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

- Tension- Tension

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

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

I. I. Newton’s first law:Newton’s first law: If no If no net forcenet force acts on a body, then the body’s velocity acts on a body, then the body’s velocity cannot change; the body cannot accelerate cannot change; the body cannot accelerate

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

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 speed of

light Einstein’s special theory of relativity.

2) The interacting bodies are on the scale of the atomic structure2) The interacting bodies are on the scale of the atomic structure Quantum mechanicsQuantum mechanics

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

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

amFnet

zznetyynetxxnet maFmaFmaF ,,, ,,

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

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

gmFg

mgW

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

mgN

- Normal force: Normal force: perpendicularperpendicular force on a body from a surface against whichforce on a body from a surface against which the body presses.the body presses.

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

-Tension: Tension: pull on a body directed away from thepull on a body directed away from thebody along a massless cord.body along a massless cord.

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

CBBC FF

QUESTIONSQUESTIONS

Q.Q. Two Two horizontal forces Fhorizontal forces F11, F, F2 2 pull a banana split across a frictionless counter.pull a banana split across a frictionless counter.

Without using a calculator, determine which of the vectors in the free body diagram Without using a calculator, determine which of the vectors in the free body diagram below best represent: a) Fbelow best represent: a) F11, b)F, b)F22. What is the net force component along (c) the . What is the net force component along (c) the

x-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) thex-axis, (d) the y-axis? Into which quadrant do (e) the net-force vector and (f) the split’s acceleration vector point?split’s acceleration vector point?

jNiNF

jNiNF

ˆ)2(ˆ)1(

ˆ)4(ˆ)3(

2

1

jNiNFFFnet ˆ)6(ˆ)2(21

Same quadrant, 4 F1

F2

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

ay≠0

a=0

ay≠0

a=0

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

Fg

)( agmTamTFF gnet

Movem

ent

(a) Increasing speed: (a) Increasing speed: vvf f >v>v0 0 a>0 a>0 T< F T< Fgg

(b) Decreasing speed: (b) Decreasing speed: vvff < v < v00 a<0 a<0 T> F T> Fgg

Q.Q. 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 F. What total mass is across a frictionless floor by force F. What total mass is accelerated to the right by (a) F, (b) cord 3 (c) cord 1? (d) Rank the accelerated to the right by (a) F, (b) cord 3 (c) cord 1? (d) Rank the blocks according to their accelerations, greatest first. (e) Rank the blocks according to their accelerations, greatest first. (e) Rank the cords according to their tension, greatest first.cords according to their tension, greatest first.

T1 T2 T3

(a) F pulls m(a) F pulls mtotaltotal= (10+3+5+2)kg = 20kg= (10+3+5+2)kg = 20kg

(b) Cord 3 (b) Cord 3 T T33 m=(10+3+5)kg = 18kg m=(10+3+5)kg = 18kg

(c) Cord 1 (c) Cord 1 T T11 m= 10kg m= 10kg

(d) F=ma(d) F=ma All tie, same acceleration All tie, same acceleration

(e) F-T(e) F-T33= 2a= 2a

TT33-T-T22= 5a= 5a

TT22-T-T11=3a =3a

TT11=10a=10a

F-TF-T33= 2a = 2a F=18a+2a=20a F=18a+2a=20a

TT33-13a= 5a -13a= 5a T T33=18a=18a

TT22-10a=3a -10a=3a T T22=13a =13a

TT11=10a=10a

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

(a) F(a) Fgg on dog house: on dog house: 44 or 5 or 5

(c) F(c) Ftoytoy from dog house: 1 from dog house: 1

(b) F(b) Fgg on toy box: 2 on toy box: 2

(d) F(d) Fdog-house dog-house from toy box: 4 or from toy box: 4 or 55

(e) F(e) Fdog-housedog-house from floor: 3 from floor: 3

(f) F(f) Ffloorfloor from dog house: 6 from dog house: 6

(g) Equal: 1=2, 1=5, 3=6(g) Equal: 1=2, 1=5, 3=6

(h) Greatest: 6,3h) Greatest: 6,3

(i) Smallest: 1,2,5(i) Smallest: 1,2,5

There are two forces on the 2 kg box in the overhead view of the figure below 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 box. Find 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 and (c) the second force (a) in unit-vector notation and as (b) magnitude and (c) direction.direction.

