4.5 Friction Forces
Discuss the tread on car tires.
Figure 4 – 1: The tread of a wheel and
how it works. How it moves water out of
the way, too. High-speed photos (Zoe).
DKX
In one form or another, friction is present in every engineering system.
Sometimes friction is a nuisance and at other times it represents a necessary part of the
workings of a system. Whether a friend or a foe, its impact on the system regularly
needs to be analyzed and understood.
Dry Friction
Dry Friction
Contact Surface
Normal Force, Friction Force
Figure 4 – 2: The force between two
bodies is divided into a normal force and
a friction force.
Maximum Static Friction Force
Kinetic Friction Force
Figure 4 – 3: The friction force F is
balanced with the applied force P until
the can begins to move.
Figure 4 – 4: The maximum static
friction force increases with the normal
force.
Dry friction arises when two solid bodies are pressed against each other, but
are otherwise unconnected. The surface between the two bodies is called the contact
surface. As shown in Fig. 4 – 2, the resultant force between the two bodies is separated
into a component perpendicular to the surface and a component tangent to the surface.
The perpendicular component is called the normal force N and the tangential
component is called the friction force F.
When the friction force is sufficiently small, the fact that the two surfaces are
not connected to each other does not come into play because the system is stationary
and sliding has not yet occurred. The reaction at the contact surface is found the same
way any other reaction is found. For example, consider Fig. 4 – 3 showing a coke can
resting on a table. Assume that you push on the can with your finger ever so slightly to
the right. A friction force F will act on the can to the left to balance with the force P by
your finger. As you press on the can to the right with greater force, a point will be
reached when the can begins to slide on the surface. Before this happens, the friction
force between the surface and the coke can is easily predicted. The friction force is
treated like any other unknown reaction. However, once sliding occurs, the nature of
the problem changes.
It’s generally difficult to predict with a high level of accuracy the maximum
static friction force Fs just before sliding takes place or the kinetic friction force Fk
during sliding. Levels of error on the order of 25% are common. The friction forces Fs
and Fk are difficult to predict because they depend on a lot of factors. The friction
forces Fs and Fk depend on the shape of the contact surface, the roughness of the two
bodies at the contact surface, the elasticity of the two bodies at the contact surface, the
presence of tiny particles between the bodies, and the temperature of the contact
surface. As an engineer, this means that friction isn’t always a desirable force to work
with, owing to its relatively poor predictability. Of course, in many engineering
problems this level of unpredictability is perfectly acceptable. In automotive belts, for
instance, the static friction force F between the belt and the disk is held below Fs by
over-design. The belt is made to be flexible, allowing it to stretch over the disk. As
external conditions change, the normal force N between the belt and the disk remains
sufficiently large to prevent it from sliding, that is to prevent the friction force F from
reaching the maximum level Fs after which sliding would occur. As a second example,
consider the disk brakes in a car. They’re based on the sliding friction force Fk. The
sliding friction force’s unpredictability is overcome through a feedback mechanism. If
more force is needed, the driver presses more on the brake pedal and if less force is
needed, the driver presses with less force.
Even though Fs and Fk depend on a lot of factors, reasonably good estimates
can be made based on their relationship with the normal force N at the contact surface.
First, consider the experimental test in which the maximum static friction force Fs is
measured repeatedly while increasing the values of N. This test produces a curve like
the one shown in Fig. 4 – 4. As shown, the maximum static friction force Fs starts out
at zero when the normal force is zero. As N increases, Fs increases in proportion to N.
The friction force Fs is related to the normal force by
Static Coefficient of Friction
Static Model of Dry Friction
Fs   s N ,
(4 – 1)
Figure 4 – 5: The kinetic friction force
increases with the normal force.
Kinetic Coefficient of Friction
Kinetic Model of Dry Friction
where the proportionality constant s is a coefficient for the maximum limit of static
friction. The coefficient is called the coefficient of statics friction and Eq. (4 – 1) is
called the static model of dry friction. Equation (4 – 1) is based on experimental
evidence. It can also be derived analytically. It is derived analytically by creating an
idealized mathematical model of a rough contact surface. Equation (4 – 1) is derived
this way in Example 4 – 5.
Next, consider the experimental test in which the sliding friction force is
measured while increasing the values of the normal force N. This test produces a curve
like the one shown in Fig. 4 – 5. The sliding friction force is related to the normal force
N by
Fk   k N ,
(4 – 2)
Table 4 – 1: Typical coefficients of static
Friction and kinetic friction (Zoe)
where the proportionality constant k is called the coefficient of kinetic friction.
Equation (4 – 2) is called the kinetic model of dry friction. You can’t help but notice
the similarity between Eqs. (4 – 1) and (4 – 2) and between Figs. (4 – 4) and (4 – 5).
The difference between them lies in the values of the coefficients of friction. Typically,
the coefficient of static friction for two bodies in contact with each other is larger than
their coefficient of kinetic friction.
The remainder of this section looks at a variety of engineering systems that
use dry friction as a primary part of how they work.
Wedges
Figure 4 – 6: A wedge is placed under a
block by applying a force P and then
removing it.
Self-Locking
The simple wedge serves a variety of purposes. It can be used to secure an
object in a given position or to adjust the object’s position. For example, when placed
under a wheel, it prevents the wheel from rolling, or it can be placed under a piece of
equipment to adjust its position. Notice that two events occur when using a wedge.
