Chapter 7. NEWTONChapter 7. NEWTON’’S SECOND LAWS SECOND LAW

A PowerPoint Presentation by

Paul E. Tippens, Professor of Physics

Southern Polytechnic State University

A PowerPoint Presentation byA PowerPoint Presentation by

Paul E. Tippens, Professor of PhysicsPaul E. Tippens, Professor of Physics

Southern Polytechnic State UniversitySouthern Polytechnic State University

© 2007

The space shuttle Endeavor lifts off for an 11-day mission in space. All of Newton’s laws of motion - the law of inertia, action- reaction, and the acceleration produced by a resultant force - are exhibited during this lift-off. Credit: NASA Marshall Space Flight Center (NASA- MSFC).

NASANASA

Objectives: After completing this Objectives: After completing this module, you should be able to:module, you should be able to:•• Write Write NewtonNewton’’s second laws second law using appropriate using appropriate

units for units for massmass, , forceforce, and , and accelerationacceleration..

•• Demonstrate your understanding of the Demonstrate your understanding of the distinction between distinction between massmass and and weightweight..

•• Draw Draw freefree--body diagramsbody diagrams for objects at rest for objects at rest and in motion.and in motion.

•• Apply Apply NewtonNewton’’s second laws second law to problems to problems involving one or more bodies in constant involving one or more bodies in constant acceleration.acceleration.

NewtonNewton’’s First Law Revieweds First Law Reviewed

Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a resultant force.

NewtonNewton’’s First Law:s First Law: An object at rest or an An object at rest or an object in motion at constant speed will remain object in motion at constant speed will remain at rest or at constant speed in the absence of a at rest or at constant speed in the absence of a resultant force.resultant force.

A glass is placed on a board and the board is jerked quickly to the right. The glass tends to remain at rest while the board is removed.

A glass is placed on a board and the board is jerked quickly to the right. The glass tends to remain at rest while the board is removed.

NewtonNewton’’s First Law (Cont.)s First Law (Cont.)

Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a resultant force.

NewtonNewton’’s First Law:s First Law: An object at rest or an An object at rest or an object in motion at constant speed will remain object in motion at constant speed will remain at rest or at constant speed in the absence of a at rest or at constant speed in the absence of a resultant force.resultant force.

Assume glass and board move together at constant speed. If the board stops suddenly, the glass tends to maintain its constant speed.

Assume glass and board move together at constant speed. If the board stops suddenly, the glass tends to maintain its constant speed.

Understanding the First Law:Understanding the First Law:

(a) The driver is forced to move forward. An object at rest tends to remain at rest.

Discuss what the driver experiences when a car accelerates from rest and then applies the brakes.

(b) Driver must resist the forward motion as brakes are applied. A moving object tends to remain in motion.

NewtonNewton’’s Second Law:s Second Law:

• Second Law: Whenever a resultant force acts on an object, it produces an acceleration: an acceleration that is directly proportional to the force and inversely proportional to the mass.

•• Second Law:Second Law: Whenever a resultant force Whenever a resultant force acts on an object, it produces an acts on an object, it produces an acceleration: an acceleration that is acceleration: an acceleration that is directly proportional to the force and directly proportional to the force and inversely proportional to the mass.inversely proportional to the mass.

Fam

Acceleration and Force With Acceleration and Force With Zero Friction ForcesZero Friction Forces

Pushing the cart with twice the force produces twice the acceleration. Three times the force triples the acceleration.

Acceleration and Mass Acceleration and Mass Again With Zero FrictionAgain With Zero Friction

F F

aa/2

Pushing two carts with same force F produces one-half the acceleration. The acceleration varies inversely with the amount of material (the mass).

Measuring Mass and ForceMeasuring Mass and ForceThe The SI unit of forceSI unit of force is the is the newtonnewton (N)(N) and and the unit for the unit for massmass is the is the kilogram (kg)kilogram (kg)..

Before presenting formal definitions of these Before presenting formal definitions of these units, however, we will conduct an experiment units, however, we will conduct an experiment by slowly increasing the force on a given object.by slowly increasing the force on a given object.

