5.5 The Gravitational 5.5 The Gravitational Force and WeightForce and Weight

The Gravitational The Gravitational Force and WeightForce and Weight

The gravitational force:The gravitational force:

FFg g ≡≡ Force that The EarthThe Earth exerts on an object This force is directed toward the centertoward the center of the

earth. Its magnitudemagnitude is called THE WEIGHTTHE WEIGHT of the object

Weight Weight ≡ ≡ |F|Fgg| | ≡≡ m mg g (5.6)(5.6) Because it is dependent on gg, the weightthe weight varies

with location gg, and therefore the weightthe weight, is less at higher altitudes

Weight is not an inherent property of the object

Gravitational Force, 2Gravitational Force, 2 Object in FREE FALLFREE FALL. Newton’s 2nd Law:

∑∑F = F = m m aa If no other forces are acting, only FFgg acts (in

vertical direction). ∑∑FFyy = = m m aayy

Or: FFgg = = mmgg (down, of course)(down, of course) (5.6)(5.6)

Where: gg = 9.8 m/s = 9.8 m/s22

SI Units:SI Units: NewtonNewton (just like any force!). If If mm = 1 kg= 1 kg FFgg = = (1kg)(9.8m/s(1kg)(9.8m/s22)) = = 9.8N9.8N

Gravitational Force, Gravitational Force, finalfinal

REMARKS!!!REMARKS!!! FFg g Depends on gg, then it varies with geographic

location. FFg g Decreases from Sea level to a higher altitude. You want loselose Weight without dietwithout diet climb a climb a

high mountain.high mountain. mm in Equation 5.6 is called the gravitational gravitational

massmass The kilogramThe kilogram is notis not a unit of WeightWeight is a unit

of MassMass. Mass and Weight are two different Mass and Weight are two different

quantitiesquantities

Gravitational & Inertial Gravitational & Inertial MassMass

In Newton’s Laws, the mass is the inertial inertial massmass and measures the resistance to a resistance to a changechange in the object’s motion

In the gravitational force, the the gravitational massgravitational mass is determining the gravitational attractiongravitational attraction between the object and the Earth

Experiments show that gravitational gravitational massmass and inertial massinertial mass have the same same valuevalue

5.6 Newton’s Third 5.6 Newton’s Third LawLaw

If two objects interact, the force FF1212 exerted byby object 1 onon object 2 is equal in magnitude and opposite in direction to the force FF2121 exerted byby object 2 onon object 1

FF1212 = = ̶̶ F F21 21 (5.7) (5.7)

Note on notation: FFABAB is the force exerted byby A onon B

Newton’s Third Law, 2Newton’s Third Law, 2 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 The action force is equal in

magnitude to the reaction force magnitude to the reaction force and opposite in directionand opposite in direction One of the forces is the action forceaction force, the

other is the reaction forcereaction force It doesn’t matter which is considered the

actionaction and which the reactionreaction

Newton’s Third Law, 3Newton’s Third Law, 3 The actionaction and reaction reaction forces must:

act on different objects.different objects. be of the same typesame type

Forces exerted BYBY a body DO NOTDO NOT (directly) influence itsits motion!!motion!!

Forces exerted ONON a body (BYBY some other body) DODO influence itsits motion!!motion!!

When discussing forces, use the words “BYBY” and “ONON” carefully.

Example: 5.1Example: 5.1 Action- Action-ReactionReaction

The force FF1212 exerted BYBY object 1 ONON object 2 is equal in magnitude and opposite in direction to FF2121 exerted BYBY object 2 ONON object 1

FF1212 = = –– FF2121

Example: 5.2Example: 5.2 Action- Action-ReactionReaction

The force FFhnhn exerted BYBY the hammer ONON the nail is equal in magnitude and opposite in direction to FFnhnh exerted BYBY the nail ONON the hammer

FFhnhn = = –– FFnhnh

Example: 5.3 Example: 5.3 Action-Action-ReactionReaction

We can walk forward because when one foot pushes backwardbackward against the ground, the ground pushes forwardforward on the foot. – – FFPGPG ≡ ≡ FFGPGP

Does the force of gravity stop? OF COURSE NOTOF COURSE NOT

But, object does not movedoes not move: 2nd Law ∑∑F = F = m m a = 0a = 0 There must be some other some other forceforce acting besides gravity (weightweight) to have ∑∑F = 0.F = 0.