First, a force is applied to place the wedge in position. The force is applied to the
wedge almost perpendicular to a resulting normal force acting on the object. Then, the
dry friction between the wedge and the surrounding objects causes the wedge to selflock, meaning that the applied force can be removed and the wedge will remain in
position. Problems involving wedges deal with a) finding the minimum force P
required to place a wedge in position, and b) determining whether or not the wedge
will stay in position, that is whether the wedge is self-locking. The following example
illustrates these ideas.
Referring to Fig. 4 – 6, wedge A is placed under block B by force P. The force
will be applied and then removed. Denote the coefficient of static friction of the surface
below the wedge by A, the coefficient of static friction of the surface between the
wedge and the block by B, and assume that the surface to the right of the block is
2
smooth. Also denote the weight of the block by W and neglect the typically small
weight of the wedge. Let’s first determine the minimum force needed to move the
wedge into position, and then we’ll determine whether or not the wedge stays in
position after the force P is removed.
Figure 4 – 7: Free Body Diagrams of the
Wedge and the Block
Figure 4 – 7 shows the free body diagrams of the wedge and the block while
the force P is being applied. Notice that the normal forces are all taken to be
compressive and that the friction forces are taken to appose the displacement of the
wedge and the block. Summing forces in the x and y directions yields
0  P  FA  FB cos  N B sin  ,
0  N A  FB sin   N B cos ,
(4 – 3a-d)
0  P  FA  N C
0  NA W
Equations (4 – 2a,b) are the resultant forces acting on the wedge in the x and y
directions and Eqs. (4 – 3c,d) are the resultant forces acting on the system in the x and y
directions. Equations (4 – 3) are 4 equations expressed in terms of the six unknowns P,
FA, FB, NB, NA, and NC. Two more equations are needed. Assuming a static model of
dry friction, the maximum friction forces FA and FB are given by
F A   A N A,
(4 – 4)
FB   B N B .
Equations (4 – 3) and (4 – 4) represent six linear algebraic equations expressed in terms
of six unknowns. The solution is
(4 – 5)

 cos  sin  
P   A  B
W , F A   AW ,
cos
   B sin  

 BW
W
FB 
, NB 
,
cos    B sin 
cos   B sin 
  cos  sin  
N A  W , NC   B
W
 cos   B sin  
In Eq.(4 – 5) we see that the six unknowns P, FA, FB, NB, NA, and NC are positive if
cos   B sin   0, that is when  B  cot  . In other words, the wedge can be
pushed under the block when  B  cot  and the minimum applied force P required to
Figure 4 – 8: Free Body Diagrams of the
Wedge and the Block when the applied
force is to the left. The friction forces
change direction and the normal force at
wall C is taken to be zero.
do this is given in Eq. (4 – 5). When  B  cot  , it requires an infinite amount of force
to push in the wedge, at least in theory.
Let’s now determine whether the wedge is self-locking. To do this, imagine
that the applied force P is applied to the left rather than to the right. The wedge now
prevents the block from moving in the opposite directions. Also, the normal force NC is
now taken to be zero. The free body diagram required for this problem is shown in Fig.
4 – 8. Summing forces in the x and y directions, we get
3
0   P  F A  FB cos   N B sin  ,
0  N A  FB sin   N B cos  ,
(4 – 6a-d)
0   P  FA
0  NA W
Equations (4 – 6) are 4 equations expressed in terms of five unknowns, P, FA, FB, and
NB, and NA. Substituting Eqs. (4 – 6c) into (4 – 6a) yields
FB  tan N B
(4 – 7)
When FB   B N B , it follows from Eq. (4 – 7) that
tan    B .
(4 – 8)
Equation (4 – 8) expresses a condition necessary for the block not to slide on the
wedge. From Eqs. (4 – 6a) and (4 – 6b), FA   A N A if
P   AW
(4 – 10)
Equation (4 – 10) is the necessary condition for the block not to slide on the bottom
surface. Since  AW is positive, this means that a positive force P is required to remove
the wedge from under the block. It follows that the block is self-locking if tan    B .
Square-Threaded Screws
Figure 4 – 9: The thread on a square
threaded screw, when unwound, makes
an angle 
Photo: DKX
Assisting Moment, Resisting Moment
Lead, Lead Angle, Mean Radius
Square threaded screws have strong threads, capable of supporting relatively
large axial loads. Square threaded screws are used in heavy machinery of all kinds.
See Fig. 4 – 9. The axial load P acting on a square threaded screw would
cause it to turn if it were not for the dry friction at the contact surface of the screw
thread. The dry friction prevents the screw from turning. In order to turn the screw an
additional moment must be applied. The moment can be applied in one of two
directions. It can be applied in the direction in which the load tends to turn the screw,
or it can be applied in the opposite direction that the load tends to turn the screw. The
minimum moment needed to turn the screw in the direction that the load tends to turn
the screw is called the assisting moment. The minimum moment needed to turn the
screw in the opposite direction that the load tends to turn the screw is called the
resisting moment. Problems involving square-threaded screws deal with a) finding the
assisting moment Ma, b) finding the resisting moment Mr and c) determining whether
the screw is self-locking.
To simplify the analysis, imagine that the thread is “unwound.” The amount h
that any point on a thread moves up or down after one revolution is called the lead of
the screw. The unwound angle  shown in the figure is called the lead angle. Notice
that the lead angle is found from
tan 
(4 – 11)
4
h
,
2R
where R denotes the mean radius of the screw.
Finding the Assisting Moment
Figure 4 – 10: Free body diagram of the
wedge.
