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5.7 Some 5.7 Some Applications of Applications of Newton’s Law, contNewton’s Law, cont
Multiple ObjectsMultiple Objects When two or more objects are When two or more objects are
connected or in contact, connected or in contact, Newton’s lawsNewton’s laws may be applied may be applied to the system as a to the system as a whole and/or whole and/or to each individual objectto each individual object
Whichever you use to solve the Whichever you use to solve the problem, the other approach problem, the other approach can be used as a checkcan be used as a check
Example 5.16Example 5.16 Multiple Multiple ObjectsObjects
First treat the system as a whole:
Apply Newton’s Laws to the individual blocks
Solve for unknown(s)Solve for unknown(s) Check: |P|P2121| = |P| = |P1212||
systemx xF m a
Example 5.17Example 5.17 Two Boxes Two Boxes Connected by a CordConnected by a Cord
Boxes A & B are connected by a cord (mass neglected). Boxes are resting on a frictionless table.
FFPP = 40.0 N = 40.0 N Find:Find:
Acceleration Acceleration (a)(a) of each box
TensionTension ((FFTT)) in the cord connecting the boxes
Example 5.17Example 5.17 Two Boxes Two Boxes Connected by a Cord, finalConnected by a Cord, final
There is only horizontal motionThere is only horizontal motion
With:With: a aAA = a = aBB = a = a Apply Newton’s Laws for box A: ΣΣFFxx = = FFP P –– FFTT = = mmAAaa (1)(1) Apply Newton’s Laws for box B: ΣΣFFxx = = FFTT = = mmBBaa (2)(2)Substituting (2)(2) into (1):(1):
FFP P –– mmBBaa = = mmAAaa FFP P == ( (mmAA + + mmBB))aa
a a = = FFPP/(/(mmAA + + mmBB) = 1.82m/s) = 1.82m/s22
Substituting aa into (2) (2)
FFTT = = mmBBa a = (12.0kg)(1.82m/s= (12.0kg)(1.82m/s22)) = 21.8N= 21.8N
Example 5.18Example 5.18 The Atwood’s The Atwood’s MachineMachine
Forces acting on the objects: TensionTension (same for both objects,
one string) Gravitational forceGravitational force
Each object has the same same accelerationacceleration since they are connected
Draw the free-body diagrams Apply Newton’s Laws Solve for the unknown(s)
Example 5.18Example 5.18 The The Atwood’s Machine, 2Atwood’s Machine, 2
The Atwood’s Machine:The Atwood’s Machine: Find:Find: aa and TT Apply Newton’s 2nd Law to each
Mass.
ΣΣFFyy = = TT –– m m11gg = m = m11 aa (1)(1)
ΣΣFFyy = = TT –– m m22gg = = –– m m22 aa (2)(2) Then:
TT = m= m11g + mg + m11 aa (3)(3)
TT = m= m22gg –– m m22 aa (4)(4)
Example 5.18Example 5.18 The The Atwood’s Machine, 3Atwood’s Machine, 3
The Atwood’s Machine:The Atwood’s Machine:
Equating: (3) = (4)Equating: (3) = (4) and Solving for aa
mm11gg + m + m1 1 aa = m = m22gg –– m m2 2 a a
mm1 1 aa + m + m2 2 aa = m = m22gg – m – m11gg
a a (m(m11 + m + m22) = (m) = (m22 – m – m11))g g
(5)(5) gmm
mma
21
12
Example 5.18Example 5.18 The The Atwood’s Machine, finalAtwood’s Machine, final
The Atwood’s Machine:The Atwood’s Machine:Substituting (5)(5) into (3)(3) or (4):(4):
TT = m= m11g + mg + m11 aa (3)(3)
2 11 1
1 2
m mT m g m g
m m
1 2 1 11
1 2
mm g m m gT m g
m m
1 1 1 2 1 2 1 1
1 2
mm g m m g m m g m m gT
m m
1 2
1 2
2mm
T gm m
Active Figure 5.14Active Figure 5.14
Example 5.19Example 5.19 Two Objects Two Objects and Incline Planeand Incline Plane
Find: aa and TT One cord:One cord: so
tension tension is the same same for both objects
Connected:Connected: so acceleration acceleration is the samesame for both objects
Apply Newton’s Laws Solve for the
unknown(s)
Example 5.19Example 5.19 Two Objects Two Objects and Incline Plane, 2and Incline Plane, 2
xyxy plane:
ΣΣFFxx = 0 = 0 & ΣΣFFyy = m = m1 1 aa
TT – m– m11g = mg = m1 1 aa
TT = m= m11g + mg + m1 1 aa (1)(1)
x’y’x’y’ plane:
ΣΣFFxx = m = m2 2 aa & ΣΣFFyy = 0 = 0
mm22ggsinsinθθ – – TT = = mm2 2 aa (2)(2)
n – mn – m22ggcoscosθθ = 0 = 0 (3)(3)
Example 5.19Example 5.19 Two Objects Two Objects and Incline Plane, Finaland Incline Plane, Final
1 22 1 1 2
1 2 1 2
sin 1sin mm gm m g m m gT
m m m m
