1. Unit 1 Our Dynamic Universe Newton’s Laws
CfE Higher Physics
Unit 1
Our Dynamic Universe
Newton’s Laws

2. Learning Intentions
To analyse the motion of an object using free body diagrams and Newton’s first and second laws.
To be able to carry out calculations using Newton’s second law (F=ma) in one direction only.

3. Newton I
“An object remains at rest or continues to move at constant speed in the same direction unless acted upon by an unbalanced force.”
Newton II
F = ma

4. Newton II
and

5. Newton II (continued)
Since the unit of force, the newton, is defined as the force that will accelerate a mass of 1 kg at 1 ms-2 we can write:
a = F m
or, more commonly:
F = ma

6. Common Forces
There are many different names used for different forces. These are some examples:
Thrust, Th - force from an engine or rocket
Weight, W - downward force due to gravity (W=mg)
Friction, Fr - force opposing motion
Tension, T - pulling force of string, rope, cable or chain
Buoyancy, B - upward force due to floatation
Lift, L - upward force generated by wings or rotors
Reaction, R - force of contact from a surface
(always at right angles to the surface)

7. Friction Friction is a force that opposes motion.
It always acts in the opposite direction of an action.
If you want to find the resultant force (the direction and force the block will move) then you simply take away the friction from the pull force:
Resultant = 6N – 2N = 4N right
Friction 2N
Pull force 6N

8. Air resistance (drag)
Terminal Velocity
In free-fall – unbalanced forces
Weight > drag
Terminal velocity is when an object is in free-fall and reaches its maximum velocity.
Before, the object in free-fall would be accelerating constantly due to gravity (-9.8 ms-2). The forces are unbalanced.
Eventually, the object reaches it’s maximum velocity. This is known as its terminal velocity.
During terminal velocity, the forces acting on the object are balanced.
Weight
Air resistance (drag)
Terminal velocity – balanced forces
Weight = drag
Weight

9. A car driving along a road
Resolving Forces
From National 5, you would have been able to use Newton’s second law linearly and vertically.
It is important in higher that you use free body diagrams for ALL forces problems. This will help you visualise the problem and the forces!
For example:
A car driving along a road
A skier on a ski tow R R T Th Fr Fr W W

10. Resolving Forces Linearly
Linearly, the forces acting on an object would be:
Friction
The Unbalanced force
Any other external force applied to the object
Example:
A block moving with a constant acceleration of 1 ms-2
Pull force
Friction
a= 1 ms-2
Unbalanced force = pull force – force of friction

11. Resolving Forces vertically - lift
Once a free body diagram has been drawn the forces acting in a particular direction can be ‘resolved’. For example:
A person in a lift
Reaction
Resolving upwards: R
- W
= ma
The rest of the problem can then be solved numerically.
Weight

12. Resolving Forces Vertically - Rocket
Thrust
Weight

13. Example
Thrust
Weight = mass x gravity

14. Example

15. Success Criteria
I can analyse the motion of an object using free body diagrams and Newton’s first and second laws.
I can carry out calculations using Newton’s second law (F=ma) in one direction only.

16. Learning Intentions
To be able to work out resultant forces for objects that are tied together and in contact.

17. Objects tied together
Below the tractor and the trailer are moving at a constant velocity.
This means that the forces of the tractor and the trailer are balanced.
This means that the Thrust produced by the engine of the tractor, is equal and opposite to the frictional forces.

18. Objects tied together
The forces on the tractor and trailer are balanced because the thrust and frictional forces are equal and opposite.

19. Objects tied together
The tractor is now accelerating at a constant rate.
Because the tractor is accelerating, the thrust of the tractor’s engine and the frictional forces are now unbalanced. There is an unbalanced force acting on the tractor and the trailer.

20. Objects tied together
IF we assume the frictional forces are the same, we can work out that the unbalanced force is equal to the thrust minus the friction.

21. Objects tied together
The trailer in this example is held by a coupling. The trailer is attached to the tractor, so it must have the same acceleration as the tractor.
There is an unbalanced force acting on the trailer through the tow bar.
The total force exerted through the tow bar is also known as the TENSION, and in this case, the tension is greater than the frictional forces acting on the trailer.
In general:
Unbalanced force = Tension – Friction
Tension = unbalanced force + friction

22. Example Problem 1
Two blocks, connected by a string, are being pulled along a horizontal surface. Their motion is uniform acceleration. The force applied to the right hand 10kg block is 10 N and the frictional force acting on each block is 2 N.
What is the acceleration of the two blocks?
What is the tension in the string?