F2

221

2

22

ˆ20

)ˆ78.20ˆ12(/)ˆ39.10ˆ6(2

/)ˆ39.10ˆ6(/)ˆ240sin12ˆ240cos12(

FiFFF

NjismjikgamF

smjismjia

T

T

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(

222

2

or

NF

NjiF

yTy

x

xTx

FNF

NF

NFNF

2

2

2

78.20

32

2012

221

ˆ20 FiFFFT

Rules to solve Dynamic problems

- Select a reference system.

- Make a drawing of the particle system.

- Isolate the particles within the system.

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

- Find the components of the forces present.

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

**. (a) A 11kg salami is supported by a cord that runs to a spring scale, which is supported by another cord from the ceiling. What is the reading on the scale, which is marked in weigh units? (b) Here the salami is supported by a cord that runs around a pulley and to a scale. The opposite end of the scale is attached by a cord to a wall. What is the reading on the scale? (c) The wall has been replaced by a second salami on the left, and the assembly is stationary. What is the reading on the scale now?

T

Fg

T

T

T NFTaa

NsmkgmgFW

g

g

8.1070)(

8.107)/8.9)(11( 2

T

T

Fg

T

NFT

ab

g 8.107

0)(

T

Fg

T

Fg

T T

NFTac g 8.1070)(

In all three cases the scale is not accelerating, which means that the two In all three cases the scale is not accelerating, which means that the two cords exert forces of equal magnitude on it. The scale reads the cords exert forces of equal magnitude on it. The scale reads the magnitude of either of these forces. In each case the tension force of the magnitude of either of these forces. In each case the tension force of the cord attached to the salami must be the same in magnitude as the weight cord attached to the salami must be the same in magnitude as the weight of the salami because the salami is not accelerating.of the salami because the salami is not accelerating.

***. 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.

nsttsmmtvd xx 4.2)/102.1(03.0 7

dx=0.03m

dyF

v0Fg

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

21431

23116

/1094.4)1011.9(

)/8.9)(1011.9(105.4

smaakgF

smkgNFFmaF

yynet

gynet

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

Ngam

FmamgF 21.7320cos

)5.0(30sin20cos

y

x

30º

F

N

Fg

20º

NNFmgN 9.109020sin30cos

****** In the figure below, m In the figure below, mblockblock=8.5kg and =8.5kg and θθ=30º. Find (a) Tension in =30º. Find (a) Tension in

the cord. (b) Normal force acting on the block. (c) If the cord is cut, the cord. (b) Normal force acting on the block. (c) If the cord is cut, find the find the magnitude of the block’s acceleration.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)(

Friction

• Related to microscopic interactions of surfaces

• Friction is related to force holding surfaces together

• Frictional forces are different depending on whether surfaces are static or moving

Kinetic Friction

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

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

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

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

Kinetic Friction

The kinetic frictional force can be written as

Where k is a constant called the coefficient of kinetic friction

The value of k depends on the surfaces involved

Nkfr FF

m = 20.0 kg

Fg = mg

FN = -mg

Fp

+

+

Ffr = FN

Static Friction

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

m

Fg = mg

FN = -mg

Static Friction

– If the person pushes with a greater force and still does not manage to move the block, the static frictional force still balances it

m

Fg = mg

FN = -mg

Ffr = FN

Fp

Static Friction

– If the person pushes hard enough, the block will move.

• The maximum force of static friction has been exceeded

m

Fg = mg

FN = -mg

Ffr = FN

Fp

Static Friction

• The maximum force of static friction is given by

• Static friction can take any value from zero to sFN

– In other words

Nsfr FF (max)

Nsfr FF

Connected Object problems

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

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

Connected masses

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

m1=10kg

m2=10kg

a2 m/s2

T1

T2

gmam

gmam

up

up

2222

12111

TF

TTF

Connected masses• Two 10 kg blocks are

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

m1=10kg

m2=10kg

a2 m/s2

T1

T2

gmam

gmam

up

up

2222

12111

TF

TTF

m1a + m2a = T1 – T2 – m1g + T2 – m2g

T1 = (m1+m2) (a+g)=(10+10)(2+9.8)=236N

T2 = m2(a+g)=10(2+9.8)=118N

Connected masses

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

• What is T*?

m2=10kg m1=1000kgTT*

*

*

22

11

TF

TTF

am

amm1a=T-m2a a = T / (m1+m2)=.99m/s2

T* = m2a=9.9N

a

Two bodies, m1= 1kg and m=2kg are connected over a massless pulley. The coefficient of kinetic friction between m2 and the incline is 0.1. The angle θ of the incline is 20º. Calculate: a) Acceleration of 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

14.283

27.684.1

:2

8.9

:1

smaNT

NTAdding

aT

amFfTBlock

aT

amTgmBlock

xg