2009 05 Dynamics Presentation 2011-08-30 6 Slides Per Page

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New Jersey Center for Teaching and Learning

Progressive Science Initiative PSI Physics

Dynamics:The Laws of motion

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Table of Contents: Dynamics

Dynamics Thought Experiment

Newton's 1st Law of Motion

Newton's 3rd Law of Motion

Free Body Diagrams

Friction

Net Force

Newton's 2nd Law of Motion

Mass, Weight, and Normal Force

Click on the topic to go to that section

Tension

General Problems Return to

Table ofContents

Intro to Dynamics -

Thought Experiment

We all have an intuition about how objects move.

Our beliefs are hard to change since they work well in

our day to day lives.

But they limit us in developing an understanding of

how the world works - we must build on our intuition

and move beyond it.

Intuitive Physics Galileo vs. Aristotle

In our experience, objects must be pushed in order to

keep moving. So force would be needed to have a

constant velocity. This is what Aristotle claimed inhis in his series of books entitled "Physics", written

2400 years ago.

But 400 years ago, another scientist and astronomer,

Galileo, proposed the following thought experimentwhich revealed another perspective.


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Imagine two perfectly smooth ramps connected

together by a perfectly smooth surface.

What Will Happen?

Thought Experiment



What Will Happen?

Thought Experiment



What Will Happen?

Thought ExperimentImagine two perfectly smooth ramps connected


What Will Happen?

Thought Experiment

If a ball rolls down one ramp, it keeps rolling up theother side until it reaches the same height.

Thought ExperimentNow repeat that experiment, but make the second

ramp less steep.

What Will Happen?

Thought Experiment


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Now repeat that experiment, but make the second

ramp less steep.

What Will Happen?

Thought Experiment


ramp less steep.

What Will Happen?

Thought Experiment


ramp less steep.

What Will Happen?

Thought ExperimentIt will still keep rolling until it reaches the same

height, but it has to roll farther!

Thought Experiment

Thought ExperimentFinally, make the ramp flat.

Now what will happen?




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Thought Experiment

It will keep rolling forever, no external force is

necessary.

Galileo vs. Aristotle

It's not that Aristotle was wrong.

In everyday life, objects do need to keep being pushed

in order to keep moving.

Push a book across the table. When you stoppushing, it stops moving. Aristotle is right in terms ofwhat we see around us every day.


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Force and Motion

It's just that Galileo, and later

Newton, imagined a world

where friction could be

eliminated.

Friction represents an external

force acting on the object, just

as your push is an externalforce.

Fapplied

Ffriction

In the absence ofall external forces, an object's velocity remains constant.

Two equal and opposite forces have the same effect, they cancel to createzero net force.

Return to

Table of

Contents

Newton's 1st Law of Motion

Newton's First Law of Motion

Galileo's observations were more fully formed in 1687 by the 'father of

physics,' Sir Isaac Newton, who called it "The First Law of Motion".

Newton's First Law of Motion:

An object at rest remains at rest, and an object in motion remainsin motion, unless acted on by a net external force.

Newton's First Law of Motion

Newton's First Law of Motion:

An object at rest remains at rest, and an object in motion remainsin motion, unless acted on by a net external force.

In other words, an object maintains its velocity (both speed and direction)

unless acted upon by a nonzero net force.

Having zero velocity, being at rest, is not special, it is just one possible

velocity...no more special than any other.

D

1 In the absence of an external force, a movingobject will

A stop immediately.

Bslow down and eventually come to astop.

C go faster and faster.

D move with constant velocity.

2 When the rocket engines on the spacecraft aresuddenly turned off, while traveling in emptyspace, the starship will

A stop immediately.

B slowly slow down, and then stop.C go faster and faster.

D move with constant speed.

D


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3 When you sit on a chair, the net external forceon you is

A zero.

B up.

C down.

D depending on your weight.

A

4 A rocket moves through empty space in a straightline with constant speed. It is far from thegravitational effect of any star or planet. Underthese conditions, the force that must be applied tothe rocket in order to sustain its motion is

A equal to its weight.

B equal to its mass.

C dependent on how fast it is moving.

D zero.

D

5 You are standing in a moving bus, facing forward,and you suddenly fall forward. You can infer fromthis that the bus's

A velocity decreased.

B velocity increased.

Cspeed remained the same, but it'sturning to the right.

Dspeed remained the same, but it'sturning to the left.

A

6 You are standing in amoving bus, facingforward, and you suddenlyfall forward as the bus

comes to an immediatestop. What force causedyou to fall forward?

A gravity

Bnormal force due to your contact withthe floor of the bus

Cforce due to friction between you andthe floor of the bus

DThere is not a force leading to yourfall.

D

Newton's laws are only valid in inertial reference frames:

An inertial reference frame is one in which Newtons first law

is valid. This excludes rotating and accelerating frames.

For instance, when your car accelerates, it is not an inertial

reference frame.

That's why a soda on the front dashboard can suddenly

seem to accelerate backwards without any force acting on it.

It's not accelerating, it's standing still.

The reference frame, the car, is accelerating underneath it.

Inertial Reference Frames

Return to

Table of

Contents

Newton's 2nd Law of Motion


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An object doesn't change its

velocity unless a force acts on it.

How does an object respond to a

force when it is applied?

Newtons Second Law of Motion

Newtons second law is the

relation between acceleration and

force.

As a net force is applied, an

object accelerates.


Newton's 2nd Law of Motion:

F = ma

*the word 'net' means overall, or total. We will discuss this in further detail later,

but for now just think of F as any force on an object

Units

Newton's 2nd Law of Motion:

F = ma

Mass is measured in kilograms (kg).