Figure 4 – 10 shows an exploded view of the screw and the surrounding
casing. The cylindrical screw shaft, the screw thread and the casing thread are shown.
The cylindrical screw is acted on by a load P and an applied moment M. The applied
moment M is either resisting or assisting depending on its direction. In the figure, the
load is downward. Looking from the top, the screw turns clockwise in the direction of
the load. Therefore, in the diagram a positive moment M assists the load P and a
negative moment M resists the load P. For now, we’ll assume that M is positive.
Figure 4 – 10, also shows the forces acting on the different bodies. The
applied load P and the assisting moment M acting on the cylindrical screw shaft are
balanced with distributed forces acting on the interior surface of the cylinder, where the
screw thread protrudes out. The resultant force and moment produced by these
distributed forces are equal an opposite to P and M, respectively. The distributed force
has an upward component, which is equal to the load P, and a lateral component Q =
M/R that produces a moment that is equal and opposite to M. Notice that Q isn’t really
a force. It’s the sum of the circumferential components of the force around the
cylinder.
Looking at the screw thread, it’s acted on by equal and opposite interior
distributed forces that produce a downward resultant force P and a lateral component
Q. The screw thread is also acted on by distributed forces at the contact surface
between the screw thread and the casing thread. The resultant force produced by these
distributed forces is resolved into a normal force N and friction component F.
Figure 4 – 11: The resultants of the
distributed forces acting on the screw
thread are transferred to a point.
To simplify the analysis, a novel approach to drawing the free body diagram
of the screw thread will be adopted. Referring to Figure 4 – 11, P, Q, N, and F have
been transferred to a point. The horizontal direction in the figure is not a direction of
force, but rather a circumferential direction around the cylinder. The resultant in the
horizontal direction therefore represents the resultant moment about the axis of the
screw divided by the radius, which has units of force – so it looks and acts like a
direction of force, although it really isn’t a force Summing forces in the vertical
direction and summing the quasi-forces in the horizontal direction, yields
0  Q  F cos  N sin  ,
0   P  F sin   N cos
(4 – 12)
The minimum moment is applied when the friction component F and the normal force
N are related to each other by the dry friction model given in Eq. (4 – 1), given by F =
sN. Substituting Eq. (4 – 1) and the relationship Q = M/R into (4 – 12) yields the
assisting moment
Ma 
(4 – 13)
 s  tan 
RP.
1   s tan 
In Eq. (4 – 13) we see that Ma could actually be negative depending on the sign of
 s  tan  . When  s  tan  is negative, the assisting moment prevents the screw from
5
turning. In other words, when  s  tan  , the coefficient of static friction is not large
enough to prevent the screw from turning, even without an applied moment M. On the
other hand, when  s  tan  , Ma in Eq. (4 – 13) truly represents the minimum moment
needed to turn the screw. Also, notice that the minimum moment required to sustain
the turning of the screw (once it has already begun to turn) is determined from Eq. (4 –
13) by replacing  s with  k .
Finding the Resisting Moment
Figure 4 – 12: Free body diagram of the
screw thread while acted upon by a
resisting moment.
The resisting moment is found the same way the assisting moment was found.
The difference lies in the signs of some of the quantities. The resisting moment is in the
opposite direction of the assisting moment, causing the friction component to change
direction, too. This means in the free body diagram shown in Fig. 4 – 11 and in Eq. (4
– 12) that the signs of Q and F need to be switched. Figure 4 – 12 shows the free body
diagram of the screw thread while it’s being acting on by the load and the resisting
moment. From Eq. (4 – 12), the governing equations change to
0  Q  F cos  N sin  ,
0   P  F sin   N cos
(4 – 14)
The minimum moment is applied when F and N are related by the dry friction model
given in Eq. (4 – 1). Substituting Eq. (4 – 1) and the relationship Q = M/R into (4 – 14)
yields the resisting moment
Mr 
(4 – 15)
 s  tan 
RP.
1   s sin 
In Eq. (4 – 15) we see that Ma could actually be positive or infinite depending on the
denominator 1   s tan  . When 1   s tan   0 the resisting moment becomes infinite,
meaning that it’s theoretically impossible to turn the screw. In order to be able to turn
the screw, the coefficient of static friction must satisfy the relationship  s  cot  . As
 s increases and approaches cot  , the moment steadily increases. Also, note that the
minimum moment required to sustain the turning of the screw is determined from Eq.
(4 – 15) by replacing  s with  k .
It’s also worth pointing out that one would expect the assisting moment to be
lower than the resisting moment. To see whether this is true, notice in Eq. (4 – 13) that
 0
 0
Ma  s
RP   s RP and in Eq. (4 – 15) that M r  s
RP   s RP. It follows
1 0
1 0
that Ma is lower than Ms as expected.
Disks and Collar Bearings
Figure 4 – 13: The dry friction on the
bottom surface of the collar bearing
resists turning motion.
Photo: DK X
Disks and collar bearings, like square threaded screws, are used in engineering
systems to support axial loads. Figure 4 – 13 shows a collar bearing or a disk being
acted on by a load P and a moment M. Figure 4 - 13 represents a disk when the inner
6
radius is zero. The problem is to find the minimum moment M needed to turn the shaft
while it’s being acted on by the load P. The analysis performed here to predict M is
similar to the analysis in the previous sub-section to predict the assisting moment Ma
and the resisting moment Mr, although there are two notable differences. First, the
contact surface of the collar bearing in not inclined whereas the screw had a lead angle
associated with an inclined thread. Secondly, the contact surface of the collar bearing is
wide whereas the contact surface of the thread was thin.
