Vector and scalar quantities

• A scalar quantity is defined only by its magnitude (or size)

for example: distance, speed, time ….

• A vector quantity is defined by stating both its magnitude

(or size) and direction for example: displacement, velocity,

force……

It is easy to combine two or more

scalar quantities e.g.

2 metres + 3 metres = 5 metres!

The direction must be taken into

account when combining two or

more vector quantities

e. 2 metres + (-3 metres) = -1 metre

Introduction to

Forces II

Sharon Tripconey

Some Basic PrinciplesNewton’s first law (N1L)

• Every particle continues in a state of rest or uniform motion in a straight line unless acted on by a resultant external force.

This means that for a particle to be in equilibrium* it must be the case that there is no resultant force acting on it (*dynamic or static equilibrium)

Newton’s second law (N2L)

• When a force acts on a particle, the change in momentum is proportional to the force. For constant mass, F = ma

Newton’s third law (N3L)

• When one object exerts a force on another there is always a reaction that is equal in magnitude and opposite in direction to the applied force.

This means that we expect forces to be found in ‘pairs’

Draw a force diagram

Assumptions:

•Each object is a particle in equilibrium

•The scales and the brush have zero mass

W

W = Weight of the person

Hint: Think about pairs of

forces (N3L)

W

F

F

S

SR

R

N

N

W Weight of the person

F Force exerted on the brush

S Contact force (brush and floor)

N Contact force (person and scales)

R Contact force (scales and floor)

W

Problem:

Explain why the reading goes down when I press on the floor

with a brush

F has the same magnitude as

the force exerted on the brush

N is the contact force (the reading on

the bathroom scales)

Consider the forces acting

on the person

F + N – W = 0+

F + N = W

W is the weight of the person (which

is constant)

Newton’s Second LawIn a given direction F = ma

Example

A toy train of mass 0.5 kg moves on a horizontal

straight track. There is a driving force of 0.4 N and

a resistance to motion of 0.35 N. Find the

acceleration.

0.4 N0.35 N

a m s -2

0.5 kg

Newton’s Second Law

The acceleration is 0.1 ms -2 in the direction of the 0.4 N force.

2 Resultant force = ma

0.4 0.35 0.5

0.05 = 0.5

0.1

N L

a

a

a

0.4 N0.35 N

a m s -2

0.5 kg

Two identical objects are connected to the ends of a light

inelastic string which passes over a fixed pulley as shown

What happens if the system is released from rest?

What happens if….

Motion in a lift

A boy of mass 50 kg is standing in a lift of mass 200 kg. The lift is raised by a vertical cable. Find the reaction of the floor on the boy and the

tension in the cable when the acceleration is 1

4𝑔 upwards.

T

R 200g

¼ g

lift and boy (250kg)boy (50kg) lift (200kg)

¼ g ¼ g

50g

R T

250g

Motion in a lift

14

50 50 so 62.5R g g R g

14

200 200 so 312.5T R g g T g

For the lift and boy, N2L ↑,

𝑇 − 250𝑔 = 250 ×1

4𝑔 so 𝑇 = 312.5𝑔

For the boy, N2L ↑

For the lift, N2L ↑,

• A model is a representation of a real situation.

A real situation will invariably contain a rich

variety of detail and any model of it will simplify

reality by extracting those features which are

considered to be most important.

• Modelling is at the heart of the subject of

Mechanics

Mathematical modelling

Modelling assumptions• Common assumptions are:

• the body is a particle

• value of g is constant

• the string is light and inextensible

• the pulley is light and smooth

• air resistance is negligible

• friction is negligible

• friction obeys the law F µ R

Modelling Cycle

Real-world

problem

Simplifying

assumptions

Mathematical

model

(equations etc)

Analysis

and solutionPredictionExperiment

How can we support students with their

understanding of mathematical modelling in

Mechanics?

• Incorporate simple practical demonstrations into our teaching;

many students quickly see something of the principle even if

their depth of understanding is not great

• As teachers, knowledge of common student misconceptions

helps us to plan lessons that will reduce the likelihood of their

development and increase the likelihood that misconceptions

already formed will be corrected.

• An approach of confronting hard things is usually more

successful than that of avoiding them for as long as possible!

About MEI• Registered charity committed to improving

mathematics education

• Independent UK curriculum development body

• We offer continuing professional development

courses, provide specialist tuition for students and

work with employers to enhance mathematical

skills in the workplace

• We also pioneer the development of innovative

teaching and learning resources