Download Friction and ramps notes

Objectives: After completing these
notes, you should be able to:
• Define and calculate the coefficients of
kinetic and static friction, and give the
relationship of friction to the normal force.
• Apply the concepts of static and kinetic
friction to problems involving constant
motion or impending motion.
Friction Forces
When two surfaces are in contact, friction forces
oppose relative motion or impending motion.
P
Friction forces are parallel to
the surfaces in contact and
oppose motion or impending
motion.
Friction and the Normal Force
12 N n
8N n
4N
n2N
4N
6N
The force required to overcome static or kinetic
friction is proportional to the normal force, Fn.
Ff ≤ msFn
Ff = mkFn
Friction forces are independent of area.
4N
4N
If the total mass pulled is constant, the same
force (4 N) is required to overcome friction
even with twice the area of contact.
For this to be true, it is essential that ALL
other variables be rigidly controlled.
Friction forces are independent of speed.
5 m/s
2 N
20 m/s
2 N
The force of kinetic friction is the same at
5 m/s as it is for 20 m/s. Again, we must
assume that there are no chemical or
mechanical changes due to speed.
Friction Forces
Friction is dependent on:
- normal force, Fn
- coefficient of the
surfaces in contact, m
Ff  mFn
Fs
Static friction opposes the intended
motion of two surfaces in contact
but at rest relative to one another.
Kinetic friction opposes motion of
two surfaces in contact that are
moving relative to one another.
PHYSICS
book pulled
Fa
Fs
wheel driven
Kinetic friction is less than static friction.
Fk
velocity
PHYSICS
book dragged
Fa
Coefficients of Friction
surfaces in contact
ms
mk
leather-soled shoes on wood
0.3
0.2
rubber-soled shoes on wood
0.9
0.7
climbing boots on rock
1.0
0.8
shoes on ice
0.1
0.05
auto tires on dry concrete
1.0
0.8
auto tires on wet concrete
0.7
0.5
auto tires on icy concrete
0.3
0.02
waxed skis on dry snow
0.08
0.04
waxed skis on wet snow
0.14
0.1
wood on wood
0.4
0.2
glass on glass
0.9
0.4
steel on steel - dry
0.6
0.4
steel on steel - greased
0.1
0.05
synovial joints in humans
0.01
0.003
The Static Friction Force
When an attempt is made to move an
object on a surface, static friction slowly
increases to a MAXIMUM value.
fs
n
P
W
Ff ≤ msFn
In this unit, when we use the following
equation, we refer only to the maximum
value of static friction and simply write:
Ff = msFn
Constant or Impending Motion
For motion that is impending and for
motion at constant speed, the resultant
force is zero and SF = 0. (Equilibrium)
Ff
F
Rest
F – Ff = 0
Ff
F
Constant Speed
F – Ff = 0
Here the weight and normal forces are
balanced and do not affect motion.
Friction and Acceleration
When F is greater than the maximum FF
the resultant force produces acceleration.
FF
a
F
Changing Speed
This case we will
have to use
Newton 2nd law
Ff = mkFn
Note that the kinetic friction force remains
constant even as the velocity increases or
decreases. Kinetic is a hard equality.
EXAMPLE 1: If mk = 0.3 and ms = 0.5,
what horizontal pull P is required to
just start a 250-N block moving?
1. Draw sketch and freebody diagram as shown.
Fn
Ff
P
+
mg
2. List givens and label
what is to be found:
mk = 0.3; ms = 0.5; W = 250 N
Find: P = ? to just start
3. Recognize for impending motion: P – Ffs = 0
EXAMPLE 1(Cont.): ms = 0.5, mg = 250 N. Find
P to overcome Ff (max). Static friction applies.
Fn
For this case: P – Ffs = 0
P
Ff
+
4. To find P we need to
know Ffs , which is:
Ff = ms Fn
250 N
5. To find Fn :
SFy = 0
mg= 250 N
Fn = ?
Fn – mg = 0
Fn = 250 N
(Continued)
EXAMPLE 1(Cont.): ms = 0.5, W = 250 N. Find P
to overcome Ff (max). Now we know Fn = 250 N.
6. Next we find Ff from:
Ff = ms
Fn = 0.5 (250 N)
7. For this case: P – Ff = 0
P = Ff = 0.5 (250 N)
P = 125 N
Fn
Ff
P
+
250 N
ms = 0.5
This force (125 N) is needed to just start motion.
Next we consider P needed for constant speed.
