ME 201Engineering Mechanics: Statics

Unit 8

Friction

Friction

A retarding force that resists the relative movement

of two bodies in contact with each other

Always acts in oppose direction of motion – actual motion

or impending motion

Always acts parallel or tangent to surface(s) in contact

Magnitude is mainly dependent on surface roughness

(other contributing factors: temperature, molecular

adhesion, electrostatic attraction, lubrication, relative

velocities)

Friction Theory

Consider a block on a horizontal surface

No Frictional resistance

N is referred to as the Normal force, perpendicular to the

contact surface

W

N

Friction Theory

Next an external force is applied

If no friction, block would immediately move

If force is applied, F = P, for equilibrium

At some point, P > F, motion occurs

W

N

P

F

Friction Theory

Ratio of F to N is called the Coefficient of Static

Friction or μ

NF

s Where

μs = coefficient of static friction (unitless)

F – Max or limiting frictional force resistance (lb, k, N)

N – Normal force, perpendicular to contact surface (lb, k, N)

Friction Theory

Stated another way:

Fs = μs N (for impending motion)

Fk = μk N (for sustaining motion, kinetic)

In general:

F ≤ μ N

Coefficient of Static Friction on Dry Surfaces

Materials Coefficient of Static Friction or μ

Metal on metal 0.15 – 0.60

Metal on wood 0.20 – 0.60

Metal on stone 0.30 – 0.70

Metal on leather 0.30 – 0.60

Wood on wood 0.25 – 0.50

Wood on leather 0.25 – 0.50

Stone on stone 0.40 – 0.70

Earth on earth 0.20 – 1.00

Rubber on concrete 0.60 – 0.90

Example Problem - PullingGiven:

W = 400 lb.

μ = 0.4

θ = 30º

Find:

Force P required to cause motion to impend

W Pθ

Solution - PullingGiven:

W = 400 lb.μ = 0.4θ = 30º

Find:P for impending motion

Solution:1-FBD2-Eqn of Equil

F=μN

W Pθ

FN

0 yF 0 xF

NF 4.

030sin400 PN

NF 4005. NP

04.866. NP

04.30cos NP

030cos FP

NF 4.

Substituting

4005. NP

04.866. NP

Solve Simult. Eqn.

W Pθ

FN

lbP 150

lbN 325

lbF 130

Friction

A retarding force that resists the relative movement

of two bodies in contact with each other

Always acts in oppose direction of motion – actual motion

or impending motion

Always acts parallel or tangent to surface(s) in contact

Magnitude is mainly dependent on surface roughness

(other contributing factors: temperature, molecular

adhesion, electrostatic attraction, lubrication, relative

velocities)

Coefficient of Static Friction on Dry Surfaces

Materials Coefficient of Static Friction or μ

Metal on metal 0.15 – 0.60

Metal on wood 0.20 – 0.60

Metal on stone 0.30 – 0.70

Metal on leather 0.30 – 0.60

Wood on wood 0.25 – 0.50

Wood on leather 0.25 – 0.50

Stone on stone 0.40 – 0.70

Earth on earth 0.20 – 1.00

Rubber on concrete 0.60 – 0.90

Friction

Two types of friction

Dry or Coulomb friction

Static – surfaces are at rest with respect to each other

Kinetic/dynamic – surfaces are moving with respect to each other

Fluid Friction

Viscosity, friction developed between layers of a fluid moving at different velocities

We’ll limit our discussion to dry friction

Friction Theory

Consider a block on a horizontal surface

No Frictional resistance

N is referred to as the Normal force, perpendicular to the

contact surface

W

N

Friction Theory

Next an external force is applied

If no friction, block would immediately move

If force is applied, F = P, for equilibrium

At some point, P > F, motion occurs

W

N

P

F

Friction Theory

Ratio of F to N is called the Coefficient of Static

Friction or μ

NF

s Where

μs = coefficient of static friction (unitless)

F – Max or limiting frictional force resistance (lb, k, N)

N – Normal force, perpendicular to contact surface (lb, k, N)

Friction Theory

Stated another way:

Fs = μs N (for impending motion)

Fk = μk N (for sustaining motion, kinetic)

In general:

F ≤ μ N

Friction

Which of the following best explains the friction

force developed in a boulder at rest vs. a boulder

being acted on by a force:

A. The friction forces are equal

B. The friction force is greater in the left boulder

C. The friction force is greater in the right boulder

D. There is no friction force in either boulder

E. None of the above

WPθ

W

Example ProblemTo Push or to Pull, That is the Question

Given:

W = 400 lb.

