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Mechanics
Module III: Friction
Lesson 12: Friction - I
All real world (macroscopic) contacts/interactions between two bodies may
be considered to consist of a normal force (along the common normal at the
point of contact) and a tangential force perpendicular to the common normal.
The tangential force at the contact is known as the frictional force. This force
always opposes relative tangential sliding or impending sliding at the contact
point.
Friction is a necessary nuisance. It is necessary in brakes, clutches, nuts
and bolts, road-tyre interface etc. It is a nuisance in bearings, gear contacts,
power screws etc. It is responsible for energy loss (primarily through heating) and wear in the contact zone. The nature of friction forces depends
on the contact conditions between the surfaces. Corresponding to the absence/presence of a fluid layer between the surfaces, we have dry/viscous
friction. These two friction modes are qualitatively different. Here, we will
restrict our discussions to dry friction, which is also known as Coulomb friction.
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Friction Model and Mechanism
Figure 1:
Consider a rigid block of weight W on a flat horizontal table, as shown in
Fig. 1. Let a horizontal force P act on the block. It is observed that, as P is
increased from zero slowly, the block initially does not move. However, as P
reaches a certain critical value, the block just about exhibits a tendency to
move. Any further attempt to increase P results in acceleration of the block,
and a steady velocity can actually be maintained by a force P lower than the
critical value of P which initiated the motion. The variation of the friction
force f (see Fig. 1 ) with the applied load is shown in Fig. 2.
The main observations are as follows:
• The friction force f can exactly follow and cancel the applied force P
up to a limit fmax. Thus, the friction force under static conditions can
have any value between zero and fmax.
• When the body starts moving, the force attains a (practically) constant
value fk .
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Figure 2:
• The force of friction in the static and sliding conditions is found to be
independent of the area of contact between the block and the surface,
but proportional to the normal force N at the contact.
Based on the above observations and laws of friction due to Amonton,
Coulomb proposed a model for the friction force which may be summarized
as
f ≤ µs N
= µk N
Static condition (equality for impending slippage)
Sliding condition
where µs is the static friction coefficient, and µk is the kinetic friction coefficient. Normally µs ≥ µk .
It is important to note that, under static conditions, unless the impending
slippage condition is imperative at a contact, the friction force and its direction may be unknown. They can be ascertained only by solving them using
the equations of equilibrium.
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The primary origin of the frictional force is the microscopic surface asperities on the contacting surfaces. Resistance to free motion is developed
because the crests of one surface must move over the crests of the other surface. Microscopic adhesion/joining at points under high pressure can also
take place in static conditions. This contributes to the higher value of the
static friction force limit.
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Classification of friction problems
Friction problems may be classified as (a) simple, and (b) compound. In
simple friction problems, the direction of friction force and the impending
slippage condition if any are all known. This is not the case in compound
friction problems where multiple possibilities of motion can exist.
Figure 3:
Problem 1
A force F is used to pull a 50 kg block, as shown in Fig. 3. Determine the
magnitude and direction of the friction force when (a) F = 0, (b) F = 200 N,
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Figure 4:
and (c) F = 250 N. What is the minimum force F require to initiate motion
of the block up the plane?
Solution
The FBD of the block is shown in Fig. 4.
(a) F = 0
Considering force equilibrium
X
X
Fy = 0 ⇒ N − 50g cos 15◦ = 0 ⇒ N = 474 N
Fx = 0 ⇒ −50g sin 15◦ − f = 0 ⇒ f = −127 N
Thus, the friction force is up the plane.
Maximum friction force possible: fmax = µs N = 118.4 N. Since fmax < f ,
the block will slide down. Hence, the force of friction is f = µk N = 94.8 N
up the plane.
(b) F = 200 N
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Considering force equilibrium
X
Fy = 0 ⇒ N − 50g cos 15◦ + 200 sin 20◦ = 0 ⇒ N = 405 N
X
Fx = 0 ⇒ −50g sin 15◦ − f + 200 cos 20◦ = 0 ⇒ f = 61 N
The friction force is down the plane.
Maximum friction force possible: fmax = µs N = 101.3 N. Since f < fmax,
the block will remain in static equilibrium condition. Hence, the force of
friction is f = µk N = 61 N down the plane.
(c) F = 250 N
Considering force equilibrium
X
Fy = 0 ⇒ N − 50g cos 15◦ + 250 sin 20◦ = 0 ⇒ N = 388 N
X
Fx = 0 ⇒ −50g sin 15◦ − f + 250 cos 20◦ = 0 ⇒ f = 108 N
Maximum friction force possible: fmax = µs N = 97 N. Since f > fmax,
the block will move up, and the force of friction is f = µk N = 77.6 N down
the plane.
(d) Minimum force to initiate motion up the plane
In this case, the applied force F should be able to overcome the static friction
force limit fmax = µs N = 0.25N .
Considering force equilibrium
X
X
Fy = 0 ⇒ N − 50g cos 15◦ + F sin 20◦ = 0
Fx = 0 ⇒ −50g sin 15◦ − 0.25N + F cos 20◦ = 0
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Solving the above equations simultaneously, F = 239 N, and N = 392 N.
Thus, the minimum force Fmin = 239 N, and the corresponding value of the
friction force is f = 0.25(392) = 98 N.
Figure 5:
Problem 2
Determine the range of mass m for which the arrangement shown in Fig. 5
is in equilibrium.
Solution
There can be two cases for static equilibrium in impending slippage condition
Figure 6:
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of the 100 kg block, as shown in Figs. 6(b) and (c)
From Fig. 6(a)
X
Fy = 0 ⇒ T = mg cos 10◦.
(1)
Case I: 100 kg block in impending slippage down the plane (Fig. 6(b))
X
X
Fy = 0 ⇒ R − 100g cos 20◦ = 0 ⇒ R = 922 N.
Fx = 0 ⇒ 2T + µs R − 100g sin 20◦ = 0 ⇒ T = 29.5 N.
Hence, from (1), m = 3.05 kg.
Case II: 100 kg block in impending slippage up the plane (Fig. 6(c))
X
X
Fy = 0 ⇒ R − 100g cos 20◦ = 0 ⇒ R = 922 N.
Fx = 0 ⇒ 2T − µs R − 100g sin 20◦ = 0 ⇒ T = 306 N.
Hence, from (1), m = 31.7 kg.
Thus, the arrangement will be in static equilibrium for 3.05 kg ≤ m ≤ 31.7
kg.
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