Acceleration is measured in meters/second2 (m/s2)

Therefore, the unit of force, the Newton, can be found fromthe second law

F = ma

N = kg*m/s2

The unit of force in the SI system is thenewton (N).

7 A 3.5 kg object experiencesan accleration of 0.5 m/s

2

.What net force does theobject feel?

F = ma

F = 3.5 * 0.5

F = 1.75 N

8 A 12 N net force acts on a36 kg object? How muchdoes it accelerate?

F= ma

a= F/m

a= 12/36

a= .33m/s2

9 How much net force isrequired to acclerate a 0.5kg toy car, initally at rest toa velocity of 2.4 m/s in 6 s?

v= V0

+at

a= v/ t= 2.4/6= .4 m/s2

F= ma

F= .5*.4= 0.2N


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We can use this equation to understand how Force, Mass, and

Acceleration are related. When two variables are 'directly proportional',

it means that if we increase one, the other will increase as well.

Newtons Second Law of MotionNewton's 2nd Law of Motion:

F = ma

F = m a

In Newton's 2nd Law, Force and Mass are directly proportional. More ma

requires more force to create the same acceleration.

Similarly, Force and Acceleration are directly proportional. To increase thacceleration of an object, more force must be applied.

We can rearrange this equation to see how mass is related to

acceleration:


a = F

m

Acceleration is directly proportional to force

(which we already knew!)

and inversely proportionalto mass.

When two variables are 'inversely proportional', it means

that if we increase one, the other will decrease!

So if we increase the mass of an object under a constant

force, the acceleration will decrease.

Think about what happens when you fill a wheelbarrow

with dirt...

10 A net force F accelerates amass m with anacceleration a. If the samenet force is applied to

mass 2m, then theacceleration will be

A 4a

B 2a

C a/2

D a/4

C

11 A constant net force acts on an object.The object moves with:

A constant accelerationB constant speedC constant velocityD increasing acceleration

A

12 A net force F acts on amass m and produces an

acceleration a. Whatacceleration results if a net

force 2F acts on mass 4m?

A a/2

B 8a

C 4a

D 2a

A

13 The acceleration of anobject is inverselyproportional to

A the net force acting on it.

B its position.

C its velocity.

D i ts mass .

D


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Return to

Table of

Contents

Net Force

FThe greek letter sigma "" means "the sum of".

Sometimes F is written as Fnet, it means the same thing.

It means you have to add up all the forces acting on an object.

Net Force

F

F = maLet's look at the left side of this equation first.

Net Force

F

The arrow above "F" reminds you that force is a vector. We won't

always write the arrow but remember it's there.

It means that when you add forces, you have to add them like

vectors: forces have direction, and they can cancel out.

Example: A 5.0 kg object is being acted on by a 20N force to

the right (F1), and a 30N force, also to the right (F 2). What is

the net force on the object?

Net Force

F

First we'll draw a free body diagram. This consists of a dot,

representing the object, and arrows representing the forces.

The direction of the arrows represents the direction of the

forces...their length is roughly proportional to their size.

Example: A 5.0 kg object is being acted on by a 20N force to

the right (F1), and a 30N force, also to the right (F 2). What is

the net force on the object?


F

The first force, F1, acts to the right with a magnitude of 20 N

F1

For example: A 5.0 kg object is being acted on by a 20N

force to the right (F 1), and a 30N force, also to the right (F 2).What is the net force on the object?


F

The second force, F2, acts to the right also, with a greater

magnitude of 30N. This is drawn slightly larger than F-1.

F1

F2


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force to the right (F 1), and a 30N force, also to the right (F 2).

What is the net force on the object?


F

F1 F2

To add vectors, move the second vector so it starts where

the first one ends. The sum is a vector which starts where

the first vector started, and ends where the last one ends.





F

F1 F2

F

To add vectors, move the second vector so it starts where

the first one ends. The sum, F, is a vector which startswhere the first vector started, and ends where the last one

ends.

These free body diagrams are critically important to ourwork. Once done, the problem can be translated into an

algebra problem.





First we will define "to the right" as positive.

Then we can interpret our diagram to read:

F = F1

+ F2

F = 20N + 30N

F = 50N to the right we get the direction from our diagram

and from our positive answer, which we

defined as meaning "to the right"

F1 F2

F

14 Two forces act on an object. One force is40N to the west and the other force is 40N tothe east. What is the net force acting on theobject?

Remember to draw a force diagram first!

F = F1

- F2

F = 40N - 40N

F = 0

15 Two forces act on an object. One force is 8.0N to the north and the other force is 6.0N to the

south. What is the net force acting on the object?

Remember to draw a force diagram first!

A 14 N

B 14 N to the north

C 14 N to the south

D 2 N to the north

E 2 N to the south

D

Now let's look at the right side of our equation.

Mass is a scalar...it does not have a direction.

But acceleration does have a direction...it is a vector.

The direction of the acceleration vector is always the

same as the direction of the net force, F, vector.


ma


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We found the net force on the object to be 50N to the right.

Now let's find its acceleration.


F = maa =F / ma = 50N / 5.0 kg

a = 10 m/s2

to the right

F1 F2

F

Force is a vector, so F = ma is true along each coordinate axis.


F1

F2

F3a = 1 m/s-

2


That means we can add up all the forces in the vertical direction and

those will equal "ma" in the vertical direction.


F1

F2

F3 a = 1 m/s-2

F1 - F2 = m*a

F1 - F2 = 0

F1

F2


That means we can add up all the forces in the vertical direction and

those will equal "ma" in the vertical direction.

And then can do the same thing in the horizontal direction.


