Chap 3 3a to 3d

Introduction

• This chapter teaches you how to deal with forces acting on an object

• You will learn how to use several formulae (inspired by Isaac Newton)

• You will learn how to model situations involving friction, particles moving on slopes and when joined over pulleys

• You will also learn laws of momentum and impulse

Dynamics of a Particle moving in a Straight Line

You can use Newton’s Laws and the formula F = ma to solve problems involving forces and acceleration

Before we start looking at question we need to go through some ‘basics’ that are essential for you to understand this

chapter…

Newton’s second law of motion“The force needed to accelerate a particle is

equal to the product of the mass of the object and the acceleration required”

F = ma

Force is measured in Newtons (N). A Newton is:

“The force that will cause a mass of 1kg to accelerate at 1ms-2”

3A

You need to understand all the forces at work in various situations…

The Normal Reaction

The normal reaction acts perpendicular to the

surface which an object is resting on

It is equal and opposite to the force exerted on

the surface by the object, which is

determined largely by gravity and the mass of

the object

R

mg (mass x gravity)

The table matches the force from the brick, which is why the brick remains still on the table (there of course would be a

maximum possible weight the table could take, but we will not worry about this for

now!




chapter…



F = ma



3A


Direction of motion

Frictional Force

Frictional Force

The frictional force opposes motion between two ‘rough’

surfaces

Although it is a force, friction does not cause movement in its own

direction. It just reduces the effect of another force

Surfaces will have a maximum level of friction where it is unable to completely prevent movement




chapter…



F = ma



3A


Tension in string

Tension

If an object is being pulled along (for example by a

string), then the force acting on the object is called the

Tension

Tension = PULLING force




chapter…



F = ma



3A


Thrust

Thrust

If an object is being pushed along (for example by a rod), then the force acting

on the object is called the Thrust (or sometimes compression)

Tension = PUSHING force




chapter…



F = ma



3A


Resistance

Any object moving through air, fluid or a solid will experience resistance caused by

the particles in the way

Gravity

Gravity is the force between any object and the earth.

The Force caused by gravity acting on an object is its weight

Remember Newton’s formula…

The Force is called the weight Mass is just mass!

The acceleration due to gravity is 9.8ms-2 (or can be left as ‘g’




chapter…



F = ma



3A


Resolving

When there are multiple forces acting on an object, we ‘resolve’ these forces in different

directions

One direction will usually be the direction of acceleration

The other will be perpendicular to this



3A

R

mg (mass x gravity)

Normal Reaction

Direction of motion

Frictional Force

Tension

Thrust


You can use Newton’s Laws and the formula F = ma to solve

problems involving forces and acceleration

Find the weight in Newtons, of a particle of mass 12kg

3A

The mass is already in kg, and use acceleration due to gravity

Calculate

As the acceleration was given to 2sf, you should give you

answer to the same accuracy

Ensure you use the exact amount in any subsequent

calculations though!




Find the acceleration when a particle of mass 1.5kg is acted on by a force of 6N

3A

Sub in F and m

Divide by 1.5




Find the values of the missing forces acting on the object in the diagram

below

3A

X

Y

2g N

4N

2ms-2

In this example you need to consider the horizontal forces and vertical forces separately (This is called

resolving)

Resolving HorizontallyTake the direction of acceleration as the positive one

2kg Resolving Vertically

Take the direction of the force Y as positive

Sub in values. You must subtract any forces acting in the opposite

direction!Calculate

Add 4

Sub in values. Acceleration is 0 as there is none in the vertical

directionCalculate

Add 2g




Find the values of the missing forces acting on the object in the diagram

below

3A

X

Y

4g N

80N

2ms-2

In this example you need to consider the horizontal forces and vertical forces separately (This is called

resolving)

Resolving HorizontallyTake the direction of acceleration as the positive one

4kg Resolving Vertically

Take the direction of the force Y as positive

Sub in values. You must subtract any forces acting in the opposite

direction!Calculate

Add X and Subtract 8

Sub in values. Acceleration is 0 as there

is none in the vertical directionCalculate

Add 20, add 4g

20N


You can solve problems involving forces by drawing a diagram

including all relevant forces, and then resolving in multiple directions

if necessary

A particle of mass 5kg is pulled along a rough horizontal table by a force of 20N,

with a frictional force of 4N acting against it. Given that the particle is initially at

rest, find:

a)The acceleration of the particleb)The distance travelled by the particle in

the first 4 secondsc)The magnitude of the normal reaction

between the particle and the table

3B

20N

5g N

R

a ms-2

5kg4N

Start by drawing a diagram

a) Resolve horizontally and sub in values. Take the direction of acceleration as positive

