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 heat-

ing) and wear in the contact zone. The nature of friction forces depends

on the contact conditions between the surfaces. Corresponding to the ab-

sence/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 fric-

tion.

1 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.

2

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 ≤ µsN Static condition (equality for impending slippage)

= µkN Sliding condition

where µs is the static friction coefficient, and µk is the kinetic friction coeffi-

cient. 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 direc-

tion may be unknown. They can be ascertained only by solving them using

the equations of equilibrium.

3

The primary origin of the frictional force is the microscopic surface as-

perities on the contacting surfaces. Resistance to free motion is developed

because the crests of one surface must move over the crests of the other sur-

face. 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.

2 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,

4

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

∑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 = µsN = 118.4 N. Since fmax < f ,

the block will slide down. Hence, the force of friction is f = µkN = 94.8 N

up the plane.

(b) F = 200 N

5

Considering force equilibrium

∑Fy = 0 ⇒ N − 50g cos 15◦ + 200 sin 20◦ = 0 ⇒ N = 405 N

∑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 = µsN = 101.3 N. Since f < fmax,

the block will remain in static equilibrium condition. Hence, the force of

friction is f = µkN = 61 N down the plane.

(c) F = 250 N

Considering force equilibrium

∑Fy = 0 ⇒ N − 50g cos 15◦ + 250 sin 20◦ = 0 ⇒ N = 388 N

∑Fx = 0 ⇒ −50g sin 15◦ − f + 250 cos 20◦ = 0 ⇒ f = 108 N

Maximum friction force possible: fmax = µsN = 97 N. Since f > fmax,

the block will move up, and the force of friction is f = µkN = 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 = µsN = 0.25N .

Considering force equilibrium

∑Fy = 0 ⇒ N − 50g cos 15◦ + F sin 20◦ = 0

∑Fx = 0 ⇒ −50g sin 15◦ − 0.25N + F cos 20◦ = 0

6

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:

7

of the 100 kg block, as shown in Figs. 6(b) and (c)

From Fig. 6(a)

∑Fy = 0 ⇒ T = mg cos 10◦. (1)

Case I: 100 kg block in impending slippage down the plane (Fig. 6(b))

∑Fy = 0 ⇒ R − 100g cos 20◦ = 0 ⇒ R = 922 N.

∑Fx = 0 ⇒ 2T + µsR − 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))

∑Fy = 0 ⇒ R − 100g cos 20◦ = 0 ⇒ R = 922 N.

∑Fx = 0 ⇒ 2T − µsR − 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.

8