SPH4UI: Lecture 2SPH4UI: Lecture 2

“Free Body Diagrams”“Free Body Diagrams”

Review: Newton's LawsReview: Newton's LawsLaw 1Law 1:: An object subject to no external forces is at An object subject to no external forces is at

rest or moves with a constant velocity if viewed rest or moves with a constant velocity if viewed from an inertial reference frame. from an inertial reference frame.

Law 2Law 2: : For any object, For any object, FFNETNET = m = maa

Where Where FFNETNET = = FF

Law 3Law 3: : Forces occur in Forces occur in action-reactionaction-reaction pairs, pairs,

FFA ,BA ,B = - = - FFB ,AB ,A. . Where Where FFA ,BA ,B is the force acting is the force acting

on object on object AA due to its interaction with object due to its interaction with object BB and vice-versa.and vice-versa.

m is “mass” of object

Honda videoHonda video

GravityGravity::Mass vs WeightMass vs Weight

What is the What is the force of gravity force of gravity exerted by the exerted by the earth on a typical physics student?earth on a typical physics student?

Typical student mass Typical student mass m = 55kgm = 55kgg = 9.81 m/sg = 9.81 m/s22..FFgg = mg = (55 kg)x(9.81 m/s = mg = (55 kg)x(9.81 m/s2 2 ))

FFgg = 540 N = WEIGHT = 540 N = WEIGHTFFE,S = -= -mg g

FFS,E = F = Fg = = mg g

Act 1:Act 1:Mass vs. WeightMass vs. Weight

An astronaut on Earth kicks a An astronaut on Earth kicks a bowling ball straight ahead and bowling ball straight ahead and hurts his foot. A year later, the hurts his foot. A year later, the same astronaut kicks a bowling same astronaut kicks a bowling ball on the moon in the same ball on the moon in the same manner with the same force. manner with the same force.

His foot hurts...His foot hurts...

(a) more

(b) less (c) the same

Ouch!

Act 1:Act 1:SolutionSolution

TheThe massesmasses of both the bowling ball of both the bowling ball and the astronaut remain the same, and the astronaut remain the same, so his foot will feel the same so his foot will feel the same resistance and hurt the resistance and hurt the same same as as before.before.

When his foot hits the ball it imparts When his foot hits the ball it imparts an “an “impulseimpulse” that virtually ” that virtually instantaneously changes the velocity instantaneously changes the velocity of the ball.of the ball.

Ouch!

time

time

accel

vel

Act 1 :Act 1 :SolutionSolution

mmballballaa = Force from foot on ball = Force from foot on ball

action - reaction pair of forcesaction - reaction pair of forces Foot on ballFoot on ball Ball back on footBall back on foot

| Force on foot | = | Force on foot | = mmballballaa

doesn’t depend on weightdoesn’t depend on weight

time

time

accel

vel

Ouch!

Act 1: Act 1: SolutionSolution

However theHowever the weightsweights of the of the bowling ball and the astronaut are bowling ball and the astronaut are less:less:

Thus it would be easier for the Thus it would be easier for the astronaut to pick up the bowling astronaut to pick up the bowling ball on the Moon than on the ball on the Moon than on the Earth.Earth.

W = mgMoon gMoon < gEarth

Wow!

That’s light.

Principle of EquivalencePrinciple of Equivalence Look at Look at F=maF=ma for gravitational force (it’s one type of force….) for gravitational force (it’s one type of force….) FF is the force e.g is the force e.g mgmg or something more complicated like or something more complicated like

GMm/RGMm/R22

mama is the consequence on motion ( is the consequence on motion (mm is the coefficient of acceleration) is the coefficient of acceleration) Newton said force proportional to accelerationNewton said force proportional to acceleration

Look, there is Look, there is mm on both sides of equation on both sides of equation MASSMASS is the property of something that couples to gravity is the property of something that couples to gravity

F=mgF=mg or or F=F=GMm/RGMm/R2 2 this is this is Gravitational massGravitational mass

And And MASSMASS quantifies how much inertia some motion has quantifies how much inertia some motion has ((mv=pmv=p),),

F=maF=ma, , (its resistance to motion), this is (its resistance to motion), this is inertial massinertial mass

These are two very different things for mass to doThese are two very different things for mass to do

Gravitational mass and intertial mass are equialent

The Free Body DiagramThe Free Body Diagram

Newton’s 2nd Law says that for an object Newton’s 2nd Law says that for an object FF = = mmaa..

