Basic Engineering Mechanics 3-1-0
S De Chowdhury 1
1 Assistant Professor
Department of Ocean Engineering and Naval Architecture
IIT Kharagpur
Lecture 13: Friction, Part II
sdip@naval.iitkgp.ac.in
Spring 2025
S De Chowdhury (OENA IIT KGP) ME11003 Spring 2025 1/4
Summary of last class
Coulomb law of friction applies when there is an impending motion
from rest for a body on ground. Friction is proportional to normal
force (F = µR, where µ is coefficient of static friction) and
independent of contact area. When the body is in motion, less force/
effort is required to overcome friction. In this case, F = µd R and µd
is coefficient of dynamic friction.
We
P can findPstatic friction
P force either by equilibrium conditions (i.e.,
Fx = 0; Fy = 0; Mo = 0, let us call it as F1 ), or by impending
motion (let be it F2 ). For non-rolling bodies, if F1 < F2 then the body
is in rest. If F1 > F2 the body moves. For two bodies in contact,
friction forces can not be calculated by imposing impending motions
on both bodies simultaneously, rather start with impending motion on
one body and find friction force on another body by equilibrium
conditions.
S De Chowdhury (OENA IIT KGP) ME11003 Spring 2025 2/4
Rolling friction
Why a roller has to rotate to move forward ?
For a roller on a shaft, the friction effect
between them can be represented by a
moment M0 . The free wheel refers to a
case when F and PH is zero (which
leads to M0 = 0.) and the roller moves
on ground without rolling. This never
happens.
In reality, the ground deforms, the point
of action of total ground reaction is
offset by a and this gives the idea of
Figure: force states for rolling motion coeffient of rolling friction.
S De Chowdhury (OENA IIT KGP) ME11003 Spring 2025 3/4
Rolling friction
What happens during rolling ?
Why a roller has to rotate to move forward ?
For a roller on a shaft, the friction effect
between them can be represented by a
moment M0 . The free wheel refers to a
P = Wa r where, P is the force required
case to F and PH is zero (which
when
move the roller when it rolls. The leads
coefficient
to M0 = 0.) and the roller moves
of rolling resistance a is in unit if lengths.
on groundIn without rolling. This never
this case, the angle ϕ is small andhappens.
so the a.
In reality, the ground deforms, the point
of action of total ground
Figure: force reaction
states for is
offset by a and this gives the
rolling motion withidea of
actual
point of actions
Figure: force states for rolling motion coeffient of rolling friction.
S De Chowdhury (OENA IIT KGP) ME11003 Spring 2025 3/4
Key points
For a roller with two contact surfaces, P can be found by putting
impending motion on one surface and using equilibrium conditions.
If P < F where F = µR if impending motion is imposed, the body will
roll without slipping/ skidding.
if P > F where F = µR from impending motion, the body will skid.
S De Chowdhury (OENA IIT KGP) ME11003 Spring 2025 4/4
Belt drives are called flexible machine elements. Flexible machine elements are used for a large
number of industrial applications, some of them are as follows.
1. Used in conveying systems
Transportation of coal, mineral ores etc. over a long distance
2. Used for transmission of power.
Mainly used for running of various industrial appliances using prime movers like electric motors, I.C.
Engine etc.
Flexible machine elements has got an inherent advantage that, it can absorb a good amount of
shock and vibration. It can take care of some degree of misalignment between the driven and the
driver machines and long distance power transmission, in comparison to other transmission systems, is
possible. For all the above reasons flexible machine elements are widely used in industrial
application.
Although we have some other flexible drives like rope drive, roller chain drives etc. we will only discuss
about belt drives.
3
Basic Principle:
A Belt Drive is shown in Fig.1. Prime movers like electric
motors, I.C. Engine etc. are used to rotate the driving pulley
through application of moment Ta as shown. The power is
β transmitted to the driven pulley through belt friction.
αL
Driven αS
Ta
pulley
Nomenclature of open Belt Drive
Driving dS
dL pulley dL - Diameter of the larger pulley
dS – Diameter of the smaller pulley
C Fig. 1
αL- Angle of wrap of the larger pulley = 180ο + 2β
αS – Angle of wrap of the smaller pulley = 180ο – 2β
Where, β = sin-1 {(dL – dS)/2C}
C- Center distance between the two pulleys
π(2
1 4
L
dS
)
2
C
(
d
dS
)
2
C
L
L
Length of open belt,
(Note that above formula is approximate, but used widely in design)
4
reaction
T1
belt motion reaction In belt drive, for any given pulley, one end
T1 T1 T1
Tr RL Driven Ta RS of the belt has a higher tension than the
T2> T1 pulley Driving T2> T1
Tt pulley other end. It is important to identify the
End opposite to
T2 T2 T2 T2 direction of friction on high tension and low tension ends of a
belt is in high tension
Fig. 2a
Fig. 2b belt.
