WarmupWarmup

2

m mx

s sm

x msm

x ss

2

2

2

m mx

s sm

x msm

x ss

ForceForce

Affects on Velocity and AccelerationAffects on Velocity and Acceleration

Force Causes Force Causes ChangeChange in Velocity in Velocity

This one time at band camp…

Ha, ha,

good one

Oh! Skeeter!

Force and MotionForce and Motion

Newton’s First LawNewton’s First Law At rest – Stays at rest (until force is applied)At rest – Stays at rest (until force is applied) In motion – Stays in motion (until force is applied)In motion – Stays in motion (until force is applied) Force causes change in velocityForce causes change in velocity Force causes accelerationForce causes acceleration Force causes a change in directionForce causes a change in direction

Warm-upWarm-up

A blue one with a nubbin was A blue one with a nubbin was moving at 5m/s straight down moving at 5m/s straight down when the problem started. The when the problem started. The difference between the bottom difference between the bottom and the top is 3 times the and the top is 3 times the height of a trans-atlantic 10m height of a trans-atlantic 10m building. This nubbin sporting building. This nubbin sporting thing is thing is earthboundearthbound. (include . (include units)units)

i

f

i

f

d

d

a

v

v

r

r

r

Types of Fundamental ForceTypes of Fundamental Force

Gravitational ForceGravitational Force Force we use in this sectionForce we use in this section

Electromagnetic ForceElectromagnetic Force Includes the contact forces we work with in this section Includes the contact forces we work with in this section

Nuclear ForceNuclear Force Weak ForceWeak Force

(summarize)(summarize)

Electromagnetic ForceElectromagnetic Force

Contact ForcesContact Forces NormalNormal FrictionFriction

Static: friction when the object is not in motionStatic: friction when the object is not in motion Sliding: friction when the object is in motionSliding: friction when the object is in motion

TensionTension SpringSpring

(Use article to organize with topic oval)(Use article to organize with topic oval)

ForceForce

Newton’s Third LawNewton’s Third Law Every action has an opposite Every action has an opposite

and equal reaction.and equal reaction.

_ _ _ _A on B B on AF Fur ur

ForceForce

Experiment with force sensors.Experiment with force sensors. equal and oppositeequal and opposite

Key VocabKey Vocab

Net force (FNet force (Fnetnet)) Vector addition/subtractionVector addition/subtraction

5 N + 7N5 N + 7N

The resultant is the net force The resultant is the net force

5N + 7N =5N + 7N =

12N12N

2

kg mN

s

g

Newton’s Second LawNewton’s Second Law

Two men pull a 50-kg box with forces 19.7 N Two men pull a 50-kg box with forces 19.7 N and 15.6 N in the directions shown below. and 15.6 N in the directions shown below. Find the net force of the box.Find the net force of the box.

19.7 N 15.4 N19.7 N 15.4 N

or or

-19.7N-19.7N

2

kg mN

s

g

ForceForce

The The pound-forcepound-force or simply or simply poundpound (abbreviations: (abbreviations: lblb, , lbflbf, or , or lbflbf) is a unit of force) is a unit of force 1N=0.225lb; 1lb=4.45N1N=0.225lb; 1lb=4.45N

Normal plus lift (FNormal plus lift (Fnetnet)) Weight from force gauge.Weight from force gauge. Upward force must be greater than gravity to have Upward force must be greater than gravity to have

upward acceleration.upward acceleration.HW: HW: 16,17, (pg 97), 16,17, (pg 97), Example problem 2 pg 99 with elevator 3 times with a = 2.50, 3.00, 15.0m/s^2 and t=1.50, 3.25, 3.00sExample problem 2 pg 99 with elevator 3 times with a = 2.50, 3.00, 15.0m/s^2 and t=1.50, 3.25, 3.00s19, 20, 22, 25 (pg 100 & 101)19, 20, 22, 25 (pg 100 & 101)

Demo

ForceForce

Newton’s Second Law………...........Newton’s Second Law………...........

Weight is a Force………………...........Weight is a Force………………...........

