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2.2. Forces

Here we are going back to the fundamentals with this bit - diving into some Newtonian mechanics. Newton developed a set of three equations describing the motion of objects. These ideas essentially underpin most of what we describe as 'Classical Physics,' i.e. Physics up to the introduction of relativity and quantum ideas. However, Newton's Laws are a pretty accurate model of most of the things we are able to observe on Earth (... unless very small, very fast or very heavy). 

 

We need to now a little bit about the idea of forces. Forces are vector quantities, so all vector mathematics we looked at in Section 1.3 will be important here. This section is also one of the IB favourites to ask questions on in exams, as well as being built upon in subsequent chapters, so it's worth making sure you are completely comfortable with it. Before we dive in, have a little watch of this clip here - a healthy dose of Physics cringe to start us off on this topic. 

The section has been split up as follows:

GodfreyKneller-IsaacNewton-1689.jpg

Big shout out to this guy. Big fan of your work.

DrR

 

Free Body Diagrams

All too often my students decide to either drawing skip these diagrams when solving Mechanics problems or to rush them, missing the key points. I can't emphasise enough just how useful these diagrams can be when solving problems, particularly when stuff starts getting harder (e.g. looking at circular motion or analysing torques on an object).

The key things you need to remember:

  • The length of the force arrows are proportional to their magnitude.

  • The force arrows should start at their point of action (e.g. from the surface for friction/ reaction force, from the CoG for the weight)

  • If an object is in equilibrium (balanced forces), then the sum of vector components in any direction (e.g. vertical and horizontal) will also be in equilibrium (more later)

Let's start with a simple example, of a ship slowing down in the sea. We can see the various forces and their relative sizes through the lengths of each arrow. A few key points about these diagrams

  • Notice that in the vertical direction, the buoyancy and weight forces are equal, therefore there is no acceleration in that direction.

  • In the horizontal direction, the drag force is greater than the thrust vector, therefore the resultant force is backwards, causing an acceleration in that direction.

  • Occasionally it can be useful to annotate a velocity (or acceleration) onto these diagrams for clarity. However, these are NOT forces so make sure they are separate.

  • The situation diagram shows the forces acting from their point of action. e.g. the thrust force acts from the propeller, the weight from the CoG etc.

  • The Free Body Diagram is often drawn as a point as in diagram 2) (thereby ignoring the geometry of the object).

  • The vector polygon allows us to clearly visualise in which direction the resultant force (and therefore acceleration) acts.

FBD.png
geogebra_logo.png

These diagrams definitely don't need to be works of art (coming from someone who is definitely not an artist)! A little sketch can make your working much clearer.

Take a look at this Geogebra simulation - it allows you to vary the forces acting on the object and to see its subsequent motion.

Video Lessons

Resources

IB Physics
Topic 2 Notes
IB-Physics.net
Chapter 2 Summary
IB Revision Notes
Isaac Physics
Forces
Mr. G
2.2 Teaching Notes
2.2 Student Notes
Physics and Maths Tutor
Motion Definitions
Motion Key Notes
Motion Detailed Notes
Mechanics Flashcards
A Level Resources - content slightly different

Questions

Cambridge University Press
Topic 2: Add Qs
Topic 2: Add Qs MS
Topic 2: MCQs
CUP Website Link
Freely available online
Dr French's Eclecticon
Statics and Moments
Link to Dr French's Site
Extension: Pre-University Material
Grade Gorilla
2.2 (Forces) MCQ
Topic 2 (Mechanics B) End Quiz
Quick IB Specific Mixed MCQs
Mr. G
2.2 Formative Assessment
Topic 2 Summary Qs
IB Specific Questions
Physics and Maths Tutor
MCQ Force, Energy, Momentum 1 (AQA 2)
MCQ Force, Energy, Momentum 1 MS (AQA 2)
MCQ Force, Energy, Momentum 2 (AQA 2)
MCQ Force, Energy, Momentum 2 MS (AQA 2)
A-Level Qs: overlapping content
Physics and Maths Tutor
Resolving & Moments (AQA 1)
Resolving & Moments MS (AQA 1)
A-Level Qs: overlapping content
Physics and Maths Tutor
Motion & Force (AQA 1)
Motion & Force MS (AQA 1)
Mechanics (Edexcel 2)
Mechanics MS (Edexcel 2)
A-Level Qs: overlapping content
Physics and Maths Tutor
Forces (OCR)
Forces MS (OCR)
Forces (Edexcel 1)
Forces MS (Edexcel 1)
A-Level Qs: overlapping content
 

Newton's 1st Law

Here we start to look at Newton's 3 Laws of Motion.  These a key idea of classical physics, so it's worth us spending a bit of time exploring these. Crash Course have a nice overview of the 3 laws, take a look.

