newton’s first law. newton’s second law. particular forces:

I.I. Newton’s first law.Newton’s first law.

II.II. Newton’s second law.Newton’s second law.

III.III. Particular forces:Particular forces: - Gravitational- Gravitational

- Weight- Weight - Normal- Normal

- Friction- Friction - Tension- Tension

IV. Newton’s third law.IV. Newton’s third law.

Chapter 5 – Force and Motion IChapter 5 – Force and Motion I

ForceForce

• Forces are what cause any change in the Forces are what cause any change in the velocity of an objectvelocity of an object– A force is that which causes an accelerationA force is that which causes an acceleration

• The The net forcenet force is the vector sum of all the forces is the vector sum of all the forces acting on an objectacting on an object– Also called total force, resultant force, or Also called total force, resultant force, or

unbalanced forceunbalanced force

Zero Net ForceZero Net Force

• When the net force is equal to zero:When the net force is equal to zero:– The acceleration is equal to zeroThe acceleration is equal to zero– The velocity is constantThe velocity is constant

• EquilibriumEquilibrium occurs when the net force is equal occurs when the net force is equal to zeroto zero– The object, if at rest, will remain at restThe object, if at rest, will remain at rest– If the object is moving, it will continue to move If the object is moving, it will continue to move

at a constant velocityat a constant velocity

Classes of ForcesClasses of Forces

• Contact forces involve Contact forces involve physical contact physical contact between two objectsbetween two objects

• Field forces act Field forces act through empty spacethrough empty space– No physical contact is No physical contact is

requiredrequired

Fundamental ForcesFundamental Forces

• Gravitational forceGravitational force– Between two objectsBetween two objects

• Electromagnetic forcesElectromagnetic forces– Between two chargesBetween two charges

• Nuclear (strong) forceNuclear (strong) force– Between subatomic particlesBetween subatomic particles

• Weak forcesWeak forces– Arise in certain radioactive decay processesArise in certain radioactive decay processes

• A spring can be used A spring can be used to calibrate the to calibrate the magnitude of a forcemagnitude of a force

• Forces are vectors, Forces are vectors, so you must use the so you must use the rules for vector rules for vector addition to find the net addition to find the net force acting on an force acting on an objectobject

I.I. Newton’s first law:Newton’s first law: If no If no net forcenet force acts on a body, then the acts on a body, then the body’s velocity cannot change; the body body’s velocity cannot change; the body cannot accelerate cannot accelerate vv = constant in = constant in magnitude and direction. magnitude and direction.

Principle of superpositionPrinciple of superposition:: when two or more forces act when two or more forces act on a body, the net force can be obtained by adding the on a body, the net force can be obtained by adding the individual forces as vectors.individual forces as vectors.

Newton mechanics laws cannot be applied when:Newton mechanics laws cannot be applied when: 1) The speed of the interacting bodies are a fraction of the 1) The speed of the interacting bodies are a fraction of the speed of light speed of light Einstein’s special theory of relativityEinstein’s special theory of relativity..2) The interacting bodies are on the scale of the atomic 2) The interacting bodies are on the scale of the atomic structure structure Quantum mechanicsQuantum mechanics

Newton’s First LawNewton’s First Law• If an object does not interact with other If an object does not interact with other

objects, it is possible to identify a objects, it is possible to identify a reference frame in which the object has reference frame in which the object has zero accelerationzero acceleration– This is also called the This is also called the law of inertialaw of inertia– It defines a special set of reference frames It defines a special set of reference frames

called called inertial framesinertial frames,,• We call this an We call this an inertial frame of referenceinertial frame of reference

Inertial reference frame:Inertial reference frame: where Newton’s laws hold.where Newton’s laws hold.

Inertial FramesInertial Frames

• Any reference frame that moves with constant Any reference frame that moves with constant velocity relative to an inertial frame is itself an velocity relative to an inertial frame is itself an inertial frameinertial frame

• A reference frame that moves with constant A reference frame that moves with constant velocity relative to the distant stars is the best velocity relative to the distant stars is the best approximation of an inertial frameapproximation of an inertial frame– We can consider the Earth to be such an inertial We can consider the Earth to be such an inertial

frame although it has a small centripetal acceleration frame although it has a small centripetal acceleration associated with its motionassociated with its motion

Newton’s First Law – Alternative StatementNewton’s First Law – Alternative Statement

• In the absence of external forces, when viewed In the absence of external forces, when viewed from an inertial reference frame, an object at rest from an inertial reference frame, an object at rest remains at rest and an object in motion remains at rest and an object in motion continues in motion with a constant velocitycontinues in motion with a constant velocity– Newton’s First Law describes what happens in the Newton’s First Law describes what happens in the

absence of a forceabsence of a force– Also tells us that when no force acts on an object, the Also tells us that when no force acts on an object, the

acceleration of the object is zeroacceleration of the object is zero

Inertia and MassInertia and Mass

• The tendency of an object to resist any The tendency of an object to resist any attempt to change its velocity is called attempt to change its velocity is called inertiainertia

• MassMass is that property of an object that is that property of an object that specifies how much resistance an object specifies how much resistance an object exhibits to changes in its velocityexhibits to changes in its velocity