Defined with the universal constant ‘G’.Defined with the universal constant ‘G’.

netFa

m

urr

gF mgur ur

1 22g

Gm mF

r

ur

11 2 26.7 10 /G x N m kg

1 lbf ≈ 4.448222 N

Force and GravityForce and Gravity

gF mgur ur

2gGMm

Fr

ur

2

GMmmg

r

2

GMg

r

11 2 26.7 10 /G x N m kg

2

kg mN

s

g

Key VocabKey Vocab

Normal ForceNormal Force Force due to gravity and mass is referred to as a Force due to gravity and mass is referred to as a

normal force or ‘the normal vector’.normal force or ‘the normal vector’. Normal vector also refers to a vector that Normal vector also refers to a vector that

intersects at 90 degrees.intersects at 90 degrees.

(also stated as: A vector that is (also stated as: A vector that is perpendicular to the tangent perpendicular to the tangent line at the interface)line at the interface)

Newton’s Second LawNewton’s Second Law

Two men pull a 50-kg box with forces 19.7 N Two men pull a 50-kg box with forces 19.7 N and 15.6 N in the directions shown below. and 15.6 N in the directions shown below. Find the resultant acceleration of the box and Find the resultant acceleration of the box and the direction in which the box moves.the direction in which the box moves.

19.7 N 15.4 N19.7 N 15.4 N

or or

-19.7N-19.7N

2

kg mN

s

g

Practice - Practice -

A large helicopter is used to lift a heat pump to the A large helicopter is used to lift a heat pump to the roof of a new building. The mass of the helicopter is roof of a new building. The mass of the helicopter is 7.0x10^3 kg and the mass of the heat pump is 1700 7.0x10^3 kg and the mass of the heat pump is 1700 kg.kg.

a.a. How much force must the air exert on the How much force must the air exert on the helicopter to lift the heat pump with an acceleration helicopter to lift the heat pump with an acceleration of 1.2 m/s^2?of 1.2 m/s^2?

b.b. Two chains connected to the load each can Two chains connected to the load each can withstand 95,000 N. Can the load be safely lifted at withstand 95,000 N. Can the load be safely lifted at 1.2 m/s^2?1.2 m/s^2?

AirAir

Drag Force: exerted by a fluid on an object : exerted by a fluid on an object moving through the fluid. moving through the fluid.

Terminal Velocity: drag force is equal to force : drag force is equal to force of gravityof gravity

FrictionFriction

Static friction FStatic friction Ff,sf,s

Friction when there is no motion between the Friction when there is no motion between the objects.objects.

FFf,s f,s <=<= u ussFFNN

Sliding friction (or kinetic friction) FSliding friction (or kinetic friction) Ff,kf,k

Friction when surfaces are rubbing against each Friction when surfaces are rubbing against each other (in motion). other (in motion).

FFf,k f,k == u ukkFFNN

Friction Coefficients: Table 5.1 pg 129

FrictionFriction

A wood block on a wood A wood block on a wood plank. plank.

m= kgm= kg uuss = 0.5 = 0.5 uukk = 0.2 = 0.2 FFNN = N = N

Normal

Friction

Review Elevator ExampleReview Elevator Example

FFnetnet = F = FEE + (-F + (-Fgg)) FFnetnet = ma = ma (represents the force on the system as a whole)(represents the force on the system as a whole)

FFE E = F= Fnetnet + F + Fgg

TensionTension

Force exerted by pulling (usually a string or Force exerted by pulling (usually a string or rope).rope).

For now, consider strings, ropes and pulleys For now, consider strings, ropes and pulleys (also called sheaves or blocks) to be massless (also called sheaves or blocks) to be massless and frictionless.and frictionless.

Provide a change in direction of force.Provide a change in direction of force.

Tension ProblemTension Problem

The blocks shown The blocks shown are placed on a are placed on a smooth horizontal smooth horizontal surface and surface and connected by a connected by a piece of string. If piece of string. If a 8.8-N force is a 8.8-N force is applied to the 8.8-applied to the 8.8-kg block, what is kg block, what is the tension in the the tension in the string?string?

6.1N6.1N

2

,19.4 19.4 2

19.4 8.8 28.2

8.80.312

28.2

19.4 *0.312 6.1

net t

t

net

t

T a

F m a

m kg kg kg

F N ma

m kg s

mF F m a kg N

s

r r

rr

r r r

TensionTension

Three blocks A, B, Three blocks A, B, and C are connected and C are connected by two massless by two massless strings passing over strings passing over smooth pulleys as smooth pulleys as shown below, with shown below, with the 3.4-kg block on a the 3.4-kg block on a smooth horizontal smooth horizontal surface. Calculate surface. Calculate the tension in the the tension in the strings connecting A strings connecting A and B, and B and C.and B, and B and C.

