TOPIC: STATICS [정역학]
Course # 1040
ARGHYA NARAYAN BANERJEE
(Prof. BANERJEE A. NARAYAN)
바너지나라얀 교수
LECTURE 11: CHAPTER 8 – FRICTION
My contact: Room # 507
Office Ph.: 053-810-2453
Mobile: 010-5717-1975
E-mail: arghya.banerjee@hotmail.com
arghya@ynu.ac.kr
APPLICATIONS
In designing a brake system for a bicycle, car,
or any other vehicle, it is important to
understand the frictional forces involved.
For an applied force on the brake pads, how
can we determine the magnitude and
direction of the resulting friction force?
The rope is used to tow the refrigerator.
In order to move the refrigerator, is it best to
pull up as shown, pull horizontally, or pull
downwards on the rope?
What physical factors affect the answer to
this question?
1
CHARACTERISTICS OF DRY FRICTION
Friction is defined as a force of resistance
acting on a body which prevents or retards
slipping of the body relative to a second body.
Experiments show that frictional forces act tangent
(parallel) to the contacting surface in a direction
opposing the relative motion or tendency for motion.
For the body shown in the figure to be in
equilibrium, the following must be true:
Fx = 0
OR F = P,
Fy = 0
OR N = W, and
MO = 0
OR W*x = P*h.
CHARACTERISTICS OF DRY FRICTION
(continued)
Static friction
Kinetic friction
To study the characteristics of the
friction force F, let us assume that
tipping does not occur (i.e., “h” is
small or “a” is large). Then we
gradually increase the magnitude of
the force P. Typically, experiments
show that the friction force F varies
with P, as shown in the right figure
above.
FA = Applied Force; Ff = Frictional force
For Static Friction: FA = Ff
For Kinetic Friction: FA > Ff
2
CHARACTERISTICS OF DRY FRICTION
(continued)
The maximum friction force is attained
just before the block begins to move (a
situation that is called “impending
motion”). The value of the force is
found using Fs = s N, where s is
called the coefficient of static friction.
The value of s depends on the two
materials in contact.
Once the block begins to move, the
frictional force typically drops and is
given by Fk = k N. The value of k
(coefficient of kinetic friction) is less
than s (i.e. s > k )
CHARACTERISTICS OF DRY FRICTION
(continued)
Static friction
FA
Kinetic friction
FA < Fs
PA
It is also very important to note that the friction force may be
less than the maximum friction force. So, just because the object
is not moving, don’t assume the friction force is at its maximum
of Fs = s N unless you are told or know motion is impending!
•
•
s depends on the two materials involved and their surface properties.
How to find the value of s ? (See next slides)
3
DETERMINING s EXPERIMENTALLY (Method – 1)
If the block just begins to slip, the
maximum friction force is Fs = s N,
where s is the coefficient of static
friction.
Thus, when the block is on the verge of
sliding, the normal force N and
frictional force Fs combine to create a
resultant Rs
From the figure,
tan s = ( Fs / N ) = (s N / N ) = s
So, Fs = s N and tan s = s
DETERMING s EXPERIMENTALLY (Method – 2)
s
A block with weight W is placed on an
inclined plane. The plane is slowly tilted
until the block just begins to slip.
The inclination, s , is noted. Analysis of
the block just before it begins to move
gives (using Fs = s N):
+  F┴ = N – W cos s
= 0
+  F║ = S N – W sin s = 0
s
W
Using these two equations, we get
s = (W sin s ) / (W cos s ) = tan s