• Mass is an inherent property of an objectMass is an inherent property of an object• Mass is independent of the object’s Mass is independent of the object’s

surroundingssurroundings• Mass is independent of the method used Mass is independent of the method used

to measure itto measure it• Mass is a scalar quantityMass is a scalar quantity• The The SI SI unit of mass is unit of mass is kgkg

Mass vs. WeightMass vs. Weight

• Mass and weight are two different Mass and weight are two different quantitiesquantities

• Weight is equal to the magnitude of the Weight is equal to the magnitude of the gravitational force exerted on the objectgravitational force exerted on the object– Weight will vary with locationWeight will vary with location

Newton’s Second LawNewton’s Second Law

• When viewed from an inertial frame, the When viewed from an inertial frame, the acceleration of an object is directly acceleration of an object is directly proportional to the net force acting on it proportional to the net force acting on it and inversely proportional to its massand inversely proportional to its mass– Force is the cause of change in motion, as Force is the cause of change in motion, as

measured by the accelerationmeasured by the acceleration• Algebraically, Algebraically, F = F = m m aa

II.II. Newton’s second law: Newton’s second law: The net force on a body is The net force on a body is equal to the product of the body’s mass and its equal to the product of the body’s mass and its acceleration.acceleration.

amFnet

zznetyynetxxnet maFmaFmaF ,,, ,,

The acceleration component along a given axis is caused The acceleration component along a given axis is caused only by the sum of the force components along the same only by the sum of the force components along the same axis, and not by force components along any other axis. axis, and not by force components along any other axis.

Units of ForceUnits of Force

System:System: collection of bodies.collection of bodies.

External force:External force: any force on the bodies inside the system.any force on the bodies inside the system.

III. Particular forces:III. Particular forces:

Gravitational:Gravitational: pull directed towards a second body, normally pull directed towards a second body, normally the Earth the Earth

gmFg

Weight:Weight: magnitude of the upward force needed to balance magnitude of the upward force needed to balance the gravitational force on the body due to an astronomical the gravitational force on the body due to an astronomical body body

mgW

mgN

Normal force:Normal force: perpendicular force on a body from a surface perpendicular force on a body from a surface against which the body presses.against which the body presses.

Frictional force:Frictional force: force on a force on a body when the body body when the body attempts to slide along a attempts to slide along a surface. It is parallel to the surface. It is parallel to the surface and opposite to the surface and opposite to the motion.motion.

Tension:Tension: pull on a body directed away from the body pull on a body directed away from the body along a massless cord.along a massless cord.

Newton’s Third LawNewton’s Third Law

• If two objects interact, the force If two objects interact, the force FF1212 exerted by object exerted by object 11 on object on object 22 is equal in is equal in magnitude and opposite in direction to the magnitude and opposite in direction to the force force FF2121 exerted by object exerted by object 22 on object on object 11

• FF1212 = - = - FF2121

– Note on notation: Note on notation: FFABAB is the force exerted by is the force exerted by AA on on BB

Newton’s Third Law, Alternative Newton’s Third Law, Alternative StatementsStatements

• Forces always occur in pairsForces always occur in pairs• A single isolated force cannot existA single isolated force cannot exist• The action force is equal in magnitude to the The action force is equal in magnitude to the

reaction force and opposite in directionreaction force and opposite in direction– One of the forces is the action force, the other is the One of the forces is the action force, the other is the

reaction forcereaction force– It doesn’t matter which is considered the action and It doesn’t matter which is considered the action and

which the reactionwhich the reaction– The action and reaction forces must act on different The action and reaction forces must act on different

objects and be of the same typeobjects and be of the same type

Action-Reaction ExamplesAction-Reaction Examples

• The force The force FF1212 exerted exerted by object by object 11 on object on object 22 is equal in is equal in magnitude and magnitude and opposite in direction opposite in direction to to FF2121 exerted by exerted by object object 22 on object on object 11

• FF1212 = - = - FF2121

Newton’s third lawNewton’s third law:: When two bodies interact, When two bodies interact, the forces on the bodies from each other are always the forces on the bodies from each other are always equal in magnitude and opposite in direction.equal in magnitude and opposite in direction.

CBBC FF

QUESTIONSQUESTIONSQ2.Q2. Two horizontal forces,Two horizontal forces,

pull a banana split across a frictionless counter. Without using pull a banana split across a frictionless counter. Without using a calculator, determine which of the vectors in the free body a calculator, determine which of the vectors in the free body diagram below best represent: a) diagram below best represent: a) FF11, , FF22;; b) What is the net b) What is the net force component along (c) the x-axis, (d) the y-axis? Into force component along (c) the x-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) the split’s which quadrant do (e) the net-force vector and (f) the split’s acceleration vector point?acceleration vector point?

jNiNFandjNiNF ˆ)2(ˆ)1(ˆ)4(ˆ)3( 21

QUESTIONSQUESTIONSQ2.Q2. Two horizontal forces,Two horizontal forces,

pull a banana split across a frictionless counter. Without using pull a banana split across a frictionless counter. Without using a calculator, determine which of the vectors in the free body a calculator, determine which of the vectors in the free body diagram below best represent: a) Fdiagram below best represent: a) F11, F, F22;; b) What is the net b) What is the net force component along (c) the x-axis, (d) the y-axis? Into force component along (c) the x-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) the split’s which quadrant do (e) the net-force vector and (f) the split’s acceleration vector point?acceleration vector point?

jNiNFandjNiNF ˆ)2(ˆ)1(ˆ)4(ˆ)3( 21

)ˆ6(ˆ)2(21 jNiNFFFnet

Same quadrant, 4Same quadrant, 4

Q1. The figure below shows overhead views of four situations Q1. The figure below shows overhead views of four situations in which forces act on a block that lies on a frictionless floor. If in which forces act on a block that lies on a frictionless floor. If the force magnitudes are chosen properly, in which situation it the force magnitudes are chosen properly, in which situation it is possible that the block is (a) stationary and (b) moving with is possible that the block is (a) stationary and (b) moving with constant velocity?constant velocity?

