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CHAPTER 6 Expected Outcome: Able to analyze and solve problems involving dry friction using the concept of equilibrium of a particle or a rigid body FRICTION

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Page 1: CHAPTER 6ocw.ump.edu.my/pluginfile.php/1155/mod_resource... · CHAPTER 6 Expected Outcome: • Able to analyze and solve problems involving dry friction using the concept of equilibrium

CHAPTER 6

Expected Outcome:

• Able to analyze and solve problems involving dry friction using the concept of equilibrium of a particle or a rigid body

FRICTION

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Application

8 - 2

Friction is both problematic and useful in many engineering applications, such as in tires and brakes.

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Introduction

8 - 3

• 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.

• Types of friction:

• dry or Coulomb friction

• fluid friction

In this chapter, only dry friction will be consider

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The Laws of Dry Friction. Coefficients of Friction

8 - 4

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

• Small horizontal force P is applied. For the 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 Fincreases as well until it reaches a maximum value Fm. S is coefficient of static friction.

NF sm

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

NF kk

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The Laws of Dry Friction. Coefficients of Friction

8 - 5

• 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

Page 6: CHAPTER 6ocw.ump.edu.my/pluginfile.php/1155/mod_resource... · CHAPTER 6 Expected Outcome: • Able to analyze and solve problems involving dry friction using the concept of equilibrium

The Laws of Dry Friction. Coefficients of Friction

8 - 6

• No friction,(Px = 0)

• No motion,(Px < F)

• Motion impending,(Px = Fm)

• Motion,(Px > Fm)

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Angles of Friction

8 - 7

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

• No friction • No motion • Motion

kk

kkk N

NNF

tan

tan

• Motion impending

ss

sms N

NN

F

tan

tan

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Angles of Friction

8 - 8

• Consider block of weight W resting on board with variable inclination angle q.

• No friction • No motion • Motion impending

• Motion

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Problems involving Dry Friction

8 - 9

• 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

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Sample Problem 8.1

8 - 10

A 100-N force acts as shown on a 300-N block placed on an inclined plane. The coefficients of friction between the block and plane are s = 0.25 and k= 0.20. Determine whether the block is in equilibrium and find the value of the friction force.

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8 - 11

1. Draw the FBD of the system2. Determine values of friction force and normal reaction force from plane required to maintain equilibrium.

:0 xF 0N 300 -N 100 53 F

N 80F

:0 yF 0N 300 54 N

N 240N3. Calculate maximum friction force and compare with friction force required for equilibrium.

N 60N 240250 .NF sm

What does the sign tell you about the assumed direction of impending motion?Answer: (-ve) sign shows that your first assumption on the force direction is wrong

The block will slide down the plane as the Fmis smaller than the force required to maintain the block in equiblirium

4. What does this solution imply about the block?

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Wedges

8 - 12

• Wedges - simple machines used to raise heavy loads.

• Force required to lift block is significantly less than block weight.

• Friction prevents wedge from sliding out.

• Want to find minimum force P to raise block.

• Block as free body

0

:00

:0

21

21

NNW

FNN

F

s

y

s

x

Fx 0 :

sN2 N3 s cos6 sin6 P 0

Fy 0 :

N2 N3 cos 6s sin6 0

• Wedge as free body

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2 - 13

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2 - 14

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References:

1. Beer, Ferdinand P.; Johnston, E. Russell; “Vector Mechanicsfor Engineers - Statics”, 8th Ed., McGraw-Hill, Singapore,2007.