Chapter 8 - Friction

Sections 8.1 - 8.4

Friction

Frictional forces resist movement of a rigid body over a rough surface.

It is assumed that the friction force acts parallel to the surface.

A distinction is made between frictional forces that act on an object that is not yet moving and an object that is moving.

Friction cont.

Static-friction force is the term for the friction force that exists before the object begins to move. This force can be counted as part of the equilibrium on a stationary rigid body.

Kinetic-friction force is the term for the friction force that exists on a rigid body that started to move.

Friction cont.

Both the static and kinetic friction forces are related to the normal force on the rigid body.

A normal force is by definition the support reaction force exerted by the supporting surface on the rigid body. It always acts perpendicular to the surface and up from the surface. (See Fig 8.1)

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Friction cont.

In equation form, the forces are: where:

Fm = sN

and

Fk = kN

•Fm is the maximum static friction force that exists just before the body begins to move.

s is the coefficient of static friction.

•N is the normal force.

Friction cont.

Fk is the kinetic friction force that exists once the object begins to move. Its value remains fairly constant as the velocity of the object increases.

k is the coefficient of kinetic friction.

Some typical values for s (coefficient of static friction) are given in Table 8.1.

Friction cont. Four possible situations for frictional forces

are shown in Fig 8.2. Of these, Case C (where motion is

impending) is probably the most important to a study of statics. For this case, an equilibrium free body diagram can be drawn because the body is not in motion. And the friction force is Fm which is equal to sN.

Angle of Friction The angle of static friction (s) is the angle

between a line perpendicular to the surface and the vector resultant of the normal force and the frictional force (see Fig’s 83(c) and 8.4(c)) when motion is impending on the rigid body.

Note from the geometry of Fig’s 8.3(c) and 8.4(c) that:

tan s = Fm/N = sN/N = μs

Angle of Friction cont.

Looking at Fig 8.4(c) it can be seen that the angle between the surface and the horizontal (angle ) is equal to s when motion is impending.

This angle is called the angle of repose. If this angle is increased any further, the block will begin to slide down the board!

Angle of Friction cont. Problems involving friction are solved by

setting up a free body diagram including all weights, external loads, normal reaction forces, and frictional forces. Break all forces into x and y components and use equilibrium equations to solve for unknowns. (ΣF = 0 + ΣM = 0). As long as there are no more than 3 unknown forces the problem should be statically determinate.

Angle of Friction cont. You will encounter 3 common types of

friction problems.(1) All applied forces are given and the coeff. of

friction s is also given. You must determine if the body will remain at rest or begin to slide! Solve the problem by setting up a FBD and solving for the frictional force and the normal force. Calculate Fm (Fm = μsN). If F is Fm, the body will not slide. If F is > Fm then the body will begin to move.

Angle of Friction cont.(2) All applied forces are given and it is stated that

motion is impending (F = Fm). You must find s. Solve the problem by finding the maximum friction force (Fm) and the normal force (N). Then

find s with Fm = sN.

(3) Given s and the fact that motion is impending (F = Fm) find the magnitude and direction of one applied force. Solve this problem with a FBD showing Fm with a sence opposite to that for impending motion.

Sample Problem 8.1A 300 lb block is on the inclined plane.

s = .25 and k = .2.

Determine whether the block is in equilibrium and the value of F.

FBD

300 lb

100 lb

y

x

F

N334

4

55

Sample Problem 8.1 cont.ΣFx = 0 = 100(4/5) - F(4/5) - N(3/5)

N = 100(4/5 x 5/3) - F(4/5 x 5/3)

N = 133.3 - 1.33 F

ΣFy = 0 = 100(3/5) - 300 + N(4/5) - F(3/5)

F = -240(5/3) + N(4/5 x 5/3)

F = -400 + 1.33 N

F = -400 + 1.33(133.3 - 1.33F)

F + 1.72 F = -400 + 173

2.73 F = -227

F = -83.1 lb

Sample Problem 8.1 cont.