Q1. The figure below shows overhead views of four situations Q1. The figure below shows overhead views of four situations in which forces act on a block that lies on a frictionless floor. If in which forces act on a block that lies on a frictionless floor. If the force magnitudes are chosen properly, in which situation it the force magnitudes are chosen properly, in which situation it is possible that the block is (a) stationary and (b) moving with is possible that the block is (a) stationary and (b) moving with constant velocity?constant velocity?

aayy≠0≠0 a=a=00

aayy≠0≠0a=a=00

Q5. In which situations does the object acceleration have Q5. In which situations does the object acceleration have (a) an x-component, (b) a y component? (c) give the direction (a) an x-component, (b) a y component? (c) give the direction of of aa..

FFnetnetFFnetnet

Q5. In which situations does the object acceleration have Q5. In which situations does the object acceleration have (a) an x-component, (b) a y component? (c) give the direction (a) an x-component, (b) a y component? (c) give the direction of a.of a.

Q. A body suspended by a rope has a weigh Q. A body suspended by a rope has a weigh WW of of 75N75N. Is . Is TT equal to, greater than, or less than equal to, greater than, or less than 75N75N when the body is when the body is moving downward at (a) increasing speed and (b) decreasing moving downward at (a) increasing speed and (b) decreasing speed?speed?

FFgg

Q. A body suspended by a rope has a weigh Q. A body suspended by a rope has a weigh WW of of 75N75N. Is . Is TT equal to, greater than, or less than equal to, greater than, or less than 75N75N when the body is when the body is moving downward at (a) increasing speed and (b) decreasing moving downward at (a) increasing speed and (b) decreasing speed?speed?

FFgg

(a) Increasing speed: (a) Increasing speed: vvf f >v>v00 a>0a>0

)( agmTamTFF gnet

T < FT < Fgg

(b) Decreasing speed:(b) Decreasing speed: vvff < v < v00 a<0a<0

)( agmTamTFF gnet

T > Fg

Q8.Q8. The figure below shows a train of four blocks being pulled The figure below shows a train of four blocks being pulled across a frictionless floor by force across a frictionless floor by force FF. What total mass is . What total mass is accelerated to the right by (a) accelerated to the right by (a) FF, (b) , (b) cord 3cord 3 (c) (c) cord 1cord 1? ? (d) Rank the blocks according to their accelerations, greatest (d) Rank the blocks according to their accelerations, greatest first. (e) Rank the cords according to their tension, greatest first. (e) Rank the cords according to their tension, greatest first.first.

Q8.Q8. The figure below shows a train of four blocks being pulled The figure below shows a train of four blocks being pulled across a frictionless floor by force across a frictionless floor by force FF. What total mass is . What total mass is accelerated to the right by (a) accelerated to the right by (a) FF, (b) , (b) cord 3cord 3 (c) (c) cord 1cord 1? ? (d) Rank the blocks according to their accelerations, greatest (d) Rank the blocks according to their accelerations, greatest first. (e) Rank the cords according to their tension, greatest first. (e) Rank the cords according to their tension, greatest first.first.

TT11 TT22 TT33

(a) (a) FF pulls pulls mmtotaltotal= (10+3+5+2)kg = 20kg= (10+3+5+2)kg = 20kg

(b) Cord (b) Cord 33 TT33 m=(10+3+5)kg = 18kgm=(10+3+5)kg = 18kg

(c) Cord (c) Cord 11 TT11 m= 10kgm= 10kg

(d) (d) F=maF=ma All tie, same acceleration All tie, same acceleration (e) T(e) T33 > T > T22 >T >T11

Q.Q. A toy box is on top of a heavier dog house, which sits on a wood floor. A toy box is on top of a heavier dog house, which sits on a wood floor. These objects are represented by dots at the corresponding heights, and six These objects are represented by dots at the corresponding heights, and six vertical vectors (not to scale) are shown. Which of the vectors best vertical vectors (not to scale) are shown. Which of the vectors best represents (a) the gravitational force on the dog house, (b) on the toy box, represents (a) the gravitational force on the dog house, (b) on the toy box, (c) the force on the toy box from the dog house, (d) the force on the dog (c) the force on the toy box from the dog house, (d) the force on the dog house from the toy box, (e) force on the dog house from the floor, (f) the house from the toy box, (e) force on the dog house from the floor, (f) the force on the floor from the dog house? (g) Which of the forces are equal in force on the floor from the dog house? (g) Which of the forces are equal in magnitude? Which are (h) greatest and (i) least in magnitude?magnitude? Which are (h) greatest and (i) least in magnitude?

