Forces

Things to recall

• A force is usually simply defined as a push or pull.

• Alt: An influence which causes motion.• A force acting on a body can cause the object

to accelerate.• When multiple forces act on a body, the

resultant (or net) force is a superposition of the forces.

• We start this discussion by recalling Newton’s 2nd law.

• Definition: The net force on a body is equal to the product of its mass and acceleration.

amF

(As an equation)

• We start this discussion by recalling Newton’s 2nd law.

• Definition: The net force on a body is equal to the product of its mass and acceleration.

• Note that we can write this law in component form for each axis.

amF

(As an equation)

zzyyxx maFmaFmaF ,,

Application of Newton’s laws

Example

Example

• A jumbotron weighing 48000N is suspended above the 3Ws field by cables shown in the diagram. Find the tension in the cables if there is equilibrium.

2020

• We start by drawing a free body diagram.

• We start by drawing a free body diagram.• A free body diagram is one where all of the

forces acting on a body are shown while all other forces and bodies are removed.

• Using a free body diagram helps you to visualise the problem!

• Identify the system of interest .– In this case the jumbotron is the object of interest.

• Identify and draw of the external forces acting on the body.

• Identify the system of interest .– In this case the jumbotron is the object of interest.

• Identify and draw of the external forces acting on the body.– The forces on the jumbotron are its weight and

the tension due to the cables.

• Write the equations for each axis.

Example

• Free body diagram:

2020

T1 T2

W=48000N

• Sol: writing Newton’s 2nd law:T1 T2

W

20 200, xxnet maF

0, yynet maF

(1)

(2)

Fnet = 0 in each case as there is no acceleration in equilibrium

• Sol: writing Newton’s 2nd law:

• Resolving components,

T1 T2

W

20 200, xxnet maF

0, yynet maF

(1)

(2)

Fnet = 0 in each case as there is no acceleration in equilibrium

020sin20sin 21 ymaWTT

020cos20cos 12 xmaTT +ve dir(3)

(4)

• From equation 3,

NT 7017120sin2

480001

020cos20cos 12 xmaTT

020sin20sin 21 ymaWTT

21 TT

4800020sin20sin 11 TT

NT 701712

Special types of Forces

Some special forces

• Normal force• Friction• Tension• Gravitational force

Friction

• A simple explanation of friction (or formally the frictional force) is the resistance to motion.

• The friction is attributed to a single force.• The friction acts in the direction opposing the

intended motion.a

Tfr

• The friction force is usually:– Proportional to the force pressing to surfaces

together (the normal force)– Depends on the “roughness” of the surfaces.

Normal force

• The normal force is the force which occurs when an object is in contact with a stable object.

• It is an example of an applied force. • It can be considered the “supporting” force.• The normal force acts perpendicular to the

surface to which the applied the force acts.

• From Newton’s third law, a book placed on a table pushes down on the table and the table pushes up on the book.

Halliday/ Resnick/ WalkerFundamentals of Physics

The forces acting on the book: The weight of the book and the normal force N. The normal force acts perpendicular to the surface of table.

N

W

Tension

• The tension is a force which is transmitted through a string, rope, cable or wire when it is pulled tight by forces acting from opposite ends.

• The tension force is directed along the length of the wire and pulls equally on the objects on the opposite ends of the wire.

Pull Tension (T)

Rope attached to a wall

Tension (T)

Newton’s 3rd law

Applications of Action-Reaction

• When an object interacts with another object they exert a force on each other.

• These forces are equal in magnitude and are called action-reaction forces.

• Definition: when two bodies interact, the forces on the bodies from each other are always equal in magnitude and opposite in direction.

• Consider the given situation:

Halliday/ Resnick/ WalkerFundamentals of Physics

FBT (The force that the table exerts on the book)

FTB (The force that the book exerts on the table)

The table-book interaction

Example

• Two blocks attached by a “massless” cord which slides over a frictionless pulley as shown below. The hanging block falls causing the large block to slide. Find the (a) acceleration of the blocks and (b) tension in the cord.

M

mFrictionlesssurface

a

kgM 4 kgm 2

Solution

• Show the forces acting on the blocks.

M

m

a

a

N

T

mg

N

T

Mg

Mg

mg

T

(The free body diagrams for each block)

Sliding block

• Consider the sliding block:

• Newton’s 2nd law for the x and y direction:

N

T

MgMaT (x-direction)

0 yMaMgN (y-direction)

a

Sliding block

• Consider the sliding block:

• Newton’s 2nd law for the x and y direction:

N

T

MgMaT (x-direction)

0 yMaMgN (y-direction)

MgN

MaT MgN

a

Falling block

• Consider the falling block:

• Newton’s 2nd law for the y direction:

T

mg

a

mamaTmg y

agmT

List of equations

MaT

MgN agmT

(1)(2)

(3)

• Using equations 1 and 3, we eliminate T:

MaT agmT

(1)

(3)

agmMa

mgamM

mM

mga

• Using equations 1 and 3, we eliminate T:

• Substituting into equation 1:

MaT agmT

(1)

(3)

agmMa

mgamM

mM

mga

mM

MmgT

• Therefore:

• and,

23.3

24

8.92

msa

NT 1.133.34

More about friction

• Here we will be considering the friction between dry solid surfaces.

