(2010-11-03) Newtonian coefficients of sliding friction
(dry friction)
The static coefficient of friction is slightly larger than the kinetic one.

When two dry solids at rest press against each other
they won't move as long as the force applied tangentially
to their common surface
of contact does not exceed a certain limit, which is
roughly proportional to the normal force that
presses them against each other.

The coefficient of proportionality is called the
static coefficient of friction and is often denoted by the
symbol m or
m_{s} (the Greek letter mu
is pronounced mew ).

As soon as such solids start moving, a lesser friction force appears in the
direction which opposes the motion. It is also proportional to the
normal force applied but the coefficient of proportionality
(called kinetic coefficient of friction or
sliding coefficient of friction) is always
less than in the static case described above.
This means that, unless the normal force changes, a constant tangential
force large enough to overcome static friction will impart motion
at a positive (nonzero) acceleration.

By definition, the "tangential force" is in the same plane as the surface
of contact and the "normal force" is perpendicular to it.

The above is a valid approximation for dry surfaces under moderate pressure.
At high pressure, seizure can occur, whereby the two solids
will not slide at all under any reasonable force.
This happens, in particular, for polished surfaces between solids consisting of
the same material: Pressure makes the interface between the two solids
effectively disappear and they may well behave as a single crystal,
with large binding forces between nearby atoms.
The solids may well break before they slide.

When a liquid lubricant is used, the above does not apply at all.
Instead, the slightest tangential force will produce sliding and the
resistive forces depend on the speed of sliding
(the normal forces are then largely irrelevant).

Gingerbread (Yahoo!
2011-03-05)
Static or kinetic regime?_{ } In the figure at right,^{ }
the table is frictionless, the rope is massless, the pulley is both.
The coefficients of static and kinetic friction between m1 and m2 are 0.539 and 0.439.
With m1 = 3.67 kg, m2 = 5.71 kg, m3 = 7.91 kg,
what are the accelerations of m1 and m2? What's the tension of the string?

Let's use the following notations:

F is the horizontal force of friction between m1 and m2.

a is the acceleration of m2.

b is the acceleration of m1 and m3

T is the tension of the string.

g = 9.80665 m/s^{2} is the gravitational field
(standard value).

Under the static regime where m1 and m2
are at rest with respect to each other, we have
a = b so that the last of the above equations
boils down to:

m3 g = (m1+m2+m3) a Therefore:
F = m2 a = g m2 m3 / (m1+m2+m3)

The coefficient of static friction must exceed the ratio of F
to the normal force (m1 g).
Thus, in the static regime, we must have:

F / (m1 g) = m2 m3 / [m1 (m1+m2+m3)] < 0.539

With the given values of the masses, that inequality doesn't hold
(since the left-hand side is more than 0.711).
We are thus faced with the kinetic regime where a and
b need not be equal.
Instead, motion is determined by the following additional equality, involving
the kinetic coefficient (u).

F = u m1 g [where u = 0.439]

Therefore, the acceleration of m2 is:

a = F/m2
= g u m1/m2 = 0.282 g = 2.767 m/s^{2}

The acceleration b of m1 (and also of m3)
is obtained as follows:

F = g u m1 = m3 (g-b) - m1 b
Therefore:
b = g (m3 - u m1) / (m1 + m3) =
0.544 g = 5.334 m/s^{2}

Finally, the tension of the string is:

T = m3 (g - b)
= g (1+u) m1 m3 / (m1+m3) = 35.4 N

Mark Taylor (2010-10-31)
Ladder leaning against a frictionless wall_{ } What's the least angle at which a ladder can rest against a frictionless wall?
When is such a ladder safe to climb?

Let m be the
static coefficient of friction
between the floor and the ladder (whose other end is resting against a
frictionless wall, consisting of something very slippery like wet ice or
glass).

Let L be the length of the ladder and
q be its inclination (the angle between the ladder
and the horizontal). Let's introduce a parameter
k that indicates the position of the center of gravity of the loaded ladder.
The combined center of gravity of the ladder and the person who climbs it
(if any) is located in the vertical line which intersects the ladder at a distance
k L from its bottom (for an ordinary ladder with nobody on it,
k = 0.5).

At equilibrium, the ladder does not move vertically or horizontally, so the sum of the horizontal
components of all existing forces is zero; so is the sum of all vertical components.

The only downward force is the weight of the whole thing (ladder + climber) and
is balanced by the upward vertical force exerted by the floor at the bottom of the ladder
(since the frictionless wall cannot exert any vertical force).

The horizontal forces exerted by the floor and the wall have the same magnitude F
(they have opposite directions).

Furthermore, the ladder doesn't rotate. Therefore, the
torques of all applied forces
cancel. The torques about the point of contact with the floor reduce to the torque of
the vertical weight (of magnitude Mg) and the purely horizontal force (F) exerted by
the frictionless wall. Therefore:

(k L cos q) Mg = (L sin q) F

The ladder will not move as long as the tangential friction force (of magnitude F) exerted
by the floor does not exceed its allowed maximum for the applied normal force (of magnitude Mg).

F < m Mg

Combining the two relations above, we obtain:

k Mg / tan q < m Mg

Which is to say that q >
arctg ( k / m )

The unloaded ladder will stay in place as long as this inequality is satisfied for
k = 0.5 but it's safe to climb to the last rung only when the
equality is satisfied when k is nearly equal to 1.

(2011-04-25) Rolling Resistance (rolling friction)
A complicated form of dry friction.

(2010-12-19) Coefficient of restitution (e).
The ratio of the initial to final closing speed.

When a ball dropped from a height h bounces off the floor to
a height r h, the dimensionless coefficient r
is called the coefficient of restitution.

Technically, the coefficient of restitution
is defined as the ratio of the initial to final closing speed
(with a change of sign that makes e positive in the common
case where the closing reverses direction after a shock).
It's thus the rato of the
speed of approach to the speed of separation.

e = 1 in an ellastic collision.

e = 0 if the colliding objects stick together after the collision.

The above values correspond to the so-called normal
coefficients of restitution, which are appropriate for head-on collision
with little or no tangential speed.
Otherwise, if the relative velocity of the two objects is not
perpendicular to their surface of contact, the ratio of the final to
the initial tangential speed is defined to be another
type of coefficient of restitution whose value is only very loosely related
to the normal coefficient of restitution discussed so far.