Figure 4 – 14: Transition diagrams for
the collar
Figure 4 – 14 shows several diagrams; the free body diagram of the collar, an
infinitesimal element of the contact surface, and a ring-shaped infinitesimal area used
in a minute for integration. As shown in the free body diagram, the collar is acted on by
a load P and a moment M on the top surface and a distributed force on the bottom
surface. The distributed force maintains the collar in equilibrium in reaction to P and
M. The normal component of the distributed force is ideally uniform when the contact
surface is dry (unlubricated). Assuming that the normal force is uniform, the part of the
normal force that acts over a surface element of area dA is
dN 
(4 – 16)
N
P
dA 
dA,
2
A
 ( R2  R12 )
where R1 and R2 are the inner and outer radii, and A   ( R22  R12 ) is the area of the
contact surface. By using a dry friction model, the friction force acting over the
element becomes
dF   s dN 
s P
 ( R22  R12 )
dA.
The resultant moment acting on the shaft by the friction force is then

M  rdF 
(4 – 17)

r s P
  (R
R2
r s P
R1
 ( R22  R12 )

2
2
 R12 )
dA
2rdr ,
where the element of area has been taken to be an infinitesimally thin disk. From Eq. (4
– 17)
2  R 3  R13
M   s P  22
3  R2  R12
(4 -18)

.


In the case of a disk, R1 = 0 so from Eq. (4 – 18) the minimum moment M to turn a disk
is
M
(4 -19)
7
2
 s RP.
3
Also, notice that the minimum moment required to sustain the spinning of a disk or
collar bearing is determined from Eq. (4 – 18) or (4 – 19) by replacing  s with  k .
Journal Bearings
Figure 4 – 15: A journal bearing
Photo: DK X
Journal bearings are designed to support lateral loads, whereas the collar
bearings treated in the previous sub-section were designed to support axial loads.
Figure 4 – 15 shows a journal bearing and a shaft. Notice that the lateral load P presses
the shaft against the journal bearing, causing the shaft to make contact with the journal
bearing at point A. The shaft is also acted on by a moment M. The problem is to find
the minimum moment M needed to turn the shaft. The location of point A is also not
yet known.
Figure 4 – 16 shows the free body diagram of the shaft. The shaft is acted on
by the lateral load P, the applied moment M, and a reaction force S that has been
resolved into a normal component N and a friction component F. By summing forces in
the direction of the load one concludes that S = P. By summing moments about the
long axis of the shaft,
Figure 4 – 16: Free body diagram of the
shaft.
0  M  RF
(4 – 20)
Assuming that the journal bearing is dry (unlubricated), the friction force is related to
the normal force by the dry friction model, so Eq. (4 – 20) reduces to
M   s RP sin 
(4 – 21)
where  is the angle of static friction. The angle of static friction is related to the
coefficient of static friction by tan    s . Using the trigonometric identity
sin   tan  1  tan 2  it follows from Eq. (4 – 21) that the minimum moment
needed to turn the shaft is
M
(4 – 22)
s
1   s2
RP ,
The angle that contact point A makes relative to the line of the load is equal to the angle
of static friction 
Friction Windings
Figure 4 – 17: Different friction
windings.
DK X
A friction winding is a winding of a belt or a belt-like object around another
object, most commonly a round object, in which the belt or the belt-like object is held
in position by friction. Belt-like objects are frequently chains, ropes, wires, and strings
and the round objects are frequently columns, disks, and pulleys. In a friction winding,
the tension on one side of the winding differs from the tension on the other side of the
winding depending on the friction force between the belt and the object. The problem
is to find the tension forces in the winding.
Figure 4 – 18 shows a typical winding. The angle w is called the winding
angle. The winding angle can be a small fraction of 360○ or it can be considerably
Winding Angle
8
Figure 4 – 18: A typical winding
larger than 360○. Also, notice that a free body diagram of a small section of the belt has
been drawn. The free body diagram shows the forces acting on the belt section just
before the belt slides in the clockwise direction. The two sides of the belt section are
acted on by tension forces that differ by a small amount dT and the round object exerts
a small force on the belt section that has been resolved into a normal component dN
and a friction component dF. Summing forces is the radial and circumferential
directions yields
0  T  dF  (T  dT ) cos d ,
,
0  dN  (T  dT ) sin d .
(4 – 23)
The angle d is infinitesimal, so cos(d ) = 1 and sin (d ) = d where dis measured
in radians. Next, assume that contact is made between the round object and the bottom
side of the belt. Belts that make contact on their bottom sides are called flat belts.
Representing the friction by the dry friction model, the friction force is related to the
normal force by dF = sdN and Eq. (4 – 23) reduces to
1
dT   s d .
T
(4 – 24)
where the term dT·d goes to zero. Equation (4 – 24) shows how the change in tension
over the belt section depends on the change in the angle. Integrating Eq. (4 – 24) from
the left contact point to the right contact point, gives

TB 1
TA
T
dT 

w
0
 s d ,
so
ln TB  ln T A   s w .
(4 – 25)
Solving Eq. (4 – 25), we get the tension on the right side expressed in terms of the
tension on the left side, the coefficient of static friction and the winding angle, given by
TB  T A e
(4 – 26)
 s w
.
Notice that the minimum tension TB to initiate clockwise sliding is larger than the
tension TB on the left, as expected. Equation (4 – 26) also shows that TB increases with
s and , which are also expected, but is independent of the radius R..