EXAMPLE 1(Cont.): If mk = 0.3 and ms = 0.5,
what horizontal pull P is required to move with
constant speed? (Overcoming kinetic friction)
SFy = may = 0
mk = 0.3
Fn
P
Ff
Fn - mg = 0
Fn = mg
Now: Ff = mk
Fn = mkmg
+
mg
P = (0.3)(250 N)
SFx = 0;
P - fk = 0
P = Ff = mkmg
P = 75.0 N
The Normal Force and Weight
The normal force is NOT always equal to
the weight. The following are examples:
P
Fn
m
Here the normal force is
less than weight due to
upward component of P.
300
mg
P
Fn
mg
Here the normal force is
equal to only the component of weight perpendicular to the plane.
Example 2. A force of 100 N drags a 300-N
block by a rope at an angle of 400 above the
horizontal surface. If uk = 0.2, what force P
will produce constant speed?
W = 300 N P
fk
n
=100
400
m
W
The force P is to be
replaced by its components Px and Py.
1. Draw and label a sketch
of the problem.
2. Draw free-body diagram.
P sin 400 Py
n
fk
W
P
400
Py
Px
P cos 400
+
Example 2 (Cont.). P = 100; W = 300 N; uk = 0.2.
3. Find components of P:
Px = P cos 400 = 0.766P
Py = P sin 400 = 0.643P
Px = 0.766P; Py = 0.643P
P sin 400
n
P
400
fk
P cos 400
mg
+
Note: Vertical forces are balanced, and for
horizontal forces are unbalanced.
F
x
0
F
y
0
Example 2 (Cont.). P = 100; W = 300 N; uk = 0.2
Px = 0.766P
Py = 0.643P
4. Apply Equilibrium conditions to vertical axis.
SFy = 0
0.643P
n
fk
300 N
P
400
0.766P
+
n + 0.643P – 300 N= 0 [Py and n are up (+)]
n = 300 N – 64.3P; Solve for n
n = 223 N
Example 2 (Cont.). P = 100; W = 300 N; uk = 0.2.
n = 223N
0.643P
n
5. Apply SFx = 0 to con-
stant horizontal motion.
SFx = 0.766P – fk = ma
fk
300 N
fk = mk n = (0.2)(223)
fk = (0.2)(223) = 44.7N
76.6 – fk = ma;
76.6-44.7= ma
P
400
0.766P
+
Example 2 (Cont.).P = 100N; W = 300 N; uk = 0
0.643P
n
P
76.6-44.7= 31.9N = ma
400
fk
0.766P
300 N
+
6. Solve for mass.
300N= mg; m=30kg
31.9N=30 a
a = 31.9/30
If P = 100 N, the
block will be
accelerated.
a = 1.06 m/s2
a = 1.06 m/s2
Example 3: What push P up the incline is
needed to move a 230-N block up the
incline at constant speed if mk = 0.3?
P
Step 1: Draw free-body
including forces, angles
and components.
y
n
W sin 600
fk
P
x
W cos 600
600
230 N
W =230 N
Step 2:
SFy = 0
n – W cos 600 = 0
n = (230 N) cos 600
n = 115 N
Example 3 (Cont.): Find P to give
move up the incline (W = 230 N).
n
y
W sin 600
x
P
fk
W cos 600
600
W
n = 115 N
W = 230 N
Step 3. Apply SFx= 0
P - fk - W sin 600 = 0
fk = mkn = 0.2(115 N)
fk = 23 N, P = ?
P - 23 N - (230 N)sin 600 = 0
P - 23 N - 199 N= 0
P = 222 N
Summary: Important Points to Consider
When Solving Friction Problems.
• The maximum force of static friction is
the force required to just start motion.
fs
n
P
W
f s  ms
n
Equilibrium exists at that instant:
SFx  0;
SFy  0
Summary: Important Points (Cont.)
• The force of kinetic friction is that force
required to maintain constant motion.
n
fk
P
W
f k  mk
n
• fk does not get larger as the speed is
increased.
Summary: Important Points (Cont.)
• Remember the normal force n is not
always equal to the weight of an object.
n
m
P
It is necessary to draw
the free-body diagram
and sum forces to solve
300
W
P
n
W
for the correct
n value.
Summary
Static Friction: No
relative motion.
Kinetic Friction:
Relative motion.
fs ≤ msn
fk = mkn
CONCLUSION: Chapter 4B
Friction and Equilibrium