This simple experiment allows us to find the
S between two materials in contact.
4
PROBLEMS INVOLVING DRY FRICTION
Steps for solving equilibrium problems involving dry friction:
1. Draw the necessary free body diagrams. Make sure that
you show the friction force in the correct direction (it
always opposes the motion or impending motion).
2. Determine the number of unknowns. Do not assume
FS = S N unless the impending motion condition is given.
3. Apply the equations-of-equilibrium (E-of-E) and
appropriate frictional equations to solve for the unknowns.
IMPENDING TIPPING versus SLIPPING
P
h
W
O
SLIPPING
TIPPING
For a given W and h of the box, how can we
determine if the block will slide or tip first? In
this case, we have four unknowns (F, N, x, and P)
and only three E-of-E (Fx = Fy = Fz = 0).
Hence, we have to make an assumption
to give us another equation (the friction
equation!). Then we can solve for the
unknowns using the three E-of-E.
Finally, we need to check if our
assumption was correct.
5
IMPENDING TIPPING versus SLIPPING
(continued)
Assume: Slipping occurs
Known: P = Fs = s N (assuming
impending motion)
Solve:
x, P, and N
Check:
0  x < b/2
Or
Assume: Tipping occurs
Known: x = b/2 (point of rotation at
the edge)
Solve:
P, N, and F
Check:
P = F < Fs = s N
PROBLEM 01
Given: A uniform ladder weighs 30 lb. The vertical wall is
smooth (no friction). The floor is rough and s = 0.2.
Find: Whether it remains in this position when it is released.
Plan:
a) Draw a FBD.
b) Determine the unknowns.
c) Make any necessary friction assumptions.
d) Apply E-of-E (and friction equations, if appropriate ) to solve for the
unknowns.
e) Check assumptions, if required.
6
PROBLEM 01 (continued)
FBD of the ladder
NB
12 ft
12 ft
26 ft
30 lb
FA
NA
5 ft 5 ft
There are three unknowns: NA, FA, NB.
  FY = NA – 30 = 0 ;
so NA = 30 lb
+  MA = 30 ( 5 ) – NB( 24 ) = 0 ;
If +   FX = NB – FA > 0
so NB = 6.25 lb
pole will not remain stationary
<0
pole will remain stationary
=0
pole will remain stationary and under ‘impending motion’
PROBLEM 01 (continued)
NB
FBD of the ladder
12 ft
12 ft
30 lb
5 ft
5 ft
NA
FA
Now check the friction force to see if the ladder slides or stays.
Fmax = s NA = 0.2 * 30 lb = 6 lb = FA
Since NB = 6.25 lb > FA
NB – FA > 0 pole will not remain stationary
the pole will not remain stationary.
< 0 pole will remain stationary
= 0 pole will remain stationary and under
It will move.
‘impending motion’
7
PROBLEM 02
Given: Refrigerator weight = 180 lb,
s = 0.25
Find: The smallest magnitude of P
that will cause impending
motion (tipping or slipping)
of the refrigerator.
Plan:
a) Draw a FBD of the refrigerator.
b) Determine the unknowns.
c) Make friction assumptions, as necessary.
d) Apply E-of-E (and friction equation as appropriate) to solve for the unknowns.
e) Check assumptions, as required.
PROBLEM 02 (continued)
FBD of the refrigerator
1.5 ft 1.5 ft
P
180 lb
4 ft
3 ft
o
F
x
N
There are four unknowns: P, N, F and x.
First, let’s assume the refrigerator slips. Then the friction
equation is F = s N = 0.25 N.
8
PROBLEM 02 (continued)
1.5
ft
P
P
1.5
ft
180 lb
4 ft
o
+   FX = P – 0.25 N = 0
N
FBD of the refrigerator
These two equations give:
and
F = 0.25 N
x
+   FY = N – 180 = 0
P = 45 lb
3 ft
N = 180 lb
+  MO = - 45 (4) + 180 (x) = 0
Check: x = 1.0 < 1.5 so OK!
Refrigerator slips as assumed at P = 45 lb
 Fy = 0
n1 + F sin = Mg
OR n1 = Mg – F sin
n1
n2
 Fy = 0
n2 = Mg + F sin
F sin 
F