•> 7

89

10

Q.Q. A toy box is on top of a heavier dog house, which sits on a wood floor. A toy box is on top of a heavier dog house, which sits on a wood floor. These objects are represented by dots at the corresponding heights, and six These objects are represented by dots at the corresponding heights, and six vertical vectors (not to scale) are shown. Which of the vectors best vertical vectors (not to scale) are shown. Which of the vectors best represents (a) the gravitational force on the dog house, (b) on the toy box, represents (a) the gravitational force on the dog house, (b) on the toy box, (c) the force on the toy box from the dog house, (d) the force on the dog (c) the force on the toy box from the dog house, (d) the force on the dog house from the toy box, (e) force on the dog house from the floor, (f) the house from the toy box, (e) force on the dog house from the floor, (f) the force on the floor from the dog house? force on the floor from the dog house?

(a)(a) F Fgg on dog house: on dog house: 44 or or 55

(c(c) ) FFtoytoy from dog house: from dog house: 1010((b) b) FFgg on toy box: on toy box: 22

(d) (d) FFdog-housedog-house from toy box:from toy box: 9 9

(e) (e) FFdog-housedog-house from floor: from floor: 88

(f) (f) FFfloorfloor from dog house: from dog house: 66

•> 7

89

10

5.5. There are two forces on the There are two forces on the 2 kg2 kg box in the overhead view of box in the overhead view of the figure below but only one is shown. The figure also shows the the figure below but only one is shown. The figure also shows the acceleration of the box. Find the second force (a) in unit-vector acceleration of the box. Find the second force (a) in unit-vector notation and as (b) magnitude and (c) direction.notation and as (b) magnitude and (c) direction.

F2

5. There are two forces on the 5. There are two forces on the 2 kg2 kg box in the overhead view of the figure box in the overhead view of the figure below but only one is shown. The figure also shows the acceleration of the below but only one is shown. The figure also shows the acceleration of the box. Find the second force (a) in unit-vector notation and as (b) magnitude box. Find the second force (a) in unit-vector notation and as (b) magnitude and (c) direction.and (c) direction.

F2

5. There are two forces on the 5. There are two forces on the 2 kg2 kg box in the overhead view of the figure box in the overhead view of the figure below but only one is shown. The figure also shows the acceleration of the below but only one is shown. The figure also shows the acceleration of the box. Find the second force (a) in unit-vector notation and as (b) magnitude box. Find the second force (a) in unit-vector notation and as (b) magnitude and (c) direction.and (c) direction.

F2

221

2

22

ˆ20

)ˆ78.20ˆ12(/)ˆ39.10ˆ6(2

/)ˆ39.10ˆ6(/)ˆ240sin12ˆ240cos12(

FiFFF

NjismjikgamF

smjismjia

net

net

5.5. There are two forces on the 2 kg box in the overhead view of the figure There are two forces on the 2 kg box in the overhead view of the figure below but only one is shown. The figure also shows the acceleration of the below but only one is shown. The figure also shows the acceleration of the box. Find the second force (a) in unit-vector notation and as (b) magnitude box. Find the second force (a) in unit-vector notation and as (b) magnitude and (c) direction.and (c) direction.

F2

213331803332

78.20tan

27.382132

)ˆ78.20ˆ32(22

2

2

or

NF

NjiF

yynet

x

xxnet

FNFNF

NFNF

2,

2

2,

78.2032

2012

221ˆ20 FiFFFnet

Rules to solve Dynamic problemsRules to solve Dynamic problems

- Select a reference system.Select a reference system.

- Make a drawing of the particle system.Make a drawing of the particle system.

- Isolate the particles within the system.Isolate the particles within the system.

- Draw the forces that act on each of the isolated bodies.Draw the forces that act on each of the isolated bodies.

- Find the components of the forces present.Find the components of the forces present.

- Apply Newton’s second law (Apply Newton’s second law (F=maF=ma) to each isolated particle.) to each isolated particle.

23. An electron with a speed of 1.2x107m/s moves horizontally into a region where a constant vertical force of 4.5x10-16N acts on it. The mass of the electron is m=9.11x10-31kg. Determine the vertical distance the electron is deflected during the time it has moved 30 mm horizontally.

ddxx=0.03m=0.03m

ddyyF

v0Fg

23. An electron with a speed of 1.2x107m/s moves horizontally into a region where a constant vertical force of 4.5x10-16N acts on it. The mass of the electron is m=9.11x10-31kg. Determine the vertical distance the electron is deflected during the time it has moved 30 mm horizontally.

ddxx=0.03m=0.03m

ddyyF

v0Fg

nsttsmmtvd xx 5.2)/102.1(03.0 7

21431

23116

/1094.4)1011.9(

)/8.9)(1011.9(105.4

smaakgF

smkgNFFmaF

yynet

gynet

mssmtatvd yoyy 0015.0)105.2()/1094.4(5.05.0 292142

13. In the figure below, 13. In the figure below, mmblockblock=8.5kg =8.5kg and and θθ=30º=30º. . Find (a) Tension Find (a) Tension in the cord. (b) Normal force acting on the block. (c) If the cord is in the cord. (b) Normal force acting on the block. (c) If the cord is cut, find the magnitude of the block’s acceleration.cut, find the magnitude of the block’s acceleration.