F1

F2

F3 a = 1 m/s-2

F1 - F2 = m*a

F1 - F2 = 0

F3 = m*a

F3 = (2kg) * (1 m/s2

)

F3 = 2 N

F1

F2

F3

a = 1 m/s-2

16 A force F1 = 50N acts to the right on a 5.0 kg object.Another force, F2 = 30N, acts to the left. Find theacceleration of the object:

17 A force F1 = 350N pushes upward on 20.0 kg object.Another force, F2 = 450N pulls downward. Find theacceleration of the object:


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18 An object accelerates downward at a rate of 4.9 m/s2

.If the downward force on the object is 500N and theupward force is 250N, what is the mass of the object?

Return to

Table of

Contents

Mass, Weight, and Normal Force

Mass is the measure of the inertia of an object, the

resistance of an object to accelerate.

In the SI system, mass is measured in kilograms.

Mass is not weight: Mass is a property of an object. It

doesn't depend on where the object is located.

Weight is the force exerted on that object by gravity.

If you go to the moon, whose gravitational acceleration is

about 1/6 g, you will weigh much less. Your mass, however,

will be the same.

Mass

Note that the pound is a unit

offorce, not of mass, and can

therefore be equated to

newtons but not to kilograms.


Units for Mass and Force

Conversion factors: 1 dyne = 10-5N;

1 lb 4.45N

System Mass Force

SI kilogram(kg) newton (N)

cgs gram (g) dyne

British slug pound (lb)

The unit of force in the SI system is thenewton (N).

Weight is the force exerted on an object by gravity.

Close to the surface of the Earth, where the

gravitational force is nearly constant, the weight is:

Weight the Force of Gravity;and the Normal Force

Near the surface of Earth, g is 9.8 m/s2

downwards.

19 Determine the Force of Gravity (weight) in Newtons ona 6.0 kg bowling ball.


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20 Determine the Force of Gravity (weight) in Newtons ona small car with a mass of 900 kg

21 Using a spring scale, you find that the weight of afriction block in the lab is around 24 N. What is themass of the block, in kilograms?

22 A 120 lb woman has a mass of about 54.5 kg.What is her weight in Newtons?

23 What is the weight of a 25kg object located near thesurface of Earth?

w= mg

w= 25*9.8

w= 245N

24 Which of the following properties of an object is likely tochange on another planet?

A Mass

B Weight

C Color

D Volume (size and shape)

25 The acceleration due togravity is lower on theMoon than on Earth.

Which of the following istrue about the mass and

weight of an astronaut onthe Moon's surface,

compared to Earth?

A Mass is less, weight is same

B Mass is same, weight is less

C Both mass and weight are less

D Both mass and weight are the same

B


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FG

An object at rest must have no net force on it. If it is sitting on a

table, the force of gravity is still there; what other force is

there?


FG

FN

An object at rest must have no net force on it. If it is sitting on a

table, the force of gravity is still there; what other force is

there?

The force exerted perpendicular

to a surface is called the normal

force.

It is exactly as large as needed to

balance the force from the object

(if the required force gets too

big, something breaks!)

The words "normal" and

"perpendicular" are synonyms.


26 A 14 N brick is sitting on atable. What is the normalforce supplied by the

table?

A 14 N upwards

B 28 N upwards

C 14 N downwards

D 28 N downwards

A

27 What normal force is supplied by a desk toa 2.0 kg box sitting on it?

(Use g = 9.8 m/s2

and round to two figures.)

F= m a = 0

FN

-m g= 0

FN

= mg= 2*9.8 20N

Return to

Table of

Contents

Newton's 3rd Law of Motion

Any time a force is exerted on an object, that

force is caused by another object.

You need two objects to have a force.

Newtons Third Law of Motion

Newtons third law:

Whenever one object exerts

a force on a second object,

the second object exerts an

equal force in the opposite

direction on the first object.

Force exerted on

cat by table

Force exerted

on table by cat


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Newton's 3rd Law of Motion:

Whenever one object exerts a force on a second object,the second object exerts an equal force in the opposite

direction on the first object.

You may have heard this, or something similar, before:

For every action, there is an equal, opposite reaction.

This is another way to state Newton's 3rd Law.

It is important to remember that the forces (or actions) are always

applied to two different object


A key to the correct application of the third law is that the

forces are exerted on different objects.

Make sure you dont use them as if they were acting on the

same object. Then they would add to zero!

Force onhands

Force onfloor

Rocket propulsion can also be explained using Newtons third law: hot

gases from combustion spew out of the tail of the rocket at highspeeds. The reaction force is what propels the rocket.


Note that the rocket

does not need

anything to push

against.

Helpful notation: the first subscript

is the object that the force is being

exerted on; the second is the

source.

Subscripts help keep your

ideas and equations clear.


Horizontal forceexerted on theground byperson's foot

FGP

Horizontal forceexerted on theperson's foot byground

FPGFGP = -FPG

28 An object of mass m sits on a flat table. TheEarth pulls on this object with force mg, whichwe will call the action force. What is the reactionforce?

AThe table pushing up on the objectwith force mg

BThe object pushing down on the tablewith force mg

CThe table pushing down on the floorwith force mg

DThe object pulling upward on theEarth with force mg

D

29 A 20-ton truck collides witha 1500-lb car and causes alot of damage to the car.

Since a lot of damage isdone on the car

Athe force on the truck is greater thenthe force on the car

Bthe force on the truck is equal to theforce on the car

Cthe force on the truck is smaller thanthe force on the car

Dthe truck did not slow down duringthe collision

B


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30 As you are sitting in a chair, you feel the chairpushing up on you. The reaction force in thissituation is:

A The chair pushing down on the ground

B Gravity pulling down on you

C You pushing down on the chair

D The ground pushing up on the chair

B

31 A student is doing push-ups in gym class. Areaction pair of forces is best described as:

A The student pushes down on the ground / the ground pushes up on the student

B Gravity is pulling the student down / The ground is pushing the student up

C Gravity is pulling the student down / The student's arms push the student up

D The student's hands push down on the ground / The students arms push the studentup

B

32 Which of Newton's lawsbest explains whymotorists should wear seatbelts?

A the first law

B the second law

C the third law

D the law of gravitation

A

33 If you blow up a balloon, and then release it, theballoon will fly away. This is an illustration of:(Note: there may be more than one answer. Be prepared to explain WHY!)