Newton's 1st Law (N1L) is quoted as follows:

"An object an object will remain at rest or in uniform motion in a straight line unless acted upon by a net external force"

I quite like the following video as an illustration of N1L - the football is in equilibrium once it leaves the astronaut's hand (N.B. equilibrium does not mean it has no forces acting on it, only that there is no net force), and will travel at a constant speed in a straight line until it is acted upon by a resultant force (e.g. it hits the wall).

The reverse is also true, i.e. if an object is travelling at a constant speed in a straight line, it will have no net force acting on it. For example, if I am driving my car at 70 mph down the motorway, the forces acting on my car will be balanced - the vector sum of the engine force, the friction, the air resistance, the weight, the normal reaction force etc will sum to zero.

FBD2.png

Video Lessons

Resources

IB Physics
Topic 2 Notes
IB-Physics.net
Chapter 2 Summary
IB Revision Notes
Isaac Physics
Equilibrium
N1L
Isaac Physics
Equilibrium
Level 1/2 applicable to this section
Mr. G
2.2 Teaching Notes
2.2 Student Notes
Physics and Maths Tutor
Motion Definitions
Motion Key Notes
Motion Detailed Notes
Mechanics Flashcards
A Level Resources - content slightly different

Questions

 

Newton's 2nd Law

Newton's 2nd Law (N2L) can be phrased in a number of ways, but is often described in terms of a rate of change of momentum:

"The rate of change of momentum of a body is directly proportional to the net force applied."

In other words, if there is a net force, the object will accelerate in the same direction as that force. You most likely are more familiar with the equation that described this relationship:

Net Force (N) = mass (kg) x acceleration (msˉ²)

ΣF = ma

You will have come across this before, but at IB the questions they ask generally require a little more thought (... or at the very least a FBD sketch!)

Worked example  - acceleration of a lift

Q. A lift of mass 2000 kg travelling downwards and accelerating at a rate of 0.5 msˉ². Calculate the tension force in the lift cables.

Now, this question is classic IB style, designed to really test your understanding of this topic. Your initial inclination might be to use F = ma to give an answer of 1000 N for the force. However, this is not the final answer!

As with most of these questions, start with a Free Body Diagram. We label all the forces acting on the object, in this case the weight of the lift and the tension force in the cables.

The lift has an acceleration downwards, which means the resultant force must act in that direction. This must mean that the Tension force acting upwards is less than the Weight force acting down.

FBD3.png

From here we know that the resultant force, ΣF = W - T. We also know through Newton's 2nd Law that ΣF = ma.

This means: 

W - T = ma

(2000 × 9.81) - T = 2000 × 0.5

T = 19 620 N - 1000 N

 ∴ T = 18 620 N

Video Lessons

Resources

IB Physics
Topic 2 Notes
IB-Physics.net
Chapter 2 Summary
IB Revision Notes
Isaac Physics
N2L
Level 4/5 Beyond IB
Mr. G
2.2 Teaching Notes
2.2 Student Notes
Physics and Maths Tutor
Motion Definitions
Motion Key Notes
Motion Detailed Notes
Mechanics Flashcards
A Level Resources - content slightly different

