Forces I

Newtons Laws

Kinematics

• The study of how objects move

Dynamics

• The study of why objects move

Newton’s Laws and Forces

• What is force?

• What are they?

Force

• A push or a pull

• Symbol is F

• Unit is N (Newton)

• One newton is the force necessary to cause

a one kilogram mass to accelerate at the rate

of 1m/s2

• 1N=1 kg m/s2

FORCEA force is any influence that can change the velocity of a body. Forces can act either through the physical contact of two objects (contact forces: push or pull) or at a distance (field forces: magnetic force, gravitational force).

Contact Forces Action-at-a-Distance Forces

Frictional Force Gravitational Force

Tensional Force Electrical Force

Normal Force Magnetic Force

Air Resistance Force

Applied Force

Spring Force

What do we mean by balanced and unbalanced forces?

The forces on the book are unbalanced

Balanced forces are EQUAL and OPPOSITE in direction

Unbalanced forces have a greater force in one direction

If all the forces are balanced, we say the object is in EQUILIBRIUM

A net force is the vector sum of the forces acting on an object

What are the net forces?FNET = Σ forces ≠ 0 N

These are free-body diagrams

Free Body Diagram Practice

1. A book is at rest on a tabletop. Diagram the forces acting on the book.

2. A picture is hanging from the ceiling by two ropes. Diagram the forces

acting on the picture. He was not injured in the experiment.

3. An egg is free-falling from a nest in a tree. Neglect air resistance. Diagram

the forces acting on the egg as it is falling.

4. A rightward force is applied to a book in order to move it across a desk

with a rightward acceleration. Consider frictional forces. Neglect air

resistance. Diagram the forces acting on the book.

5. A rightward force is applied to a book in order to move it across a desk at

constant velocity. Consider frictional forces. Neglect air resistance.

Diagram the forces acting on the book.

6. A college student rests a backpack upon his shoulder. The pack is

suspended motionless by one strap from one shoulder. Diagram the

vertical forces acting on the backpack.

7. A skydiver is descending with a constant velocity. Consider air resistance.

Diagram the forces acting upon the skydiver.

Newton Three Laws of Motion

• His laws explain why objects move (or

don’t move) as they do

Newton's First Law of Motion

(Law of Inertia) or [Law of Balanced

Forces]

Newton's First Law of Motion says:

an object at rest stays at rest and an

object in motion stays in motion

unless the object is acted on by an

unbalanced force.

“Things keep doing what they are

doing!”

Constant Velocity

It can be Zero or it can be greater than

zero…..It just can’t change!!!!!!

What is required to change the

constant velocity?

• An unbalanced force

Diagram from the Physics Classroom

http://www.physicsclassroom.com/mmedia/newtlaws/efff.html

Diagram from the Physics Classroom

http://www.physicsclassroom.com/mmedia/newtlaws/efff.html

Other Examples to Consider:

• Blood rushes from your head to your feet

while quickly stopping when riding on a

descending elevator.

• The head of a hammer can be tightened

onto the wooden handle by banging the

bottom of the handle against a hard surface.

Inertia and Mass

• Inertia: the resistance an object has to a

change in its state of motion

• The more mass the more inertia

• As mass increases the more an object resists

change in its motion

• Mass (kg) is the measurement of Inertia

Newton’s First Law

Remember an object in constant

motion has balanced forces working

on it. This is the first law described

mathmaticallyΣF = 0

• Where Σ (sigma) is sum

• F is force

FREE BODY DIAGRAM:Identify the missing

force marked with a (?) in the diagram below.

The car is moving at a constant velocity of 25

m/s eastward.

Car20N

15N

Air Friction

Ground

Friction

Ground Friction

?

Force pushing car

FREE BODY DIAGRAM:Identify the missing

force marked with a (?) in the diagram below.

The car is accelerating eastward with a net

force of 50 N.

Car20N

15N

Air Friction

Ground

Friction

Ground Friction

?

Force pushing car

Assign Fun With Forces

Forces WS I A

• Work an example on top and on bottom

.