The Normal ForceThe Normal Force : FFNN = n = n NormalNormal is math term for

PerpendicularPerpendicular (())

FFNN is to the surface & opposite

to the weightweight (in this simple in this simple case onlycase only)

Example: 5.4Example: 5.4 Action- Action-Reaction (Normal Force)Reaction (Normal Force)

The normal force:The normal force: nn (table on monitor) is the reaction of the force the monitor exerts ONON the table

The actionThe action (FFgg, Earth on , Earth on monitormonitor) force is equal in magnitude and opposite in direction to the reactionthe reaction force, the force the monitor exerts ONON the Earth

Caution: Caution: The normal force is The normal force is not always = & opposite to not always = & opposite to the weight!!the weight!! As we’ll see!

Example: 5.4Example: 5.4 Normal Normal Force, 2Force, 2

Free Body DiagramFree Body Diagram In a free body

diagram, you want the forces acting ONON a particular object

The normal forceThe normal force and the force of force of gravitygravity are the forces that act ONON the monitor

Normal ForceNormal Force Where does the Normal ForceNormal Force

come from? From the From the other bodyother body!!!!!! Does the normal forcenormal force ALWAYSALWAYS

equal to the weight weight ?

NO!!!NO!!!Weight and Normal Force are not Weight and Normal Force are not

Action-ReactionAction-Reaction Pairs!!! Pairs!!!

Example 5.5 Example 5.5 Normal Normal ForceForce

mm = 10 kg = 10 kgWeight: FWeight: Fgg = = mmg = g = 98.0N98.0NThe normalThe normalforceforce is equalis equal to the weight!! weight!!Only this caseOnly this caseFFNN = n = = n = mmg = 98.0Ng = 98.0N

Example 5.6 Example 5.6 Normal Normal Force, 2Force, 2

mm = 10 kg = 10 kgWeight: FWeight: Fgg = = mmg = 98.0Ng = 98.0NPushing Force = 40.0NPushing Force = 40.0NThe normal forceThe normal force is NOTNOTALWAYSALWAYS equal to the weight!!weight!!

FFNN = n = 40.0N mg= = n = 40.0N mg= 138.0N138.0N

Example 5.7 Example 5.7 Normal Force, Normal Force, finalfinal

mm = 10 kg = 10 kgWeight: FWeight: Fgg = = mmg = 98.0Ng = 98.0NPuling Force = 40.0NPuling Force = 40.0NThe normal forceThe normal force is NOTNOTALWAYSALWAYS equal to the weight!! weight!!

FFNN = n = 98.0N = n = 98.0N –– 40.0N= 40.0N= 58.0N58.0N

Example 5.8Example 5.8 Accelerating the boxAccelerating the box

mm = 10 kg = 10 kg F Fgg = 98.0N = 98.0N

From Newton’s 2nd Law:∑∑F = F = mmaa

FFPP – – mmg = g = m m aa m m aa = 2.0N= 2.0NThe box accelerates accelerates upwards upwards because

FFPP > > m m gg

Example 5.9Example 5.9 Weight LossWeight Loss (Example 5.2 Text Book)Example 5.2 Text Book)

Apparent weight lossApparent weight loss. The lady weights

65kg = 640N, the elevator descends with

a =a = 0.2m/s2. What does the scale read (FFNN)?

From Newton’s 2nd law: ∑∑F = F = mmaa

FFNN – – mmg = g = – – m m aa FFNN = = mmg g – – m m aa FFNN = 640N – 13N = 627N = 52kg52kg

Upwards!Upwards!