Questions

Cambridge University Press
Topic 2: Add Qs
Topic 2: Add Qs MS
Topic 2: MCQs
CUP Website Link
Freely available online
Dr French's Eclecticon
Force and Acceleration
Force and Acceleration Solutions
Link to Dr French's Site
Extension: Pre-University Material
Grade Gorilla
2.2 (Newton's Laws) MCQ
Topic 2 (Mechanics B) End Quiz
Quick IB Specific Mixed MCQs
Isaac Physics
Force and Momentum
Conservation of Momentum
Mixed Questions, including Momentum
Mr. G
2.2 Formative Assessment
Topic 2 Summary Qs
IB Specific Questions
Physics and Maths Tutor
Newton's Laws (AQA 2)
Newton's Laws MS (AQA 2)
A-Level Qs: overlapping content
Physics and Maths Tutor
Motion & Force (AQA 1)
Motion & Force MS (AQA 1)
Mechanics (Edexcel 2)
Mechanics MS (Edexcel 2)
A-Level Qs: overlapping content
Physics and Maths Tutor
Forces (OCR)
Forces MS (OCR)
Forces (Edexcel 1)
Forces MS (Edexcel 1)
A-Level Qs: overlapping content
Physics and Maths Tutor
MCQ Force, Energy, Momentum 1 (AQA 2)
MCQ Force, Energy, Momentum 1 MS (AQA 2)
MCQ Force, Energy, Momentum 2 (AQA 2)
MCQ Force, Energy, Momentum 2 MS (AQA 2)
A-Level Qs: overlapping content
 

Newton's Third Law

At GCSE you may have learnt Newton's Third Law as something about every action having an equal and opposite reaction. I want you to forget that definition right now - seriously. This is one of those areas in the Physics syllabus that is full of misconceptions.

 

Picture1.png

Here is why: as you are sitting in your chair right now, you are not accelerating, therefore the forces acting on you are balanced. Your weight force acting down equals the normal reaction force acting up, equal and opposite right?

 

This is not, I repeat, this is NOT Newton's Third Law.

 

Yes they are equal and opposite, but N3L pairs must act on different objects.

Normal Reaction

Weight

Instead I'd like you to learn the following definition:

"If Object A exerts a force on Object B, then Object B will exert an equal and opposite force (of the same type) on Object A."

(admittedly it's not as catchy, but it is much better at helping you understand what is going on)

Key thing to get to grips with is any Newton's 3rd Law pairs of forces (what you previously thought of as action and reaction) must act on different objects.

The following video shows a nice demonstration of this effect on the International Space Station. 

Let's apply this to the above example, now we need to think about what object are exerting that normal reaction and weight force on you. The normal reaction force is the force the chair exerts upwards on you. Well, N3L says we must exert an equal and opposite force down on the chair. My weight comes from the Earth pulling me down with a force of 800 N. N3L says I must be pulling the Earth upwards with a force of 800 N.

These are what are known as Newton's 3rd Law pairs. These pairs have been colour coded below. 

Picture1.png

Normal Reaction

(of chair pushing

up on you)

Weight

(of Earth pulling

down on you)

Chair.png

Normal Reaction

(of you pushing

down on chair)

earth.png

Weight

(of you pulling

up on Earth)

Video Lessons

Chris Doner
Newton's 3rd Law
IB Specific
Khan Academy
Newton's 3rd Law
More Newton's 3rd
Physics Online
Newton's 3rd Law
Study Nova
Forces and Newton's Laws (N3L)
Forces/ FBDs (Lecture 4)

Resources

IB Physics
Topic 2 Notes
IB-Physics.net
Chapter 2 Summary
IB Revision Notes
Isaac Physics
N3L
Mr. G
2.2 Teaching Notes
2.2 Student Notes
Physics and Maths Tutor
Motion Definitions
Motion Key Notes
Motion Detailed Notes
Mechanics Flashcards
A Level Resources - content slightly different

Questions

 

Friction

Friction is a force that acts to oppose motion. At IB we need to consider both static friction (when stuff is not moving), and dynamic friction (when stuff is moving).

The equation for static friction is given in your formula book as:

F ≤ μs R

Why the inequality? Well, if I push a chair with a force of 10 N, it will push back with a frictgional force of 10 N. If I push with 20 N, it pushes back with 20 N. If I keep increasing the force it will push back with that same frictional force until it starts to slide

Dynamic friction is very similar, except the coefficient might be slightly less (as it's easier to keep things sliding once you've started) i.e. μD > μs.

F = μD R

Crash Course have an overview of this section here.

Let's start with an example of a 100 N block on a flat horizontal surface of μ = 0.5 (let's say both dynamic and static coefficients are equal for this example).