Rolling Resistance
Figure 4 – 19: The contact surface under
a rolling object deforms ever so slightly.
Ideally, a round wheel will roll forever, assuming that the contact force is
perfectly horizontal, that the wheel is perfectly round, that the wheel and the contact
surface are rigid, and that no other forces act on the wheel. In reality, the situation is
different. When a wheel rolls on a surface, a load P acts on the contact surface, causing
the surface to deform, ever so slightly. As shown in Fig. 4 – 20, contact is made over a
deformed surface rather than at a point. Figure 4 – 20 also shows the line of the
resultant force S going through point C. A force Q on the wheel is needed in order to
maintain the wheel in translational equilibrium (rolling at a constant speed). The
9
problem is to find the minimum rolling force Q needed to maintain the wheel in
equilibrium.
Minimum Rolling Force
Coefficient of Rolling Resistance
The line of the contact force needs to go through point C in order to maintain
the wheel in rotational equilibrium. Let the resultant force make contact with the wheel
at point A. Point A is a distance a from the vertical axis of the wheel. The quantity a,
which is called the coefficient of rolling resistance, is regarded as a property of the
wheel and the contact surface.
Assuming that the angle  in Fig. 4 – 20 is small, the approximation cos =
1, and sin =  can be used. It follows that a = R, where R is the radius of the wheel.
Summing forces in the x and y directions,
Figure 4 – 20: To maintain rotational
equilibrium, the contact force S acts
through the wheel’s mass center.
0  Q  S sin  ,
0  P  S cos .
(4 – 27)
Using the small angle approximation again,
Q  Pa / R
.
(4 – 28)
Equation (4 – 28) expresses the minimum rolling force needed in terms of the load P,
the coefficient of rolling resistance a, and the radius of the wheel R.
Key Terms
Assisting Moment, Coefficient of Kinetic Friction, Coefficient of Static Friction, Collar Bearings, Disks, Dry
Friction, Friction Force, Friction Winding, Journal Bearings, Kinetic Friction Force, Kinetic Model of Dry Friction,
Lead, Lead Angle, Maximum Static Friction Force, Mean Radius, Minimum Rolling force, Normal Force, Resisting
Moment, Rolling Resistance, Roughness Angle, Self-Locking, Square-Threaded Screw, Static Model of Dry Friction,
Wedge, Winding Angle
Review Questions
1. Describe the name of this section.
2. What are the unique elements of this template?
Examples
Fig. 1a
4 – 1 A 20 ft ladder is leaned up against a wall in a house. The bottom of the ladder
rests on a small carpet. The coefficient of static friction between the carpet and the
floor is s = 0.5. The wall is assumed to be smooth. Determine whether the ladder slips
when a man climbs up it. If it does slip, how far up the ladder has the man climbed
before it slips? Neglect the weight of the ladder.
10
Fig. 1b
Solution: The problem is best set up by regarding the system as one planar rigid body
consisting of the man, the ladder, and the small carpet. To transition to equations, the
free body diagram is drawn. Notice in Fig. 1b that the unknown weight of the man was
denoted by W. One can only hope that the man’s weight does not affect the answer.
Otherwise, the problem can’t be solved. Also notice that the friction force was taken to
be positive to the right. If the friction force had been a simple reaction, like the normal
force at B, the direction of the force taken in the free body diagram would not have
been very important. For example, if the wrong direction is assumed for the normal
force NB then in the answer step we’d find that NB is negative, and the analysis would
still be correct. On the other hand, the friction force is not a simple reaction. We’ll see
momentarily that the directions of the friction force and the corresponding normal
force in the free body diagram need to be correct in order to get the correct
answer.
Now that the free body diagram has been drawn, let’s move on to the equation
step where the equations are listed. Summing forces in the x and y directions, and
summing moments about point A yields
Set-up – The ladder and the man are
treated as one planar rigid body.
Transition – In the free-body-diagram,
the friction force needed to be taken in
the correct direction.
Equation – The equations were found
from the equilibrium conditions and the
static model of dry friction.
Answer – The equations are not linear
because the unknown W is multiplied by
the unknown x.
Knowledge – The ladder slid to the left
when the man was 2/3 of the way up the
ladder – independent of his weight.
Would this result make sense if the
man’s weight was small compared to the
weight of the ladder?
0  F  NB,
0  W  N A ,
(a)
0  Wx  16 N B
Equations (a) are three equations in terms of the four unknowns F, NB, NA, and x.
Notice that W is not being counted as an unknown because we’re hoping that it cancels
out of the equations before we’re done. We still need an additional equation. The
additional equation comes from the static model of dry friction, given by
F  s N A
(b)
The reason that the direction of the friction force needs to be correct in the free body
diagram is explained by looking at Eq. (b). In Eq. (b), we see that taking the wrong
direction for the friction force causes Eq. (b) to be wrong, in turn causing the answer to
be wrong. This does not happen with simple reactions.
Equations (a) and (b) are four algebraic equations in terms of four unknowns. The
solution is
F  0.5W , N B  0.5W ,
(c)
N A  W , x  8 ft.
Notice that W cancelled out of the calculation for x. We now see that the ladder slips
when the man is 8 ft to the right of point A, which is about 10.7 ft above point A or 2/3
of the way up the ladder.
Fig. 2a
4 – 2 You have a new roommate moving in and you’re helping him move his empty
100 lb dresser across a moderately smooth floor. If the coefficient of static friction
between the floor and the dresser is s = 0.25, find the minimum force P needed to
move the dresser. Does the dresser slide or tip over?