Substituting (1)(1) in (2)(2) gives:
mm22ggsinsinθθ – (– (mm11g + mg + m1 1 aa) = ) = mm22 aa
mm22ggsinsinθθ – – mm11g g –– m m1 1 aa = = mm22 a a
a a ((mm11 + m + m22) = ) = mm22ggsinsinθθ – – mm11gg Substituting aa in (1)(1) we’ll get:
TT = m= m11g + mg + m1 1 aa (1)(1)
2 1
1 2
sinm ma g
m m
2 1
1 11 2
sinm mT m g m g
m m
1 1 2 1 1 2 1 1
1 2
sinmm g m m g m m g m m gT
m m
Problem-Solving Hints Problem-Solving Hints Newton’s LawsNewton’s Laws
Conceptualize the problemConceptualize the problem – draw a diagram
Categorize the problemCategorize the problem Equilibrium (Equilibrium (F = 0)F = 0) or Newton’s Second Law (Newton’s Second Law (F = F = m m a)a)
AnalyzeAnalyze Draw free-bodyfree-body diagrams for each
object Include only forces acting on the object
Problem-Solving Hints Problem-Solving Hints Newton’s Laws, contNewton’s Laws, cont
Analyze, cont.Analyze, cont. Establish coordinate system Be sure units are consistent Apply the appropriate equation(s) in component
form Solve for the unknowns.
This always requires Kinder Garden Kinder Garden Algebra (KGA).Algebra (KGA). Like solving two linear equations with two unknowns
FinalizeFinalize Check your results for consistency with your free-
body diagram Check extreme values
5.8 Forces of 5.8 Forces of FrictionFriction
When an object is in motion on a surface or through a viscous medium, there will be a resistance to the a resistance to the motionmotion This is due to the interactionsinteractions between
the object and its environment This resistance is called the Force of Force of
FrictionFriction
Forces of Friction, 2Forces of Friction, 2 FrictionFriction exists between any 2 sliding surfaces. Two types of friction:Two types of friction:
StaticStatic (no motion) friction KineticKinetic (motion) friction
The size of the friction forceThe size of the friction force depends on: The microscopic details of 2 sliding surfaces. The materials they are made of Are the surfaces smooth or rough? Are they wet or dry?
Forces of Friction, 3Forces of Friction, 3 FrictionFriction is proportional to the normal normal
forceforce ƒƒss µµss nn (5.8)(5.8) and ƒƒk k = = µµkk n n (5.9)(5.9) These equations relate the magnitudesmagnitudes of
the forces, THEY ARE NOTTHEY ARE NOT vector equations The force of static frictionforce of static friction
(maximum) (maximum) is generally greater than the force of kinetic frictionforce of kinetic friction ƒƒss >> ƒƒk k
The coefficients of friction (The coefficients of friction (µµk,sk,s)) depends on the surfaces in contact
Forces of Friction, Forces of Friction, finalfinal
The direction of the frictional The direction of the frictional force force is oppositeis opposite the direction the direction of motion and of motion and parallel parallel to the to the surfaces in contactsurfaces in contact
The coefficients of friction The coefficients of friction ((µµk,sk,s))
are nearly independent of the are nearly independent of the area of contactarea of contact
Static FrictionStatic Friction Static frictionStatic friction acts to
keep the object from moving: ƒƒss = = F F
If FF increases, so does ƒƒss If FF decreases, so does ƒƒss ƒƒss µ µss n n where the
equality holdsequality holds when the surfaces surfaces are on the verge of slippingverge of slipping Called impending impending
motionmotion
Static Friction, contStatic Friction, cont Experiments determine the relation
used to compute friction forces. The friction force ƒƒss exists ║║ to the
surfaces, even if there is no motion. Consider the applied force F F ∑∑F = ma = 0F = ma = 0 & also v = 0v = 0
There must be a friction forceThere must be a friction force ƒƒss to oppose F F
FF – ƒƒss = 0= 0 ƒƒss == FF
Kinetic FrictionKinetic Friction The force of kinetic frictionkinetic friction (ƒƒkk )
acts when the object is in motion Friction force ƒƒkk is proportional to
the magnitude ofmagnitude of the normal force nn between 2 sliding surfaces.