 

Our equation for static friction says that our maximum frictional force before slipping is:

F = μsR

= 0.5 × 100 

= 50 N 

 This means that any applied force below (or equal to) 50 N will not be enough to cause slipping, and the frictional force will be equal to the applied force. Above 50 N and the block will begin to slide. 

  1. No force applied means no frictional forces acting

  2. An applied force of 20N is applied, which means a frictional force of 20N acts to oppose this.

  3. An applied force of 50N  is applied, which is the maximum frictional force permissible. This block is at the point of slipping. Any additional force will cause an acceleration.

  4. An applied force of 70N is greater than the frictional force of 50N, therefore there will be a resultant force (of 20N) and an acceleration to the right. (Note: In reality, the frictional force may decrease once it starts to slide, as the dynamic coefficient of friction is less than the static coefficient).

Friction.png

Angled Slopes

The tricky bit with this topic is having an object on an angled slope, as this draws together all the different aspects of section 2.2. Free body diagrams are key. Remember that the reaction force always acts perpendicular to the surface, and the frictional force always acts along the surface. 

Q. A 20 N block is placed on an angled surface at an angle of 28° to the horizontal. If the block is at the point of slipping find the static coefficient of friction.

As with most of these problems, start with a Free Body Diagram. We have three forces acting, the weight acting directly downwards, the Reaction force out of the slope and the frictional force up the slope. The block is in equilibrium, so these forces are balanced.

blockonslope1.png

These vectors are currently not all perpendicular to one another, so we will need to resolve these. Previously, we have resolved in the vertical and horizontal directions. Here, it actually makes more sense to resolve into the slope and along the slope (rotating our x- and y- coordinate system). That means we only need to resolve our single weight vector that does not fit into this new coordinate system. 

blockonslope2.png

We know that the block is in equilibrium, which means the forces in the x- and y- directions are balanced.

 

(in the y-direction - i.e. out of the slope)

R = W cosθ

 = 20 × cos 28

∴ R = 17.6 N

Now that we have worked out the reaction force, we can substitute this into our equation for static friction to work out our coefficient of friction. 

(in the x-direction - i.e. along the slope)

Ff = μR = W sinθ

μ × 17.6 = 20 × sin 28

∴ μ = 0.53

geogebra_logo.png

A couple of Geogebra simulations to summarise this subsection:

  1. Friction on a slope : See what happens to the forces and the coefficient of friction as we vary the angle.

  2. This simulation ties together some of this resolving along a slope with the Newton's 2nd Law stuff. The sliders allow you to change the mass, the angle, the frictional force and the tension in a string pulling up the slope. By resolving these forces you can work out the resulting acceleration. 

Video Lessons

Resources

IB Physics
Topic 2 Notes
IB-Physics.net
Chapter 2 Summary
IB Revision Notes
Isaac Physics
Friction
Mr. G
2.2 Teaching Notes
2.2 Student Notes
Physics and Maths Tutor
Motion Definitions
Motion Key Notes
Motion Detailed Notes
Mechanics Flashcards
A Level Resources - content slightly different

Questions

Cambridge University Press
Topic 2: Add Qs
Topic 2: Add Qs MS
Topic 2: MCQs
CUP Website Link
Freely available online
Grade Gorilla
2.2 (Forces) MCQ
Topic 2 (Mechanics B) End Quiz
Quick IB Specific Mixed MCQs
Mr. G
2.2 Formative Assessment
Topic 2 Summary Qs
IB Specific Questions

And to finish, another classic Physics joke:

Two cats of the same size slide down a roof at the same time, but one falls off first. Which one?

The one with the smaller “mew.”

 

Additional Resources

Quizlet_Logo.png

Definitions and Key Words : Chapter 2

A set of Quizlet flashcards of the key words and definitions for this chapter is provided here. 

IB Questions

A question by question breakdown of the IB papers by year is shown below to allow you to filter questions by topic. Hopefully you have access to many of these papers through your school. If available, there may be some links to online sources of questions, though please be patient if the links are broken! (DrR: If you do find some broken links, please contact me through the site)

 

Questions on this topic (Section 2) are shown in red.