μ = 0.4

θ = 30ºWP

θ

Find Force P required to cause motion to impend

W Pθ

Example Problem - Pushing

Given:

W = 400 lb.

μ = 0.4

θ = 30º

Find:

Force P required to cause motion to impend

WP

θ

PushingGiven:

W = 400 lb.

μ = 0.4

θ = 30º

Find:

Force P required to cause motion to impend

WPθ

Solution - PushingGiven:

W = 400 lb.μ = 0.4θ = 30º

Find:P for impending motion

Solution:1-FBD2-Eqn of Equil

F=μN

WPθ

0 yF

030cos FP

NF 4.

030sin400 PN

0 xF

NF 4005. NP

NF 4.04.30cos NP

04.866. NP

4005. NP

04.866. NP

Substituting

Solve Simult. Eqn.

WP

θ

FN

lbP 240

lbN 520

lbF 208

Example Problem - PullingGiven:

W = 400 lb.

μ = 0.4

θ = 30º

Find:

Force P required to cause motion to impend

W Pθ

PullingGiven:

W = 400 lb.

μ = 0.4

θ = 30º

Find:

Force P required to cause motion to impend

W Pθ

Solution - PullingGiven:

W = 400 lb.μ = 0.4θ = 30º

Find:P for impending motion

Solution:1-FBD2-Eqn of Equil

F=μN

W Pθ

FN

0 yF 0 xF

NF 4.

030sin400 PN

NF 4005. NP

04.866. NP

04.30cos NP

030cos FP

NF 4.

Substituting

4005. NP

04.866. NP

Solve Simult. Eqn.

W Pθ

FN

lbP 150

lbN 325

lbF 130

In Class Exercise – Up an InclineGiven:

W = 400 lb.

μ = 0.4

θ = 30º

Find:Force P required to cause motion to impend

W P

θ

lbP 6.338

Solution - Up an InclineGiven:

W = 400 lb.μ = 0.4θ = 30º

Find:P for impending motion

Solution:1-FBD2-Eqn of Equil

F=μN

400 lb P

θ

F

N

x’

y’

0' yF

030sin400 FP

θ

0' xF

030sin4006.138 P

NF 4.

030cos400 N

4.346*4.F

W P

θ

lbN 4.364

lbP 6.338

lbF 6.138

NF

In Class Exercise - Down an Incline

Given:

W = 400 lb.

μ = 0.4

θ = 30º

Find:

Force P required to cause motion to impend

W

Pθ

lbP 4.61

Solution - Down an InclineGiven:

W = 400 lb.μ = 0.4θ = 30º

Find:P for impending motion

Solution:1-FBD2-Eqn of Equil

F=μN

400 lb

Pθ

F

N

0' yF

x’

y’

θ

What does the “-” mean?

A “push” of at least 61.4 lb required

to keep block from sliding

0' xF

NF

030sin400 PF

NF 4.

030cos400 N

030sin4006.138 P

4.346*4.F

W

Pθ

lbN 4.346

lbF 6.138

lbP 4.61

In Class Exercise – Angle of Impending Motion

Given:

W = 400 lb.

μ = 0.4

Find:

θ such that it causes motion to impend

W

θ

8.21

Solution - Impending Motion

Given:μ = 0.4

Find:θ for impending motion

Solution:1-FBD2-Eqn of Equil

F=μN

W

θ

F

N

0' yFN

F

x’

y’

θ

Note that θ is independent of weight, dependent only on μ

sinWF NF

cosWN

0sin WF

4.tan

cos

sin

W

W

N

F

0cos WN

0' xF

tan

cos

sin

tan

Note the following

relationships:Dividing/Substituting:

8.21

W

θ