• Here we will be considering the friction between dry solid surfaces.

• We recall that there are two types of friction for this case: kinetic and static friction.

• Here we will be considering the friction between dry solid surfaces.

• We recall that there are two types of friction for this case: kinetic and static friction.

• Consider the following example when each case occurs.

• Consider a book resting on a table. The forces acting on it are shown below.

Halliday/ Resnick/ WalkerFundamentals of Physics

FN

W

• An external force F is applied to the book. Again the forces acting on it are shown below.

Halliday/ Resnick/ WalkerFundamentals of Physics

FN

W

fsF (Stationary)

The applied force is not large enough to overcome friction

• The friction at this point is static friction.

• As the magnitude of the external force F is increased so does the friction.

• NB: The applied force must balance the friction.

Halliday/ Resnick/ WalkerFundamentals of Physics

FN

W

fsF (Stationary)

The applied force is still not large enough to overcome friction

• The friction increases with the applied force until it reach a maximum.

• At this point the book will move when a greater external force is applied.

• When the external force is greater than the friction the book is moving. At this point the friction acting on the book is kinetic friction.

Halliday/ Resnick/ WalkerFundamentals of Physics

FN

W

fkF (Moving)

The book begins to accelerate when the friction is overcome.

a

Definition

• Static Friction: the friction that occurs between the two surface when the two surfaces are at rest relative to each other.

• Kinetic Friction: when there is relative motion between surfaces.

Properties of Friction

• For static friction, the static frictional force balances the component of the net external force parallel to the surface.

• When the static friction reaches a maximum:

• Where is the coefficient of static friction.– a measure of the relative amount of adhesion

between the surfaces

Nf ss max,

s

Properties of Friction

• When the body is moving the kinetic friction is given by:

• Where is the coefficient of kinetic friction.

Nf kk

k

Example

• A 75kg roller is pulled at angle of 42° along a cricket pitch at constant velocity. If the coefficient of friction between the roller and pitch is 0.1, find the tension T in the handle.

• The free body diagram:

TN

mg

fr

42

• The free body diagram:

TN

mg

fr

42T cos42

T sin42

• Newton’s 2nd for each axis:

042cos mmafT xr

TN

mg

fr

42T cos42

T sin42

042sin mmamgNT y

• Newton’s 2nd for each axis:

042cos mmafT xr

TN

mg

fr

42T cos42

T sin42

042sin mmamgNT y

42sinTmgN rfT 42cos

• Newton’s 2nd for each axis:

• For a moving body,

042cos mmafT xr

TN

mg

fr

42T cos42

T sin42

042sin mmamgNT y

42sinTmgN rfT 42cos

Nf kr

• Newton’s 2nd for each axis:

• For a moving body,

042cos mmafT xr

TN

mg

fr

42T cos42

T sin42

042sin mmamgNT y

42sinTmgN rfT 42cos

Nf kr

042cos NT k042sin mgNT

042cos NT k

• Solving for T:T

N

mg

fr

42T cos42

T sin42

042sin mgNT042cos NT k

042sin42cos TmgT k

42sin42cos k

kmgT

42sin1.042cos

8.9751.0

T

N91

Application of friction

Uniform circular motion

• Michael Schumacher travels in his Ferrari of mass 600kg travels around a bend of radius 100m. The coefficient of static friction is 0.75. (Note that a negative lift helps to keep the car on the track.) Find the negative lift if when v=28.6m/s it is about to slide out of the turn.

• Diagram,

v

R

• Diagram and free body diagram for the car:

v

R

fsfs

N

mg

FL

a

• Newton’s 2nd for each axis,

fs

N

mg

FL

a

r

mvmaf xs

2

0 yL maFmgN

• Newton’s 2nd for each axis,

• For a body on the verge of sliding

fs

N

mg

FL

ar

mvmaf xs

2

0 yL maFmgN

Nf ss

• Newton’s 2nd for each axis,

• For a body on the verge of sliding

fs

N

mg

FL

ar

mvmaf xs

2

0 yL maFmgN

Nf ss

r

mvNs

2

LFmgN r

mvNs

2

• Substituting for N,fs

N

mg

FL

a

r

mvFmg Ls

2

LFmgN r

mvNs

2

• Substituting for N,

• Solving for the negative lift,

fs

N

mg

FL

a

r

mvFmg Ls

2

LFmgN r

mvNs

2

ssL mgrmv

F

2

gr

vmF

sL

2

• Substituting for N,

• Solving for the negative lift,

fs

N

mg

FL

a

r

mvFmg Ls

2

LFmgN r

mvNs

2

ssL mgrmv

F

2

gr

vmF

sL

2

N6648.910075.0

6.28600

2