A Newton's first law

B Newton's second law

C Newton's third law

D Galileo's law of inertia

C

Return to

Table of

Contents

Free Body Diagrams

Free Body Diagrams

1. Draw and label a dot to represent the first object.

2. Draw an arrow from the dot pointing in the direction of one of the forces

that is acting on that object. Label that arrow with the name of the force.

3. Repeat for every force that is acting on the object. Try to draw each ofthe arrows to roughly the same scale, bigger forces getting bigger arrows.

4. Once you have finished your free body diagram, recheck it to make surethat you have drawn and labeled an arrow for every force. This is no timeto forget a force.

5. Draw a separate arrow next to your free body diagram indicating thelikely direction of the acceleration of the object. This will help you use yourfree body diagram effectively.

6. Repeat this process for every object in your sketch.


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A hot air balloon hovers in the sky.

Draw it's free body diagram:

Determine F in the x and y directions

FB

mg

a = 0

Fx= ma

x= 0

Fy= ma

y= 0

Fy= F

B- mg = 0

FB

= mg

A boy pulls a toy along (with constant velocity) on a string.

Draw the toy's free body diagram:


mg

FN

FT

a = 0 (const. velocity)

Fx = max = 0

Fx = FT = 0

Fy = may = 0

Fy = FN - mg = 0

FN = mg

An object on a crane descends with constant velocity.

Draw the object's free body diagram:


FT

mg

a = 0

Fx= ma

x= 0

Fy= ma

y= 0

Fy= F

T- mg = 0

FT

= mgReturn toTable of

Contents

Friction

There are many different types of forces that occur in

nature, but perhaps none is more familiar to us than the

force of friction (F f).

Friction is a resistive force that opposes the motion ofan object.

Friction is the reason objects stop rolling or sliding

along a surface. It is the reason it is difficult to start

pushing a heavy box along the floor.

There are many different types of friction:

Friction between solid objects and air is often called air

resistance. Friction between two fluids is called

viscosity, and so on.

Friction - A Resistive ForceWhere does friction come from?

On a microscopic scale, most surfaces are rough.

This leads to complex interactions between them that wedon't need to consider yet, but the force can be modeled in

a simple way.

Kinetic Friction Force

v


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Friction that acts on an object that is already in motion

is called kinetic friction.

Forkinetic sliding friction, we write:

Kinetic friction is the product of two things:

k is called the coefficient ofkinetic friction, and isdifferent for every pair of surfaces.

FN is simply the Normal Force, which, on flat

surfaces, is equal to the weight of the object.


A larger coefficient of friction means a greater frictional force.

Notice the friction that occurs between different materials in the

table below:

Fy = may

FN - mg = 0

FN = mg

A man accelerates a crate along a rough surface.

Draw the crate's free body diagram:

DetermineF in the x and y directions

FN

mg

Fapp

Ffr

a

Fx= maxFapp - F fr = max

34 A 4.0kg brick is sliding ona surface. The coefficientof kinetic friction between

the surfaces is 0.25. What

it the size of the force offriction?

Ffr= kFNF

N= mg

Ffr= kmg

Ffr= (0.25)(4)(9.8) = 9.8 N

35 A brick is sliding to the righton a horizontal surface.

What are the directions of thetwo surface forces: the frictionforce and the normal force?

A right, down

B right, up

C left, down

D left, up

D

Static friction is the frictional force between two surfaces

that are not moving along each other.

Static friction keeps objects from moving when a force is

first applied.

Static Friction Force

Fapplied

v = 0

**


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sis the coefficient ofstatic friction,and is different for every pair of

surfaces.

Static Friction Force

Fapplied

v = 0

** Static Friction Force**Note the symbol in this equation.

Imagine pushing on a box until it moves. You can apply a small force...nothing happens.

You apply more and more force until the box finally starts moving - this is

the maximum amount of static friction.

The friction can be LESS than the maximum amount or EQUAL to the

maximum amount, but never greater.

The force of friction is equal to sFNat the instant when the object startsmove.

Then what happens?

Friction ForceThe static frictional

force increases as the

applied force increases,

always equal to the net

applied force.

Until it reaches its

maximum, sFN.

Then the object starts to

move, and the kinetic

frictional force takes

over, KFN .

10

30

20

50

40

10 20 30 40 50 60 70

Frictionforce,f

Applied force, FA

f =S FN

S FN0

nomotion

sliding

** Friction Force

Surface

Coefficient

of Static

Friction

Coefficient of

Kinetic

Friction

Wood on wood 0.4 0.2

Ice on ice 0.1 0.03

Metal on metal ( lubr icated) 0.15 0.07

Steel on steel (unlubricated) 0.7 0.6

Rubber on dry concrete 1.0 0.8

Rubber on wet concrete 0.7 0.5

Rubber on other solidsurfaces

1-4 1

Teflon on Teflon in air 0.04 0.04

Joints in human limbs 0.01 0.01

**The table below shows values for both static and

kinetic coefficients of friction.

Notice that static friction is greater than kinetic friction.Once an object is in motion, it is easier to keep it in motion.