Calculate a



including all relevant forces, and then resolving in multiple directions if

necessary


with a frictional force of 4N acting against it. Given that the particle is initially at rest,

find:

a)The acceleration of the particle – 3.2ms-2

b)The distance travelled by the particle in the first 4 seconds

c)The magnitude of the normal reaction between the particle and the table

3B

20N

5g N

R

3.2ms-2

5kg4N


Use SUVAT

Sub in values

Calculate

b)



including all relevant forces, and then resolving in multiple directions if

necessary


with a frictional force of 4N acting against it. Given that the particle is initially at rest,

find:

a)The acceleration of the particle – 3.2ms-2

b)The distance travelled by the particle in the first 4 seconds – 25.6m

c)The magnitude of the normal reaction between the particle and the table

3B

20N

5g N

R

3.2ms-2

5kg4N


c)

Calculate

Resolve vertically, taking R as the positive direction



including all relevant forces, and then resolving in multiple

directions if necessary

A small pebble of mass 500g is attached to the lower end of a light string. Find the tension in the string

when the pebble is:

a)Moving upwards with an acceleration of 2ms-2

b)Moving downwards with a deceleration of 4ms-2

3B


0.5g N

T

2ms-2 0.5kg

a) Resolve vertically, taking the direction of the

acceleration as positive

Calculate T





A small pebble of mass 500g is attached to the lower end of a light string. Find the tension in the string

when the pebble is:

a)Moving upwards with an acceleration of 2ms-2 – 5.9N

b)Moving downwards with a deceleration of 4ms-2

3B


0.5g N

T

-4ms-2 0.5kg

b) Resolve vertically, taking the direction of movement

as positive

Calculate T

Even though the pebble is moving downwards, there is more tension in the string as the pebble is decelerating – the

string is working against gravity!

In this case, the pebble is moving downwards at a

decreasing rate, so you can put the acceleration on as

negative





A particle of mass 3kg is projected at an initial speed of 10ms-1 in the

horizontal direction. As it travels, it meets a constant resistance of magnitude 6N. Calculate the

deceleration of the particle and the distance travelled by the time it

comes to rest.

3B

3kg

R

3g N

6N

a ms-

2 Start by drawing a diagram

It is important to note that the initial projection speed is NOT a force, there are actually no forces acting in the positive direction

Deceleration

Take the direction of movement as positive – remember to

include 0 as the positive force!

Calculate a

So the deceleration is 2ms-

2





A particle of mass 3kg is projected at an initial speed of 10ms-1 in the

horizontal direction. As it travels, it meets a constant resistance of magnitude 6N. Calculate the

deceleration of the particle and the distance travelled by the time it

comes to rest.

Deceleration = 2ms-2

3B

3kg

R

3g N

6N

a ms-

2 Start by drawing a diagram

It is important to note that the initial projection speed is NOT a force, there are actually no forces acting in the positive direction

Distance travelled

Sub in values

Work through to calculate s


If a force at applied at an angle to the direction of motion you can resolve it to find the component

of the force acting in the direction of motion

3C

A horizontal force has no effect on the object in the vertical direction

10N

A vertical force has no effect on the

object in the horizontal direction

10N

10N

20°

OppHyp

Adj

SO

H CA

H TO

A

Opp = Sinθ x Hyp

Opp = Sin20 x 10

10sin20

10cos20

Adj = Cosθ x Hyp

Adj = Cos20 x 10

So a force can be split into its horizontal and vertical components using Trigonometry!

However, a force at an angle will have some

effect in BOTH the horizontal and vertical

directions!




Find the component of each force in the x and y-directions

3C

9N

x

y

40°

9Cos40

9Sin40

Force in the x-direction Force in the y-direction

= 9Cos40

= 6.89N

= 9Sin40

= 5.79N




Find the component of each force in the x and y-directions

3C

12N

x

y

23°12Cos23

12Sin23

Force in the x-direction Force in the y-direction

= 12Cos23

= 11.05N

= 12Sin23

= 4.69N

= -11.05N

(This will be negative as it is the opposite direction to x!)