Newton's Second

Law of Motion

(Law of Acceleration)

[Law of Unbalanced

Forces]

SECOND LAW OF MOTION

According to Newton's Second Law of Motion, the net force acting on a body equals the product of the mass and the acceleration of the body. The direction of the force is the same as that of the acceleration. In equation form:

F = ma

F = force applied to an object

[N = Newtons = kg ∙m/s2]

m = object’s mass [kg]a = acceleration [m/s2]

What is the relationship between

Force, Mass, and Acceleration?

According to Newton's second law of motion,

if mass is constant what is the relationship

between force and acceleration?

Direct

According to Newton's second law of motion,

if force is constant what is the relationship

between mass and acceleration?

Inverse

• What happens to acceleration of an object if

force is doubled? (mass is constant)

• Force Tripled?

• Cut in half?

• What happens to acceleration if force is

constant but the mass is doubled?

• Mass Tripled?

• Mass cut in half?

2nd Law: F = m·a

EX A• A 2000 N net force acts horizontally to the

right on a 50 kg object. Draw the free body

diagram. Calculate the acceleration.

amF

akgN 502000

2/40 sma

Are weight and mass the same

thing?

WEIGHT (of a body or object) aka Force of Weight

The gravitational force with which the Earth attracts the body. Causes it to accelerated downward with the acceleration of gravity g.

Vector quantity Measured in Newtons

Varies with its location near the Earth (or other astronomical body)

ForceWeight = mass x gravity

FW = m x g

Newton = kg x m/s2

1 N = 1 kgm/s2

Mass

A scalar quantity

Same everywhere in the universe. It doesn’t change

Measured in kg

EX B: What is the weight of a person

whose mass at sea level is 72 kg? Draw

the free body diagram. g = 9.80 m/s2

m = 72 kg

F = mg

Fw = mg

Fw = (72 kg)(9.80 m/s2)

Fw = 706 kg m/s2

Fw = 706 N

Ex C: What is the mass of a box (in

grams) if it weighs 625N? Answer

to the nearest tenth. Draw the free

body diagram.

F = ma

m = F/a

m = 625N/9.8 m/s2

m = 63.8 kg

m = 63800 grams

What is the third law of motion

• For every action force there is an equal and opposite reaction force

• Two different objects and two different forces (equal but opposite)

• Forces always occur in pairs

• Also known as The Law of Force Pairs

• Are the effects of the force the same?

• NO

• They are not always acting on the same amount of mass

• Examples:

• Bug hitting windshield, recoil of shot gun, pushing off of on a raft

Third Law Mathmatically

F = F

Which can be written as:

m1a1= m2a2

Forces occurs in pairs…

What forces cause the swimmer to

move forward?

• Her push on the wall?

Or

• The wall’s push on her?

Ex D

• A 2500 kg car hits a 0.001 kg bug with a

force of 500N. According to the third law

of motion, the bug hits the car with how

many Newton's of force? Why isn’t the car

damaged? Calculate the acceleration of the

bug due to the force. Calculate the

acceleration of the car due to the force.

A 2500 kg car hits a 0.001 kg bug with a force of 500N. According to the

third law of motion, the bug hits the car with how many Newton's of

force? Why isn’t the car damaged? Calculate the acceleration of the bug

due to the force. Calculate the acceleration of the car due to the force.

acceleration of the bug due to impact:

a = F/m

a = 500N/.001 kg = 500,000 m/s2 !

acceleration of the car due to impact :

a = F/m

a = 500N/2500kg = 0.2 m/s2

Third Law of Motion

• The girl pushes the boy with 10N of force.

How much does the boy push back with?

Force and Kinematics

Net Force (Fnet )

• Fnet in simple situations (only 1 force

implied): Fnet = ma

• Fnet and acceleration are always in the same

direction

• How is net force on an object determined

with multiple?

• It is the sum of all forces

• Fnet = ΣF

• Remember force is a vector, therefore it has

direction (+,-, N,S etc)

Net force (F) = maThe second law of motion is the key to understanding the behavior of moving bodies since it links cause(force) and effect (acceleration) in a definite way.

