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