BELT AND WRAP FRICTION
Rob Wendland, driving Mike Troxel's
Federal-Mogul Dragster In-N-Out Burger
A Capstan-Style Winch
Belt and Wrap Friction
When flat belts, V-belts, band brakes, and line-wrap around
capstans are used in any form of product, the frictional forces
developed must be determined. All forms of wrap friction are
analyzed similarly. Let's first look at flat belts


θ
r
T2
T1
2

A flat belt has a wrap angle around a
pulley of θ radians. Normal forces
develop at each finite point of contact
with a resulting frictional force acting
tangent to the pulley and opposite the
direction of rotation. This reduces belt
tension around the pully such that
tension T1<T2
Furthermore, since belt tension changes
continuously, so do the incremental
normal and frictional forces
Belt and Wrap Friction
To help determine how frictional forces affect belt tension,
consider a finite element over angle ∆θ. Acting on this
element is slackside tension T and tightside tension T + ∆T


y
T T

2
 F=⋅ N

2
T
N
Incremental forces ∆N and ∆F develop
as a result of the applied incremental
tensions. By performing an equilibrium
analysis:

x
 Fx = 0
= T cos
 
 


 ⋅ N −  T   T ⋅cos
2
2
Fy= 0
=  N −  T   T ⋅sin
3
 
 


− T⋅sin
2
2
Belt and Wrap Friction
y
T T

2
 F=⋅ N

2
T
 
N
x
Substituting  N
=>
T = ⋅T⋅
Since this expression is developed over a very small angle,
this would have to be summed around the wrap angle
  T i =   ⋅T⋅ i 
i
4
 



=
and cos
=1
2
2
2
Substituting:
 F x =>
T = ⋅ N
 F y =>
 N = T⋅ 


For small , sin
i
Belt and Wrap Friction
y
T T

2
 F=⋅ N


2
T
N


x
Since this expression is developed
over a very small angle, applying such
a summation would be impractical
However, if we replaced the finite
quantities of ∆T and ∆θ with infinitely
small values of dT and dθ, we could
integrate over θ as follows:
T2
∫ dTT
T1
5

= ∫ d 
0
T 2 = T 1 e 

This evaluates to:

It's important to remember T2 is always
larger than T1 and θ is in radians
V-Belt Friction
α

The friction developed by a V-belt can
be developed in a similar fashion. The
relationship between tightside and
slackside tension for a V-belt is:
T 2 = T 1 e /sin  / 2

6
Angle α must be expressed in radians
Example 18
The accessory package for the engine shown below requires a
torque of 30 ft-lbf. The pulley in contact with the belt has a
diameter of 8” and a coefficient of static friction of 0.30.
Determine the tension in each part of the belt if the belt is not
to slip and:



40o
60o
7
the system uses a flat belt
the system uses a V-belt with a 38o V
and the same coefficient of friction
Example 18
First determine angle of wrap. Draw
a construction line at the base of
T
vector TB and parallel to vector TA.
Angle α is the difference between
angles of the two vectors and is equal
θ
α
α
to 20o. This results in a wrap angle of
200o or 1.11π radians
 Since there are two unknowns, we
must develop two mathematical relationships to solve for the
unknown tensions. These are the moment about the center of
the pulley and the friction equation for a flat belt. Since
moment is applied clockwise, tension TB is tightside tension;
we will let T2 = TB and T1 = TA
TA

40o
B
60o
8
Example 18
 M o = T B⋅4 in − T A⋅4 in −  30 ft⋅lb f ⋅12 in / ft
.
. . T B = T A  90
(Moments)
-and0.3⋅1.11 
T B = T A⋅e
(Friction)
TA
40o
TB
60o
θ
Substituting:
0.3⋅1.11 
T A  90 = T A⋅e
= 2.85⋅T A
.
. . T A = 48.6 lb f
and T B = 139 lb f
α
α
o
The application of a V-belt changes only the friction equation. The 38 V
is 0.211 radians. Modifying the friction equation changes the solution to:
0.3⋅1.11  /sin 0.106 
T A  90 = T A⋅e
.
. . T A = 3.82 lb f
and
= 24.5⋅T A
T B = 98.4 lb f
Notice the efficiency increase of a V-belt over that of a flat belt. The reduced
tensions help increase bearing life
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Frictional Forces
Real Life Mechanisms

T1
45o
90o
T1
A
135o
T1
T2
B
µ = 0.40
C
(A)
T2
(A)
o
 = 45
45⋅
=
180
= 0.25 rad
0.40⋅0.25 
T 2 = T 1⋅e
= 1.37⋅T 1
10
Since area does not affect
the developed friction for a
belt, the formula for flat
belts is applicable to round
line. Only angle of wrap
and coefficient of friction
govern developed tension
T2
(B)
o
 = 90
90⋅
=
180
= 0.50  rad
0.40⋅0.50  
T 2 = T 1⋅e
= 1.87⋅T 1
(C)
o
 = 135
135⋅
=
180
= 0.75 rad
0.40⋅0.75 
T 2 = T 1⋅e
= 2.56⋅T 1
Example 19

300 lbf
11
F
You are lifting an engine out of your
friends car. In your front yard is a sturdy
oak tree. You wrap a rope around a
tree branch to lift the engine. The
engine weighs 300 lbf, the rope has a
180o wrap, and µ = 0.1
 How much force does it take to lift the
engine from the vehicle?
 How much force must you apply to
lower the engine back into place?
Example 19

Line wrap is 180o or π radians. When
lifting, force F is the tightside of the line
(T2). To solve for T2:

T 2 = T 1e
T 2 = 300⋅e 0.1⋅ = 411 lb f

300 lbf
12
F
To lower the engine, force F becomes
the slack side of the line (T1). To solve
for T1:
T 1 = T 2 e−
−0.1⋅
T 1 = 300⋅e
= 219 lb f
Frictional Forces
Real Life Mechanisms

Mechanisms using wrap friction are often critical life-safety devices.
Such devices are used in rappelling, rock climbing, sailing, and
rigging of equipment
From left to right are: RescueTECH rappell rack; PMI brake tube for lowering long distances; and
the Wichard gyb`easy to control the boom of a sailboat during a gybe. All devices must meet
stringent NFPA, OSHA, ABYC, or Coast Guard requirements depending upon their application

Some devices, such as the rack, have adjustable bars. Others, like
the brake tube, rely on the number of wraps around the tube, while
the anti-gybe device relies on how many passes of line through the
non-adjustable bars
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Example 20

(A)
(A)
(B)
o
 = 800
= 4.44  rad
T 2 = T 1⋅e 0.25⋅4.44 
= 32.8⋅T 1
14
(C)
(B)
Consider the adjustable rack
shown. The leftmost configuration
has closely spaced bars resulting
in 120o of contact on each turn.
The other two are spaced to
provide contact arcs as shown.
Determine the tension ratio for
each configuration if µ = 0.25
o
 = 560
= 3.11 rad
T 2 = T 1⋅e 0.25⋅3.11 
= 11.5⋅T 1
(C)
o
 = 315
= 1.75  rad
T 2 = T 1⋅e 0.25⋅1.75 
= 3.95⋅T 1
Building the Great Pyramids...
Friction Was a Factor
µ=0.5
roll-rope
µ=0.04
bearings
µ=0.04
sledge-tracks
µ=0.2
static friction

The Egyptions and Mayans responsible for building the pyramids on
their respective continents were able to reduce friction where it wasn't
wanted and maximize friction where it was useful and desired. We
still do this today in equipment design
15