Key phrase here isKey phrase here is for an objectfor an object.. Object has mass and experiences forcesObject has mass and experiences forces

So before we can apply So before we can apply FF = m = maa to any given object we to any given object we isolate the forces acting on this object:isolate the forces acting on this object:

The Free Body Diagram...The Free Body Diagram...

Consider the following case as an example of this….Consider the following case as an example of this…. What are the forces acting What are the forces acting on the plank on the plank ?? Other forces act on Other forces act on FF, , WW and and E E. . focus on plankfocus on plank

PP = plank = plank

FF = floor = floor

WW = wall= wall

E E = earth = earth

FFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P

The Free Body Diagram...The Free Body Diagram...

Consider the following caseConsider the following caseWhat are the forces acting What are the forces acting on the plank on the plank ??

Isolate the plank from

the rest of the world.

FFW,P

FFP,W

FFP,F FFP,E

FFF,P

FFE,P

The Free Body Diagram...The Free Body Diagram...

The forces acting on the plank should The forces acting on the plank should reveal themselves...reveal themselves...

FFP,W

FFP,F FFP,E

Aside...Aside... In this example the plank is not moving...In this example the plank is not moving...

It is certainly not accelerating!It is certainly not accelerating! So So FFNETNET = m = maa becomes becomes FFNET NET = 0= 0

This is the basic idea behind This is the basic idea behind staticsstatics, which we will , which we will discuss later.discuss later.

FFP,W + FFP,F + FFP,E = 0

FFP,W

FFP,F FFP,E

The Normal ForceThe Normal Force

When person is holding the bag

above the table he must supply a force.

When the bag is placed on the table, the table supplies

the force that holds the bag on it

That force is perpendicular or

normal to the surface of table

ExampleExample

Example dynamics problem:Example dynamics problem:

A box of mass A box of mass m = 2 kg m = 2 kg slides on a horizontal frictionless slides on a horizontal frictionless floor. A force floor. A force FFx x = 10 N = 10 N pushes on it in the pushes on it in the xx direction. direction.

What is the acceleration of the box?What is the acceleration of the box?

FF = Fx ii aa = ?

m

y y

x x

Example...Example... Draw a picture showing all of the forcesDraw a picture showing all of the forces

FFFFB,F

FFF,BFFB,E

FFE,B

y y

x x

Example...Example... Draw a picture showing all of the forces.Draw a picture showing all of the forces. Isolate the forces acting on the block.Isolate the forces acting on the block.

FFFFB,F

FFF,BFFB,E = mgg

FFE,B

y y

x x

Example...Example... Draw a picture showing all of the forces.Draw a picture showing all of the forces. Isolate the forces acting on the block.Isolate the forces acting on the block. Draw a free body diagram.Draw a free body diagram.

FFFFB,F

mgg

y y

x x

Example...Example... Draw a picture showing all of the forces.Draw a picture showing all of the forces. Isolate the forces acting on the block.Isolate the forces acting on the block. Draw a free body diagram.Draw a free body diagram. Solve Newton’s equations for each component.Solve Newton’s equations for each component.

FFXX = ma = maXX FFB,FB,F - mg = ma - mg = maYY

FFFFB,F

mgg

y y

x x

Example...Example... FF XX = ma = ma

XX So So aa XX = F = F

X X / m = (10 N)/(2 kg) = 5 m/s / m = (10 N)/(2 kg) = 5 m/s 22..

FF B,FB,F - mg = ma - mg = maYY

But But aa YY = 0 = 0 So So FF B,F B,F = mg = mg ..

The vertical component of the forceThe vertical component of the forceof the floor on the object of the floor on the object ((FF B,F B,F ) is) isoften called the often called the Normal Force Normal Force ((NN)) ..

Since Since aa YY = 0 = 0 , , N = mgN = mg in this case. in this case.

FX

N

mg

y y

x x

Example RecapExample Recap

FX

N = mg

mg

aX = FX / m y y

x x

• Analyzing forces

• Free body diagram

• Tension in a rope = magnitude of the force that the rope exerts on object.