36 A 4.0 kg brick is sitting on atable. The coefficient of staticfriction between the surfaces is0.45. What is the largest forcethat can be applied horizontallyto the brick before it begins toslide?

**

Ffr=sFNFN = mg

Ffr

=smg

Ffr= (.45)(4)(9.8)= 17.64N

Fapp = 17.64N

37 A 4.0kg brick is sitting on atable. The coefficient ofstatic friction between thesurfaces is 0.45. If a 10 Nhorizontal force is appliedto the brick, what will be theforce of friction?

**

Ffr

= sF

N

FN = mg

Ffr= smg

Ffr= (.45)(4)(9.8)= 17.64N

Fapp < Ffr (a= 0)

Ffr=10N


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Return to

Table of

Contents

Tension FT

mg

a

When a cord or rope pulls on

an object, it is said to be under

tension, and the force it exerts

is called a tension force, F T.

Any object that is hanging or

suspended is considered to

have tension acting upward.

In some cases, tension can be

applied in other directions as

well.

Tension Force

FT

mg

a

There is no special formula to

find the force of tension.

We need to use force diagrams

and net force equations to solve

for it!

Tension Force 38 A 25 kg lamp is hangingfrom a rope. What is thetension force being suppliedby the rope?

F= ma= 0

FT

-mg= 0

FT

= mg

FT

= (25)(9.8)= 245 N

39 A crane is lifting a 60 kgload at a constant velocity.Determine the tensionforce in the cable.

F= ma = 0

FT

- mg = 0

FT

= mg

FT

= (60)(9.8) = 588 N

40 A 90 kg climber rappels from the top of a cliff withan acceleration of 1 m/s

2

.

Determine the Tension in the climber's rope.

F= ma

FT - mg = -90 N

FT = 900 N - 90N

FT = 810N


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41 A crane lifts a 400 kg crate upward with anacceleration of 3 m/s

2

.

Determine the Tension in the crane.

F= ma

FT - mg = 1200 N

FT = 4000N + 1200N

FT = 5200N

The tension in a rope is the

same everywhere in the rope.

If two masses hang down from

either side of a cable, for

instance, the tension in both

sides must be the same.

Tension Force

20 kg

50 kg

T1 = T2

Tension Force

A 50 kg mass hangs from one end of a rope that passes over a smallfrictionless pulley. A 20 kg weight is suspended from the other end ofthe rope.

Which way will the 50 kg mass accelerate?

Which way will the 20 kg mass accelerate?

Draw a Free Body Diagram for each mass:

50 kg20 kg

This system is sometimes called an "Atwood Machine".

Tension Force

50 kg20 kg

50 kg 20 kg

FT

m1 g

FT

m2 g

Remember the Tension in the rope is the same everywhere, so

FT is the same for both masses. The direction of acceleration is

also different. What about the magnitude of acceleration?

aa

Since the two masses are connected to each other, they must

accelerate at the same rate. Otherwise, the rope would have to

continuously get longer as the masses got further apart!

Slide to reveal

Tension Force

50 kg 20 kg

FT

m1 g

FT

m2 g

a a

Remember there is no special equation for tension. We need to

use net force to find the tension.

Below each diagram, write the Net Force equation for each mass:

F = m1am1g - FT = m1a

F = m2aFT - m2g = m2a

Tension Force

50 kg 20 kg

FT

m1 g

FT

m2 g

a a

F = m1am1g - FT = m1a

F = m2aFT - m2g = m2a

What is different about each formula?Why aren't they the same?


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August 30, 20

Tension Force

m1g - FT = m1a FT - m2g = m2a

We have two equations (one for each mass) and two

unknowns (FT and a). This means we can combine the

equations together to solve for each variable!

First we'll solve each one for F T:

m1g - FT = m1am1g = m1a + FT

m1g - m1a = FT

FT - m2g = m2aFT = m2g + m2a

Now we can set them equal to one another:

m1g - m1a = FT FT = m2g + m2a

m1g - m1a = m2g + m2a

Tension Force

There is only one unknown (a) here. Solve for a:

m1g - m1a = m2g + m2a

m1g = m2g + m2a + m1a

m1g - m2g = m2a + m1a

m1g - m2g = a (m2 + m1)

m1g - m2g = a(m2 + m1)

Add m1a to both sides

subtract m2g from both sides

factor out 'a'

(remember factoring is just theopposite of distributing!)

divide by (m1 + m2)

m1g - m1a = m2g + m2a

Tension Force

Substitute and solve:

m1g - m2g = a(m2 + m1)

a = 50(9.8) - 20(9.8)70

a = 4. 2 m/s2

50 kg 20 kg

Remember: this is theacceleration for both m 1 and m2.

Tension Force

50 kg20 kg

Now we can use either equation to solve for Tension:

m1g - m1a = FT

FT = m1g - m1a

FT = 50(9.8) - 50 (4.2)

FT = 280 N

FT = m2g + m2a

FT = 20(9.8) + 20(4.2)

FT = 280 N

We get the same answer either way, sincethe Tension is the same in both ropes!

Return to

Table of

Contents

General Problems

Putting it all together...

An 1800 kg elevator moves up and down on a cable. Calculate the

tension force in the cable for the following cases:

a) the elevator moves at a constant speed upward.

b) the elevator moves at a constant speed downward.c) the elevator accelerates upward at a rate of 2.4 m/s

2

.

d) the elevator accelerates downward at a rate of 2.4 m/s2

.


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August 30, 20


An 1800 kg elevator moves up and down on a cable. Calculate the

tension force in the cable for the following cases:


a = 0

mg

T

Given:m = 1800 kg

g = 9.8 m/s

2

a = 0T = ?

