Belt friction lecture_new

11/29/2013
1
Applications
Engineering Mechanics: Belt Friction
Band brakes
Torque
Transmission
Capstan
(Use less smaller force to
hold loads)
Pulley
(Raising or lowering
loads)
Theory
• Relate T1 and T2 when belt is about to slide to right.

11/29/2013
2
2 1T / T e s 

Condition for impendingslip for
belt about to slide toward right
(T2 > T1 )
T2 = Tension in the belt that pulls
T1 =Tension in the belt that resist
Theory
Angle of contact shall be expressed in
radian
The formula can be applied for
impendingslip condition for
problems involving:
1. Flat belt passing over a fixed
cylindrical drum
2. Ropes/Belt wrapped around a
post or capstan or pulleys
3. Belt drives, both pulley and
belt rotate
2 1T / T e k 
If the belt or rope is actually
slipping (relative slip between
belt/rope and drum/pulley)
Problem-1
A cord is wrapped twice around a
capstan A and three times around
second capstan B. Finally cord goes
over a half barrel section and supports
a mass M of 500 kg. What is tension T
required to maintain this load? Take
coefficient of friction 0.1 for all
surfaces of contact.
Maximum value of tension can be
found out by upward impending
motion of mass ‘M’ for which
Mg<T.
Minimum value of tension can be
found out by the impending
downward motion of ‘M’ for
which Mg>T .
Solution:

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3
Problem-1
For 181.15 <T<132811.04 N, the load will be
held in place .
For impending downward motion of M ,
Mg > T2>T1> T
M=500 Kg, βA=4π, βB=6π, βpulley=0.5 πSOLUTION:
0.20s  0.15k 
Acable passes around three 2-in.-radius
pulleys and supports two blocks as
shown. Two of the pulleys ( D and E) are
locked to prevent rotation, while the third
pulley is rotated slowly at a constant
speed. Knowing that the coefficients of
friction between the cable and the pulleys
are
determine the largest weight which can
be raised if pulley C is rotated clockwise
.
Problem-2
Solution:
Impending slip at C
Slips at D and E

11/29/2013
4
Note: 1
sin 30 rad,
2 6
r
r

 
   
2
,
2 6 3
C D E
  
       
2 ,s C
AW T e 

 1 2 1, 16 lbk D k ET T e T e   
 
   
     5 2
3 3
0.15 0.20
16 lb 16 lb 11.09 lbk E D s C
AW e e e e
 
      
   
C is rotated clockwise slowly
Problem-2
Belt Drives
Driver Pulley
T2 >T1
Driven pulley
M1
Connected to electric motor (say).
An external torque (M) is applied
Connected to a machine tool (say).
Torque transferred to the pulley by the
machine tool is M1
Torque Transmitted
M1= (T2-T1)*0.2

11/29/2013
5
Problem-3
A flat belt connects pulley A, which
drives the a machine tool, to pulley B
,which is attached to shaft of an electric
motor. The coefficients of friction are s
= 0.25 and k = 0.20 between both
pulleys and the belt. Knowing that the
maximum allowable tension in the belt is
600 N, determine the largest torque
which can be exerted by the belt on
pulley A.
Solution:
Since slipping depends upon angle of
contact and coefficient of friction ,pulley
‘B’ will slip first.(Since it is having same
coefficient of friction but smaller angle of
contact than pulley ‘A’).
For the maximum tension in one side of
belt of pulley ‘B’, resistance tension in
other side should be found out.
From moment about center of pulley ‘A’
,maximum allowable torque can be found
out.
For no slip at ‘A’ , µs < (µs) max condition
should be satisfied at pulley ‘A’.
Problem-3
.
Pulley-B
SOLUTION:
 
 
0 25
240 4 3
120 2 3
s
o
A
o
B
.
 
 

 
 

11/29/2013
6
Problem-3
Here couple MA is applied to the pulley by the
machine tool to which it is attached and is equal
and opposite to the torque exerted by the belt.
Pulley-A
Problem-4
A differential band brake is used to control
the speed of a drum which rotates at a
constant speed. Knowing that the coefficient
of kinetic friction between the belt and the
drum is 0.30 and that a couple of magnitude
125 lb-ft is applied to the drum, determine
the corresponding magnitude of the force P
that is exerted on end D of the lever when
the drum is rotating (a) clockwise, (b)
counter clockwise.
Solution
• Relative slip between belt and
drum
• Moment equilibrium of drum
about center of drum will give
tension generated in both side of
the drum.
• Again moment equilibrium of
lever about point B will give the
force ‘P’

11/29/2013
7
Problem-4
SOLUTION:
Problem-4
Lever

Belt friction lecture_new

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