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Friction myth buster #3: A fixed pulley (“drum”) CAN change the tension in a cable! OK then, which one is larger, F 1 or F 2 ? ANSWER: It depends… SMOOTH (no friction) F 1 = F 2 F 1 F 2 F 1 F 2 ROUGH (friction) F 1 = F 2

Friction myth buster #3: A fixed pulley (“drum”) CAN change ...Friction myth buster #3: A fixed pulley (“drum”) CAN change the tension in a cable! OK then, which one is larger,

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Page 1: Friction myth buster #3: A fixed pulley (“drum”) CAN change ...Friction myth buster #3: A fixed pulley (“drum”) CAN change the tension in a cable! OK then, which one is larger,

Frictionmythbuster#3:Afixedpulley(“drum”)CANchangethetensioninacable!

OKthen,whichoneislarger,F1orF2?

ANSWER:Itdepends…

SMOOTH(nofriction)

F1 = F2 F1

F2

F1

F2

ROUGH(friction)

F1 = F2

Page 2: Friction myth buster #3: A fixed pulley (“drum”) CAN change ...Friction myth buster #3: A fixed pulley (“drum”) CAN change the tension in a cable! OK then, which one is larger,

PROBLEMAbeltiswrappedaroundaroughstationarydrum,withcoefficientsofstaticandkineticfriction,μSandμK.

Lecture19summary:beltfriction

me270-cmk

FUNDAMENTALEQUATIONS:•  Forimpendingslippingofbeltondrum:

•  Forslippingofbeltondrum:

NOTES:•  T1≥T2,always.Youmustknowaprioriwhichofthetwo

tensionsisthelargest.Generally,youneedtoconsiderthedirectionofimpendingslippingconsiderationstodeterminethelargesttension.

•  βistheangleofwrapforthebeltinradians.•  Theaboveequationsarealsovalidforapulleyonwhicha

torqueMacts.

Chapter6

D. Belt friction

Whenever a belt, cable, or rope is wrapped around a rough-surfaced object, friction develops at theinterface. Up to this point in the course, we have ignored the influence of friction for cables beingpulled over pulleys. With the assumption that the pulley is smooth, we have concluded that thepulley does not alter the tension in the cable. Here we will now address the question as to how thisfriction produces a tension di↵erence as the cable is wrapped around a non-ideal drum. In orderto determine the relationship for this tension di↵erence, it is necessary to build up a mathematicalmodel that utilizes di↵erential elements.

Consider the situation shown in the figure below left where we have a section of a belt pulled overa rough, stationary circular drum. The tensions at the ends of the belt are T1 at ✓ = 0 and T2 at✓ = �, where � is the total angle of wrap for the belt. For the sake of the mathematical developmentthat follows, it is assumed that T1 > T2.

Since we have assumed T1 > T2, the impending motion of the belt on the drum is toward the highertension end where the tension is T1. The friction force on the belt by the drum will oppose thedirection of impending slip; therefore, the distributed friction force on the belt will be as shown inthe figure below right.

A free body diagram for a di↵erential section of the belt is shown in the following figure. Thisfigure shows four forces acting on the section of the belt: the tensions at ✓ and ✓ + d✓, the normalforce dN and the friction force df = µdN . It is important to note that the direction of the frictionforce is drawn to oppose the direction of impending sliding of the section of the belt.

Summing forces acting on the belt section gives:X

Fx = T (✓ + d✓)cos(d✓/2)� T (✓)cos(d✓/2) + µdN = 0

XFy = �T (✓ + d✓)sin(d✓/2)� T (✓)sin(d✓/2) + dN = 0

Since d✓ is of di↵erential (small) size, sin(d✓/2) ! d✓/2 and cos(d✓/2) ! 1. Therefore, the above

T1T2

= eµSβ ≥1

T1T2

= eµKβ ≥1

T1

T2

M

β

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Skill Building Problem SB.6.1

HomeworkProblemH6.D.7Given: BlocksAandB,eachhavingaweightofW,aresupportedbythecable/fixed-

drumsystemshownbelow.BlockCissupportedatthejunctionofthetwocablessupportingAandB.Thecoefficientsofstaticfrictionforthetwofixeddrumsareasindicatedinthefigure.

Find: DeterminethemaximumweightofblockCforwhichthesystemcanremain

inequilibrium.ExpressyouranswerintermsofW.Usethefollowing:μs=0.3andθ=60°.

µS

g

fixeddrum

fixeddrum

µS

A

C

B

horizontalθ

D

E

Page 4: Friction myth buster #3: A fixed pulley (“drum”) CAN change ...Friction myth buster #3: A fixed pulley (“drum”) CAN change the tension in a cable! OK then, which one is larger,

Skill Building Problem SB.6.2

HomeworkProblemH6.D.9Given: Thebandbrakesystemshownbelowhasabandwrappedaroundadrum,

withacounter-clockwisetorqueMbeingappliedtothedrumandthedrumrotatingataconstantrate.ThebandistensionedbyarmAC,withtheloadPactingatendCofthearm.ThecoefficientofkineticfrictionbetweenthebandandthedrumisμK.ConsidertheweightofarmACtobenegligiblecomparedtoP.

Find: DeterminethecoupleMactingonthedrum.Writeyouranswerintermsof

P,RandμK.

R

A

2R 2R

B C

O M

R

P

D

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ME 270 – Fall 2020 ©Freeform 2020 38

Homework H19.A

Given: A belt is wrapped around two pulleys, each of radius R. The ends of the belt are attached to rigid bar OH at points A and B, as shown. A rigid arm is welded to the disk on the right, with a force P acting at end E of the arm. A second force F acts at end H of arm OH. Let μS represent the coefficient of static friction between the belt and the drums.

Find: Determine the smallest value of F that is required to prevent slipping between

the belt and either drum. Express your answer in terms of the load P. Use the following in your analysis: μS = 0.4. P

R R

smooth bearings

weld

R 5R R

F A O B

C D E

H

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ME 270 – Fall 2020 ©Freeform 2020 39

Homework H19.B

Given: Block C, having a weight of W, is supported by cable CD that is pulled over a

pair of rough, fixed cylinders, where μs is the coefficient of static friction between the cable and cylinders. Note that section AB is horizontal.

Find: Determine the range of values for the force F applied to end D of the cable for

which block C remains in static equilibrium. Provide your answer in terms of the block’s weight W.

For this problem, use the following parameter: μs = 0.3.

g

F

W

4

3

C

B

A

D

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