F = ma

T - mg = ma and a =0,T - mg = 0T = mgT = (1800 kg)(9.8 m/s2)T = 17,640 N


An 1800 kg elevator moves up and down on a cable. Calculate thetension force in the cable for the following cases:

b) the elevator moves at a constant speed downward.

a = 0

mg

T

Given:m = 1800 kgg = 9.8 m/s

2

a = 0T = ?

F = maT - mg = ma and a =0,T - mg = 0T = mgT = (1800 kg)(9.8 m/s2)T = 17,640 N

No different than if the constant speed is upward!



c) the elevator accelerates upward at a rate of 2.4 m/s2

.

a

mg

T

Given:m = 1800 kgg = 9.8 m/s2

a = 2.4 m/s2

T = ?

F = ma

T - mg = ma

T = ma + mgT = m(a + g)

T = (1800 kg)(2.4 m/s2 + 9.8 m/s2)

T = 21,960 N



d) the elevator accelerates downward at a rate of 2.4 m/s2.

Given:m = 1800 kgg = 9.8 m/s

2

a = -2.4 m/s2

T = ?

F = ma

T - mg = ma

T = ma + mg

T = m(a + g)

T = (1800 kg)(-2.4 m/s2 + 9.8 m/s2)T = 13,320 N

T

mg

a

A 50 kg man stands on a scale inside an elevator. State the scale

measurement for the following cases:




.

d) the elevator accelerates downward at a rate of 1.4 m/s

2

.

A 50 kg man stands on a scale inside an elevator. State the scalemeasurement for the following cases:


Given:m = 50 kgg = 9.8 m/s2

a = 0

FN

= ?

F = ma

FN

- mg = ma and a =0,

FN

- mg = 0

FN

= mg

FN

= (50 kg)(9.8 m/s2)

FN

= 490 N

FN

mg

a = 0


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August 30, 20



Given:m = 50 kg

g = 9.8 m/s2

a = 0

FN

= ?

F = ma

FN

- mg = ma and a =0,

FN

- mg = 0

FN = mgFN

= (50 kg)(9.8 m/s2)

FN

= 490 N

FN

mg

a = 0

A 50 kg man stands on a scale inside an elevator. State the scale

measurement for the following cases:


.

Given:

m = 50 kgg = 9.8 m/s2

a = 1.4 m/s2

FN = ?

F = ma

FN

- mg = ma

FN

= ma + mg

FN

= m(a + g)

FN = (50 kg)(1.4 m/s2 + 9.8 m/s2)F

N= 560 N

FN

mg

a


d) the elevator accelerates downward at a rate of 1.4 m/s2

.

Given:m = 50 kgg = 9.8 m/s2

a = -1.4 m/s2

FN

= ?

F = ma

FN

- mg = ma

FN

= ma + mg

FN

= m(a + g)

FN

= (50 kg)(-1.4 m/s2 + 9.8 m/s2)

FN

= 420 N

FN

mg

a

Read the problem carefully; then read it again.

Draw a sketch, and then a free-body diagram.

Choose a convenient coordinate system .

List the known and unknown quantities;

Find relationships between the knowns and unknowns.

Estimate the answer.

Solve the problem without numbers, a lgebraically.

Then put the numbers in and solve for a numerical answer.

Keep track of dimensions .

Make sure your answer is reasonable .

Problem Solving A General Approach

#75. A train with a mass of 25000 kg increases its speed from 36 km/h

(10 m/s) to 90 km/h (25 m/s) in 20 seconds. Assume that the

acceleration is constant and that you can neglect friction.

a. Find the acceleration of the train;

b. Find the distance traveled during this 20 s?

c. Draw a free- body diagram for the train;d. Find the average net force supplied by the locomotive.

A train with a mass of 25000 kg increases its speed from 36 km/h

(10 m/s) to 90 km/h (25 m/s) in 20 seconds. Assume that theacceleration is constant and that you can neglect friction.

a. Find the acceleration of the train

Given:m = 25000 kg

V0

= 36 km/h (10 m/s)v = 90 km/h (25 m/s)t = 20 sa = ?

v = V0

+ at

a = (v - V0)/t

a = (25 - 10)/20

a = .75 m/s2


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b. Find the distance traveled during this 20 s?

Given:m = 25000 kg

V0

= 36 km/h (10 m/s)v = 90 km/h (25 m/s)

t = 20 sa = .75 m/s2

x = x0

+ V0t + (1/2at2) andx

0= 0

x = (10*20) + (.5*.75*202)

x = 350 m




c. Draw a free- body diagram for the train

Given:

m = 25000 kg

V0

= 36 km/h (10 m/s)v = 90 km/h (25 m/s)

t = 20 sa = .75 m/s2

mg

FN

a

FApplied




d. Find the average net force supplied by the locomotive.

Given:

m = 25000 kg

V0

= 36 km/h (10 m/s)v = 90 km/h (25 m/s)

t = 20 sa = .75 m/s2

F = ma

F = (25000 kg)(.75 m/s2)

F = 18750 N

#77. Two blocks, with masses m1

= 400 g and m2

= 600 g,are connected by a string and lie on a frictionless

tabletop. A force F = 3.5 N is applied to block m2.

a. Draw a free-body diagram for each block showing all

applied forces to scale. Next to each diagram show the

direction of the acceleration of that object.

b. Find the acceleration of each object.

c. Find the tension force in the string between twoobjects.

Two blocks, with masses m1

= 400 g and m2

= 600 g, are connected by a string and lie on a

frictionless tabletop. A force F = 3.5 N is applied to block m2.

a. Draw a free-body diagram for each block showing all applied forcesto scale. Next to each diagram show the direction of the acceleration of

that object.m

1F

N

m1g

FT

a

m2

FN

FT F

app

m2g

a


= 400 g and m2



b. Find the acceleration of each object.