FFNN is the force the scale exerts on the person, and is equal and opposite to the force she exerts on the scale.

Example 5.9Example 5.9 Weight Loss, 2Weight Loss, 2

What does the scale read when the elevator descends at a constant constant speedspeed of 2.0m/s?

From Newton’s 2nd law: ∑∑F = F = 00

FFNN – – mmg = 0g = 0 FFNN = = mmg =g = 640N = 65kg65kg

The scale reads her true mass!

NOTE:NOTE: In the first case the scale reads an “apparent mass”“apparent mass” but her mass does notdoes not change as a result of the acceleration: it stays at 65 kg65 kg

5.7 Some Applications 5.7 Some Applications of Newton’s Lawof Newton’s Law

OBJECTS IN EQUILIBRIUMOBJECTS IN EQUILIBRIUM If the accelerationacceleration of an object that

can be modeled as a particle is zerozero, the object is said to be in EQUILIBRIUM!!EQUILIBRIUM!!

Mathematically, the net forcenet force acting on the object is zerozero0

0 and 0x y

F

F F

Example 5.10 Example 5.10 EquilibriumEquilibrium

A lamp is suspended from a chain of negligible mass

The forces acting on the lamp are Force of gravityForce of gravity (F(Fgg)) Tension in the chain (Tension in the chain (TT))

Equilibrium gives0 0y gF T F gT F

Example 5.10 Example 5.10 Equilibrium, Equilibrium, finalfinal

The forces acting on the chain are TT’’ and TT””

TT”” is the force exerted by the ceiling

TT’’ is the force exerted by the lamp

TT’’ is the reaction force to TT Only TT is in the free body

diagram of the lamp, since TT’’ and TT”” do not act on the lamp

Example 5.11 Example 5.11 Traffic Traffic Light at RestLight at Rest

Example 5.4 Example 5.4 (Text Book)(Text Book) This is an equilibrium

problem No movementNo movement, so a = 0a = 0

Upper cables are not strong as the lower cable. They will break if the tension exceeds 100N100N.

Will the light remain or Will the light remain or will one of the cables will one of the cables break?break?

mmgg =122N =122N

Example 5.11 Example 5.11 Traffic Traffic Light at Rest, 2Light at Rest, 2

Apply Equilibrium Apply Equilibrium Conditions:Conditions:

ΣΣFFyy = 0= 0 TT3 3 – F– Fg g = 0 = 0 TT3 3 = F= Fg g = 122N= 122N Find Components:Find Components:

TT1x1x = = –– TT11cos37 cos37 TT1y1y = = TT11sin37sin37

TT2x2x = = TT22cos53 cos53 TT2y2y = = TT22sin53sin53

TT3x3x = 0 = 0 TT3y3y = = –– 122N 122N Apply Newton’s 2Apply Newton’s 2ndnd Law: Law:

ΣΣFFx x = 0 = 0 ΣΣFFy y = 0= 0

-T-T1x1x TT2x2x

TT2y2y

TT1y1y

Example 5.11 Example 5.11 Traffic Light Traffic Light at Rest, finalat Rest, final

ΣΣFFx x = –= – TT11cos37 + cos37 + TT22cos53 = cos53 = 0 0 (1)(1)

ΣΣFFy y = = TT11sin37 + sin37 + TT22sin53 – 122N = 0 sin53 – 122N = 0 (2)(2) Solving for Solving for TT11 or or TT22::

From EqnFrom Eqn (1) (1) solve for solve for TT22

TT22 = (cos37/cos53)= (cos37/cos53)TT11 ==1.331.33TT11

Substituting this value into Eqn (2)(2)

TT11sin37 + (sin37 + (1.331.33TT11)sin53 =122N )sin53 =122N

TT11 = 74.4 N = 74.4 N and and TT22 = 97.4N= 97.4N

Both values are less than 100N, soBoth values are less than 100N, so

the cables will not break!!!the cables will not break!!!