Although the force in Although the force in newtonsnewtons will become will become our standard, we begin by using the more our standard, we begin by using the more familiar unit of forcefamiliar unit of force----the pound (lb).the pound (lb).

Force and AccelerationForce and Acceleration

4 lbF

a = 2 ft/s2

8 lb a = 4 ft/s2F

12 lb a = 6 ft/s2F

Acceleration Acceleration aa is directly proportional to force is directly proportional to force F F and is in the direction of the force. Friction and is in the direction of the force. Friction

forces are ignored in this experiment.forces are ignored in this experiment.

Force and AccelerationForce and AccelerationFF

aa

FF

aa

FF

aa= Constant= Constant

8 lb8 lb

4 ft/s4 ft/s22= = 22

lblb

ft/sft/s22

Inertia or mass ofInertia or mass of 1 slug = 1 lb/(ft/s1 slug = 1 lb/(ft/s22))

Mass m = Mass m = 22 slugsslugs

MASS: A Measure of InertiaMASS: A Measure of Inertia

a = 6 ft/s26 lb1 slug

a = 3 ft/s26 lb2 slugs

a = 2 ft/s26 lb3 slugs

One One slugslug is that mass on which a constant force is that mass on which a constant force of of 11 lblb will produce an acceleration of will produce an acceleration of 1 ft/s1 ft/s22. . Friction forces are ignored in this experiment.Friction forces are ignored in this experiment.

Two Systems of UnitsTwo Systems of Units

USCU system:USCU system: Accept Accept lblb as unit of force, as unit of force, ftft as unit of length, and as unit of length, and ss as unit of time. as unit of time. Derive new unit of mass, the Derive new unit of mass, the slugslug.

F (lb) = m (slugs) a (ft/s2)F (lb) = m (slugs) a (ft/s2)

SI system:SI system: Accept Accept kgkg as unit of mass, as unit of mass, mm as as unit of length, and unit of length, and ss as unit of time. Derive as unit of time. Derive new unit of force, the new unit of force, the newtonnewton (N)(N)..

F (N) = m (kg) a (m/s2)F (N) = m (kg) a (m/s2)

Newton: The Unit of ForceNewton: The Unit of ForceOne One newtonnewton is that resultant force which imparts is that resultant force which imparts an acceleration of an acceleration of 11 m/sm/s22 to a mass ofto a mass of 1 kg1 kg..

F (N) = m (kg) a (m/s2)F (N) = m (kg) a (m/s2)

What resultant force will give a 3 kg mass an What resultant force will give a 3 kg mass an acceleration of 4 m/sacceleration of 4 m/s22? ?

2(3 kg)(4 m/s )F

F = 12 NF = 12 N

Remember F = m aF = ?

a = 4 m/s2

3 kg

Comparing the Comparing the Newton to the PoundNewton to the Pound

1 N = 0.225 lb

1 lb = 4.45 N 1 lb1 lb 4.454.45 NN

A 160A 160--lb person weighs about 712 Nlb person weighs about 712 N

A 10A 10--N hammer weighs about 2.25 lbN hammer weighs about 2.25 lb

Example 1:Example 1: What resultant force What resultant force FF is required is required to give a to give a 6 kg6 kg block an acceleration of block an acceleration of 2 m/s2 m/s22??

F = ?6 kg

a = 2 m/s2

F = ma = (6 kg)(2 m/s2)

F = 12 NF = 12 N

Remember consistent units for Remember consistent units for forceforce, , massmass, , and and accelerationacceleration in all problems.in all problems.

Example 2:Example 2: A A 4040--lblb resultant force causes a resultant force causes a block to accelerate at block to accelerate at 5 ft/s5 ft/s22. What is the mass?. What is the mass?

F = F = 40 lb40 lbm=?m=?

aa = 5 ft/s= 5 ft/s22

m = 8 slugsm = 8 slugs

or FF ma ma

2

40 lb5 ft/s

Fma

You must You must recallrecall that the that the slugslug is the appropriate is the appropriate mass unit when mass unit when FF is in is in lblb and and aa is in is in ft/sft/s22..

Example 3Example 3. . A net force of A net force of 4.2 x 104.2 x 1044 NN acts on acts on a a 3.2 x 103.2 x 1044 kgkg airplane during takeoff. What airplane during takeoff. What is the force on the planeis the force on the plane’’s s 7575--kgkg pilot?pilot?