Example E

A 2000 N net force (horizontally to the right) acts

on a 50 kg object. Draw the free body diagram.

Calculate the acceleration. If the object started

from rest, how far would it move in 5 seconds?

amF akgN 502000

2/40 sma

Example EA 2000 N net force (horizontally to the right) acts

on a 50 kg object. Draw the free body diagram.

Calculate the acceleration. If the object started

from rest, how far would it move in 5 seconds?

d = vit + .5at2

d = (.5)(40 m/s2)(5sec)2

d = 500m

Could we determine the velocity at 5 seconds? How?

a = (vf – vi)/t

vf = vi + at

vf = 0 m/s + (40 m/s2)(5sec) = 200 m/s

Ex F A 1000 kg car goes from 10 to 20 m/s in 5 s. What force is

acting on it?

m = 1000 kg

vi = 10 m/s

vf = 20 m/s

t = 5 s

F = ma

a = (vf – vi)/t a = (20m/s-10 m/s) / 5s

a = 2 m/s2

F = (1000 kg) (2 m/s2) = 2000 N

Ex G-I as time permits

Ex G A 60-g tennis ball approaches a racket at 15 m/s, is in contact

with the racket for 0.005 s, and then rebounds at 20 m/s. Find the

average force exerted by the racket. How can we draw this?

Ex G A 60-g tennis ball approaches a racket at 15 m/s, is in contact

with the racket for 0.005 s, and then rebounds at 20 m/s. Find the

average force exerted by the racket.

m = 0.06 kg

vi = 15 m/s

t = 0.005 s

vf = - 20 m/s a = (vf – vi)/t a = (-20m/s-15 m/s) / 0.005s

a = -7000 m/s2

F = maF = (0.06 kg) (-7000 m/s2) = -420 N

Ex H The brakes of a 1000-kg car exert 3000 N.

a. How long will it take the car to come to a stop from a velocity of

30 m/s? How can we draw this?

Ex H The brakes of a 1000-kg car exert 3000 N.

a. How long will it take the car to come to a stop from a velocity of

30 m/s?m = 1000 kg

F = -3000 N

vi = 30 m/s

vf = 0 m/sm

Fa

1000

3000 = - 3 m/s2

a

vvt

if

3

300

= 10 s

b. How far will the car travel during this time?

d = vit+ .5at2

= 30(10)+ .5 (-3)(10)2

= 150 m

If you got 450m you

did not use -3 m/s2

Ex H The brakes of a 1000-kg car exert 3000 N.

a. How long will it take the car to come to a stop from a velocity of

30 m/s?m = 1000 kg

F = -3000 N

vi = 30 m/s

vf = 0 m/sm

Fa

1000

3000 = - 3 m/s2

a

vvt

if

3

300

= 10 s

b. How far will the car travel during this time?

d = vit+ .5at2

= 30(10)+ .5 (-3)(10)2

= 150 m

If you got 450m you

did not use -3 m/s2

Ex I A net horizontal force of 4000 N is applied to a car at rest

whose weight is 10,000 N. What will the car's speed be after 8 s?

FA = 4000 N

Fw = 10,000 N

t = 8sg

w

a

Fm

8.9

10000 = 1020.4 kg

m

Fa

4.1020

4000 = 3.92 m/s2

vf = vi + at

= 0 +3.92(8)

= 31.36 m/s

Forces Part II

Friction

Friction

• Is a force that opposes motion.

When surfaces are pressed

together, we can identify

four forces

• Friction Force: FF

• Force opposing motion

• Measured in Newtons (N)

• Applied Force: FA

• The push or pull applied to the object

• Measured in Newtons (N)

• Fw Force of Weight or Gravity

• (Mass in kg) (Acceleration due to Gravity)

• Kg x 9.8 m/s2

• Measured in Newtons (N)

• Normal Force: FN

• Usually a 3rd Law reaction to gravity, that is

equal and opposite of Force of Weight (Fw)

• Perpendicular to the surface.

