EQUILIBRIUM CASE 3: Static Friction
EQUILIBRIUM1 CASE3
Static friction is the force of friction developed between surfaces that are at rest.
Static frictions, as well as other types of friction, are degrading forces that
always oppose motion. Degrading forces take away energy and one must do work
to overcome frictional forces.
We show frictional forces with a lower case f or script upper case F , and static
friction as fs or Fs. When one attempts
to slide and object across a surface (and
F
it requires a force F to move the object)
Fs
friction fs will always act in the opposite
direction.
Frictional force acts here
between the surfaces.
F
OBJECT
Surfaces have microscopic hills and valleys.
The smoother the surface,
the smaller the hills and
valleys. The rougher the
surface, the larger the hills and valleys.
friction, fs
Several factors cause static friction:
♠ Hills and valleys interact. They snag tending to hold the object back.
♠ Electrical interaction between atoms and molecules contribute.
♠ Micro-weld effect – pressure tends to fuse or weld coinciding high spots.
Centuries ago, da Vinci discovered that the force of static friction was related to
the Normal force that the surface exerts on the object. The static frictional force
on a flat surface is proportional to the Normal:
fs  N
To make the proportion an equality, a constant is used. The constant here is
called the coefficient of static friction and we use the Greek letter mu, , to
symbolize it. For static friction this coefficient is written as: s. The general
equation for mechanical friction is:
f = N
and in a case with static
equilibrium where the system is at rest, the equation is written:
fs = sN
where s is the coefficient of static friction. Several sources, including the text
book will have a list of some common values for the static coefficient of some
surfaces. These values are determined experimentally and for each set of
surfaces in contact there is a specific coefficient. The values are pure numbers –
no units are assigned to them.
Examples:
wood on wood:
steel on steel:
metal on wood
rubber on dry concrete:
s
s
s
s
=
=
=
=
0.70
0.20
0.25
0.90
Note that: 0  s  1
Static frictional forces will have a maximum value for any set of surfaces. Beyond
that point, motion will occur and the object can no longer be in static equilibrium.
The maximum static frictional force is written as:
fs (max)  sN
This is the definition of static friction we will use. The maximum force of static
friction is equal to or greater than the product of the coefficient of static friction
for the surfaces involved and the Normal force exerted by the surface on the
object.
Layout of vectors around a block:
We lift the Normal vector to the
cg. We
y
slide the
forces F
N
and fs to
the cg.
y
N
F
-x
fs
x
-x
fs
F
x
forward pull on block
friction holds
back
w
w
-y
-y
Normal arises from the surface,
Weight vector from the cg.
If we collapse a body down to
Its cg, all the forces act at cg.
Let’s look at an example of the same object in different circumstances. Consider
a 100 lb crate sitting on a level floor. The coefficient of (static) friction for the two
surfaces – the floor and the underside of the crate – is 0.6. We will examine
several situations in which we increase the pull (F) on the crate.
First we will pull with a force of 10 lbs.
What happens? y
N
-x
fs
F = 10 lbs
w = 100 lbs
-y
x
Fy = 0 = N + (-)w  N = w = 100 lbs
The maximum frictional force is:
fs(max) = sN
Then fs(max) = 0.6(100 lbs) = 60 lbs, the
greatest frictional force that can exist
at 100 lbs.
To break free from friction we must pull
forward with a force greater than 60 lbs.
We pull with a force of 10 lbs and the
force of static friction is 10 lbs. Why?
By Newton’s 3rd Law and because the
crate is in equilibrium and Fx = 0.
Fx = F – fs = 0,  F = fs = 10 lbs.
y
N
F = 20 lbs
-x
x
fs
w
-y
Next we pull forward with a force of 20 lbs. Friction pulls against the motion.
Still in equilibrium we see: Fx = F – fs = 0, 
F = fs = 20 lbs. Bump the force
of the pull to 50 lbs. This is still less
N x
than the 60 lb maximum static frictional
force for the surfaces. No motion takes
fs
F = 50 lbs
place. Equilibrium still exists and
-x
x
F = fs = 50 lbs. y
N
fs
w
F = 60 lbs
-x
-y
x
Finally we match the maximum static
frictional force with a pull of 60 lbs.
w
This is a special case when F = fs(max).
-y
At F = 60 lbs, the crate just wants to
move. The crate is now in a state of
impending motion. This state can only exist at fs(max). The crate is ready to
“break away” from friction’s grip, but does not yet move. If we exert any force
greater than fs(max), motion occurs and we have a resulting net force  0. If we
have a Resultant  0, we are no longer in equilibrium. Now we must use F = ma
(Newton’s 2nd Law) to solve the problem. If we are not in equilibrium, some
unbalanced force will equal mass times acceleration. Let’s pull with a force of
70 lbs. The Fy = 0, since the crate is not moving up or down, but the Fx = ma.
acceleration!
y
N = 100 lbs
-x
fs = 60 lbs
F = 70 lbs
x
w = 100 lbs
-y
R = F – fs = 70 lbs – 60 lbs = 10 lbs.
We have Fnet = 10 lbs = ma. (m = w/g)
No more equilibrium, so now we solve for the acceleration:
F = ma, a = F/m = F/w/g = 10 lbs/100lbs/32 ft/s2 = 3.2 ft/s2
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