Free body diagram

System of InterestSystem of Interest

System of InterestSystem of Interest

f - All forces opposing

the motion

System 1: Acceleration of the professor and the cart

System 2: Force the professor exerts on the cart

A traffic light weighing hangs from a cable tied to two other cables fastened to a support as shown in the figure.

Example

Example

A crate of mass m is placed on a frictionless plane of incline

(a) Determine the acceleration of the crate.

Example

Attwood’s machine.

Two objects of mass m1 and m2 are hung over a pulley.

(a) Determine the magnitude of the acceleration of the two objects and the tension in the cord.

W

AB

C

W

BC

BA

Free Body Diagram

Question Question Consider a person standing in an elevator that is accelerating

upward. The upward normal force N exerted by the elevator floor on the person is

a) larger than

b) identical to

c) less than

the downward weight W of the person.

Person is accelerating upwards - net upwards force is non zero

mg

N

Free Body Diagram of the person:

Act 2 : Act 2 : Normal ForceNormal Force

A block of mass A block of mass mm rests on the floor of an elevator that is rests on the floor of an elevator that is accelerating upward. What is the relationship between accelerating upward. What is the relationship between the force due to gravity and the normal force on the the force due to gravity and the normal force on the block? block?

m

(a)(a) N > mgN > mg

(b)(b) N = mgN = mg

(c)(c) N < mgN < mg

a

Act 2 : Act 2 : SolutionSolution

m

N

mg

All forces are acting in the y direction, so use:

Ftotal = ma

N - mg = ma

N = ma + mg

therefore N > mg

a

Understanding Understanding You are driving a car up a hill with You are driving a car up a hill with constant velocityconstant velocity. On a piece of . On a piece of paper, draw a Free Body Diagram (FBD) for the car. paper, draw a Free Body Diagram (FBD) for the car.

How many forces are acting on the car? How many forces are acting on the car? 1122334455

weight/gravity (W)normal (FN)

engine/motor (Fcar_on_road(action) ==> (reaction) Froad

on car)

W

Froad on car

FN V

correct True if accelerating also

UnderstandingUnderstanding

The net force on the same car isThe net force on the same car is

1. Zero 1. Zero

2. Pointing up the hill2. Pointing up the hill

3. Pointing down the hill3. Pointing down the hill

4. Pointing vertically downward4. Pointing vertically downward

5. Pointing vertically upward 5. Pointing vertically upward

W

Froad on car

FN V

W

Froad on car

FN F = ma = 0

correct

What is the tension in the string?What is the tension in the string?

A) T<WA) T<W

B) T=WB) T=W

C) W<T<2WC) W<T<2W

D) T=2WD) T=2W

Pulley IPulley I

WW

Pull withforce = WWSame answer

W

T

W

Look at Free Body Diagram: T=W

Net Force = 0 = acceleration

Pulley IIPulley II

What is the tension in the string?What is the tension in the string?

A) T<WA) T<W

B) T=WB) T=W

C) W<T<2WC) W<T<2W

D) T=2WD) T=2W W2Wa a

2Wa

T

2W

T<2W: Wa

T

W

W<T:

Look at FreeBody Diagrams:

Tools: Ropes & StringsTools: Ropes & Strings Can be used to pull from a distance.Can be used to pull from a distance. TensionTension (T)(T) at a certain position in a rope is the magnitude of the at a certain position in a rope is the magnitude of the

force acting across a cross-section of the rope at that position.force acting across a cross-section of the rope at that position. The force you would feel if you cut the rope and grabbed the The force you would feel if you cut the rope and grabbed the

ends.ends. An action-reaction pair.An action-reaction pair.

cut

TT

T

L

R

T = FR,LT = FL,R

Tension doesn’t have a directionWhen you hook up a wire to an object the direction is determined by geometry of the hook up

Tools: Ropes & Strings...Tools: Ropes & Strings... Consider a horizontal segment of rope having massConsider a horizontal segment of rope having mass

mm:: Draw a free-body diagram (ignore gravity).Draw a free-body diagram (ignore gravity).

