Ch. 4: Friction

4.0 OutlineIntroductionTypes of FrictionDry Friction

4.0 Outline

Ch. 4: Friction

4.1 Introduction

4.1 Introduction

In real situation, the forces of action and reactionbetween contacting surfaces have their componentsboth in the tangential and normal directions to thecontacting surface. Tangential forces are known asFriction forces. Whenever a tendency exists for onecontacting surface to slide along another surface,the friction forces developed are always in a directionto oppose this tendency.In some systems, friction is undesirable since it normallyspoils the required behavior. But in many situations,friction functions the systems.In real case where sliding motion between parts occurs,the friction forces result in a loss of energy.

Ch. 4: Friction

4.2 Types of Friction

4.2 Types of Friction

a) Dry (Coulomb) friction when unlubricated surfacesare in contact under a condition of slidingor tendency to slide.Friction force tangent to the surfaces of contact isdeveloped both during the interval leading up toimpending slippage and while slippage takes place.Its direction always opposes the motion or impendingmotion which would occur if no friction were present.b) Fluid frictionc) Internal friction

Ch. 4: Friction

4.3 Dry Friction

4.3 Dry Friction

Mechanism of friction

Ch. 4: Friction

maxF F<

max sF F Nμ= =

4.3 Dry Friction

Three regions of static motion transition

a) No motion is the region up to the point of slippageor impending motion. Friction force is determinedby the equations of equilibrium becausethe system is in equilibrium. When the motion is notimpending,

b) Impending motion is the moment where the bodyis on the verge of slipping. Static friction force reachesthe max value. For a given pair of mating surfaces,

.c) Motion The body starts moving in the direction

of the applied force. Here, friction force drops to alower value called kinetic friction .It will drop further with higher velocity.

kF Nμ=

Ch. 4: Friction

tan F/Nα =s stanφ μ=

k ktanφ μ=

4.3 Dry Friction

Friction cone Friction coefficient reflects the roughnessof a pair of mating surfaces. The smaller thecoefficient value, the smoother the surfaces.Direction of resultant R is specified by .When the friction force reaches max value, .When slippage occurs, .

The friction angle defines the limiting position ofthe total reaction force R. The friction cone ofvertex angle represents the locus of possiblepositions for the reaction force R.

Friction force is independent of the apparent or projected area of contact.

s k, φ φ

s k2 , 2φ φ

Ch. 4: Friction

4.3 Dry Friction

Friction cone

Ch. 4: Friction

sF Nμ=

4.3 Dry Friction

Types of dry friction problems First step is to identifywhich of these categories applies.

1) Condition of impending motion is known to existThe body is in equilibrium and on the verge of slipping.Friction force is the max static friction

2) Relative motion is known to existFriction force is the kinetic friction

3) Unknown status of the problemAssume static equilibrium and solve for the requiredfriction force F. Then check and conclude the status.

kF Nμ=

Ch. 4: Friction

sNμsF Nμ>

sF Nμ=

4.3 Dry Friction

Possible outcomes

a) friction force for the assumed equilibriumcan be provided and so the body is in static equilibrium.

b) max friction force is required for thestatic equilibrium condition and so motion impends.

c) surfaces cannot support more friction than. So the equilibrium assumption is invalid

and motion occurs instead. Friction forceis the kinetic friction . Even with the correctkinetic friction substituted, equilibrium equationsare still not hold accelerated motion

sF Nμ<

kF Nμ=

Ch. 4: Friction

4.3 Dry Friction

SP. 6/1 Determine the max angle θwhich the adjustableincline may have with the horizontal before theblock of mass m begins to slip. The coefficientof static friction between the block and theinclined surface is μs.

Ch. 4: Friction

y

x s

1s s

F 0 N mgcos 0

F 0 N mgsin 0

tan or tan

θ

μ θ

μ θ θ μ−

⎡ ⎤= − =⎣ ⎦⎡ ⎤= − =⎣ ⎦

= =

∑∑

4.3 Dry Friction

SP. 6/1

at the moment of slipping, friction is upward sF Nμ=

when the friction force reaches max value, s stanφ μ=by equilibrium, R = W and sφ θ= 1

stanθ μ−∴ =

sφ

R

Ch. 4: Friction

4.3 Dry Friction

SP. 6/2 Determine the range of values which the massmo may have so that the 100 kg block shownin the figure will neither start moving up theplane nor slip down the plane. The coefficientof static friction for the contact surface is 0.30.