F cos 
F cos 


When the body is taller
than the person, it is
easier to push than pull.

F sin 

Fr1
Fr2
W = Mg
Fr1 = s n1 = s (Mg – F sin)
 Less friction: easy to move

W = Mg
Fr2 = s n2 = s (Mg + F sin)
 More friction: hard to move
n2 + F sin = Mg
OR n2 = Mg – F sin
 Fy = 0

 Fy = 0
n1 = Mg + F sin
n2
n1

Q. But for grass-cutter we push it.
Why?
A. This will create more friction,
so that more grass can be cut.
Fr1

F cos 
F sin  F
W = Mg
Fr1 = s n1 = s (Mg + F sin)
 More friction: hard to move
When the body is shorter
than the person, it is
easier to pull than push.

F sin 

Fr1 < Fr2
F cos 

Fr2
W = Mg
Fr2 = s n2 = s (Mg - F sin)
 Less friction: easy to move
Fr1 > Fr2
9
QUIZ
1. A friction force always acts _____ to the contact surface.
A) Normal
B) At 45°
C) Parallel
D) At the angle of static friction
2. If a block is stationary, then the friction force acting on it is
________ .
A)  s N
B) = s N
C)  s N
D) = k N
Static friction
FA
Kinetic friction
FA < Fs
PA
QUIZ
3. A 100 lb box with a wide base is pulled by a force P
and s = 0.4. Which force orientation requires the
least force to begin sliding?
100 lb
A) P(A)
B) P(B)
C) P(C)
D) Can not be determined
P||
FA
NA
P(A)
P(B)
P(C)
NA = 100 lb
P(A) = FA = s NA
= s 100
= 0.4 x 100 = 40 lb
FB
NB
P┴
P(A) < P(B)
4. A ladder is positioned as shown. Please indicate
the direction of the friction force on the ladder at
B.
A) 
B) 
C)
D)
NB = 100 + P┴
P(B) = FB = s NB
= s (100 + P┴)
> 40 lb
B
A
10
QUIZ
5. A 10 lb block is in equilibrium. What is
the magnitude of the friction force
between this block and the surface?
 S = 0.3
2 lb
A) 0 lb B) 1 lb C) 2 lb D) 3 lb
N = 10 lb
FS = s N = 0.3 x 10
= 3 lb > 2 lb
So, FA = 2 lb
So the motion is not
impending motion.
It is below the
impending motion.
6. The ladder AB is positioned as shown. What is the
direction of the friction force on the ladder at B.
A)
B)
C) 
D) 
B
A
WEDGES AND FRICTIONAL FORCES ON FLAT BELTS
Wedges are used
adjust the elevation
provide stability
heavy objects such
this large steel pipe.
APPLICATIONS:
to
or
for
as
How can we determine the force required
to pull the wedge out?
When there are no applied forces on the
wedge, will it stay in place (i.e., be selflocking) or will it come out on its own?
Under what physical conditions will it
come out?
11
APPLICATIONS
(continued)
Belt drives are commonly used for
transmitting the torque developed by a
motor to a wheel attached to a pump,
fan or blower.
How can we decide if the belts will
function properly, i.e., without
slipping or breaking?
APPLICATIONS (continued)
In the design of a hand brake, it
is essential to analyze the
frictional forces acting on the
band (which acts like a belt).
How can you determine the
tensions in the cable pulling on
the band?
Also from a design perspective,
how are the belt tension, the
applied force P and the torque M,
related?
12
ANALYSIS OF A WEDGE
W
A wedge is a simple machine in which a small
force P is used to lift a large weight W.
To determine the force required to push the wedge
in or out, it is necessary to draw FBDs of the wedge
and the object on top of it.
It is easier to start with a FBD of the wedge since
you know the direction of its motion.
Note that:
a) the friction forces are always in the direction
opposite to the motion, or impending motion, of
the wedge;
b) the friction forces are along the contacting
surfaces; and,
c) the normal forces are perpendicular to the
contacting surfaces.
Forces on the wedge
ANALYSIS OF A WEDGE (continued)
Forces on the weight (W)
Next, a FBD of the object on top of the wedge
is drawn. Please note that:
a) at the contacting surfaces between the
wedge and the object the forces are equal in
magnitude and opposite in direction to those
on the wedge; and, b) all other forces acting on
the object should be shown.
To determine the unknowns, we must apply E-of-E,  Fx = 0 and
 Fy = 0, to the wedge and the object as well as the impending
motion frictional equation, FS = S N.
13
ANALYSIS OF A WEDGE (continued)
Now of the two FBDs, which one should we start analyzing first?
