5.5 the gravitational force and weight. the gravitational force and weight the gravitational force:...
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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
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?
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