F = 4.2 x 104 N

m = 3.2 x 104 kg

++F = ma

4

4

4.2 x 10 N3.2 x 10 kg

Fam

a = 1.31 m/s2

To find To find FF on on 7878--kgkg pilot, assume same acceleration:pilot, assume same acceleration:

F = ma = (75 kg)(1.31 m/s2); F = 98.4 N

First we find the First we find the acceleration acceleration aa of of plane.plane.

A Word About Consistent UnitsA Word About Consistent UnitsNow that we haveNow that we have derivedderived units ofunits of newtonsnewtons andand slugsslugs, , we can no longer use units that we can no longer use units that are inconsistent with those definitionsare inconsistent with those definitions..

Acceptable measures of Acceptable measures of LENGTHLENGTH are:are:

SI units: meter meter (m)(m)

USCU units: foot foot (ft)(ft)

UnacceptableUnacceptable units are: centimeters (units are: centimeters (cmcm); ); millimeters (millimeters (mmmm); kilometers (); kilometers (kmkm); ); yards (yards (ydyd); inches (); inches (in.in.); miles (); miles (mimi))

Consistent Units (Continued . . .)Consistent Units (Continued . . .)

UnacceptableUnacceptable units are: grams (units are: grams (gmgm); ); milligrams (milligrams (mgmg); newtons (); newtons (NN); ); pounds (pounds (lblb); ounces (); ounces (ozoz))

Acceptable measures of Acceptable measures of MASSMASS are:are:

SI units: kilogram kilogram (kg)(kg)

USCU units: slug slug (slug)(slug)

The last three unacceptable units are actually units of force instead of mass.

Consistent Units (Continued . . .)Consistent Units (Continued . . .)

UnacceptableUnacceptable units are: units are: kilonewtonskilonewtons ((kNkN); tons (); tons (tonstons); ounces (); ounces (ozoz); ); kilograms (kilograms (kgkg); slugs (); slugs (slugslug))

The last two unacceptable units are not force units—they are units for mass.

Acceptable measures of Acceptable measures of FORCEFORCE are:are:

SI units: newtonnewton (N)(N)

USCU units: pound pound (lb)(lb)

Consistent Units (Cont.)Consistent Units (Cont.)

When we say that the acceptable units for force and mass are the newton and the kilogram, we are referring to their use in physical formulas. ( Such as F = m a)

The centimeter, the millimeter, the milligram, the mile, and the inch may be useful occasionally in describing quantities. But they should not be used in formulas.

Problem Solving StrategyProblem Solving Strategy (For the Simpler Problems.)(For the Simpler Problems.)

• Read problem; draw and label sketch.

• List all given quantities and state what is to be found.

• Make sure all given units are consistent with Newton’s second law of motion (F = m a).

• Determine two of the three parameters in Newton’s law, then solve for the unknown.

•• Read problem; draw and label sketch.Read problem; draw and label sketch.

•• List all given quantities and state what is to List all given quantities and state what is to be found.be found.

•• Make sure all given units are consistent with Make sure all given units are consistent with NewtonNewton’’s second law of motion (s second law of motion (F = m F = m aa).).

•• Determine two of the three parameters in Determine two of the three parameters in NewtonNewton’’s law, then solve for the unknown.s law, then solve for the unknown.

Example 4.Example 4. A A 5454--gmgm tennis ball is in contact tennis ball is in contact with the racket for a distance of with the racket for a distance of 40 cm40 cm as it as it leaves with a velocity of leaves with a velocity of 48 m/s48 m/s. What is the . What is the average force on the ball?average force on the ball?

Given: Given: vvoo = 0; = 0; vvff = 48 m/s = 48 m/s xx = 0.40 m; = 0.40 m; m = 0.0540 km; m = 0.0540 km; a a = ?= ?

First, draw sketch and list First, draw sketch and list given quantities:given quantities:

Given: Given: vvoo = 0; = 0; vvff = 48 m/s = 48 m/s xx = 40 cm; = 40 cm; mm = 54 gm = 54 gm a a = ?= ?