Solution: The dresser is set up as a planar rigid body. In order to help transition to
equations, the free body diagram was drawn, which is shown in Fig. 2b. The forces
acting at the contact surface between the dresser and the floor could be distributed in
any of a number of ways. As shown in Fig. 2b, if the dresser had legs, forces would act
on the dresser at points A and B. If the bottom of the dresser had been flat, contact
could be made over the entire bottom surface. It really doesn’t matter, though, since the
11
dresser is a rigid body. It’s the resultant force and the resultant moment acting at the
contact surface that affect the dresser’s equilibrium.
Fig. 2b
Summing force in the x and y directions, and summing moments about point A,
yields
0  P  F,
Set-up – The dresser is a planar rigid
body.
0  100  N ,
0  5P  1.5(100 )  (1.5  x) N
(a)
Transition – In the free-body-diagram,
the force acting on the contact was
treated as a single force acting at an
unknown location x.
Equations (a) are three equations in terms of the four unknowns P, F, N, and x. We
need an additional equation, which comes from the static model of dry friction
Equation – The equations were found
from the equilibrium conditions and the
static model of dry friction.
The solution to (a) and (b) is
Answer – The equations are not linear
because the unknown N is multiplied by
the unknown x.
Knowledge – The dresser did not tip
over. It slid when the applied force was
25 N, which is 25% of the dresser’s
weight.
Fig. 3a
Fig. 3b
F  0.25 N
(b)
P  25 N, F  25 N,
N  100 N, x  1.25 ft.
(c)
Since x is smaller than the 1.5 ft, we see that the dresser slides before it tips over. (The
dresser tips just after x = 1.5). Note that this problem could also have been solved by
assuming that the dresser had legs and that the reactions at the contact surface were
concentrated at points A and B. If you do this, you’d see that the two approaches are
related to each other by N = NA + NB, and F = FA + FB, in which FA  0.25 N A and
FB  0.25 N B . Checking that x is less than 1.5 is equivalent to checking that NA and NB
are both positive. The dresser would have tipped if x = 1.5 or, equivalently, if NA = 0.
4 – 3 Two blocks A and B are resting on each other when a force P is applied, as
shown. Find the minimum force P needed to cause either of the blocks to move. Let the
coefficient of static friction between the two blocks be 0.9 and let the coefficient
of static friction between block B and the ground be B0.2. Also, assume that the
blocks weigh WA = 10 lb and WB = 50 lb.
Solution: This system consists of two planar rigid bodies. To transition to equations,
free body diagrams of each block were drawn (See Fig. 3b). Notice that there are two
possibilities. Either a) block A slides and block B remains at rest, or b) both blocks
move together as a single body. We don’t yet know which of the possibilities occurs
when the minimum force P is applied. So, we can guess either one of the possibilities
and determine whether it makes sense or leads to a contradiction. For no particular
reason, we’ll guess the first possibility.
Let’s now list the equations. From the free body diagrams, summing forces in the x
and y directions yields
Set-up – The system is two planar rigid
bodies – the two blocks.
Transition – Remember that the friction
forces need to be assumed in the correct
directions.
(a)
For Block A:
(b)
For Block B:
0  P  FA , 0  N A  10
0   FA  FB , 0   N A  N B  50
and from the static model of dry friction for the surface between the blocks
FA  0.9 N A .
(c)
12
Equations (a), (b), and (c) are five linear algebraic equations in terms of the five
unknowns P, NA, FA, FB, and NB. The solution is
(d)
Equation – There were two possible
sets of equations depending on the
impending motion. The equations
were found from the equilibrium
conditions and the static model of dry
friction between the blocks.
Answer – Five linear
equations were solved.
algebraic
Knowledge – The top block slid and
the bottom block remained stationary
even though the coefficient of static
friction of the top block was 4.5 times
greater than the the coefficient of static
friction of the bottom block.
Fig. 4a
Fig. 4b
P  9 N, N A  10 N, FA  9 N, FB  9 N, N B  60 ft.
Next, since no impending motion is assumed at point B, FB needs to be less than FB =
BNB = 0.2(60) =12 lb. Indeed FB = 9 lb which is less than 12 lb so the guess that block
A would slide and block B would remain at rest was correct. Had it been incorrect, the
solution would have needed to be reworked assuming that both blocks move together
as a single body.
In this example, notice that the top block slipped relative to the bottom block and
that the bottom block remained stationary even though the coefficient of static friction
of the top block was much greater than the static coefficient of the bottom block.
This example also illustrates an interesting feature of many dry friction problems. In
many dry friction problems there is more than one contact surface and the nature of the
impending motion is not known in advance. When this happens several possibilities
will arise, necessitating the different possibilities to be examined. In a properly
formulated problem, one of the possibilities will lead to no contradictions and the
others will all lead to contradictions.
4 – 4 Three cement pipes were stacked on top of each other, as shown. The interest lies
in gaining a wide-ranging understanding of the limitations of this stacking, given that
the coefficient of static friction between any two pipes is A and between any pipe and
the ground is C. The question arises for which values of A and C is the stack stable.