Using Newton’s 2nd law (in Using Newton’s 2nd law (in xx direction): direction): FFNETNET = T = T22 - T - T11 = ma = ma

So if So if m = 0m = 0 (i.e. the rope is light) then (i.e. the rope is light) then TT11 = =TT22 And if And if a = 0 a = 0 then then TT11 = =TT22 Otherwise, Otherwise, TT is a function of position is a function of position

T1 T2

m

a x x

Tools: Ropes & Strings...Tools: Ropes & Strings... An ideal (An ideal (masslessmassless) rope has constant tension ) rope has constant tension

along the rope.along the rope.

If a rope has mass, the tension can vary along the If a rope has mass, the tension can vary along the roperope For example, a heavy rope For example, a heavy rope

hanging from the ceiling... hanging from the ceiling...

We will deal mostly with ideal massless ropes.We will deal mostly with ideal massless ropes.

T = Tg

T = 0

T T

Tools: Ropes & Strings...Tools: Ropes & Strings... What is force acting on mass – isolate the mass!!!What is force acting on mass – isolate the mass!!! The direction of the force provided by a rope is along the The direction of the force provided by a rope is along the

direction of the rope:direction of the rope: (this is a massless rope)(this is a massless rope)

mg

T

m

Since ay = 0 (box not moving),

T = mg

Act 3 : Act 3 : Force and accelerationForce and acceleration A fish is being yanked A fish is being yanked upwardupward out of the water using a fishing line that breaks when the tension reaches out of the water using a fishing line that breaks when the tension reaches

180 N180 N. The string snaps when the acceleration of the fish is observed to be is . The string snaps when the acceleration of the fish is observed to be is 12.2 m/s12.2 m/s22. What is the . What is the mass of the fish?mass of the fish?

m = ?a = 12.2 m/s2

snap ! (a) 14.8 kg

(b) 18.4 kg

(c) 8.2 kg

Act 3 : Act 3 : Solution:Solution:

Draw a Free Body Diagram!!Draw a Free Body Diagram!!

T

mg

m = ?a = 12.2 m/s2

Use Newton’s 2nd lawin the upward direction:

FTOT = ma

T - mg = ma

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

mT

g a

kg28

sm21289

N180m

2.

..

Tools: Pegs & PulleysTools: Pegs & Pulleys Used to change the direction of forcesUsed to change the direction of forces

An ideal massless pulley or ideal smooth peg will change the An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:direction of an applied force without altering the magnitude:

FF1 ideal peg

or pulley

FF2

| FF1 | = | FF2 |

Tools: Pegs & PulleysTools: Pegs & Pulleys Used to change the direction of forcesUsed to change the direction of forces

An ideal massless pulley or ideal smooth peg will change the An ideal massless pulley or ideal smooth peg will change the direction of an applied force without altering the magnitude:direction of an applied force without altering the magnitude:

mg

T

m T = mg

FW,S = mg

Scales:Scales: Springs can be calibrated to tell us the applied force.Springs can be calibrated to tell us the applied force.

We can calibrate scales to read We can calibrate scales to read NewtonsNewtons, or..., or... Fishing scales usually read Fishing scales usually read

weight in weight in kgkg or or lbslbs..

02468

1 lb = 4.45 N

How many Newtons is 1 kg?

m m m

(a)(a) 0 lbs. (b)(b) 4 lbs. (c)(c) 8 lbs.

(1) (2)

?

Act 4 : Act 4 : Force and accelerationForce and acceleration

A block weighing A block weighing 4 lbs4 lbs is hung from a rope attached to a scale. The is hung from a rope attached to a scale. The scale is then attached to a wall and reads scale is then attached to a wall and reads 4 lbs4 lbs. What will the scale . What will the scale read when it is instead attached to another block weighing read when it is instead attached to another block weighing 4 lbs4 lbs??

Act 4 : Act 4 : Solution:Solution:

Draw a Free Body Diagram of one of the blocks!!Draw a Free Body Diagram of one of the blocks!!

Use Newton’s 2nd Lawin the y direction:

FTOT = 0

T - mg = 0

T = mg = 4 lbs.

mg

T

m T = mg

a = 0 since the blocks are stationary

Act 4 : Act 4 : Solution:Solution:

The scale reads the tension in the The scale reads the tension in the rope, which is rope, which is T = 4 lbs T = 4 lbs in both cases!in both cases!

m m m

T T T T

TTT