Consistent units require converting Consistent units require converting gramsgrams to to kilogramskilograms and and centimeterscentimeters to to metersmeters::

Cont. . .

Example 4 (Cont).Example 4 (Cont). A A 5454--gmgm tennis ball is in tennis ball is in contact with the racket for a distance of contact with the racket for a distance of 40 40 cmcm as it leaves with a velocity of as it leaves with a velocity of 48 m/s48 m/s. . What is the average force on the ball?What is the average force on the ball?

2 202 ;fax v v

00

22(48 m/s) ; 2880 m/s

2(0.40 m)a a

F= F= (0.054 kg)(2880 m/s(0.054 kg)(2880 m/s22);); F = 156 N

F = maF = ma

2

2fv

ax

Knowing that Knowing that F = m F = m aa, we need , we need first to find acceleration first to find acceleration aa::

Weight and MassWeight and Mass

•• WeightWeight is the force due to gravity. It is is the force due to gravity. It is directed downward and it varies from directed downward and it varies from location to location.location to location.

•• MassMass is a universal constant which is a is a universal constant which is a measure of the inertia of a body.measure of the inertia of a body.

F = m F = m aa so that:so that: W = mgW = mg and and m =m =WW

gg

Weight and Mass: ExamplesWeight and Mass: Examples

• What is the mass of a What is the mass of a 6464--lblb block?block?

W = mgW = mg

64 lb64 lb3232 ft/sft/s22

• What is the weight of a What is the weight of a 1010--kgkg block?block?

9.89.8 mm/s/s22 WW

mm10 kg10 kg W = mg = W = mg = (10 kg)(9.8 m/s(10 kg)(9.8 m/s22))

W = 98 NW = 98 N

2

64 lb32 ft/

2 s ss

lugm

Mass is Constant; W Varies.Mass is Constant; W Varies.

EarthEarth

9898 NN 9.8 m/s9.8 m/s22

49 N49 N 4.9 m/s4.9 m/s22 32 lb32 lb 16 ft/s16 ft/s22

EarthEarth

64 lb64 lb 32 ft/s32 ft/s22

m = = m = = 10 kg10 kgWW

ggm = = m = = 22 slugsslugs

WW

gg

Description of ObjectsDescription of Objects• Objects described by mass or weight:Objects described by mass or weight:

• Conversions made by NewtonConversions made by Newton’’s 2nd Law:s 2nd Law:

W W (N) = (N) = mm (kg) x 9.8 m/s(kg) x 9.8 m/s22

W = mgW = mg m =m =WW

gg

W W (lb) = (lb) = mm (slugs) x 32 ft/s(slugs) x 32 ft/s22

Inconsistent Common UsageInconsistent Common UsageIn the In the United StatesUnited States, objects are often , objects are often referred to by their weight at a point where referred to by their weight at a point where gravity is equal to 32 ft/sgravity is equal to 32 ft/s22. .

W = 3200 lb

800 lbYou might hear:You might hear: ““An An 800800--lb force pulls a lb force pulls a 32003200--lb car.lb car.””

This car should be called a called a 100100--slugslug car.car.

Thus, when an object is described as a _?_-lb object, we remember to divide by g to get mass.

Inconsistent Usage (Cont.)Inconsistent Usage (Cont.)Even Even metric units metric units are used inconsistently. Mass are used inconsistently. Mass in in kgkg is often treated as if it were weight (is often treated as if it were weight (N)N). . This is sometimes called the kilogramThis is sometimes called the kilogram--force. force.

The kilogram is a mass - never a force - and it doesn’t have direction or vary with gravity.

A chemist might be asked A chemist might be asked to to weighweigh out 200 g of a out 200 g of a certain element. Also, you certain element. Also, you hear about a 10hear about a 10--kg kg loadload as as if it were if it were weightweight. .

F

10 kg

Always Remember!!Always Remember!!In Physics, the use of Newton’s second law and many other applications makes it absolutely necessary to distinguish between mass and weight. Use the correct units!

Metric SI units: Mass is in kg; weight is in N.

USCU units: Mass is in slugs; weight is in lb.

Always give preference to the SI units.

