[email protected] ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx 1 Bruce Mayer, PE Engineering-36:...

[email protected] • ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx1

Bruce Mayer, PE Engineering-36: Engineering Mechanics - Statics

Bruce Mayer, PELicensed Electrical & Mechanical Engineer

[email protected]

Engineering 36

Ch08: Wedge &

Belt Friction



Outline - Friction

The Laws of Dry Friction• Coefficient of Static Friction• Coefficient of Kinetic (Dynamic) Friction

Angles of Friction• Angle of static friction• Angle of kinetic friction• Angle of Repose

Wedge & Belt Friction• Self-Locking & Contact-Angle



Basic Friction - Review The Static Friction Force Is The force that Resists

Lateral Motion. It reaches a Maximum Value Just Prior to movement. It is Directly Proportional to Normal Force:

NF sm After Motion Commences The Friction Force Drops

to Its “Kinetic” Value NF kk



Wedge Friction Consider the

System Below

Find the Minimum Push, P, to move-in the Wedge

The Wedge is of negligible Weight

Then the FBD of the Two Blocks using Newton’s 3rd Law



Wedge Friction

For Equilibrium of the Heavy Block

Solve for FA,n

For Equilibrium of the Wt-Less Wedge

sincos0 ,, nAsnAy FFWF

sincos,s

nA

WF

cossin0

sincos0

,,,

,,,

nAnAsnCy

nAnAsnCsx

FFFF

FFFPF



Wedge Friction

In the last 2-Eqns Sub Out FA,n

Eliminating FC,n from the 2-Eqns yields an Expression for Pmin:

cossincos

sinsincos

0

sinsincos

cossincos

0

,

,

sssnCy

sssnCsx

WWFF

WWFPF

cos2sin1sincos

2min ss

s

µW

P



Wedge Friction

MATLAB Plots for P when W = 100 lbs

0 2 4 6 8 10 12 14 16 18 2040

45

50

55

60

65

70

75

80

85

(°)

P (

lbs)

W = 100 lbs, µ = 0.2

0 5 10 15 20 25 3010

20

30

40

50

60

70

80

90

µ (%)

P (

lbs)

W = 100 lbs, = 10°



MATLAB Code% Bruce Mayer, PE% ENGR36 * 22Jul12% ENGR36_Wedge_Friction_1207.m%u = 0.2W = 100a = linspace(0,20);P = W*((1-u*u)*sind(a) +2*u*cosd(a))./(cosd(a)-u*sind(a))plot(a,P, 'LineWidth',3), grid, xlabel('\alpha (°)'), ylabel('P (lbs)'), title('W = 100 lbs, µ = 0.2')disp('showing 1st plot - Hit Any Key to Continue')pause%a = 10;u = linspace(0,0.3);P = W*((1-u.*u)*sind(a) +2*u*cosd(a))./(cosd(a)-u*sind(a));plot(100*u,P, 'LineWidth',3), grid, xlabel('µ (%)'), ylabel('P (lbs)'), title('W = 100 lbs, \alpha = 10°')disp('showing 2nd plot - Hit Any Key to Continue')pause%u = linspace(0, .50);aSL =atand (2*u./(1-u.^2));plot(100*u,aSL, 'LineWidth',3), grid, xlabel('µ (%)'), ylabel('\alpha (°)'), title('Self-Locking Wedge Angle')disp('showing LAST plot')



Wedge Friction Now What Happens

upon Removing P

The Wedge can• Be PUSHED OUT• STAY in Place

– SelfLocking condition

Then the FBD When P is Removed• Note that the

Direction of the Friction forces are REVERSED



Wedge Friction

For Equilibrium of the Heavy Block

Solve for FA,n

For Equilibrium of the Wt-Less Wedge

sincos0 ,, nAsnAy FFWF

cossin0

sincos0

,,,

,,,

nAnAsnCy

nAnAsnCsx

FFFF

FFFF

KW

Fs

nA

sincos,



Wedge Friction

To Save Writing sub K for FA,n

Eliminate FC,n

Now Divide Last Eqn by Kcosα

0sincos

0sincos

,

,

KKF

KKF

snC

snCs

01sincos20

0sincos

0sincos

2,

,

ss

ssnC

snCs

µKK

KKF

KKF



Wedge Friction

Dividing by Kcosα

Recognize sinu/cosu = tanu

01cos

sin2

cos

01sincos2

2

2

ss

ss

µ

K

µKK

222

2

1

2

1

2

1

2tan

21tan

s

s

s

s

s

s

ss

µµµ

µ



Wedge Friction

After all That AlgebraFind The Maximum α to Maintain the Block in the Static Location

Since Large angles Produce a Large Push-Out Forces, and a ZERO Angle Produces NO Push-Out Force, the Criteria for Self-Locking

2max 1

2arctan

s

s

21

2arctan

s

sSL



Wedge Push-Out

SMALL PushOut Force• Likely SelfLocking

LARGE PushOut Force• Likely NOT SelfLocking

21

2arctan

s

sSL



Wedge Friction

0 5 10 15 20 25 30 35 40 45 500

10

20

30

40

50

60

µ (%)

(

°)

Self-Locking Wedge Angle



Belt Friction Consider The Belt Wrapped

Around a Drum with Contact angle .

The Drum is NOT Free-Wheeling, and So Friction Forces Result in DIFFERENT Values for T1 and T2

To Derive the Relationship Between T1 and T2 Examine a Differential Element of the Belt that Subtends an Angle • The Diagram At Right Shows

the Free Body Diagram



Belt Friction cont Write the Equilibrium Eqns for

Belt Element PP’ if T2>T1

2sin

2sin0

2cos

2cos0

TTTNF

NTTTF

y

sx

Eliminate N from the Equations

2sin

2sin

2sin

2cos

2cos

2cos0

2sin

2sin

2cos

2cos0

2sin

2sin

TTTTTT

TTTTTTF

TTTNF

s

sx

y



Belt Friction cont.1 Combining Terms

2

sin22

cos0

TTT s

Divide Both Sides by

2

2sin2

2cos0

TTT

s

Now Recall From Trig And Calculus

d

dLimLim

00

1sin

10cos

So in the Above Eqn Let: /2 →0; Which Yields

TdTTTd

dTdTT

d

dTss 2 as20



Belt Friction cont.2 The Belt Friction Differential Eqn

Integrate the Variables-Separated Eqn within Limits• T( = 0) = T1

• T( = ) = T2

From Calculus

Now Take EXP{of the above Eqn}

ddTT

Td

dTss

1Vars Sep

12120lnlnln

12

1

TTTTddTT ss

T

T

ss eTTee TT 12ln 12



Belt Friction Illustrated This is a VERY

POWERFUL Relationship

Condsider the Case at Right. Assume• A ship Pulls on the Taut

Side With A force of 4 kip (2 TONS!)

• The Wrap-Angle = Three Revolutions, or 6

• µs = 0.3

seT

T

1

2

The Tension, T1, Applied by the Worker

lbe

kip

e

TT

s14

463.0

21



WhiteBoard Work

Let’s WorkThese NiceProblems



Bruce Mayer, PERegistered Electrical & Mechanical Engineer

[email protected]

Engineering 36

Appendix 00

sinhT

µs

T

µx

dx

dy



WhiteBoard Work

Let’s WorkThis NiceProblem



Wedge Push-Out

SMALL PushOut Force• Likely SelfLocking

LARGE PushOut Force• Likely NOT SelfLocking

[email protected] ENGR-36_Lec-22_Wedge-n-Belt_Friction.pptx 1 Bruce Mayer, PE Engineering-36:...

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