You can calculate the magnitude of a frictional force using the coefficient of

friction

Friction is a force which opposes movement between two ‘rough’ surfaces.

It is dependent on two things:

1) The normal reaction between the two surfaces

2) The coefficient of friction between the two surfaces

The maximum frictional force is calculated as follows:

If a surface is described as ‘smooth’, the implication is that the coefficient of friction is

0. 3D



friction


A block of mass 5kg is lying at rest on rough horizontal ground. The coefficient of friction between the block and the ground is 0.4. A rough horizontal force, P, is applied to the block. Find the magnitude of the frictional

force acting on the block and its acceleration when:

a)P = 10Nb)P = 19.6Nc)P = 30N

3D

10N

F

R

5kg

5g N

Draw a diagrama ms-

2

We need to find the maximum possible frictional force

To do this we need R, the normal reaction

Resolve vertically

Calculate R

Now we can calculate the maximum possible frictional force

Sub in values

Calculate FMAX



friction




a)P = 10Nb)P = 19.6Nc)P = 30N

3D

10N

F

R

5kg

5g N

Draw a diagrama ms-

2

The maximum frictional force is 19.6 N

Any force will be opposed by friction up to this value

For part a), the force is only 10NTherefore, the frictional force will match this at 10N, preventing movementHence, there is also no acceleration

10N



friction




a)P = 10Nb)P = 19.6Nc)P = 30N

3D

19.6N

F

R

5kg

5g N

Draw a diagrama ms-

2

The maximum frictional force is 19.6 N

Any force will be opposed by friction up to this value

For part b), the force is only 19.6NTherefore, the frictional force will match this at 19.6N, preventing movementHence, there is also no acceleration

This situation is called ‘limiting equilibrium’, as the object is on the point of movement

19.6N



friction




a)P = 10Nb)P = 19.6Nc)P = 30N

3D

30NF

R

5kg

5g N

Draw a diagrama ms-

2

For part c), the force is 30NThe frictional force will oppose 19.6N of this, but no more.Hence, the object will accelerate…Resolve horizontally!

19.6N

Sub in values and resolve horizontally

Calculate

Divide by 5

So the acceleration will be 2.08ms-

2



friction


A 5g box lies at rest on a rough horizontal floor. The coefficient of friction between the

box and the floor is 0.5. A force P is applied to the box. Calculate the value of P required to

cause the box to accelerate if:

a)P is applied horizontallyb)P is applied at an angle of θ above the

horizontal, where tanθ = 3/4

3D

PF 5kg

5g N

R Draw a diagram

Resolve vertically to find the normal reaction

Sub in values and resolve vertically

Calculate

Now find the maximum frictional force

Sub in values

Calculate

So P will have to exceed 24.5N to make the object move!



friction





a)P is applied horizontally – 24.5Nb)P is applied at an angle of θ above the


3D

P

F 5kg

5g N

RDraw a diagram

θ

Pcosθ

Psinθ

θ

SO

H CA

H TO

ATanθ = 3/4

So Opp = 3And Adj = 4

3

4

5Opp

Adj

Hyp

We need to find the values of Cosθ and Sinθ. The ratio for Tanθ can be used to find these!

We can find the hypotenuse using

Pythagoras’ Theorem!

Sinθ = 3/5 Cosθ = 4/5

Sinθ = 0.6 Cosθ = 0.8

0.8P

0.6P



friction







3D

P

F 5kg

5g N

RDraw a diagram

θ

0.8P

0.6P

Resolve vertically to find the normal reaction

Sub in values and resolve vertically

We find the normal reaction in terms of

P

Now find the maximum frictional force

Sub in values

Find Fmax in terms

of PSo, 0.8P will have to exceed this if the box is

to move…



friction







3D

P

F 5kg

5g N

R

Draw a diagram

θ

0.8P

0.6P24.5 – 0.3P

We need to find the value for P for which the box is in ‘limiting equilibrium’ – that is, so the horizontal forces cancel each other

out…Resolve

horizontally…Sub in values and

resolve horizontallyCareful with the bracket!

Rearrange and solve

P must exceed 22N, which is less than when P was horizontal

The reason is because some of the force is upwards, this alleviates some of the friction between the

surfaces…

Chap 3 3a to 3d

Education

Transcript of Chap 3 3a to 3d