TN

Fg

y

x

13. In the figure below, 13. In the figure below, mmblockblock=8.5kg=8.5kg and and θθ=30º.=30º. Find (a) Tension Find (a) Tension in the cord. (b) Normal force acting on the block. (c) If the cord is in the cord. (b) Normal force acting on the block. (c) If the cord is cut, find the magnitude of the block’s acceleration.cut, find the magnitude of the block’s acceleration.

TN

Fg

y

x

2/9.45.865.410)( smaaNmaFTc gx

NsmkgmgFTaa gx 65.415.0)/8.9)(5.8(30sin0)( 2

NmgFNb gy 14.7230sincos)(

FrictionFriction

• Related to microscopic interactions of surfacesRelated to microscopic interactions of surfaces• Friction is related to force holding surfaces Friction is related to force holding surfaces

togethertogether• Frictional forces are different depending on Frictional forces are different depending on

whether surfaces are static or movingwhether surfaces are static or moving

Kinetic FrictionKinetic Friction

• When a body slides over a surface there is a When a body slides over a surface there is a forceforce exerted by the surface on the body in the exerted by the surface on the body in the opposite directionopposite direction to the motion of the body to the motion of the body

• This force is called This force is called kinetic frictionkinetic friction• The magnitude of the force depends on the The magnitude of the force depends on the

nature of the two touching surfacesnature of the two touching surfaces• For a given pair of surfaces, the magnitude of For a given pair of surfaces, the magnitude of

the kinetic frictional force is proportional to the the kinetic frictional force is proportional to the normal force exerted by the surface on the body. normal force exerted by the surface on the body.

Kinetic FrictionKinetic Friction The kinetic frictional force can be written as

Where k is a constant called the coefficient of kinetic frictionThe value of k depends on the surfaces involved

Nkfr FF

m = 20.0 kg

FFgg = -mg = -mg

FFNN = mg = mg

FFpp

+

+

FFfrfr = = FFNN

Static FrictionStatic Friction

A frictional force can arise even if the body A frictional force can arise even if the body remains stationaryremains stationary– A block on the floor - no frictional forceA block on the floor - no frictional force

m

FFgg = mg = mg

FFNN = -mg = -mg


– If someone pushes the desk (but it does not If someone pushes the desk (but it does not move) then a static frictional force is exerted move) then a static frictional force is exerted by the floor on the desk to balance the force by the floor on the desk to balance the force of the person on the deskof the person on the desk

m

FFgg = mg = mg

FFNN = -mg = -mg

FFfrfr = = FFNN

Fp

Static FrictionStatic Friction– If the person pushes with a greater force and If the person pushes with a greater force and

still does not manage to move the block, the still does not manage to move the block, the static frictional force still balances itstatic frictional force still balances it

m

FFgg = mg = mg

FFNN = -mg = -mg

FFfrfr = = FFNN

Fp


– If the person pushes hard enough, the block If the person pushes hard enough, the block will move.will move.• The The maximum force of static frictionmaximum force of static friction has been has been

exceededexceeded

mm

FFgg = mg = mg

FFNN = -mg = -mg

FFfrfr = = FFNN

FFpp


• The maximum force of static friction is The maximum force of static friction is given bygiven by

• Static friction can take any value from zero Static friction can take any value from zero to to ssFFNN

– In other wordsIn other words

Nsfr FF (max)

Nsfr FF

Forces on an inclineForces on an incline

• Often when solving Often when solving problems involving problems involving Newton’s laws we Newton’s laws we will need to deal will need to deal with resolving with resolving acceleration due to acceleration due to gravity on an gravity on an inclined surfaceinclined surface


xx

yy

W = mg

mgsinmgsin

mgcos

What normal forceWhat normal forcedoes the surface exert?does the surface exert?


cos

sin

mgN

mg

y

x

F

F


0cos

sin

mgN

mmg

y

x

F

aF

EquilibriumEquilibrium


• If the car is If the car is justjust stationary on the stationary on the incline what is the incline what is the (max) coefficient of (max) coefficient of static friction?static friction?

0cos

0sin

mgN

mNmg

y

sx

F

aF

tancossin

cossin

s

ss mgNmg

‘Equilibrium’ problems (Equilibrium’ problems (F = 0)F = 0)

053sin37sin

037cos53cos

30

20

1

01

02

TTy

x

TF

TTF

Connected Object problemsConnected Object problems

• One problem often posed is how to work One problem often posed is how to work out acceleration of a system of masses out acceleration of a system of masses connected via strings and/or pulleysconnected via strings and/or pulleys

– for example - Two blocks are fastened to the for example - Two blocks are fastened to the ceiling of an elevator. ceiling of an elevator.