Given:

m1

= 400 g (.4 kg)

m2

= 600 g (.6 kg)

Fapp

= 3.5 Ng = 9.8 m/s2

a = ?

to the right

F = m a

F1 = m1a F2 = m2aT = m1a Fapp - T = m 2a

Fapp - m1a = m2aFapp = m2a + m1aFapp= a (m2 + m1)

a = Fapp

(m 2 + m1)a = 3.5N

(0.4kg + 0.6kg)

a = 3.5 m/s2


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August 30, 20


= 400 g and m2



c. Find the tension force in the string between two objects.

Given:

m1

= 400 g (.4 kg)

m2

= 600 g (.6 kg)

Fapp = 3.5 Ng = 9.8 m/s2

a = 3.5m/s2

FT= ?

F = m1a

FT

= m1a

FT

= (.4 kg)(3.5 m/s2)

FT

= 1.4 N

# 79. A 12 kg load hangs from one end of a rope that passes over a smallfrictionless pulley. A 15 kg counterweight is suspended from the other end ofthe rope. The system is released from rest.

a. Draw a free-body diagram for each object showing all appliedforces in relative scale. Next to each diagram show the direction ofthe acceleration of that object.

b. Find the acceleration of each mass.c. What is the tension force in the rope?

d. What distance does the 12 kg load move in the first 3 s?e. What is the velocity of 15 kg mass at the end of 5 s?

12 kg

15 kg

A 12 kg load hangs from one end of a rope that passes over a small frict ionlesspulley. A 15 kg counterweight is suspended from the other end of the rope. Thesystem is released from rest.

a. Draw a free-body diagram for each object showing all applied forces inrelative scale. Next to each diagram show the direction ofthe acceleration of that object.

m1

(12 kg)

FT

m1g

a

m2

(15 kg)

FT

m2g

a

A 12 kg load hangs from one end of a rope that passes over a small frictionlesspulley. A 15 kg counterweight is suspended from the other end of the rope. Thesystem is released from rest.

b. Find the acceleration of each mass.

Given:

m1

= 12 kg

m2

= 15 kg

g = 9.8 m/s2

a = ?

m1 m2

F = m1a

T - m1g = m1a

T = m1a + m

1g

F = m2a

T - m2g = - m

2a

T = m2g - m

2a

m1a + m1g = m2g - m2a

a(m1+ m2) = g(m2 - m1)

a = 1.1 m/s2


c. What is the tension force in the rope?

Given:

m1

= 12 kg

m2

= 15 kg

g = 9.8 m/s2

a = 1.1 m/s2

T = ?

F = m1a

T - m1g = m1a

T = m1a + m

1g

T = 130.8 N


d. What distance does the 12 kg load move in the first 3 s?

andx0 = 0, V0 = 0

Given:

m1 = 12 kgm

2= 15 kg

g = 9.8 m/s2

a = 1.1 m/s2

t = 3 sx = ?


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August 30, 20


e. What is the velocity of 15 kg mass at the end of 5 s?

v = V0

+ at (andv0

= 0)

v = at

v = 5.5 m/s

Given:

m1

= 12 kg

m2

= 15 kg

g = 9.8 m/s2

a = 1.1 m/s2

t = 5 s

x = ?

#81. A 500 g block lies on a horizontal tabletop. The coefficient of kineticfriction between the block and the surface is 0.25. The block is connected by amassless string to the second block with a mass of 300 g. The string passesover a light frictionless pulley as shown above. The system is released fromrest.

a. Draw clearly labeled free-body diagrams for each of the 500 g and the300g masses. Include all forces and draw them to relative scale. Draw

the expected direction of acceleration next to each free-body diagram.b. Use Newtons Second Law to write an equation for the 500 g mass.c. Use Newtons Second Law to write an equation for the 300 g mass.d. Find the acceleration of the system by simultaneously solving the

system of two equations.e. What is the tension force in the string?

500 g

300 g

A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction betweenthe block and the surface is 0.25. The block is connected by a massless string to the

second block with a mass of 300 g. The string passes over a light frictionless pulleyas shown above. The system is released from rest.

a. Draw clearly labeled free-body diagrams for each of the 500 g and the300g masses. Include all forces and draw them to relative scale. Drawthe expected direction of acceleration next to each free-body diagram.

m1

(.5 kg)

FN

m1g

aF

T

m2 (.3 kg)

FT

m2g

aFfr



b. Use Newtons Second Law to write an equation for the 500 g mass.

x - direction

F = m1a

T - Ffr

= m1a

T = m1a + F

fr

y - direction

F = 0

FN

-m

1g = 0

FN

= m1g

Ffr

= FN

Ffr

= m1g

T = m1a +m

1g


second block with a mass of 300 g. The string passes over a light frictionless pulley

as shown above. The system is released from rest.

y - direction

F = - m2a

T - m2g = - m

2a

T = m2g - m

2a

c. Use Newtons Second Law to write an equation for the 300 g mass.

A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between

the block and the surface is 0.25. The block is connected by a massless string to the


d. Find the acceleration of the system by simultaneously solving the system oftwo equations.

Given:m

1= .5 kg

m2

= .3 kg

g = 9.8 m/s2

= .25a = ?

T = m1a +m1g T = m2g - m2a

m1a + m1g = m2g - m2a

a(m1+ m2) = g(m2 - m1)a = g(m2 - m1)

(m1+ m2)a = 2.14 m/s

2


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August 30, 20

A 500 g block lies on a horizontal tabletop. The coefficient of kinetic friction between

the block and the surface is 0.25. The block is connected by a massless string to thesecond block with a mass of 300 g. The string passes over a light frictionless pulley

as shown above. The system is released from rest.

e. What is the tension force in the string?