Ch. 4: Friction

yF 0 N 100gcos20 0, N 922 N⎡ ⎤= − = =⎣ ⎦∑

4.3 Dry Friction

SP. 6/2

bounded mo values block start moving

Case I: max mo, start moving up, friction downward

Case II: min mo, start moving down, friction upward

sF Nμ=

x o s oF 0 m g N 100gsin20 0, m 62.4 kgμ⎡ ⎤= − − = =⎣ ⎦∑

x o s o

o max

F 0 m g N 100gsin20 0, m 6.0 kg

6.0 m 62.4 kg and F F 277 N up/downward

μ⎡ ⎤= + − = =⎣ ⎦∴ ≤ ≤ ≤ =∑

Ch. 4: Friction

4.3 Dry Friction

SP. 6/3 Determine the magnitude and direction of thefriction force acting on the 100 kg block shownif, first, P = 500 N and, second, P = 100 N. Thecoefficient of static friction is 0.20, and thecoefficient of kinetic friction is 0.17. The forceare applied with the block initially at rest.

Ch. 4: Friction

y

s

x s

F 0 N 500sin 20 100gcos20 0, N 1092.85 N

max supportable friction N 218.6 N

F 0 500cos20 F 100gsin20 0, F 134.3 N N

the assumption is valid

μ

μ

⎡ ⎤= − − = =⎣ ⎦= =

⎡ ⎤= − − = = <⎣ ⎦∴

∑

∑

4.3 Dry Friction

SP. 6/3

don’t know if the block is impending or is moving assume static equilibriumP = 500 N: assume the block tends to move up friction downward

P = 100 N: assume the block tends to slide down friction upwardy

s

x s

F 0 N 100sin 20 100gcos20 0, N 956.04 N

max supportable friction N 191.21 N

F 0 F 100cos 20 100gsin20 0, F 241.55 N N

the assumption is invalid, block is moving downwardkinetic f

μ

μ

⎡ ⎤= − − = =⎣ ⎦= =

⎡ ⎤= + − = = >⎣ ⎦∴

∑

∑

kriction upward N 162.5 Nμ= =

Ch. 4: Friction

4.3 Dry Friction

SP. 6/4 The homogeneous rectangular block of mass m, width b, andheight H is placed on the horizontal surface and subjected to ahorizontal force P which moves the block along the surface witha constant velocity. The coefficient of kinetic friction betweenthe block and the surface is μk. Determine (a) the greatestvalue that h may have so that the block will slide without tippingover and (b) the location of a point C on the bottom face of theblock through which the resultant of the friction and normalforces acts if h = H/2.

Ch. 4: Friction

ktanθ μ=

4.3 Dry Friction

SP. 6/4

a) On the verge of tipping over, reaction acts at the corner AWhen slippage occurs,Block moves w/ const. velocity equilibriumThree-force member: reaction at A must pass through B

( )k ktan b/2h, h b/ 2θ μ μ= = ∴ =

b) When slippage occurs,Block moves w/ const. velocity equilibriumThree-force member: reaction at C must pass through G

ktanθ μ=

( )k ktan x/ H/2 , x H/2θ μ μ= = ∴ =

Ch. 4: Friction

4.3 Dry Friction

SP. 6/5 The three flat blocks are positioned on the 30°incline as shown,and a force P parallel to the incline is applied to the middle block.The upper block is prevented from moving by a wire whichattaches it to the fixed support. The coefficient of static frictionfor each of the three pairs of mating surfaces is shown. Determinethe maximum value which P may have before any slippingtakes place.

Ch. 4: Friction

4.3 Dry Friction

SP. 6/5

Ch. 4: Friction

y 1 1

2 1 2

3 2 3

F 0 N 30gcos30 0, N 254.87 N

N N 50gcos30 0, N 679.66 N N N 40gcos30 0, N 1019.5 N

⎡ ⎤= − = =⎣ ⎦− − = =− − = =

∑

1 s 1 2 s 2

3max s 3

2 3 3 3max

F N 76.46 N, F N 271.86 NF N 458.8 Nblock #3: F F 40gsin30 0, F 468.06 N F

block #3 cannot stay still -- the assumption is invalid

μ μμ

= = = =

= =

− + = = >∴

4.3 Dry Friction

SP. 6/5

Since 30 kg-block cannot slide and 50 kg-block is pulled, 50 kg-block tendsto move and only 2 cases are possible. Either 50 kg-block alone or50&40 kg-blocks move together.

50 kg-block tends to move alone F1 & F2 max (either one alone will not slip)

50&40 kg-blocks tend to move together F1 & F3 max (either one alone will not slip)

2 3 2 2max

1 2

block #3: F F 40gsin30 0, F 262.6 N Fblock #2 & #3 does not slip relative to each other

block #2: P F F 50gsin30 0, P 93.8 N

− + = = <∴

− − + = =

Ch. 4: Friction

4.3 Dry Friction

P. 6/7 The light bar is used to support the 50 kg blockin its vertical guides. If the coefficient of staticfriction is 0.30 at the upper and 0.40 at the lowerend of the bar, find the friction force acting ateach end for x = 75 mm. Also find the maximumvalue of x for which the bar will not slip.