ƒƒkk nn ƒƒkk kk nn (magnitudes)(magnitudes) k k Coefficient of kinetic frictionCoefficient of kinetic friction
kk : depends on the surfaces & their conditions
kk : is dimensionless & < 1
Static & Kinetic Static & Kinetic FrictionFriction
Experiments find that the Maximum Static Maximum Static Friction ForceFriction Force ƒƒs,maxs,max
is proportional to the magnitude (sizemagnitude (size)) of the normal force nn betweenbetween the 2 surfaces.
DIRECTIONS:DIRECTIONS: ƒƒs,maxs,max nn
Then: ƒƒs,maxs,max == ssnn
(magnitudes)(magnitudes)
Static & Kinetic FrictionStatic & Kinetic Friction
s s Coefficient of static frictionCoefficient of static friction ss : depends on the surfaces & their conditions ss : is dimensionless & < 1
Always:Always: ƒƒs,maxs,max >> ƒƒk k s s nn >> kk n n (Cancel nn) ss >> kk ƒƒss ƒƒs,maxs,max = µµssnn ƒƒss µ µss n n
Some Coefficients of Some Coefficients of FrictionFriction
Active Figure 5.16Active Figure 5.16
Friction in Newton’s Friction in Newton’s Laws ProblemsLaws Problems
Friction Friction is a force force, so it simply is included in the Net ForceNet Force (FF ) in Newton’s Laws
The rulesThe rules of friction allow you to determine the direction the direction and magnitudemagnitude of the force offorce of frictionfriction
Example 5.20 Example 5.20 Pulling Pulling Against FrictionAgainst Friction
Assume:Assume: mmg = 98.0N g = 98.0N n = 98.0 N, n = 98.0 N, ss == 0.40, 0.40, kk = = 0.30 0.30
ƒƒs,maxs,max = = ssnn = 0.40(98N)= 0.40(98N) = 39N= 39N Find Force of Friction if the force Find Force of Friction if the force
applied applied FFAA is:is:
A.A. FFAA = 0= 0 ƒƒss == FFAA = 0= 0 ƒƒss == 00
Box does not move!!Box does not move!!
B.B. FFAA = 10N 10N FFAA < ƒƒs,max s,max or (10N10N < < 39N39N) ƒƒss –– FFAA = 0= 0 ƒƒss == FFAA == 10N10N
The box still does not move!!The box still does not move!!
nn
ƒƒs,ks,k
Example 5.20 Example 5.20 Pulling Pulling Against Friction, 2Against Friction, 2
C.C. FFAA = 38N 38N < ƒƒs,maxs,max ƒƒss –– FFAA = 0= 0
ƒƒss == FFAA == 38N38NThis force is still not quite large This force is still not quite large enough to move the box!!!enough to move the box!!!
D.D. FFAA = 40N40N > ƒƒs,maxs,max kinetic friction.kinetic friction. This one will start moving the box!!!This one will start moving the box!!!
ƒƒkk kk nn = 0.30(98N) == 0.30(98N) = 29N.29N.The net force on the box is:The net force on the box is:
∑∑FF = = mmaaxx 40N – 29N = m40N – 29N = maaxx 11N11N = m = maaxx aaxx = = 11 kg.m/s11 kg.m/s22/10kg =/10kg = 1.10 m/s1.10 m/s22
ƒƒs,ks,k
nn
Example 5.20 Example 5.20 Pulling Pulling Against Friction, finalAgainst Friction, final
ƒƒ s,ks,k
ƒƒkk== 29N29N
ƒƒ ss
µ µ ss n n
ƒƒs,maxs,max = 39N= 39N
Example 5.21Example 5.21 To Push or Pull To Push or Pull a Sleda Sled
Similar to Quiz 5.14Similar to Quiz 5.14 Will you exerts less
force if you push or pull the girl?
