CHAPTER 8

FRICTION

• In preceding chapters, it was assumed that surfaces in contact were either frictionless (surfaces could move freely with respect to each other) or rough (tangential forces prevent relative motion between surfaces).

• Actually, no perfectly frictionless surface exists. For two surfaces in contact, tangential forces, called friction forces, will develop if one attempts to move one relative to the other.

• However, the friction forces are limited in magnitude and will not prevent motion if sufficiently large forces are applied.

• The distinction between frictionless and rough is, therefore, a matter of degree.

• There are two types of friction: dry or Coulomb friction and fluid friction. Fluid friction applies to lubricated mechanisms. The present discussion is limited to dry friction between nonlubricated surfaces.

8.1 Introduction

Example of friction

Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body.

Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion.

Dry Friction

For the body shown in the figure to be in equilibrium, the following must be true:

F = P, N = W, and Wx = Ph.

Impending motion

FS = sN Fk = kN

Motion

coefficient of kinetic frictioncoefficient of static friction

The maximum friction force is attained just before the block begins to move (a situation that is called “impending motion”). The value of the force is found using Fs = s N, where s is called the coefficient of static friction. The value of s depends on the

materials in contact.

• Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N.

• Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force.

• As P increases, the static-friction force F increases as well until it reaches a maximum value Fm.

NF sm

• Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk.

NF kk

8.2 The Law of Dry Friction, Coefficients of Friction

• Maximum static-friction force:NF sm

• Kinetic-friction force:

sk

kk NF

75.0

• Maximum static-friction force and kinetic-friction force are:

- proportional to normal force

- dependent on type and condition of contact surfaces

- independent of contact area

• Four situations can occur when a rigid body is in contact with a horizontal surface:

• No friction,(Px = 0)

• No motion,(Px < Fm)

• Motion impending,(Px = Fm)

• Motion,(Px > Fm)

• It is sometimes convenient to replace normal force N and friction force F by their resultant R:

• No friction • Motion impending• No motion

ss

sms N

N

N

F

tan

tan

• Motion

kk

kkk N

N

N

F

tan

tan

8.3 Angles of Friction

• No friction

• No motion • Motion impending

• Motion

Another examples will show how the angle of friction can be used to advantage in the analysis of certain types of problems.

• All applied forces known

• Coefficient of static friction is known

• Determine whether body will remain at rest or slide

• All applied forces known

• Motion is impending

• Determine value of coefficient of static friction.

• Coefficient of static friction is known

• Motion is impending

• Determine magnitude or direction of one of the applied forces

8.4 Problems Involving Dry Friction

Pushing the uniform crate that has a weight W and sits on the rough surface.

Slip, tip over, or static?

Example

Verge of slipping Tip over

Relation with moment

Relation with moment

80 N

200 N

N = 200 N

μs = 0.3

Fmax = (0.3) (200 N)

= 60 [N]

F

+Fx = 0

80N – F = 0 F = 80 [N]

+ Fy = 0

-200 N + N = 0 N = 200 [N]

80 N

200 N

N = 200 N

F

x

0.2 m

0.4 m

O

+

0)x(200)4.0(80

;0

OM

x = 0.16 m

Relation with moment

80 N

200 N

N = 200 N

μs = 0.3

Fmax = (0.3) (200 N)

= 60 [N]

F

+Fx = 0

80N – F = 0 F = 80 [N]

+ Fy = 0

-200 N + N = 0 N = 200 [N]

80 N

200 N

N = 200 N

F

x

0.2 m

0.6 m

O

+

0)x(200)6.0(80

;0

OM

x = 0.24 m

tip over

Sample problem 8.1

1350N

450 N

A 450 N force acts as shown on a 1350N block placed on a inclined plane.

The coefficients of friction between the block and the plane are μs = 0.25 and

μk = 0.20.Determine whether the block is in equilibrium, and find the value of the frictiom force.

Solution

1350N

450N

Force required for equilibrium.

We first determine the value of the friction force required to maintain equilibrium. Assuming that F is directed down and to the left, we draw the free-body diagram of the block and write

The force F required to maintan equilibrium is an 360N force directed up and to the right ; the tendency of the block is thus to move down the plane.

1350N

450 N

1350N

450N

Maximum Friction force.

The magnitude of the maximum friction force can be developed is

Since the value of the force required to maintain equilbrium (360N)is larger than the maximum value which cn be obtained (270 N), equilibrium will not be maintained and the block will slide down the plane.

1350N

450 N

1350 N

450 N

N = 1080 N

F = 216 N

Actual Value of Friction Force.

The magnitude of the actual friction force is obtained as follows :

The sence of this force is opposite to the sense of motion ;the force is thus directed up and to the right :

Factual = 216 N

It should be noted that the forcee is opposite to the sense of motion ;the force

1350N

450 N

Sample problem 8.2

A support block is acted upon by two forces as shown. Knowing that the coefficient of friction

between the block and the incline are μs = 0.35 and μk = 0.25, determine

(a) the force P for which motion of the block up the incline is impending,(b) the friction force when the block is moving up(c) the smallest force P required to prevent the block from sliding down

Free Body Diagram.

For each part of the problem we draw a free-body diagram of the block and a force triangle includingthe 800N vertical force, the horizontal force P, and the force R exerted on the block by the incline.The direction of R must be determine in each separate case. We note that since P is perpendicular to the 800N force, the force triangle is a right triangle, which can easily be solved for P. In most other problems, however, the force triangle will be an oblique triangle and should be solved by applying the law of sines.

Solution

a. Force P for impending Motion of the Block Up the Incline

P = (800 N) tan 44.29º

= 780 N

b. Friction Force F when the Block Moves Up the incline

R = (800 N) / cos 39.04º = 1029.9 N

F = R sin Φk = (1029.9 N) sin 14.04º

F = 250 N

c. Force P for impending Motion of the Block Down the Incline

P = (800 N) tan 5.71º

= 80 N

Sample problem 8.3

150mm

75mm

The moveable bracket shown may be placed at any height on the 75mm diameter pipe.

If the coefficient of friction between the pipe and bracket is 0.25,

determine the minimum distance x at which the load can be supported.

Neglect the weight of the bracket.

SOLUTION:• When W is placed at minimum x, the bracket is about to slip and friction

forces in upper and lower collars are at maximum value.

BBsB

AAsA

NNF

NNF

25.0

25.0

• Apply conditions for static equilibrium to find minimum x. :0 xF 0 AB NN AB NN

:0 yF

WN

WNN

WFF

A

BA

BA

5.0

025.025.0

0

WNN BA 2

:0 BM

05.1275.026

05.125.036

0in.5.1in.3in.6

xWWW

xWNN

xWFN

AA

AA

in.12x

150 mm

75 mm

150 mm

75 mm

37.5 mm

NA (150 mm) – FA (75mm) – W (x - 37.5 mm) = 0150 NA – 75 ( 0.25NA ) – Wx + 37.5W = 0150 (2W) – 18.75 (2W) – Wx + 37.5W = 0

x = 300 mm

PROBLEMS

Problem 8.1

Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ= 30º and P = 200 N

Free Body Diagram

Solution

Problem 8.4

Determine whether the 9 kg block shown is in equilibrium, and findthe magnitude and direction of the friction force when P = 60 N and θ = 15º

SolutionFree Body Diagram

Problem 8.11 and 8.12

The coefficients of friction are μs = 0.40 and μk = 0.30between all surface of contact.Determine the forces P for which motion of the 27kg block is impending if cable AB

a) is attached as shownb) is removed

18kg

27kg

18kg

27kg

18kg

27kg

18kg

27kg

THE END