54N48N

TensionTension

2 2

2

, ,

, ,

,

(6.9 *9.8 ) (4.1 *9.8 ) 68 40 28

6.9 3.4 4.1 14.4

281.9

14.4

(

net t

net A B A B

t A B C

netnet t

t

a BC B T AB

T AB B a BC

a BC B C

F m a

m mF F F m g m g kg kg N N N

s sm m m m kg kg kg kg

F N mF m a a

m kg s

F F F

F F F

F m m

r r

r r r

rr r r

r r

r r r

r2

,

) (3.4 4.1)( 1.9 ) 14

40 ( 14 ) 54T AB

ma N

s

F N N N

r

r

Tension PracticeTension Practice

, ,

,

T BC A a AB

B

a AB

F F F

a

F

F

r r r

r

r

r

Sketch ProblemSketch Problem

A CB68N 40N

ABC68N 40N

Sketch ProblemSketch Problem

AB C68N 40N

A68N 40N

BC

A CB68N 40N

FFnetnet

FFnetnet = = sum of all forces acting on a system = msum of all forces acting on a system = maa FFnetnet = = FFf,kf,k + + FFTT + + FFpush + push + FFgravitygravity = = mmaa

ForceForce

Newton’s Third LawNewton’s Third Law Every action has an opposite Every action has an opposite

and equal reaction.and equal reaction.

_ _ _ _A on B B on AF Fur ur

ForceForce

Newton’s Second Law………...........Newton’s Second Law………........... netFa

m

urr

ForceForce

Newton’s first lawNewton’s first law When the net forces are zero, an object at rest remains When the net forces are zero, an object at rest remains

at rest and an object in motion remains in motion in the at rest and an object in motion remains in motion in the same direction at the same speed. same direction at the same speed.

For FFor Fnetnet = 0 = 0 velocity is constant: vvelocity is constant: v11 = v = v22 acceleration is zeroacceleration is zero

FrictionFriction

Static friction FStatic friction Ff,sf,s

Friction when there is no motion between the Friction when there is no motion between the objects.objects.

FFf,s f,s <=<= u ussFFNN

Sliding friction (or kinetic friction) FSliding friction (or kinetic friction) Ff,kf,k

Friction when surfaces are rubbing against each Friction when surfaces are rubbing against each other (in motion). other (in motion).

FFf,k f,k == u ukkFFNN

Friction Coefficients: Table 5.1 pg 129

FrictionFriction

Static friction FStatic friction Ff,sf,s

Friction when there is no motion between the Friction when there is no motion between the objects.objects.

FFf,s f,s <=<= u ussFFNN

FFf,s f,s <=<= u ussmgmg

5kg10N

Review Elevator ExampleReview Elevator Example

FFnetnet = F = FEE + (-F + (-Fgg)) FFnetnet = ma = ma (represents the force on the system as a whole)(represents the force on the system as a whole)

FFE E = F= Fnetnet + F + Fgg

Setting up the ProblemSetting up the Problem

Connected by a massless string, pulled along a Connected by a massless string, pulled along a surface with a coefficient of frictionsurface with a coefficient of friction 0.75k

100N

StepsSteps

What forces are present?What forces are present?

3 2 1100net t k k kF m a N m g m g m g r r

0.75k

100N

What is the value of each force?What is the value of each force?

100 44 29 15 12net tF m a N N N N N r r

Draw the free body diagramDraw the free body diagram

StepsSteps

List known variables, solve for unknownList known variables, solve for unknown

2

12

2 4 6 12

121

12

net t

t

net

t

F m a N

m kg kg kg kg

F N ma

m kg s

r r

rr

Break apartBreak apart

Find the tension in each stringFind the tension in each string First, identify the forces of each partFirst, identify the forces of each part

, , 1for m1, 15f k mF Nr

100N

, , 2for m2, 29f k mF Nr

, , 3for m3, 44f k mF Nr

Consider each section as a systemConsider each section as a system

Draw a free body diagram for this systemDraw a free body diagram for this system

, , 1for m1, 15f k mF Nr

, , 2for m2, 29f k mF Nr

net tF m a r r

Consider each section as a systemConsider each section as a system

Draw a free body diagram for this systemDraw a free body diagram for this system

, , 1for m1, 15f k mF Nr

net tF m ar r

Break apartBreak apart

Draw a sketch (free body diagram) for each Draw a sketch (free body diagram) for each stringstring