θ is the same in both cases
Newton’s 2Newton’s 2ndnd Law: Law:
∑∑FF = = mmaa
PushingPushing
PullingPulling
Example 5.21Example 5.21 To Push or Pull To Push or Pull a Sled, 2a Sled, 2
x direction:x direction: ∑∑FFxx = m= maaxx FFxx – – ƒƒs,max s,max = = mmaaxx PushingPushing y direction:y direction: ∑∑FFyy = 0 = 0
n n – – mmgg – – FFyy = 0 = 0
n n = m= mg + Fg + Fyy
ƒƒs,max s,max == μμssnn
ƒƒs,max s,max == μμs s (m(mgg + + FFyy ) ) PushingPushing
nn
ƒƒs,maxs,maxFFxx
FFyy
Example 5.21Example 5.21 To Push or Pull To Push or Pull a Sled, finala Sled, final
PullingPulling y direction:y direction: ∑∑FFyy = 0 = 0
n n + FFyy – – mmgg = 0 = 0 n n = m= mg g – – FFyy
ƒƒs,max s,max == μμssn n
ƒƒs,max s,max == μμs s (m(mgg –– FFyy ) ) NOTE:NOTE: ƒƒs,max s,max (Pushing)(Pushing) > > ƒƒs,max s,max (Pulling)(Pulling)
Friction Force would be lessFriction Force would be lessif you pull than push!!!if you pull than push!!!
PullingPulling
FFxx
FFyynn
ƒƒs,maxs,max
Conceptual Example 5.22Conceptual Example 5.22 Why Why Does the Sled Move? Does the Sled Move? (Example 5.11 Text (Example 5.11 Text Book)Book)
To determine if the horse (sled)horse (sled) moves: consider only the horizontal forces exerted ONON the horse horse (sled)(sled) , then apply 2nd Newton’s Law: ΣΣF = F = m m a.a.
Horse:Horse: TT : tension exerted by the sled. : tension exerted by the sled. ffhorse horse : reaction exerted by the Earth.: reaction exerted by the Earth.
Sled:Sled: TT : tension exerted by the horse. : tension exerted by the horse. ffsled sled : friction between sled and snow.: friction between sled and snow.
Conceptual Example 5.22Conceptual Example 5.22 Why Why Does the Sled Move? Does the Sled Move? finalfinal
Horse:Horse: If If ffhorse horse > T> T , the horse accelerates to the right. , the horse accelerates to the right.
Sled:Sled: If If TT > f> fsled sled , the sled accelerates to the right. , the sled accelerates to the right.
The forces that accelerates the systemThe forces that accelerates the system (horse-sled) (horse-sled) is the net force is the net force ffhorse horse f fsled sled
If If f fhorse horse = f= fsled sled the system will move with the system will move with constant constant velocity.velocity.
Example 5.23Example 5.23 Sliding Sliding Hockey PuckHockey Puck
Example 5.13Example 5.13 (Text Book)
Draw the free-body free-body diagram, including the force of kinetic frictionkinetic friction Opposes the motion Is parallel to the surfaces in
contact Continue with the solution
as with any Newton’s Law problem
Example 5.23Example 5.23 Sliding Sliding Hockey Puck, 2Hockey Puck, 2
Given: vvxixi = 20.0 m/s = 20.0 m/s v vxfxf = = 0, 0, xxii = 0, = 0, xxff = 115 m = 115 m
Find μμkk?? y directiony direction:: ((aayy = 0) = 0)
∑ ∑FFyy = 0 = 0
nn – m – mgg = 0 = 0 nn = = mmgg (1)(1) x directionx direction:: ∑∑FFx x = m= maaxx
– – μμkknn = = mmaaxx (2)(2) Substituting (1) in (2) : – – μμkk(m(mgg)) = = mmaaxx
aaxx = = –– μμk k gg
Example 5.23Example 5.23 Sliding Sliding Hockey Puck, finalHockey Puck, final
aaxx = = –– μμk k gg To the left (slowing down) & To the left (slowing down) &
independent of the mass!!independent of the mass!! Replacing aaxx in the Equation:
vvff22 = v = vii
22 + 2 + 2aaxx(x(xff – x – xii) )
0 = (20.0m/s)0 = (20.0m/s)22 + 2( + 2(–– μμk k gg)(115m))(115m)
μμk k 2(9.80m/s(9.80m/s22)(115m))(115m) = 400(m400(m22/s/s22) )
μμk k == 400(m400(m22/s/s22) / (2254m) / (2254m22/s/s22) )
μμk k = 0.177= 0.177
Example 5.24Example 5.24 Two Objects Two Objects Connected with FrictionConnected with Friction
Example 5.13Example 5.13 (Text Book) Known: ƒƒkk kknn Find: aa
Mass 1Mass 1:: (Block)(Block) y direction:y direction: ∑∑FFyy = 0 = 0, aayy = 0= 0
n + n + FsinFsinθθ – m– m11g = 0g = 0
n = mn = m11g – Fg – Fsinsinθθ (1)(1) x direction:x direction: ∑∑FFx x = m= m11aa
FFcoscosθθ – – TT –– ƒƒkk = = mm11aa
FFcoscosθθ – – TT –– kknn = = mm11aa
T T = F= Fcoscosθθ –– kknn – m1aa (2)(2)
Example 5.24Example 5.24 Two Objects Two Objects Connected with Friction, 2Connected with Friction, 2
Mass 2:Mass 2: (Ball)(Ball) y direction:y direction: ∑∑FyFy = m = m22aa
TT – m– m22g = mg = m22aa
TT = m= m22g + mg + m22aa (3)(3) x direction:x direction: ∑∑FFxx = 0 = 0, aaxx = 0= 0
n = mn = m11g – Fg – Fsinsinθθ (1)(1)
T T = F= Fcoscosθθ –– kknn – m1aa (2)(2) Substitute (1)(1) into (2)(2)::
T = FT = Fcoscosθθ – – kk(m1g – F(m1g – Fsinsinθθ )) – m – m11aa (4)(4) Equate: (3) = (4)(3) = (4) and solve for a:a:
Example 5.24Example 5.24 Two Objects Two Objects Connected with Friction, finalConnected with Friction, final
21
21kk
mmgmmμsinθμcosθF
a
gmmFamm
gmFgmFamam
amFgmFamgm
kk
kk
k
2121
2121
1122
sincos
sincos
)sin(cos
Inclined Plane Inclined Plane ProblemsProblems
Tilted coordinate systemTilted coordinate system: Convenient, but not necessary.