Connected massesConnected masses

• Two Two 10 kg10 kg blocks are blocks are strung from an elevator strung from an elevator roof, which is accelerating roof, which is accelerating up at up at 2 m/s2 m/s22.. m1=10kg

m2=10kg

aa2 m/s2 m/s22

TT11

TT22


• Two Two 10 kg10 kg blocks are blocks are strung from an elevator strung from an elevator roof, which is accelerating roof, which is accelerating up at up at 2 m/s2 m/s22.. m1=10kg

m2=10kg

aa2 m/s2 m/s22

TT11

TT22

gmam

gmam

up

up

2222

12111

TF

TTF

Connected massesConnected masses• Two Two 10 kg10 kg blocks are blocks are

strung from an elevator strung from an elevator roof, which is accelerating roof, which is accelerating up at up at 2 m/s2 m/s22..

m1=10kg

m2=10kg

aa2 m/s2 m/s22

TT11

TT22gmam

gmam

up

up

2222

12111

TF

TTF

mm11a + ma + m22a = Ta = T1 1 – T– T22 – m – m11g + Tg + T22 – m – m22gg

TT11 = (m = (m11+m+m22) (a+g)=(10+10)(2+9.8)=) (a+g)=(10+10)(2+9.8)=236N236N

TT22 = m = m22(a+g)=10(2+9.8)=(a+g)=10(2+9.8)=118N118N


• What is the acceleration of the system What is the acceleration of the system below, if below, if TT is is 1000 N1000 N??

• What is What is T*T*??

mm22=10kg=10kg mm11=1000kg=1000kgTTT*T*

aa


• What is the acceleration of the system What is the acceleration of the system below, if below, if TT is is 1000 N1000 N??

• What is What is T*T*??

mm22=10kg=10kg mm11=1000kg=1000kgTTT*T*

*

*

22

11

TF

TTF

am

ammm11a=a=TT-m-m22aa a = T / (ma = T / (m11+m+m22)=)=.99m/s.99m/s22

T* = T* = mm22a=a=9.9N9.9N

aa


• What are What are T*T* and and aa now? now?

m 2=10kg

m 1=1000kg

TT

T*T*

aa


• What are What are T*T* and and aa now? now?

m 2=10kg

m 1=1000kg

TT

T*T*

sin*

sin*

222

111

gmam

gmam

TF

TTFaa

Connected massesConnected masses• Add in friction: Add in friction: μ = 0.4μ = 0.4• What are What are T*T* and and aa now? now?

m 2=10kg

m 1=1000kg

TT

T*T*

aa

Connected massesConnected masses• Add in friction: Add in friction: μ = 0.4μ = 0.4• What are What are T*T* and and aa now? now?

m 2=10kg

m 1=1000kg

TT

T*T*

cossin*

cossin*

2222

1111

gmgmam

gmgmam

k

k

TF

TTF aa

Motion over pulleysMotion over pulleys

• We know that the We know that the magnitude and magnitude and sensesense of the of the acceleration should acceleration should be the same for the be the same for the whole systemwhole system

1000kg

10kg

aa

aa TT

Θ = 30Θ = 3000

μ = 0.4μ = 0.4

Motion over pulleysMotion over pulleys

• We know that the We know that the magnitude and magnitude and sensesense of the of the acceleration should acceleration should be the same for the be the same for the whole systemwhole system

1000kg

10kg

TF

TF

gmam

gmgmam k

222

1111 cossin

aa

aa TT

Θ = 30Θ = 3000

μ = 0.4μ = 0.4

55. The figure below gives as a function of time. The figure below gives as a function of time t t, the force , the force component component FFxx that acts on a that acts on a 3kg3kg ice block, which can move only along ice block, which can move only along the the x x axis. Ataxis. At t=0 t=0, the block is moving in the positive direction of the , the block is moving in the positive direction of the axis, with a speed of axis, with a speed of 3m/s3m/s. What are (a) its speed and (b) direction of . What are (a) its speed and (b) direction of travel at travel at t=11st=11s? ?

55. The figure below gives as a function of time. The figure below gives as a function of time t t, the force , the force component component FFxx that acts on a that acts on a 3kg3kg ice block, which can move only along ice block, which can move only along the the x x axis. Ataxis. At t=0 t=0, the block is moving in the positive direction of the , the block is moving in the positive direction of the axis, with a speed of axis, with a speed of 3m/s3m/s. What are (a) its speed and (b) direction of . What are (a) its speed and (b) direction of travel at travel at t=11st=11s? ?

sx

f

txxx

x

f

dtmFvvdt

dtdv

dtdv

mFa

vstsmvt

11

00

0

0

?11/30

smsmkg

skgmv

kgsmvmvvdtFNsareagraphTotal

f

f

s

fx

/8/33

/15

3)/3()(1511

00

Two bodies, Two bodies, mm11= 1kg= 1kg and and mm22=2kg=2kg are connected over a are connected over a massless pulley. The coefficient of kinetic friction between massless pulley. The coefficient of kinetic friction between mm22 and the incline is and the incline is 0.10.1. The angle . The angle θ θ of the incline is of the incline is 20º20º.. Calculate: a) Acceleration of the blocks. (b) Tension of the Calculate: a) Acceleration of the blocks. (b) Tension of the cord.cord.

T

N

m1

m2

20ºm1g

T

m2g

f

Two bodies, Two bodies, mm11= 1kg= 1kg and and mm22=2kg=2kg are connected over a massless are connected over a massless pulley. The coefficient of kinetic friction between pulley. The coefficient of kinetic friction between mm22 and the incline and the incline isis 0.1 0.1. The angle . The angle θ ofθ of the incline is the incline is 20º.20º. Calculate: a) Acceleration of Calculate: a) Acceleration of the blocks. (b) Tension of the cord.the blocks. (b) Tension of the cord.