-T-T1x1x

TT1y1y

TT2y2y

TT2x2x

Objects Experiencing a Objects Experiencing a Net Force, ExamplesNet Force, Examples

If an object that can be modeled as a particle experiences an experiences an acceleration (a acceleration (a ≠ 0)≠ 0),, there must be a nonzero net force (F nonzero net force (F ≠ 0)≠ 0) acting on it.

Draw a free-bodyfree-body diagram Apply Newton’s Second LawApply Newton’s Second Law in

component form

Example 5.12Example 5.12 Net Net ForceForce

45)1(1tan1

21tan

1412100210022

21

F

F

NFFRF

Forces on The CrateForces on The Crate Forces acting on the Forces acting on the

crate:crate: A tension, the

magnitude of force TT The gravitational force,

FFgg

The normal force, nn, exerted by the floor

Forces on The Crate, Forces on The Crate, 22

Apply Newton’s Second Law in Apply Newton’s Second Law in component form:component form:

ONLYONLY in this case nn == F Fgg

Solve for the unknown(s) If TT is constant, then aa is constant and

the kinematic equations can be used to more fully describe the motion of the crate

x xF T ma 0y g gF n F n F

Example 5.13Example 5.13 Normal Normal ForceForce

The normal force, The normal force, FFNN is is NOT NOT always always equal to equal to the the weight!!weight!!m = 10.0 kgm = 10.0 kg FFgg = = 98.0N98.0NFind: aFind: axx ≠ 0? ≠ 0? FFNN??if a ayy = 0 = 0

m=10kgm=10kg

Example 5.13Example 5.13 Normal Normal Force, finalForce, final

FFPyPy = F = FPPsin(30) = 20.0N sin(30) = 20.0N

FFPxPx = F = FPPcos(30) = 34.6Ncos(30) = 34.6N

ΣΣFFx x = F= FPx Px = m= m aaxx aaxx = 34.6N/10.0kg= 34.6N/10.0kg

aaxx == 3.46m/s3.46m/s22

ΣΣFFyy = F = FNN + F + FPy Py – mg = m– mg = m aayy FFNN + F + FPyPy – mg = 0 – mg = 0 FFNN = mg – F= mg – FPyPy= 98.0N – 20.0N= 98.0N – 20.0N

FFNN = 78.0N = 78.0N

FFPxPx

FFPyPy

Example 5.14Example 5.14 Conceptual Conceptual Example: The Hockey PuckExample: The Hockey Puck

Moving at constant velocity,constant velocity, with NO NO friction.friction.Which free-body diagram is correct?Which free-body diagram is correct?

(b)(b)

Inclined PlanesInclined Planes Choose the coordinate Choose the coordinate

systemsystem with x along the x along the inclineincline and y y perpendicular to the perpendicular to the inclineincline

Forces acting on the Forces acting on the object:object: The normal force, nn, acts

perpendicular to the plane The gravitational force, FFgg ,

acts straight down

Example 5.15 Example 5.15 The The Runaway Car Runaway Car (Example 5.6 (Example 5.6 Text Book)Text Book)

Replace the force of force of gravitygravity with its components:

FFgxgx = = mgmgsinsin FFgy gy = mg= mgcoscosWith: a ayy = 0 = 0 & a ax x ≠ 0≠ 0

(A). (A). Find aax x

Using Newton’s 2nd Law: y-Directiony-Direction

ΣΣFy = n – Fy = n – mgmgcoscos = ma= mayy = = 0 0 n =n = mgmgcoscos

Example 5.15 Example 5.15 The The Runaway Car, finalRunaway Car, final

x-directionx-direction

ΣΣFFxx = = mmgsingsin = m = maax x aaxx = g = gsinsin

Independent of m!!Independent of m!!(B) (B) How long does it take the front of the car to reach

the bottom?

(C).(C). What is the car’s speed at the bottom?

212

212

2 2

sin

f i xi x

xx

x x v t a t

d dd a t t t

a g

2 2 22 2( sin )

2 sin

xf xi x xf

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v v a d v g d

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