K-Trigonometry:K-Trigonometry:FFggxx= F= Fggsinsinθθ = mg = mgsinsinθθFFggyy= F= Fggcoscosθθ = = –– mg mgcoscosθθ
Understand:Understand: ∑∑F = F = m m a , a , ƒƒkk kk nn
aax x ≠ 0≠ 0 aay y = 0= 0 y directiony direction:: ∑∑FFyy = 0 = 0
n –– mg mgcoscosθθ = 0= 0 n n == mg mgcoscosθθ (1)(1)
x directionx direction:: ∑∑FFx x = m= maaxx mgmgsinsinθθ – – ƒƒ = = mmaaxx (2)(2)
aaxx
Is the normal force Is the normal force nn equalequal& opposite to the weight & opposite to the weight FFg g ??
NO!!!!NO!!!!
Experimental Experimental Determination of Determination of µµs s and and µµkk
The block is sliding down the plane, so friction acts up the plane
This setup can be used to experimentally determine the coefficient of friction µµs,ks,k
µµs,ks,k = tan = tan s,ks,k
For µµs s use the angle where the block just slips
For µµk k use the angle where the block slides down at a constant speed
Active Figure 5.19Active Figure 5.19
Example 5.25 Example 5.25 The The SkierSkier
Assuming: Assuming: FFGG = mg , a= mg , ay y = = 00
ƒƒkk kkn,n, kk = 0.10= 0.10 Find: Find: aaxx Components:Components:
FFGGxx= F= FGGsinsin3030oo = mg = mgsinsin3030oo
FFGGyy= F= FGGcoscos3030oo = = –– mg mgcoscos3030oo
Newton’s 2Newton’s 2ndnd Law Law y directiony direction:: ∑∑FFyy = 0 = 0
n –– mg mgcoscos3030oo = 0= 0 n n == mg mgcoscos3030oo (1)(1)
x directionx direction:: ∑∑FFx x = m= maaxx mgmgsinsin3030oo – – ƒƒkk = = mmaaxx (2)(2)
aaxx
ƒƒkk
kk nn
nn
aaxx
Example 5.25 Example 5.25 The The SkierSkier
n n == mg mgcoscos3030oo (1)(1)
mgmgsinsin3030oo – – ƒƒkk = = mmaaxx (2)(2) Replacing ƒƒkk kkn n in (2)(2)
mgmgsinsin3030oo – – kkn n = = mmaax x (3)(3) SubstitutingSubstituting (1) (1) intointo (3) (3)
mgmgsinsin3030oo – – kkmgmgcoscos3030oo = = mmaaxx
aaxx == ggsinsin3030oo –– μμkkggcoscos 3030oo
aaxx == g(0.5)g(0.5) –– 0.10g(0.87)0.10g(0.87)
aaxx == 0.41g0.41g
aaxx = = 4.00m/s4.00m/s22
aaxx
ƒƒkk
kk nn
nn
aaxx
Examples to Read!!!Examples to Read!!! Example 5.2Example 5.2 (Page 120) Example 5.3Example 5.3 (Page 122) Example 5.12Example 5.12 (Page 134)
Homework to be solved in Homework to be solved in Class!!!Class!!! NONENONE
Material for the Material for the MidtermMidterm