EQUILIBRIUM CASE 4: Kinetic Friction
EQUILIBRIUM1 CASE4
In equilibrium there can never be a resultant other than zero, R = 0, or F = 0.
The 1st Condition of Equilibrium always holds true, x = 0 and y = 0. Whenever
there is a Resultant that is not zero, or any type of net Force (Fnet ≠ 0), one must
determine a solution using Newton’s 2nd Law:
F(net) = ma or F(net) = w/g • a (when weight is the given)
Back to equilibrium: Recall Newton’s 1st Law of Motion:
If an object is at rest, it remains at rest, or if an object is in motion, it will
travel in a straight line at a constant speed (aka vc), unless it is acted upon
by an outside force. (This outside force is aka F(net).)
Because of the 1st Law, equilibrium can and will exist at rest or in motion:
1. All bodies at rest are in a state of Static Equilibrium.
2. All bodies moving at a constant speed in a straight line (vc) are in a
state of Translatory Equilibrium.
Translatory means traveling in a straight line at a constant speed. Static means
at rest. (Dynamic indicates a net Force is applied, therefore not in equilibrium.)
Kinetic is a word that means “to move through space”. Kinetic friction is the form
of mechanical friction that concerns itself with objects that are moving.
Sometimes kinetic friction is referred to as sliding friction. If an object moves
across a level surface at a constant velocity, the object develops kinetic friction
between the surface interface – the boundary between the surfaces.
For now we will associate kinetic friction with translatory motion. We will define
kinetic friction as:
fk = μkN
For every set of surfaces there will be a μk and as a general rule for any given set
of surfaces:
μk < μs
Keep in mind:
 Systems at rest involving friction involve static friction.
 Systems moving at constant velocity involve kinetic, or sliding friction.
 Translatory equilibrium – moving at constant speed in a straight line.
 Static equilibrium – systems at rest.
 Dynamic systems are not in equilibrium and must use F(net) = ma. This
accelerated motion still must consider the effects of kinetic friction.
Let’s look at a couple of examples. A 100 lb crate is pulled across a level floor at
a constant velocity. If the coefficient of kinetic friction is 0.4, find the force (pull
or push) required to move the crate.
y
We can use y = 0 to
find N, in order to find
fk and then F:
w = 100 lbs
μk = 0.4
N
F=?
-x
Fy = 0 = N – w, so
N = w = 100 lbs
x
fk
w
fk = μkN = (.4)(100 lbs) = 40 lbs
Fx = 0 = -fk + F, so F = fk = 40 lbs
-y
Force required
y
N
fs
Now we have a box pulled across a
level floor at a constant velocity by a
force of 60 lbs. If μk = 0.5, find the
weight of the box.
F = 60 lbs
x
-x
w=?
N=?
μk = 0.5
w
-y
F = 60 lbs
fk = ?
Since equilibrium is indicated, the following relationships hold true:
Fx = 0 = -fk + F, so F = fk = 60 lbs
and
Fy = 0 = N – w plus fk = μkN
so,
N = fk/μk = 60 lbs/0.5 = 120 lbs
If we exert any force greater than 60 lbs, acceleration occurs and we have a
resulting net force  0, so we must use F = ma. 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 = 120 lbs
fs = 60 lbs
F = 70 lbs
-x
x
w = 120 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/120lbs/32 ft/s2 = 2.67 ft/s2
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