1. 8.5 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
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

2. 8.5 Frictional Forces on Flat Belts
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
T1
Obviously T2 > T1

3. 8.5 Frictional Forces on Flat Belts
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 direction
Due to this unknown force distribution, analysis
the problem by studying the
forces acting on a differential
element of the belt

4. 8.5 Frictional Forces on Flat Belts
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
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

5. 8.5 Frictional Forces on Flat Belts
Frictional Analysis
Applying equilibrium equations
∑ Fx = 0;
 dθ   dθ 
T cos  + µdN − (T + dT ) cos =0
 2   2 
∑ Fy = 0;
 dθ   dθ 
dN − (T + dT ) sin   − T sin  =0
 2   2 
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

6. 8.5 Frictional Forces on Flat Belts
Frictional Analysis
µdN = dT
dN = Tdθ
dT
= µdθ
T
Solving
T = T1 , θ = 0, T = T2 , θ = β
T2 dT β
∫T1 T = µ ∫0 dθ
T
In 2 = µβ
T1
T2 = T1e µβ

7. 8.5 Frictional Forces on Flat Belts
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
any shape of contacting surface
For application, it is valid only when
impending motion or motion occurs

8. 8.5 Frictional Forces on Flat Belts
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.
Assume that the force F applied at the end of the cord is
directed vertically downward.

9. 8.5 Frictional Forces on Flat Belts
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 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

10. 8.5 Frictional Forces on Flat Belts
Solution
T2 = T1e µ s β ;
500 N = T1e 0.25[(3 / 4 )π ]
500 N 500 N
T1 = = = 277.4 N
e 0.25[(3 / 4 )π ] 1.80
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

11. 8.5 Frictional Forces on Flat Belts
Solution
For section of the cord passing over the drum at C
W < 277.4N
T2 = T1e µ s β ;
277 n = We0.25[(3 / 4 )π ]
W = 153.9 N
W 153.9 N
m= = 2
= 15.7 kg
g 9.81m / s