Example 5.Example 5. A A resultantresultant forceforce of of 40 N40 N gives a block an gives a block an accelerationacceleration of of 8 m/s8 m/s22. . What is the What is the weightweight of the block near of the block near the surface of the Earth?the surface of the Earth?

W=?W=?

F = F = 40 N40 Naa 8 m/s8 m/s22

To find weight, we must first To find weight, we must first find the mass of the block:find the mass of the block:

; FF ma ma

2

4 0 N 5 k g8 m /s

m

Now find weight of a 5-kg mass on earth. Now find weight of a 5-kg mass on earth.

W = mgW = mg

= = (5 kg)(9.8 m/s(5 kg)(9.8 m/s22))

W = 49.0 NW = 49.0 N

NewtonNewton’’s Third Law (Reviewed):s Third Law (Reviewed):

• Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs.

•• Third Law:Third Law: For every action force, there For every action force, there must be an equal and opposite reaction must be an equal and opposite reaction force. Forces occur in pairs.force. Forces occur in pairs.

Action

Reaction ActionReaction

Acting and Reacting ForcesActing and Reacting Forces• Use the words by and on to study

action/reaction forces below as they relate to the hand and the bar:

•• Use the words Use the words byby and and onon to study to study action/reaction forces below as they action/reaction forces below as they relate to the relate to the handhand and the and the barbar::

The The actionaction force is exerted force is exerted by by the _____ the _____ onon the _____.the _____.

The reaction force is exerted The reaction force is exerted by by the _____ the _____ onon the _____.the _____.barbar

handshands barbar

handshands

Action

Reaction

Example 6:Example 6: A A 6060--kgkg athlete exerts a force on athlete exerts a force on a a 1010--kgkg skateboard. If she receives an skateboard. If she receives an acceleration of acceleration of 4 m/s4 m/s22, what is the , what is the acceleration of the skateboard?acceleration of the skateboard?

Force on runner = Force on runner = --(Force on board)(Force on board)

mm rr aarr = = --mm bb aabb

(60 kg)(4 m/s(60 kg)(4 m/s22) = ) = --(10 kg) (10 kg) aabb

a = - 24 m/s2a = - 24 m/s2

Force on Force on RunnerRunner

Force on Force on BoardBoard

2(60 kg)(4 m/s) 24 m/s-(10 kg)

a

Review of FreeReview of Free--body Diagrams:body Diagrams:

• Read problem; draw and label sketch.

• Construct force diagram for each object, vectors at origin of x,y axes.

• Dot in rectangles and label x and y compo- nents opposite and adjacent to angles.

• Label all components; choose positive direction.

•• Read problem; draw and label sketch.Read problem; draw and label sketch.

•• Construct force diagram for each object, Construct force diagram for each object, vectors at origin of x,y axes.vectors at origin of x,y axes.

•• Dot in rectangles and label x and y compoDot in rectangles and label x and y compo-- nentsnents opposite and adjacent to angles.opposite and adjacent to angles.

•• Label all components; choose positive Label all components; choose positive direction.direction.

Example of FreeExample of Free--body Diagrambody Diagram

300 600

4 kg

A ABB

W = mg

300 600

Bx

By

Ax

Ay

1. Draw and label sketch.1. Draw and label sketch.2. Draw and label vector force diagram.2. Draw and label vector force diagram.3. Dot in rectangles and label x and y compo3. Dot in rectangles and label x and y compo--

nentsnents opposite and adjacent to angles.opposite and adjacent to angles.

Applying NewtonApplying Newton’’s Second Laws Second Law

• Read, draw, and label problem.

• Draw free-body diagram for each body.

• Choose x or y-axis along motion and choose direction of motion as positive.

• Write Newton’s law for both axes:

F x = m a x F y = m a y

• Solve for unknown quantities.

•• Read, draw, and label problem.Read, draw, and label problem.

•• Draw freeDraw free--body diagram for each body.body diagram for each body.

•• Choose x or yChoose x or y--axis along motion and choose axis along motion and choose direction of motion as positive.direction of motion as positive.

•• Write NewtonWrite Newton’’s law for both axes:s law for both axes:

FF xx = m = m a xx FF yy = m = m a yy•• Solve for unknown quantities. Solve for unknown quantities.