Let us consider the driving pulley (Fig.2b). In this case the pulley rotates in the direction of applied
torque Ta (anticlockwise). The belt will oppose the motion of the pulley and the friction forces on the
pulley and belt are as shown in the Fig.2b. A motion will be imparted to the belt in the direction of
friction on the belt. Equilibrium of the belt segment suggests that T2 is higher than T1.
The driven pulley in the initial stage is not rotating. The basic nature of friction again suggests that the
driven pulley opposes the motion of the belt. The directions of friction on the belt and the driven pulley
are shown the Fig.2a. The frictional force on the driven pulley will create a motion in the clockwise
direction. Equilibrium of the belt segment for driven pulley again suggests that T2 is higher than T1. The
torque Tt =(T2-T1)x dL/2 transmitted to the driven pulley operates the load attached to it, say a pump,
which is Tr . For the driving pulley, Ta = (T2-T1)x ds/2. The power transmitted, P = Ta x ωS , where, ωS is the
angular speed of the driving pulley.
5
Relationship between T1 and T2 :
n
dθ/2 Let us consider a small segment of belt (Fig.3) from the driven pulley
t
T
shown in Fig.2a. In the direction n, only normal force dN is considered.
T + dT F
dN The effect of centrifugal force on the belt is not considered. At the
dθ impending slippage of the belt, F=μ dN.
Fig. 3 t- direction: T cosdθ/2 + μ dN – (T + dT) cosdθ/2 = 0
or, μ dN = dT [cos dθ/2 ≈ 1]
n- direction: - T sindθ/2 + dN - (T+dT) sindθ/2 = 0
or, dN = 2T sin dθ/2 + dT sin dθ/2
≈ 2T dθ/2 + dT dθ/2 ≈ T dθ [sin dθ/2 ≈ dθ/2 and 2nd term neglected]
From above equilibrium equations we get, dT/T = μ dθ, Integrating between corresponding limits yields,
µdθ, 𝑜𝑟 ln µα, or T2 = T1 eµα, whereαis the angle of wrap in radians. … (A)
Remember that:
(i) For belt drives, to use equation (A), select either (µ𝜶)driving pulleyor (µ𝜶)driven pulley, whichever is lower.
(ii) α, the angle of wrap is in radians.
(iii) Belt end opposite to the direction of friction on belt is in high tension.
6
Problem Statement: An idler pulley is used to
T2
1
increase the angle of wrap for the pulleys shown.
4 The smaller pulley transmits 16 kW power at 900
2
rpm clockwise rotation. Identify the slack and tight
side of the drive. Find the belt tensions and the
3 T1
torques that can be transmitted by the pulleys?
Take µ=0.3
For the smaller pulley, ωS = 2π x 900/60 = 94.25 r=3ad/s. For the given direction of rotation of the pulley,
friction on the belt is in the clockwise direction. Therefore, belt in the top of the drive is tight side (high tension,
T2) and the lower side is slack side.( low tension, T1). Now, Ps =Ts x ωS, or 16000 = Ts x 94.25. Therefore,
Ts = 169.76 Nm. Ts = (T2 – T1) x rs, or (T2 – T1) = 1358 N. For larger pulley, TL = (T2 – T1) x rL, TL = 339.25 Nm.
Angle of wrap for the larger pulley, αL = segments(1+2+3) = 300 + 1800 + 150 = 2250
Angle of wrap for the smaller pulley, αs = segment 4 = 1800. As μαs < μαL , μαs should be chosen.
μαs = 0.3 x 1800 =0.3 x π = 0.9425. Therefore, T2 = T1 x e 0.9425 = 2.566 x T1 or T1 = 1358 / 1.566 N.
Therefore, T1 = 867 N and T2 = 2225 N
7
Problem statement: A 1000 N load hangs from a cable passing
over a curved surface. Find T to prevent the load from slipping
down for μ =0.3?