We should start analyzing the FBD in which the number of unknowns
are less than or equal to the number of E-of-E and frictional equations.
W
NOTE:
If the object is to be lowered, then the wedge needs
to be pulled out. If the value of the force P needed
to remove the wedge is positive, then the wedge is
self-locking, i.e., it will not come out on its own.
BELT ANALYSIS
Consider a flat belt passing over a fixed
curved surface with the total angle of
contact equal to  radians.
If the belt slips or is just about to slip,
then T2 must be larger than T1 and the
motion resisting friction forces. Hence,
T2 must be greater than T1.
Detailed analysis shows that T2 = T1 e   where  is the
coefficient of static friction between the belt and the surface.
Be sure to use radians when using this formula!!
14
EXAMPLE
Given: The crate weighs 300 lb and S at
all contacting surfaces is 0.3.
Assume the wedges have
negligible weight.
Find: The smallest force P needed to
pull out the wedge.
Plan:
1. Draw a FBD of the crate. Why do the crate first?
2. Draw a FBD of the wedge.
3. Apply the E-of-E to the crate.
4. Apply the E-of-E to wedge.
EXAMPLE (continued)
Equal and opposite
FB=0.3NB
NC
300 lb
FC=0.3NC
P
NB
NC
15º
15º
FC=0.3NC
FBD of Crate
FD=0.3ND
15º
ND
FBD of Wedge
The FBDs of crate and wedge are shown in the figures. Applying
the E-of-E to the crate, we get
+  FX =
NB – 0.3NC = 0
+  FY = NC – 300 + 0.3 NB = 0
Solving the above two equations, we get
NB = 82.57 lb = 82.6 lb,
NC = 275.3 lb = 275 lb
15
EXAMPLE (continued)
FB=0.3NB
NC
FC=0.3NC
300 lb
P
NB = 82.6 lb
Fd sin15
15º
15º
15º
FC=0.3NC
Nd sin15
NC = 275 lb
FBD of Crate
FD=0.3ND
Fd cos15
Nd cos15
ND
FBD of Wedge
Applying the E-of-E to the wedge, we get
+  FY = ND cos 15 + 0.3 ND sin 15 – 275= 0;
ND = 263.7 lb = 264 lb
+  FX = 0.3(275) + 0.3(264) cos 15 – (264) sin 15 – P = 0;
P = 90.7 lb
PROBLEM
Given: Blocks A and B weigh 50 lb and
30 lb, respectively.
Find: The smallest weight of cylinder D
which will cause the loss of static
equilibrium.
16
GROUP PROBLEM SOLVING (continued)
Plan:
1. Consider two cases: a) both blocks slide together, and,
b) block B slides over the block A.
2. For each case, draw a FBD of the block(s).
3. For each case, apply the E-of-E to find the force needed to
cause sliding.
4. Choose the smaller P value from the two cases.
5. Use belt friction theory to find the weight of block D.
GROUP PROBLEM SOLVING
(continued)
Case a (both blocks sliding together):
 +  FY = N – 80 = 0
N = 80 lb
+  FX = 0.4 (80) – P = 0
P = 32 lb
P
B
A
30 lb
50 lb
F=0.4 N
N
17
GROUP PROBLEM SOLVING (continued)
30 lb
Case b (block B slides over A):
0.6 N sin20
0.6 N
P
0.6 N cos20
20º
N sin20
 +  Fy = N cos 20 + 0.6 N sin 20 – 30 = 0
N cos20
N
N = 26.20 lb
 +  Fx = – P + 0.6 ( 26.2 ) cos 20 – 26.2 sin 20 = 0
P = 5.812 lb
 = /2 (rad)
Case b has the lowest P (Case a was 32 lb) and this will
occur first. Next, using a frictional force analysis of belt,
we get
P = T1
s = 0.5
T2 > T1
WD > P
WD = P e   = 5.812 e 0.5 ( 0.5  ) = 12.7 lb
The Block D weighing 12.7 lb will cause the block B to
slide over the block A.
WD = T2
T2 = T1 e  
WD = P e  
QUIZ
1. A wedge allows a ______ force P to lift
a _________ weight W.
A) (large, large)
B) (small, small)
C) (small, large)
D) (large, small)
W
2. Considering friction forces and the
indicated motion of the belt, how are belt
tensions T1 and T2 related?
A) T1 > T2
B) T1 = T2
C) T1 < T2
D) T1 = T2 e
18
QUIZ
1. Determine the direction of the friction force on object B
at the contact point between A and B.
A) 
B) 
C)
D)
2. The boy (hanging) in the picture weighs 100 lb and the
woman weighs 150 lb. The coefficient of static friction
between her shoes and the ground is 0.6. The boy will
______ ?
A) Be lifted up
B) Slide down
C) Not be lifted up
D) Not slide down
Force P
Force on A
Force on B
N
P = 90 lb
< 100 lb
Fs = 0.6 x 150
= 90 lb
3. In the analysis of frictional forces on a flat belt, T2 = T1 e  .
In this equation,  equals ______ .
A) Angle of contact in degrees
P
150 lb
100 lb
B) Angle of contact in radians
C) Coefficient of static friction D) Coefficient of kinetic friction
19