the same body!