NgmNf

NgmFN

NgmF

kk

yg

xg

84.120cos

42.1820cos

7.620sin

22

2,22

2,2

T

N

m1

m2

20ºm1g

T

m2g

f

2

2,2

11

/42.0,38.9

14.28327.684.1

:28.9

:1

smaNT

NTAddingaT

amFfTBlockaT

amTgmBlock

xg

The three blocks in the figure below are connected by massless cords The three blocks in the figure below are connected by massless cords and pulleys. Data: and pulleys. Data: mm11=5kg=5kg, , mm22=3kg=3kg, , mm33=2kg=2kg. Assume that the incline plane . Assume that the incline plane is frictionless.is frictionless. (i) Show all the forces that act on each block.(i) Show all the forces that act on each block. (ii) Calculate the acceleration of (ii) Calculate the acceleration of mm11, , mm22, , mm33..

m1

m2

m330º

m3g

T2

T1

m2g

T2

T1

N

Fg2x

Fg2y

(iii) Calculate the tensions on the cords. (iii) Calculate the tensions on the cords. (iv) Calculate the normal force acting on (iv) Calculate the normal force acting on mm22

m1g

The three blocks in the figure below are connected by massless cords The three blocks in the figure below are connected by massless cords and pulleys. Data: and pulleys. Data: mm11=5kg=5kg, , mm22=3kg=3kg, , mm33=2kg=2kg. Assume that the incline plane . Assume that the incline plane is frictionless.is frictionless. (i) Show all the forces that act on each block.(i) Show all the forces that act on each block. (ii) Calculate the acceleration of (ii) Calculate the acceleration of mm11, , mm22, , mm33..

m1

m2

m330º

m3g

T2

T1

m2g

T2

T1

N

Fg2x

Fg2y


Fg2y=m2gcosFg2x=m2gsin30º

m1g

The three blocks in the figure below are connected by massless cords The three blocks in the figure below are connected by massless cords and pulleys. Data: and pulleys. Data: m1=5kgm1=5kg, , m2=3kgm2=3kg, , m3=2kgm3=2kg. Assume that the incline . Assume that the incline plane is frictionless.plane is frictionless. (i) Show all the forces that act on each block.(i) Show all the forces that act on each block. (ii) Calculate the acceleration of (ii) Calculate the acceleration of mm1, 1, m2m2, , m3m3..

m1

m2

m330º

m3g

T2

T1

m2g

T2

T1

N

Fg2x

Fg2y


m1gBlock 1: m1g-T1=m1aBlock 2: m2g(sin30º) +T1-T2=m2aBlock 3: T2-m3g = m3aAdding (1)+(2)+(3): g(m1+0.5m2-m3)=a(m1+m2+m3) a=4.41m/s2

(ii) T1=m1(g-a)= 5kg(9.8 m/s2-4.41 m/s2) = 26.95N

(iii) T2=m3(g+a)= 2kg(9.8 m/s2+4.41 m/s2)= 28.42N

(iv) N2= Fg2y= m2gcos30º = 25.46N

1B. 1B. (a) What should be the magnitude of (a) What should be the magnitude of FF in the figure below if the body of in the figure below if the body of mass mass m=10kg m=10kg is to slide up along a frictionless incline plane with constant is to slide up along a frictionless incline plane with constant acceleration acceleration a=1.98 m/sa=1.98 m/s22? (b) What is the ? (b) What is the magnitude of the Normal force?magnitude of the Normal force?

yyxx

30º

FN

Fg

20º

1B. 1B. (a) What should be the magnitude of (a) What should be the magnitude of FF in the figure below if the body of in the figure below if the body of mass mass m=10kg m=10kg is to slide up along a frictionless incline plane with constant is to slide up along a frictionless incline plane with constant acceleration acceleration a=1.98 m/sa=1.98 m/s22? (b) What is the ? (b) What is the magnitude of the Normal force?magnitude of the Normal force?

NgamFmamgF 21.7320cos

)5.0(30sin20cos

yyxx

30º

FN

Fg

20º

NNFmgN 9.109020sin30cos

Three forces, given by , , Three forces, given by , , and , act on an object to give it an acceleration of and , act on an object to give it an acceleration of magnitude magnitude 3.75 m/s3.75 m/s22. (a) What is the direction of the acceleration? . (a) What is the direction of the acceleration? (b) What is the mass of the object? (c) If the object is initially at rest, (b) What is the mass of the object? (c) If the object is initially at rest, what is its speed after what is its speed after 10.0 s10.0 s? (d) What are the velocity components ? (d) What are the velocity components of the object after of the object after 10.0 s10.0 s??

F1 2.00 i 2.00 j N F2 5.00 i 3.00 j N

F3 45.0 i N


F1 2.00 i 2.00 j N F2 5.00 i 3.00 j N

F3 45.0 i N

mF a 2ˆ ˆ ˆ ˆ ˆ ˆ2.00 2.00 5.00 3.00 45.0 N 3.75 m sm i j i j i awhere where a represents the direction of represents the direction of aa

2ˆ ˆ ˆ42.0 1.00 N 3.75 m sm i j a

2 242.0 1.00 N F 1 1.00tan42.0

atat

2 ˆ42.0 N at 181 3.75 m sm F a


F1 2.00 i 2.00 j N F2 5.00 i 3.00 j N

F3 45.0 i N

(a)(a) ˆ is at 181 a counterclockwise from the counterclockwise from the xx-axis -axis

(b)(b) 242.0 N 11.2 kg

3.75 m sm

(d)(d) 20 3.75 m s at 181 10.0 sfi t v v a

ˆ ˆ37.5 0.893 m sf v i j

(c)(c) 2 237.5 0.893 m s 37.5 m sf v

A bag of cement of weight A bag of cement of weight 325 N325 N hangs from three wires as hangs from three wires as suggested in Figure. Two of the wires make angles suggested in Figure. Two of the wires make angles 11 = 60.0 = 60.0 and and 2 2 = 25.0= 25.0 with the horizontal. If the system is in with the horizontal. If the system is in equilibrium, find the tensions equilibrium, find the tensions TT11, , TT22, and , and TT33 in the wires. in the wires.