23. Worked Answer

24. Worked Answer

25. Worked Example 2
A 2,000 kg tractor is towing a trailer of mass 1,200 kg. The tractor can generate an engine force of 6,500 N and the frictional forces on the tractor and trailer are 800 N and 600 N respectively.
a) Calculate the maximum acceleration of the tractor and trailer? R
Resolving horizontally: Th
- Frtractor
- Frtrailer
= ma
Frtractor Th
6, = (2, ,200) a
Frtrailer
5,100 = 3,200 a W
a = 1.59 ms-2

26. Worked Example 2 (continued)
b) Calculate the tension in the coupling between the tractor and trailer?
Trailer only
Resolving horizontally: T
- Frtrailer
= ma R
T = 1,200(1.59)
T = 1, T
Frtrailer
T = 2,508 N W

27. Objects in contact
For two blocks in contact, a similar approach is used.
There are only two acting on the 8 kg block, but 3 acting on the 4 kg block.
Determine the force exerted BY the 4 kg block on the 8 kg block.
Find the acceleration of the whole system.
Consider the forces acting on the 8 kg block ONLY.

28. Objects in Contact example
(a) Find the force exerted by the 4 kg block on the 8 kg block.
(b) Find the acceleration of the blocks.
(a)On 8 kg block, the forces are:
Friction total = 3N
Force of 4 kg on 8 kg = push force – force of friction = 6N

29. Objects in Contact example

30. Objects in Contact tips
Always consider the forces on the mass that is furthest away from the push or pull force. This is because there are only two forces acting on it – The force of friction and the pull or push force.
For the other block, there are three forces acting on it – the force of friction, the push force and the unbalanced force of the other mass on it.
Always get the total mass of the system when finding the acceleration.
Always take away friction from the push or pull force to get the resultant.

31. Success Criteria
I am able to work out resultant forces for objects that are tied together and in contact.

32. Resultant Forces at an angle
When objects are pulled at an angle, like below, they can be resolved into horizontal and vertical components – just like projectiles:

33. Resultant Forces at an angle
If the whole pull force was to act in the horizontal direction, there would be no vertical force. i.e. if you decrease the angle horizontally, the horizontal component of force would equal the pull force.

34. Resultant Forces at an angle
If the whole pull force was to act in the vertical direction, there would be no horizontal force. i.e. if you increase the angle vertically, the vertical component of force would equal the pull force.

35. Resultant Forces at an angle
But what do the horizontal and vertical forces depend on?
The angle – theta, which is the angle to the horizontal.
For example, below we have made theta = 30 degrees and the force being pulled 100N:
Resolve the vector from tip to tail!

36. Resultant Forces at and angle

37. Resultant Force at an Angle
The bigger the angle, the smaller the horizontal force, but the bigger the vertical force.

38. Resultant Force at an Angle
The bigger the angle, the smaller the horizontal force, but the bigger the vertical force.

39. Learning Intentions To be able to:
Resolve the forces for a lift at rest, accelerating and decelerating
Resolve the forces for people in a lift that is at rest, accelerating and decelerating.
Resolve the forces for a person in a lift on newton scales that is at rest, accelerating and decelerating.

40. Forces on a Lift
When a lift is or isn’t in operation, there are two forces that act on it:
Tension in the cable
Weight

41. Forces on a Lift Tension = weight
When the lift is at rest or travelling at a constant velocity, the forces of the tension and weight are balanced:
Tension in the cable
Tension = weight
Weight

42. Forces on a Lift
When the lift is accelerating upwards or decelerating down, you can simply treat the lift as if it is a rocket (the unbalanced force must overcome gravity so must be bigger):
Tension in the cable
Weight

43. Forces on a Lift
When the lift is accelerating down or decelerating up, you can simply treat the lift as if it is a rocket (the unbalanced force is small because it is going in the same direction as gravity):
Tension in the cable
Weight

44. Forces on a Lift
Going UP: The unbalanced force must be greater than the tension and the weight as it is going against gravity.
Going DOWN: The unbalanced force is smaller, because it is going down with gravity.
Tension in the cable
Weight

45. Forces on a Lift – person in lift
When the lift has a person or object in it and is at rest or travelling at a constant velocity, the forces of the tension and weight of both the person and lift are balanced:
Tension in the cable
Tension = weight of lift + weight of person in lift
Weight

46. Forces on a Lift – person in lift
When the lift is accelerating upwards/decelerating down:
Tension in the cable
Tension > weight of lift + weight of person in lift
Weight

47. Forces on a Lift – person in lift
When the lift is accelerating down/decelerating up:
Tension in the cable
Tension < weight of lift + weight of person in lift
Weight

48. Forces on a Lift – person in lift on scales
When the lift is at a constant velocity or not moving, the reading on the scale:
Weight of person = Reaction force
Reaction
Weight

49. Forces on a Lift – person in lift on Newton scales
When the lift is accelerating up or decelerating down:
Reaction force > Weight
Reading on scales is smaller
Reaction
Weight

50. Forces on a Lift – person in lift on Newton scales
When the lift is accelerating down or decelerating up:
Reaction force < Weight
Reading on scales is bigger
Reaction
Weight

51. Learning Intentions I am able to:
Resolve the forces for a lift at rest, accelerating and decelerating
Resolve the forces for people in a lift that is at rest, accelerating and decelerating.
Resolve the forces for a person in a lift on newton scales that is at rest, accelerating and decelerating.