Solution: Notice that no numbers have been supplied, so the interest lies in expressing
the results in terms of the system parameters. The system is three pipes, which are each
planar rigid bodies. Also, notice that this problem possesses a symmetry that can be
used to simplify the problem. Since the lower pipes are identical to each other, we
could imagine looking at the pipes from the front or the back and the problem would be
the same. This means that the reactions on the left and right pipes are the same, as the
three free body diagrams in Fig. 4b show. Also, because the impending motion of pipe
C is downward and the impending motion of the lower pipes is outward, the friction
forces acting on pipe C are always upward, and the friction forces acting on the lower
pipes are always inward. Because of symmetry, it turns out that it is sufficient to
consider the equilibrium of the lower left pipe and the vertical equilibrium of the whole
system. The other equilibrium conditions are redundant, meaning that they’ve already
been considered, in effect, through the symmetry. From Fig. 4b,
0  FA  FC sin   N C cos ,
(a1-3)
For Pipe A:
0  N A  W  N C sin   FC cos
0  RF A  RFC
(b)
Set-up – The system is three planar rigid
bodies.
Transition – The symmetry in the
system simplified the problem. Only the
equilibrium of the left pipe and the
equilibrium of the system in the vertical
direction needed to be considered.
For the System: 0  2 N A  3W .
The values of A and C for which the system is stable are found by first finding the
values of A and C where the system transitions from being stable to being unstable.
There are two possibilities. The transition takes place a) when the top pipe slides down
and the two bottom pipes roll outward or b) when the top pipe rolls down and the two
bottom pipes slide outward. From the static model of dry friction for the surfaces
13
FC   C N C or FA   A N A.
(c1-2)
Equations (a), (b), and (c1) or (c2) are six equations.
Equation – The equations were
nonlinear because of the unknown
angle. There were two possibilities
to consider depending on the
impending motion.
Let’s first assume that the transition takes place when the top pipe slides down
and the bottom pipes roll outward. Substituting Eqs. (b), (c1), and (a3) into (a1) yields
Answer – In addition to solving
equations, inequalities needed to be
checked.
In Eq. (d) we see that  depends on the coefficient of static friction and, perhaps more
interestingly, does not depend on the weight or the radius of any pipe (Can you provide
a purely logical reason for this by looking at the system?) Also, referring back to Fig.
4a, notice that the angle  can’t be larger than when the bottom two pipes touch, which
cos
is when  = 60○. When 0    60 o , we find in Eq. (d) that 0.268 
 0.5 .
1  sin 
When C > 0.5, it follows from the equations that FC > CNC (can you show this?) so
the top pipe doesn’t ever slide when C > 0.5. Similarly, when C < 0.268 it can be
shown that FC > CNC, so the top pipe always slides when C < 0.268. Substituting (d)
and (b) into (a) yields the remaining unknowns when 0.268   C  0.5 , given by
1
1
1
3
(e1-4)
FA   CW , FC  CW , N C  W , N A  W .
2
2
2
2
Knowledge – We found two
mechanisms of “failure,” depending
on the relative sizes of the
coefficients of static friction. The
weight and the radius did not affect
the
answer.
The
stability
characteristics of the system were
summarized in the graph below.
C 
(d)
cos
.
1  sin 
Next, let’s look at the friction force at A to check whether it has exceeded the limit FA =
ANA. Comparing the limiting value of FA with the value in (e), we see that the friction
limit is not exceeded if
 C  3 A .
(f)
Fig. 4c
The angle at which the top pipe slides down is found by solving Eq. (d). This is done
using the substitution x = cos The angle is found to be
 2 C
1  2
C

  cos 1 
(g)

.


Next, let’s consider the possibility that the transition takes place when the top
pipe rolls downward and the two bottom pipes slide outward. Substituting Eqs.(b),
(c1), and (a3) into (a1), we get
3 A 
(h)
cos
.
1  sin 
cos
 0.5 When 3A > 0.5,
1  sin 
it follows from the equations that FA > ANA so the bottom pipes never slide when 3A
> 0.5. Similarly, when 3A < 0.268 it can be shown that FA > ANA, so the bottom pipes
always slide when 3A < 0.268. The angle at which the bottom pipes slide outward is
found by solving Eq. (h). The angle is
Since 0    60 o , we again see in Eq. (h) that 0.268 
 6 A 
.
 1  9 2 
A 

  cos 1 
(i)
14
This analysis illustrated several things about friction problems in general in
addition to the characteristics of pipes stacked on each other. First, with respect to dry
friction problems in general, notice that there were a fair number of possibilities and
conditions to keep track of. This is characteristic of dry friction problems with multiple
friction surfaces causing multiple mechanisms of failure. Secondly, with respect to
pipes stacked on each other, we found that the weight and radius of any pipe did not
affect stability. Stability was influenced solely by the values of A and C. To help
visualize the stability conditions of the stacked pipes, the A – C plane was graphed, as
shown in Fig. 4c. This graph displays all of the conditions found in the analysis
succinctly giving the engineer a greater understanding of the problem.
4 – 5 Derive the static model of dry friction, Eq. (4 – 1), by creating an idealized rough
surface between two solid bodies.
Fig. 5a
Solution: A rough surface between two bodies is a series of smooth bumps and
grooves that follow an irregular pattern. To idealize a rough surface, the smooth bumps
and groves can be given a regular, repeating pattern and the two surfaces can then be
“mated.” As shown in Fig. 5a, the rough surface between the two bodies is represented
by repeating triangular bumps that are up against each other. The roughness angle is
denoted by s. The free body diagram of a section of the top body is shown in Fig. 5b.