Ch. 4: Friction

1 1A Btan 21.8 , tan 16.7φ μ φ μ− −= = ° = = °

( )1B Asin 75 / 300 14.5θ φ φ−= = ° < <

4.3 Dry Friction

P. 6/7 Bar is a two-force member.Assume the system is in equilibrium.Hence the reaction forces at both endsact along the axial direction.

at x = 75 mm:

limitation of the reaction force on each endy

R inside the static friction cone, system is in equilibrium and F Ntan 126.6 Nθ= =

yF 0 N 50g 0, N 490.5 N⎡ ⎤= − = =⎣ ⎦∑

max x before slipping when the bar angle = that of small friction cone

Bx/300 sin , x 86.2 mmφ= =

R

RN

N

F

F

Ch. 4: Friction

4.3 Dry Friction

P. 6/32 Find the tension in the cable and force Pthat makes the 15 kg lower block(a) to start sliding downward(b) to start sliding upward

Ch. 4: Friction

1

2 1 2

1max 1

2max 2

N 8gcos20 73.75 NN N 15gcos20 0, N 212 NF 0.3N 22.12 NF 0.4N 84.81 N

= =− − = == == =

1max 2max

1max

P F F 15gsin20 0, P 56.6 NF 8gsin20 T 0, T 49 N− − + = =

+ − = =

4.3 Dry Friction

P. 6/32

a) pulling down, 15 kg block impends to slide downward

15g

8g

F1

F1

N1

N1

T

P

F2N2

Ch. 4: Friction

1

2 1 2

1max 1

2max 2

N 8gcos20 73.75 NN N 15gcos20 0, N 212 NF 0.3N 22.12 NF 0.4N 84.81 N

= =− − = == == =

4.3 Dry Friction

P. 6/32

b) pushing up, assume 15&8 kg blocks impends to slide upward togetherthe cable slacks T=0

1 1 1max

1max 2max

1max

8 kg block: 8gsin20 F 0, F 26.84 N F15 kg block impends to slide upward aloneP F F 15gsin20 0, P 157.3 NT F 8gsin20 0, T 4.72 N

− = = >

∴− + + + = =

− − + = =15g

8g

F1

F1N1

N1

T

PF2

N2

Ch. 4: Friction

4.3 Dry Friction

P. 6/34 The uniform slender rod of mass m and length Lis initially at rest in a centered horizontal positionon the fixed circular surface of radius R = 0.6L.If a force P normal to the bar is gradually appliedto its end until the bar begins to slip at the angleθ= 20°, determine the coefficient of static friction.

Ch. 4: Friction

[ ]a/r a 20 R R/9180πθ π⎛ ⎞= = =⎜ ⎟

⎝ ⎠i

4.3 Dry Friction

P. 6/34

no slip until θ=20°distance on bar = length on curve

( )( )s

L/2 R/9tan F/N 0.211

L/ 2tan20π

μ α−

= = = =

20°L/(2tan20)

πR/9

α

α

Ch. 4: Friction

4.3 Dry Friction

P. 6/37 The three identical rollers are stacked on ahorizontal surface as shown. If the coefficientof static friction μs is the same for all pairsof contacting surfaces, find the minimum valueof μs for which the rollers will not slip.

Ch. 4: Friction

O A B

A B

Amax Bmax A B

A Amax B

M 0 F F

from the figure, N NF F so F reaches the limit value before Fslipping does occur first at contact AF F and F determined by equilibrium equation

⎡ ⎤= =⎣ ⎦<

∴ <

∴∴ =

∑

4.3 Dry Friction

P. 6/37

lower roller: three-force memberreaction force at A must pass through contact Bfrom geometry,

Lower roller tends rolling out at upper contactwhile tends to slide out at lower contact

condition: one or more contacts impend to slip

FBD: lower left roller (three-force member)

Amax sF F , tan tan15 0.268α μ= ∴ = = =∵

A

BNB

NAFA

FB

mg

O

Rα

30° 15°

r

r A

B

O

Ch. 4: Friction

4.3 Dry Friction

P. 6/43 The industrial truck is used to move the solid1200 kg roll of paper up the 30°incline. If thecoefficients of static and kinetic friction betweenthe roll and the vertical barrier of the truck andbetween the roll and the incline are both 0.40,compute the required tractive force P betweenthe tires of the truck and the horizontal surface.

Ch. 4: Friction

4.3 Dry Friction

P. 6/43 To move the paper roll, 3 possibilities1) A and B both slip2) only B slips3) only A slipsafter calculation, only case 3) is viable

A

O B A

x A B B B A

y A B B

A B B B

slipping at A, F 0.4N

M 0 F 0.4N

F 0 N F cos30 N sin 30 0, N 1.307N

F 0 0.4N 1200g F sin30 N cos30 0

N 22.1 kN, N 28.9 kN, F 8853 N 0.4N

=

⎡ ⎤= =⎣ ⎦⎡ ⎤= − − = =⎣ ⎦⎡ ⎤= − − − + =⎣ ⎦

= = = <

∑∑∑A O

B

NB

FBNA

0.4NA

1200g