• Measured in Newtons (N)

• FN is NOT FNET

• Don’t confuse them just because they begin

with N!

For Examples J-M we will look

at the same object under different

conditionsEx J: FNET No FF

Ex K: FF No FNET

Ex L: FNET & FF

Ex M: FNET & FF

• A 50kg object is moving

horizontally at a constant velocity.

Is there acceleration? Is there a net

force?

FW

FN

FAFF

Ex J No Friction

• Fnet = ma A 50 kg object experiences an applied force of 400 N,

what is the Fnet? (Disregard friction) What is the acceleration?:

• Fnet = FA

• Fnet = Net Force, results in acceleration

• a = Fnet /m = (400N )/ (50 kg) = 8 m/s2

Example J

FA

FW

FN

Ex K Friction and

Constant velocity. Is there a net

force?

• Fnet = 0

• FA+ -FF = ma or FA + - FF = Fnet

A 50 kg object moves at a constant velocity when acted upon by an applied force of 400 N. What is the FF? What is the Fnet? What is the acceleration?:

• FF = FA

• Fnet = FA + - FF = 0 N

• If Fnet = 0 then acceleration = ?

• 0

How does friction affect net force?• Fnet = ma

• ma = FA+ FF

Where:

• Fnet = Net Force

• FF = Friction Force

• FA = Applied Force

• You are actually subtracting FF from FA, since Ff is in the opposite direction

• Friction will reduce the net force

Example K

Friction and constant velocity

FA

FW

FF

FN

Ex K What would have to change to

have acceleration?

• Do all surfaces provide the same

amount of friction? How is this

described?

FA

FW

FF

FN

FF = Force of Friction

What does it depend on?

• Depends on the surface of the

materials.

• Depends on how tightly the surfaces

are pressed together.

Coefficient of Friction

• The coefficient of friction is a measure of how

difficult it is to slide a material of one kind over

another; the coefficient of friction applies to a pair

of materials, and not simply to one object by itself

• Coefficient of Friction Reference Table -

Engineer's Handbook

When Surfaces are Pressed

Together• Coefficent of Friction µ can be calculated

• It is a ratio of FF and FN

N

F

F

F

NF FF

Coefficient of friction

• What is it equal to?

• What is the unit for Coefficient of friction?

• If Coefficient of friction is small, what does

that mean about the FF?

• If Coefficient of friction is large, what does

that mean about the FF?

• The higher the coefficient of friction, the

more difficult to slide

What forces cause the swimmer to

move forward?

• Her push on the wall?

Or

• The wall’s push on her?

I call these FAWN problems

Example L• A 400 N force is applied

horizontally to a 50kg object.

Calculate the acceleration of the

object if = 0.3.

FA

FW

FF

FN

Example Cont’d.

• FA = 400 N

• Fg = m·g (50kg)(9.8m/s2)

Fg = 490 N

• FN = 490 N also

• FF = (0.3)(490N) = 147 N

Example cont’d.

FA = 400N

FW = 490N

FF = 147N

FN = 490N

To solve for Acceleration must

calculate Net Force

• FNET = ma= 400N – 147N

• ma = 253N

FNET = ma = FA+ -FF

Now use Net Force and mass of

object in F=ma formula

a = F/m

a = 253N/50kg

a =5.06 m/s2

Ex M

Putting it all togetherA 50 kg object accelerates horizontally at 0.2 m/s2 from

rest for 5 seconds. If the coefficient of friction is 0.01, what is the Fnet? What is the FF ? What is the FA ?How far does it travel in 5 seconds?:

• Fnet = (50 kg) (0.2 m/s2) = 10 N

• FF = μ FN

• FN = (50kg)(9.8 m/s2) = 490 N

• FF = (0.01) (490 N) =4.9N

• FA = Fnet –-FF

• FA = 10 N + 4.9 N = 14.9 N

Ex M

A 50 kg object accelerates horizontally at 0.2 m/s2 from rest for 5 seconds. If the coefficient of friction is 0.01, what is the Fnet? What is the FF ? What is the FA ?How far does it travel in 5 seconds?:

• Fnet = (50 kg) (0.2 m/s2) = 10 N

• Set up DVVAT

• d = ?