Trig IdentitiesTrig Identities

SOHCAHTOASOHCAHTOA Adjacent Leg = Hypotenuse * cosAdjacent Leg = Hypotenuse * cos Opposite Leg = Hypotenuse * sinOpposite Leg = Hypotenuse * sin Opposite Leg = Adjacent Leg * tanOpposite Leg = Adjacent Leg * tan

Force componentsForce components

The magnitude of the force 1 is 87 N, of The magnitude of the force 1 is 87 N, of force 2 is 87 N, and of 3 is 87 N. The angles force 2 is 87 N, and of 3 is 87 N. The angles θ θ 1 and 1 and θ θ 2 are 60° each. Use the 2 are 60° each. Use the Pythagorean theorem and trig identities to Pythagorean theorem and trig identities to find the resultant of the forces 1, 2 , and 3 .find the resultant of the forces 1, 2 , and 3 .

Drag ForceDrag Force

Terminal velocity is when an object in free fall Terminal velocity is when an object in free fall has reached equilibrium. That is, the drag has reached equilibrium. That is, the drag force is equal to the force of gravity. force is equal to the force of gravity.

The FThe Fnetnet for a system in equilibrium is always for a system in equilibrium is always zero.zero.

Friction with Force componentsFriction with Force components

The system shown below is in equilibrium. The system shown below is in equilibrium. Calculate the force of friction acting on the Calculate the force of friction acting on the block A. The mass of block A is 7.10 kg and block A. The mass of block A is 7.10 kg and that of block B is 7.30 kg. The angle that of block B is 7.30 kg. The angle θ θ is is 48.0°.48.0°.

Break ApartBreak Apart

The system shown below is in equilibriumThe system shown below is in equilibrium: This : This means Fnet=0, since Fnet=ma, then ma=0 and a=0means Fnet=0, since Fnet=ma, then ma=0 and a=0

In order for the net force to be zero on block A (no In order for the net force to be zero on block A (no acceleration) the tension in the rope pulling block A acceleration) the tension in the rope pulling block A has to be matched by the friction between block A has to be matched by the friction between block A and the surface.and the surface.

The friction here must be equal tothe tension pulling A

The tension here must be equal tothe friction resisting motion

Break ApartBreak Apart

There is a force from gravity acting on block There is a force from gravity acting on block B. B.

Convert this force to its x and y components. Convert this force to its x and y components. Using the concept of equilibrium, realize that Using the concept of equilibrium, realize that

the sum of x components is 0the sum of x components is 0

7.30kg

θ θ is 48.0°.is 48.0°.

Two wire tensionTwo wire tension

Familiar ProblemFamiliar Problem

Three blocks A, B, and C are connected by two Three blocks A, B, and C are connected by two massless strings passing over smooth pulleys as massless strings passing over smooth pulleys as shown below, with the B block on a smooth shown below, with the B block on a smooth horizontal surface.horizontal surface.

A=2kg, B=3kg, C=5kgA=2kg, B=3kg, C=5kg

Redraw the problemRedraw the problem

The problem may be redrawn to help visualize The problem may be redrawn to help visualize which direction the forces are acting and which direction the forces are acting and which direction is positive.which direction is positive.

A CBForce of gravity on A Force of gravity on C

Breaking the problem downBreaking the problem down

Find the total mass of the system ABC.Find the total mass of the system ABC. Find the net force of the system ABC.Find the net force of the system ABC. Find the acceleration of the system ABC.Find the acceleration of the system ABC. Find the acceleration of the system C.Find the acceleration of the system C. Find the total mass of the system C.Find the total mass of the system C. Calculate the force due to the acceleration of system C.Calculate the force due to the acceleration of system C. Calculate the force due to gravity on system C.Calculate the force due to gravity on system C. Calculate the tension in the strings connecting B and C.Calculate the tension in the strings connecting B and C.

Spring (restoring force)Spring (restoring force)

F=-kxF=-kx k=constantk=constant x=displacementx=displacement

ForceForce

Experiment with force sensors.Experiment with force sensors. equal and oppositeequal and opposite normal plus lift (Fnormal plus lift (Fnetnet)) friction (static and sliding)friction (static and sliding)