Example 7:Example 7: A cart and driver have a mass of A cart and driver have a mass of 120 kg120 kg. What force . What force FF is required to give an is required to give an acceleration of acceleration of 6 m/s6 m/s22 with no friction?with no friction?

1. Read problem and draw a sketch.1. Read problem and draw a sketch.

2. Draw a vector force diagram and label forces.2. Draw a vector force diagram and label forces.

Diagram for Cart:

n

W

F

3. Choose x3. Choose x--axis along motion and indicate the axis along motion and indicate the right direction as positive (+).right direction as positive (+).

x+

Example 7 (Cont.)Example 7 (Cont.) What force What force FF is required is required to give an acceleration of to give an acceleration of 6 m/s6 m/s22? ?

Fy = 0; n - W = 0

The normal force n is equal to weight W

Fx = max ; F = ma

F F = (120 kg)(6 m/s= (120 kg)(6 m/s22))

F = 720 N

Diagram for cart:

n

W

Fx

+

m =m = 120 kg120 kg

4. Write Newton's Law equation for both axes.4. Write Newton's Law equation for both axes.

ay = 0

Example 8:Example 8: What is the tension What is the tension TT in the in the rope below if the block accelerates upward rope below if the block accelerates upward at at 4 m/s4 m/s22? (Draw sketch and free? (Draw sketch and free--body.)body.)

1010 kgkg

aa = = +4 m/s+4 m/s22

TT aa

TT

mgmg

++

FF xx = m = m aaxx = 0 (No info)= 0 (No info)

FF yy = m = m aayy = m = m aa

T T -- mg = m mg = m aa

mg = mg = (10 kg)(9.8 m/s) = 98 N(10 kg)(9.8 m/s) = 98 N

m m aa= = (10 kg)(4 m/s) = 40 N(10 kg)(4 m/s) = 40 N

T -- 98 N98 N = = 40 N40 N T = 138 NT = 138 N

Example 9:Example 9: In the absence of friction, what In the absence of friction, what is the acceleration down the is the acceleration down the 303000 incline?incline?

303000

mgmg

nn

606000

nn

WW

mg mg coscos 606000mg mg sin 60sin 6000

++

FF xx = m = m aaxx

mg mg cos 60cos 6000 = m = m aa

aa = g = g cos 60cos 6000

aa = = (9.8 m/s(9.8 m/s22) cos 60) cos 6000

a = 4.9 m/s2a = 4.9 m/s2

Example 10.Example 10. TwoTwo--Body Problem: Find tension in the Body Problem: Find tension in the connecting rope if there is no friction on the surfaces.connecting rope if there is no friction on the surfaces.

2 kg2 kg 4 kg4 kg12 N12 N Find acceleration of Find acceleration of

system and tension system and tension in connecting cord.in connecting cord.

First apply First apply F = mF = maa to entire system (to entire system (both massesboth masses).).

12 N12 Nnn

((mm 22 + m+ m 44 ))gg

FF xx = = ((mm 22 + m+ m 44 ) ) aa

12 N = (6 kg) 12 N = (6 kg) aa

aa ==1212 NN

6 kg6 kg a = 2 m/s2a = 2 m/s2

Example 10 (Cont.)Example 10 (Cont.) The twoThe two--body problem.body problem.

2 kg 4 kg12 N

Now find tension T Now find tension T in connecting cord.in connecting cord.

Apply F = m Apply F = m aa to the to the 22 kgkg mass where mass where aa = = 2 2 m/sm/s22..

TTnn

mm 22 gg

FF xx = = mm 2 2 aa

TT = (2 kg)(2 m/s= (2 kg)(2 m/s22))

T = 4 NT = 4 N

Example 10 (Cont.)Example 10 (Cont.) The twoThe two--body problem.body problem.

2 kg2 kg 4 kg4 kg12 N12 N Same answer for Same answer for TT

results from focusing results from focusing on on 44--kgkg by itself.by itself.

Apply Apply F = m F = m aa to the to the 4 kg4 kg mass where mass where aa = 2 m/s= 2 m/s22..