A bag of cement of weight A bag of cement of weight 325 N325 N hangs from three wires as hangs from three wires as suggested in Figure. Two of the wires make angles suggested in Figure. Two of the wires make angles 11 = 60.0 = 60.0 and and 2 2 = 25.0= 25.0 with the horizontal. If the system is in with the horizontal. If the system is in equilibrium, find the tensions equilibrium, find the tensions TT11, , TT22, and , and TT33 in the wires. in the wires.

1

1 2

F

2

T 3

T 2 T 1

g

3 gT F

1 1 2 2sin sin gT T F

1 1 2 2cos cosT T

2 2

11 2 1 2 1 2

3

1

12 1

2

cos cossin cos cos sin sin

325 N

cos25.0 296 Nsin85.0cos cos60.0296 N 163 Ncos cos25.0

g g

g

g

F FT

T F

T F

T T

A block is given an initial velocity of A block is given an initial velocity of 5.00 m/s5.00 m/s up a up a frictionless frictionless 20.0°20.0° incline. How far up the incline does the incline. How far up the incline does the block slide before coming to rest?block slide before coming to rest?

A block is given an initial velocity of A block is given an initial velocity of 5.00 m/s5.00 m/s up a up a frictionless frictionless 20.0°20.0° incline. How far up the incline does incline. How far up the incline does the block slide before coming to rest?the block slide before coming to rest?

After it leaves your hand, the After it leaves your hand, the block’s speed changes only block’s speed changes only because of one component of because of one component of its weight:its weight:

2 2

sin20.0

2 .x x

fi fi

F ma mg ma

v v a x x

0fv 5.00 m siv sin 20.0a g

20 5.00 2 9.80 sin 20.0 0fx

25.0 3.73 m

2 9.80 sin 20.0fx

A woman at an airport is towing her A woman at an airport is towing her 20.0-kg20.0-kg suitcase at suitcase at constant speed by pulling on a strap at an angle constant speed by pulling on a strap at an angle above above the horizontal. She pulls on the strap with a the horizontal. She pulls on the strap with a 35.0-N35.0-N force, force, and the friction force on the suitcase is and the friction force on the suitcase is 20.0 N20.0 N. Draw a . Draw a free-body diagram of the suitcase. (a) What angle does free-body diagram of the suitcase. (a) What angle does the strap make with the horizontal? (b) What normal force the strap make with the horizontal? (b) What normal force does the ground exert on the suitcase?does the ground exert on the suitcase?


suitcase 20.0 kgm 35.0 NF

: 20.0 N cos 0: sin 0

x x

y y g

F ma FF ma n F F

(a)(a) cos 20.0 N

20.0 Ncos 0.57135.0 N

55.2

F


suitcase 20.0 kgm 35.0 NF

: 20.0 N cos 0: sin 0

x x

y y g

F ma FF ma n F F

sin 196 35.0 0.821 Ngn F F

(b)(b)

167 Nn

Three objects are connected on the table as shown in Three objects are connected on the table as shown in Figure. The table is rough and has a coefficient of kinetic Figure. The table is rough and has a coefficient of kinetic friction of friction of 0.3500.350. The objects have masses . The objects have masses 4.00 kg4.00 kg, , 1.00 1.00 kgkg and and 2.00 kg2.00 kg, as shown, and the pulleys are frictionless. , as shown, and the pulleys are frictionless. Draw free-body diagrams of each of the objects. (a) Draw free-body diagrams of each of the objects. (a) Determine the acceleration of each object and their Determine the acceleration of each object and their directions. (b) Determine the tensions in the two cords.directions. (b) Determine the tensions in the two cords.

Three objects are connected on the table as shown in Figure. The Three objects are connected on the table as shown in Figure. The table is rough and has a coefficient of kinetic friction of table is rough and has a coefficient of kinetic friction of 0.3500.350. The . The objects have masses objects have masses 4.00 kg4.00 kg, , 1.00 kg1.00 kg and and 2.00 kg2.00 kg, as shown, and the , as shown, and the pulleys are frictionless. Draw free-body diagrams of each of the pulleys are frictionless. Draw free-body diagrams of each of the objects. (a) Determine the acceleration of each object and their objects. (a) Determine the acceleration of each object and their directions. (b) Determine the tensions in the two cords.directions. (b) Determine the tensions in the two cords.

n

T12 T23

m g 2

f = n k

m g 1

T12

m g 3

T23

For m1

1m,

y yF ma12 1 1T mg ma

newton’s first law. newton’s second law. particular forces:

Documents

zerothe object

net force acts

accelerationthe net

resultant force

total force

objecta force

inertial framewe

constant velocity relative