T = m1a + m

1g

T = 2.3 N

Given:

m1

= .5 kg

m2

= .3 kg

g = 9.8 m/s2

= .25

a = 2.14 m/s2

T = ?

An elevator (mass 4850 kg) is to be designed so that themaximum acceleration is 0.0680g. What are the maximumand minimum forces the motor should exert on thesupporting cable?

**

An elevator (mass 4850 kg) is to be designed so that the maximum acceleration is 0.0680g. What are themaximum and minimum forces the motor should exert on the supporting cable?

Given:m = 4850 kg

g = 9.8 m/s2

a = 0.0680g = .67 m/s2

Fmax

= ?

Fmin

= ?

(when "a"is up)

F = ma

T - mg = ma

T = ma + mg

T = 50799.5 N (max)

(when "a"is down)

F = - ma

T - mg = - ma

T = mg - ma

T = 44280.5 N (min)

Two boxes are connected by a cord. A person pullshorizontally on box A with force F = 40.0 N. The boxeshave masses of 10 kg and 12 kg. Ignore friction betweenthe boxes and the tabletop.

a) Show the free-body diagram of the box B.

b) Show the free-body diagram of the box A.

c) Find the acceleration of the system.

d) Find the tension in the cord.

FT = 40 NmA =10 kg

mB =12 kg

Two boxes are connected by a cord. A person pulls horizontall y on box A with force F = 40.0 N. Theboxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop.

a) Show the free-body diagram of the box B.

mB

(12 kg)

FN

mBg

aF

T

Two boxes are connected by a cord. A person pulls hor izontally on box A with force F = 40.0 N. Th eboxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop.

b) Show the free-body diagram of the box A.

mA

(10 kg)

FN

aF

appFT

mAg


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August 30, 20

Two boxes are connected by a cord. A person pulls horizontally on box A with force F = 40.0 N. Theboxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop.

c) Find the acceleration of the system.

Given:

mB

= 12 kg

mA

= 10 kg

Fapp

= 40 N

g = 9.8 m/s2

a = ?

to the right

Two boxes are connected by a cord. A person pulls hor izontally on box A with force F = 40.0 N. Th eboxes have masses of 10 kg and 12 kg. Ignore friction between the boxes and the tabletop.

d) Find the tension in the cord.

Given:

mB

= 12 kg

mA

= 10 kg

Fapp

= 40 N

g = 9.8 m/s2

a = 1.82 m/s2

F = mBa

T = mBa

T = 21.84 N

Two masses are suspended over a pulleyby a cable, as shown on the diagram. Letthe mass of the elevator be 1150 kg andthe counterweight 1000 kg.

a) Show the free-body diagram of theelevator

b) Show the free-body diagram of thecounterweight

c) Calculate the acceleration of thesystem

d) Calculate the tension in the cable

Elevatorcar

Counterweight

mC

= 1000 kg

mE

=1150 kg

aE

aC

Two masses are suspended over a pulley by a cable, as shown on the diagram.Let the mass of the elevator be 1150 kg and the counterweight 1000 kg.

a) Show the free-body diagram of the elevator

mE

(1150 kg)

FT

mEg

a


b) Show the free-body diagram of the counterweight

mC (1000 kg)

FT

mCg

a


c) Calculate the acceleration of the system

mc

me

Given:m

c= 1000 kg

me

= 1150 kg

g = 9.8 m/s2

a = ?


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August 30, 20


d) Calculate the tension in the cable

Given:

mc

= 1000 kg

me

= 1150 kg

g = 9.8 m/s2

a = .68 m/s2

T = ?

F = mca

T = mca + m

cg

T = 10480 N

Two boxes are connected by acord running over a pulley. Thecoefficient of kinetic frictionbetween box A and the table is 0.2

a) Show the free-body diagramsof box A and box B

b) Find the acceleration of thesystem of two boxes

c) Find the tension in the cord

A

B

5.0 kg

2.0 kg

A FT

mAg

Ffr

FN

A

B

5.0 kg

2.0 kgB

FT

mBg

Two boxes are connected by a cord running over a pulley. The coefficient ofkinetic friction between box A and the table is 0 .2

a) Show the free-body diagrams of box A and box B

Two boxes are connected by a cord running over a pulley. The coefficient ofkinetic friction between box A and the table is 0 .2

A

B

5.0 kg

2.0 kg

b) Find the acceleration of the system of two boxes

x - direction

F = mAa

T - Ffr

= mAa

T = mAa + F

fr

y - direction

F = 0

FN

-m

Ag = 0

FN

= mAg

Ffr

= FN

Ffr

= mAg

T = mAa +m

Ag

mA

y - direction

F = - mBa

T - mBg = - m

Ba

T = mBg - m

Ba

mB

mAa + mAg = mBg - mBaa(mA+ mB) = g(mB - mA)

a = g(mB - mA)(mA+ mB)

a = 1.4 m/s2

Given:

mA

= 5 kg

mB

= 2 kg

g = 9.8 m/s2

= .2a = ?

Two boxes are connected by a cord running over a pulley. The coefficient ofkinetic friction between box A and the table is 0.2

c) Find the tension in the cord

F = - mBa

T - mBg = - m

Ba

T = mBg - m

Ba

T = 16.8 N

Given:

mA

= 5 kg

mB

= 2 kg

g = 9.8 m/s2

= .2

a = 1.4 m/s2

T = ?

2009 05 Dynamics Presentation 2011-08-30 6 Slides Per Page

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