N = 133.3 - 1.33(-83.1)

N = 133.3 + 110.6

N = 243.8 lb

Fm = sN

= .25(2438)

= 62 lb

since F > Fm the block will slide and

Factual =kN = .2(243.8) = 48.8 lb

Sample Problem 8.2If s = 0.35 and k = 0.25, determine the force P needed to (a) start the block moving up the incline, (b) to keep it moving, and (c) to prevent it from sliding down.

25°

25°

N

F

P

25°

800 N

y

x

Fm = sNFm = .35N

Sample Problem 8.2 cont.ΣFx = 0 = - P + F cos 25° + N sin 25°

P = .35 N(cos 25°) + N sin 25°

P = .317 N + .423 N

P = .74 N

ΣFy = 0 = -800 + N cos 25° - F sin 25°

800 LB = .906 N - (.35 N) x .42

N = 800/.758 = 1055 N

P = .74(1055) = 781 N

So when P exceeds 781 N the block will start to move.

Sample Problem 8.2 cont.Once the block starts moving:

Use the same equations, but F = kN = .25 N

ΣFx = 0 = - P + F cos 25° + N sin 25°

P = .25 N cos 25° + N sin 25°

P = .65 N

ΣFy = 0 = -800 N + N cos 25° - F sin 25°

800 = N cos 25° - .25 N sin 25°

800 = .8 N

N = 1000 N

P = .65 x 1000 = 650 N

Sample Problem 8.2 cont.To prevent the block from sliding down:

N

P

25°

w

F

Fm = sN

ΣFx = 0 = - P - F cos 25° + N sin 25°

P = -(.35 N)cos 25° + N sin 25°

P = -.317 N + .423 N

P = .106 N

ΣFy = 0 = -800 + N cos 25° + F sin 25°

800 = .906 N + .423(.35 N)

800 = 1.05 N

N = 759 N

P = .106(759)

P = 80.5 N

Solution 8.1

P

250 lb

s = 0.30k = 0.20

Given: = 30°, P = 50 lb

Find: Friction Force acting on block.

Solution 8.1 cont. 250 lb

xy

FN

50 lb

30°

30°

Assume Equilibrium

+ ΣFy = 0:

N - (250 lb)cos 30° - (50 lb)sin 30° = 0

N = +241.5 lb N = 241.5 lb

+ ΣFx = 0:

F - (250 lb)sin 30° + (50 lb)cos 30° = 0

F = +81.7 lb F = 81.7 lb

Solution 8.1 cont.

Maximum Friction Force:

Fm = sN = (0.30)(241.5 lb) = 72.5 lb

Since F > Fm, Block moves down

Friction Force: F = kN = (0.20)(241.5 lb) = 48.3 lb

F = 48.3 lb

Solution 8.2

P

250 lb

s = 0.30k = 0.20

Given: = 35°, P = 100 lb

Find: Friction Force acting on block.

Solution 8.2 cont. 250 lb

xy

FN

100 lb

35°

35°

Assume Equilibrium

+ ΣFy = 0:

N - (250 lb)cos 35° - (100 lb)sin 35° = 0

N = +262.15 lb N = 262.15 lb

+ ΣFx = 0:

F - (250 lb)sin 35° + (100 lb)cos 35° = 0

F = +61.48 lb F = 61.48 lb

Solution 8.2 cont.

Maximum Friction Force:

Fm = sN = (0.30)(262.15 lb) = 78.64 lb

Since F < Fm, Block is in equilibrium

Friction Force: F = 61.5 lb

Solution 8.3

800 N

25°

s = 0.20

k = 0.15P

Given: = 40°, P = 400 N

Find: Friction Force acting on block.

Solution 8.3 cont.400 N

800 N

NF

x

y

40°

25°

15°

25°

Assume Equilibrium

+ ΣFy = 0:

N - (800 N)cos 25° + (400 N)sin 15° = 0

N = +621.5 N N = 621.5 N

+ ΣFx = 0:

- F + (800 N)sin 25° - (400 N)cos 15° = 0

F = -48.28 N F = 48.28 N

Solution 8.3 cont.

Maximum Friction Force:

Fm = sN = (0.20)(621.5 N) = 124.3 N

Since F < Fm, Block is in equilibrium

F = 48.3 N