Lecture 8

Static friction and Kinetic or Sliding friction

Friction

Friction is caused by the “microscopic” interactions between the two surfaces

Friction...

Force of friction acts to oppose motion: Parallel to a surface Perpendicular to a NNormal force.

maFF

ffF mgg

NN

ii

j j

Static and Kinetic Friction

Friction exists between objects and its behavior has been modeled.

At Static Equilibrium: A block, mass m, with a horizontal force F applied,

Direction: A force vector to the normal force vector N N and the vector is opposite to the direction of acceleration if were 0.

Magnitude: f is proportional to the applied forces such that

fs ≤ s N

s called the “coefficient of static friction”

Friction: Static frictionStatic equilibrium: A block with a horizontal force F applied,

As F increases so does fs

Fm

1

FBD

fs

N

mg

Fx = 0 = -F + fs fs = F

Fy = 0 = - N + mg N = mg

Static friction, at maximum (just before slipping)

Equilibrium: A block, mass m, with a horizontal force F applied,

Direction: A force vector to the normal force vector N N and the vector is opposite to the direction of acceleration if were 0.

Magnitude: fS is proportional to the magnitude of N

fs = s N F

m fs

N

mg

Kinetic or Sliding friction (fk < fs)

Dynamic equilibrium, moving but acceleration is still zero

As F increases fk remains nearly constant

(but now there acceleration is acceleration)

Fm

1

FBD

fk

N

mg

Fx = 0 = -F + fk fk = F

Fy = 0 = - N + mg N = mg v

fk = k N

Sliding Friction: Quantitatively

Direction: A force vector to the normal force vector N N and the vector is opposite to the velocity.

Magnitude: ffk is proportional to the magnitude of N N

ffk = k N N ( = Kmg g in the previous example)

The constant k is called the “coefficient of kinetic friction”

Logic dictates that S > K for any system

Static Friction with a bicycle wheel

You are pedaling hard and the bicycle is speeding up.

What is the direction of the frictional force?

You are breaking and the bicycle is slowing down

What is the direction of the frictional force?

Coefficients of Friction

Material on Material s = static friction k = kinetic friction

steel / steel 0.6 0.4

add grease to steel 0.1 0.05

metal / ice 0.022 0.02

brake lining / iron 0.4 0.3

tire / dry pavement 0.9 0.8

tire / wet pavement 0.8 0.7

An experiment

Two blocks are connected on the table as shown. The

table has unknown static and kinetic friction coefficients.

Design an experiment to find S

Static equilibrium: Set m2 and add mass to m1 to reach the breaking point.

Requires two FBDs :

m1

m2

m2g

N

m1g

T

T

Mass 2

Fx = 0 = -T + fs = -T + S N

Fy = 0 = N – m2g

fS

Mass 1

Fy = 0 = T – m1g

T = m1g = S m2g S = m1/m2

A 2nd experiment

Two blocks are connected on the table as shown. The

table has unknown static and kinetic friction coefficients.

Design an experiment to find K.

Dynamic equilibrium: Set m2 and adjust m1 to find place when

a = 0 and v ≠ 0

Requires two FBDs

m1

m2

m2g

N

m1g

T

T

Mass 2

Fx = 0 = -T + ff = -T + k N

Fy = 0 = N – m2g

fk

Mass 1

Fy = 0 = T – m1g

T = m1g = k m2g k = m1/m2

An experiment (with a ≠ 0)

Two blocks are connected on the table as shown. The

table has unknown static and kinetic friction coefficients.

Design an experiment to find K.

Non-equilibrium: Set m2 and adjust m1 to find regime where a ≠ 0

Requires two FBDs

T

Mass 2

Fx = m2a = -T + fk = -T + k N

Fy = 0 = N – m2g

m1

m2

m2g

N

m1g

T

fk

Mass 1

Fy = m1a = T – m1g

T = m1g + m1a = k m2g – m2a k = (m1 (g + a)+m2a)/m2g

Inclined plane with “Normal” and Frictional Forces

Weight of block is mg

NormalForce

Friction ForceSliding Down

“Normal” meansperpendicular

Note: If frictional Force = Normal Force (coefficient of friction)Ffriction = Fnormal = mg sin then zero acceleration

1. At first the velocity is v up along the slide

2. Can we draw a velocity time plot?

3. What the acceleration versus time?

v

mg sin

fk Sliding

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