• vi = 0 m/s

• vf = ?

• a =0.2 m/s2

• d = (.5)(.2)(5)2

• d = 2.5m

What does a small coefficient of

friction imply about FF?

TENSION aka FT

• is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.

• It is the opposite of compression. It is a “response force”

• That is to say, if one pulls on the rope, the rope fights back by resisting being stretched

• Ropes, strings, and cables can only pull. They cannot push because they bend.

• is measured in newtons

• is always measured parallel to the string on which it applies.

• What does the rope provide?

• A lift (vertical force) and a pull (horizontal force)

• If there was no angle, would there be any vertical

force?

• No

• If the angle was at 90°, how would that affect the

force components?

• Force would only be in the vertical plane

• How would you calculate the horizontal and vertical

force components if the angle of the rope with the

floor was 57° and the Force of tension (FA) in the

rope was 400 N?

Ex N. This crate is be pulled with a rope across a friction

based horizontal surface at a constant velocity. The rope

exerts a tension of 400 N at an angle of 57°. What is the

coefficient of friction?

50 kg

57°

A box is pulled into motion with a rope across a horizontal surface.

The rope makes an angle of 57° to the floor. The Force of tension

(FA) in the rope is 400N

50 kg57º

FAX

FAY

400 N

A. Determine FAX or FHoriz

= cos (57) (400N)

= 217.86 N

B. Determine FAY or FVert

= sin (57) (400N)

= 335.47 N

I work in a circle• Determine FW

• Determine FAX

• Determine FAY

• Determine FN

• The FAY supplies part of the upward force. The total

upward force is FAY + FN and together these are equal but

opposite the FW. FN = FW – FAY

• Determine FF

• Determine coefficient of friction

50 kg57°

FAX 217.86 N

FAY 335.47 N

400 N

Determine Fw

= (50 kg)(9.8 m/s2)

= 490 N

Determine FN

= FW- FAY

= 490 N – 335.47 N

= 154.53 N

50 kg57°

FAX 217.86 N

FAY 335.47 N

400 N

Determine FF

FF= Fax (Constant Velocity)

FF= 217.86N

Determine µ= FF/ FFN

= 217.86N/ 154.53 N

= 1.41

Force III Statics

TENSION aka FT

• is the magnitude of the pulling force exerted by a string, cable, chain, or similar object on another object.

• Ropes, strings, and cables can only pull. They cannot push because they bend.

• is measured in newtons

• is always measured parallel to the string on which it applies.

Example O: What is the tension in

the cable supporting the sign?

Mass of sign 5.86 kg

Example P: What is the tension in

each cable supporting the sign?

Mass of sign 5.86 kg

Example Q: What is the tension in

each cable supporting the sign?

Angle A and Angle B = 35o

Mass of sign 5.86 kg

Example Q What do we know?Object not moving, therefore FW = FT

FW = 57.4 N

FT is shared between the two cables.

FT represents the vertical component

Do you see triangles?

Angle A and Angle B = 35o

Mass of sign 5.86 kg

Example Q SolutionObject not moving, therefore FW = FT

FW = 57.4 N

FT is shared between the two cables.

FT represents the vertical component

Do you see triangles?

Angle A and Angle B = 35o

Mass of sign 5.86 kg

Solution

S = O/H

FT = 28.7 N/ sin 35°

FT = 50 N

How would this change as angle changes?

Try with 60º, 15º

Angle A and Angle B = 35o

Mass of sign 5.86 kg

35

°

28.7 N

FT

A word about angle location:

35º is in reference to the horizontal

55º is in reference to the vertical

55°

28.7 N

FT

35°

Ex R

A 7.5 g sign is suspended

by two equal strings.

The angle of the string

to the vertical is 35°.

What is the tension in

each string?

0.045 NA 7.5 g sign is suspended

by two equal strings.

The angle of the string

to the vertical is 35°.

What is the tension in

each string?