FF xx = = mm 4 4 aa

12 N 12 N -- TT = (4 kg)(2 m/s= (4 kg)(2 m/s22))

T = 4 NT = 4 N

1212 NNnn

mm 22 gg

TT

Example 11Example 11 Find acceleration of system and Find acceleration of system and tension in cord for the arrangement shown.tension in cord for the arrangement shown.

First apply First apply F = m F = m aa to entire to entire system system alongalong the line of motion.the line of motion.

FF xx = = ((mm 22 + m+ m 44 ) ) aa

a = 6.53 m/s2a = 6.53 m/s2

nn

mm 22 gg

TT

mm 44 gg

TT

+ + aa

Note Note mm 22 gg is balanced by is balanced by nn..mm 44 g = g = ((mm 22 + m+ m 44 ) ) aa

(4 kg)(9.8 m/s(4 kg)(9.8 m/s22))

2 2 kgkg + + 4 kg4 kgaa = == =

mm 44 gg

mm 22 + m+ m 44

22 kgkg

44 kgkg

Example 11 (Cont.)Example 11 (Cont.) Now find the tension Now find the tension TT given given that the acceleration is that the acceleration is aa = 6.53 m/s= 6.53 m/s22..

To find To find TT, apply , apply F = m F = m aa to just to just the the 2 kg2 kg mass, ignoring 4 kg.mass, ignoring 4 kg.

T = T = (2 kg)(6.53 m/s(2 kg)(6.53 m/s22))

T = 13.1 NT = 13.1 N

Same answer if using 4 kg.Same answer if using 4 kg.

mm 44 g g -- T = mT = m 4 4 aaT = mT = m 44 (g (g -- aa)) = = 13.1 N13.1 N

nn

mm 22 gg

TT

mm 44 gg

TT

+ + aa

22 kgkg

4 kg4 kg2 2 or xF m a T m a

Example 11.Example 11. Find the acceleration of the system Find the acceleration of the system shown below. (The Atwood machine.)shown below. (The Atwood machine.)

First apply F = ma to entire system along the line of motion.

FF xx = = ((mm 22 + m+ m 55 ) ) aa

a = 4.20 m/s2a = 4.20 m/s2

TT

mm 22 gg mm 55 gg

TT

+a+a

2 kg2 kg 5 kg5 kg

25 2

2 5

(5 kg 2 kg)(9.8 m/s )2 kg + 5 kg

m g m gam m

5 2 2 5( )m g m g m m a

SummarySummary

Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

Newton’s Second Law: A resultant force produces an acceleration in the direction of the force that is directly proportional to the force and inversely proportional to the mass.

Newton’s First Law: An object at rest or an object in motion at constant speed will remain at rest or at constant speed in the absence of a resultant force.

NewtonNewton’’s First Law:s First Law: An object at rest or an An object at rest or an object in motion at constant speed will remain at object in motion at constant speed will remain at rest or at constant speed in the absence of a rest or at constant speed in the absence of a resultant force.resultant force.

Newton’s Third Law: For every action force, there must be an equal and opposite reaction force. Forces occur in pairs.

NewtonNewton’’s Third Law:s Third Law: For every action force, For every action force, there must be an equal and opposite reaction there must be an equal and opposite reaction force. Forces occur in pairs.force. Forces occur in pairs.

Summary: ProcedureSummary: Procedure

• Read, draw and label problem.

• Draw free-body diagram for each body.

• Choose x or y-axis along motion and choose direction of motion as positive.

• Write Newton’s law for both axes:

F x = m a x F y = m a y

• Solve for unknown quantities.

•• Read, draw and label problem.Read, draw and label problem.

•• Draw freeDraw free--body diagram for each body.body diagram for each body.

•• Choose x or yChoose x or y--axis along motion and choose axis along motion and choose direction of motion as positive.direction of motion as positive.

•• Write NewtonWrite Newton’’s law for both axes:s law for both axes:

FF xx = m = m a xx FF yy = m = m a yy•• Solve for unknown quantities. Solve for unknown quantities.

; RR

FF ma am

N = (kg)(m/s2)N = (kg)(m/s2)

CONCLUSION: Chapter 7CONCLUSION: Chapter 7 NewtonNewton’’s Second Law of Motions Second Law of Motion