6161103 8.5 frictional forces on flat belts

8.5 Frictional Forces on Flat Belts8.5 Frictional Forces on Flat Belts

� Whenever belt drives or hand brakes are designed, it is necessary to determine the frictional forces developed between the belt and its contacting surfaces

� Consider the flat belt which passes over a � Consider the flat belt which passes over a

fixed curved surface such that

the total angle of belt to surface

contact in radians is β and the

coefficient of friction between

the two surfaces is µ


� Determine the tension T2 in the belt which is needed to pull the belt CCW over the surface and overcome both the frictional forces at the surface of contact and the known tension the surface of contact and the known tension T1

� Obviously T2 > T1


Frictional Analysis� Consider FBD of the belt segment in contact with

the surface � Normal force N and the frictional force F, acting at

different points on the belt, vary both in magnitude and directiondifferent points on the belt, vary both in magnitude and direction

� Due to this unknown force distribution, analysis the problem by studying the forces acting on a differentialelement of the belt


Frictional Analysis

� Consider FBD of an element having a length ds

� Assuming either impending motion or motion of the belt, the magnitude of the frictional force the belt, the magnitude of the frictional force

dF = µ dN

� This force opposes the sliding

motion of the belt and thereby

increases the magnitude of the

tensile force acting in the belt by dT


Frictional Analysis

� Applying equilibrium equations

02

cos)(2

cos

;0

=

+−+

=∑

θµ

θ ddTTdN

dT

Fx

� Since dθ is of infinitesimal size, sin(θ/2) and cos (θ/2) can be replaced by dθ/2 and 1 respectively

� Product of the two infinitesimals dT and dθ/2 may be neglected when compared to infinitesimals of the first order

02

sin2

sin)(

;0

02

cos)(2

cos

=

−

+−

=∑

θθ dT

ddTTdN

F

dTTdNT

y


Frictional Analysis

θµ

θµ

ddT

TddN

dTdN

=

=

=

Solving

µβ

β

µβ

θµ

βθθ

θµ

eTT

T

TIn

dT

dT

TTTT

dT

T

T

12

1

2

0

21

2

1

,,0,

=

=

=

====

=

∫ ∫


Frictional Analysis� T2 is independent of the radius of the drum

and instead, it is a function of the angle of belt to surface contact, β

� This equation is valid for flat belts placed on � This equation is valid for flat belts placed on any shape of contacting surface

� For application, it is valid only when impending motion or motion occurs


Example 8.9

The maximum tension that can be developed In the cord is

500N. If the pulley at A is free to rotate and the coefficient of

static friction at fixed drums B and C is µs = 0.25, determine

the largest mass of cylinder that can be lifted by the cord. the largest mass of cylinder that can be lifted by the cord.

Assume that the force F applied at the end of the cord is

directed vertically downward.


Solution� Lifting the cylinder, which has a weight of W =

mg, causes the cord to move CCW over the drums at B and C, hence, the maximum tension T2 in the cord occur at DT2 in the cord occur at D

� Thus, T2 = 500N� For section of the cord passing

over the drum at B � 180° = π rad, angle of contact

between the drum and the cord β = (135°/180°)π = 3/4π rad


Solution

( )[ ]eTN

eTT s

500

;4/325.0

1

12

=

=π

βµ

� Since the pulley at A is free to rotate, equilibrium requires that the tension in the cord remains the same on both sides of the pulley

( )[ ] NN

e

NT

eTN

4.27780.1

500500

500

4/325.01

1

===

=

π


Solution

� For section of the cord passing over the drum at C

W < 277.4N

( )[ ]

kgsm

N

g

Wm

NW

Wen

eTT s

7.15/81.9

9.153

9.153

277

;

2

4/325.0

12

===

=

=

=π

βµ

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6161103 8.5 frictional forces on flat belts