For the impending slippage of the load 1000N downwards, the friction on the belt is towards
the load T. Therefore, belt end attached to the load 1000N is in high tension. For noncircular
section, the angle of warp depends on total angle of contact, not on belt length.
Therefore, α = 1800 = π radians
T2 = 1000 = T x e0.3 x π . Hence, T = 389.7 N
(To move the load up T must be greater than 1000N as T will be at higher tension)
8
Screws: These are used for fastenings and for transmitting power or motion.
The action of screw depends on the friction developed on the mating
W
M
threads. For fastenings V-threads are mostly used but for transmitting power
Linear
motion
or motion square threads are more efficient than other types of threads. We
α will study about analysis of square threads. Fig.4a shows a view of a right
pitch hand square bolt within a nut. The bolt top carries a load W. When the bolt
Nut is given a moment M in the manner shown, it moves up sliding over the
threads of the nut. The lower face of the bolt thread is in contact with the
upper face of the nut. In practice the force given on the wrench handle to
rO rim
produce M may be considered as a force P acting at the mean radius, rm .
Screw area Nomenclature of screw: (i) Pitch (p) is the distance between two
consecutive threads. ii) Lead is the distance by which screw advances in
Fig. 4a
P = M / rm
one full turn, L = np , where, n is the number of starts, in case of multiple
threads. For n=1, L=p. (iii) ro , ri = outer and inner radius of screw.
(iv) rm = (ro + ri)/2, (v) Lead angle, α = tan-1 [ L /(2π rm )]
9
Analysis of screw:
A representative portion of screw and nut threads (blue and yellow
colour respectively) in a stretched manner is shown in Fig. 4b. The load W,
friction force F, and reaction R are assumed to be distributed uniformly
over the entire contact portion of screw and nut threads. In this figure the
summed up forces are shown. For upward motion of the screw, the
friction force on the screw thread is down the plane. The reaction R acts
at an angle φ (friction angle, tan-1μ) with the normal N, which is inclined
at α with the vertical. The resistive moment about the screw axis, MR = rm
x R x sin (α+φ). Therefore, external moment required to overcome the
frictional resistance for winding the screw upward is, MW = MR. The sense
of moment Mw or force P is shown in Fig. 4b.
W=R x cos (α+φ), therefore, R=W/cos (α+φ),
Hence, MW = rm x W x tan (α+φ).
10
For downward motion of the screw, the friction force on the screw
thread is up the plane, as shown in Fig.4c. The reaction R acts on left of N
at an angle φ. Here, (φ-α) is the angle of R with the vertical. The resistive
moment about the screw axis, MR = rm x R x sin (φ-α). Therefore, moment
required for unwinding the screw downward, Mun = MR. This means, the
screw will unwind only if an external moment Mun or force P is applied to
the screw, as shown in Fig.4c, otherwise the screw will remain in place.
This is called self-locking of screw. The condition for self-locking should be
that, φ > α. At the verge of slipping, φ = α. Here, Mun = rm x W x tan(φ-α)
For the condition, φ < α, from the Fig.4d, we can observe that R is on the
right of the vertical. Friction F acts up the plane and the resistive moment
about the screw axis, MR = rm x R x sin (α-φ). Direction of F and MR
indicates that the screw itself will unwind. Hence, to prevent unwinding,
always an external moment M or force P should act on the screw as
shown. This situation is undesirable for any screw.
11
Problem statement: The braking mechanism consists of two pined arms
and a square threaded screw with left and right-hand threads. If pitch
of the screw is 4 mm, mean diameter 12mm, and µ=0.35, determine
the tension in the screw when a torque of 3Nm is applied to the screw.
If the coefficient of friction between the brake pads A and B and the
circular shaft is 0.5, What is the maximum torque M the shaft can resist.
FBD 1
T FBD 1 shows the tensions in the screw and the moment shown is responsible to
create the tensions, but it is not applied after loading the screw.
μAX
AX M=(2T) rm tan (α+φ), α = tan-1 [ p /(2π rm )] = tan-1 (4 /12π ) = 6.060 ,
φ= tan-1 μ = 19.30 , Here, M = 3 Nm, Hence, T= 527.45 N
CX
FBD 2
FBD 2: we consider impending condition to get maximum friction force which each
CY
brake shoe can hold. Moment about C gives, Tx600 = Ax x300, Hence, Ax = 1055 N
FBD 3: Ax = Bx = 1055 N,
Considering moment about O, M= 200x0.5x(Ax + Bx ) or M= 211 Nm
FBD 3
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