Summing forces along the x and y directions yields
0  F  ( R1  R2 ) sin  s ,
(a)
0   N  ( R1  R2 ) cos s .
Fig. 5b
Equation (a) are two equations expressed in terms of the two unknown reactions R1 and
R2. Solving for R1 and R2 gives
(b)
R1 
F cos s  N sin  s
,
sin 2 s
(c)
R2 
 F cos s  N sin  s
.
sin 2 s
The two reactions need to be positive or zero, that is R1  0 and R2  0. In (b) we see
that R1  0 when F > 0 and N > 0 although in (c) we see that R2 could be less than zero
depending on s. Sliding is initiated when R2 = 0, so from (c) we get the static model of
dry friction
Fs   s N , in which  s  tan  s .
(d)
We also see in (d) that the static coefficient of friction is equal to the tangent of the
roughness angle.
Fig. 6a
4 – 6 A 20 ft ladder is leaned up against a wall in a house with a 200 lb man standing
on it, as shown. A 15○ wedge is then placed under the bottom of the ladder at point A.
The coefficients of static friction between the wedge and the ladder and between the
wedge and the floor are each s = 0.5. The wall is assumed to be smooth. First, find the
15
minimum force P needed to place the wedge in position. Then determine whether the
wedge is self-locking. Neglect the weight of the ladder and the wedge.
Solution: This system is best set up as two rigid bodies; one represents the man and
the ladder and the other represents the wedge. To transition to equations, the
corresponding free body diagrams were drawn, as shown in Fig. 6b.
Fig. 6b
When the wedge is being placed in position, the impending motion of the wedge
is the right, so the friction forces acting on the wedge are to the left, as shown.
Summing forces in the x and y directions, and summing moments about point C, we get
(a)
For the wedge:
(b)
For the entire system:
Equation – The equations were found
from the equilibrium conditions and the
static model of dry friction.
Answer – The equations were linear
algebraic equations. There were solved
twice; when the wedge was put in and
when it was taken out.
Knowledge – It required 175 lb to put in
the wedge. It required +25 lb to remove
the wedge, so the wedge is self-locking.
0  N C  N A cos   F A sin 
0  P  FC  N B ,
Set-up – The system consists of two
planar rigid bodies.
Transition – In the free-body-diagram,
take note of the directions of the friction
forces.
0  P  F A cos   FC
0  N C  200 ,
0  16 N B  6(200 ).
Equations (a) and (b) are five linear algebraic equations in terms of the six unknowns
P, FA, FC, NC, NA, and NB. The additional equation comes from the static model of dry
friction at point C, given by
FC  0.5 N C
(c)
Equations (a), (b), and (c) are six linear algebraic equations expressed in terms of six
unknowns. The solution is
P  175 lb , FA  77 .6 lb , FC  100 lb ,
(c)
N C  200 lb, N A  186 lb, N B  75 lb
Notice that FA <A NA, so no sliding takes place between the wedge and the ladder.
Next, the question arises whether the wedge is self-locking. To answer this question,
imagine that the applied force P is to the left, that is, that we’re trying to remove the
wedge. This switches the directions of the friction forces since the impending motion
of the wedge has switched. Switching the signs of P, FA, and FC in Eqs. (a) and (b), the
solution to the new equations becomes
(d)
P  25 lb , F A  77 .6 lb , FC  100 lb ,
N C  200 lb, N A  228 lb,
N B  75 lb
Notice that | FA | <A NA, so there is no sliding between the wedge and the ladder. Also,
since P is positive, so the wedge is self-locking.
Fig. 7a
4 – 7 A 25 lb wheel rests against a smooth wall. Find the minimum force needed to
place a 15○ wedge under the wheel, so that the wedge is not touching the floor. The
coefficient of static friction between the wedge and the wheel and between the wedge
and the floor are each s = 0.2. Is the wedge self-locking? Neglect the weight of the
wedge.
16
Solution: This system consists of two rigid bodies; the wheel and the wedge. The
corresponding free body diagrams are shown in Fig. 7b.
Fig. 7b
When the wedge is being placed under the wheel, the impending motion of the
wedge is the left, so the friction forces on the wedge are to the right, as shown.
Summing forces in the x and y directions, and summing moments about point C,
(a)
For the wedge:
(b)
For the entire system:
0  N C  F A sin 15  N A cos15
0   P  FC  N B ,
Set-up – The system consists of two
planar rigid bodies.
Transition – Two sets of free body
diagrams are considered – when the
applied force is to the right and when the
applied force is to the left. .
0   P  FC  F A cos15  N A sin 15
0  N C  25,
For the wheel, summing moments about the center of the wheel:
0  RFA ,
(c)
Equation – The equations are linear
algebraic equations.
where R is the radius of the wheel. Equations (a), (b) and (c) are five linear algebraic
equations in terms of the six unknowns P, FC, FA, NA, NC, and NB. The additional
equation comes from the static model of dry friction at surface C, given by
Answer – Since the equations are linear,
they’re solved by standard methods.
(c)
Knowledge – It required 11.5 lb to place
the wedge under the 25 lb wheel. It
required –1.70 lb to remove it, so the
wedge is not self-locking.
Equations (a), (b), and (c) are six linear algebraic equations expressed in terms of six
unknowns. The solution is
P  11 .7 lb , FC  5 lb , F A  0 lb ,
(c)
N A  25 .9 lb, N C  25 lb, N B  6.70 lb
FC  0.2 N C
Notice that FA <A NA, so no sliding takes place between the wedge and the wheel.
Next, let’s determine whether the wedge is self-locking. To do this, imagine that F acts
to the right instead of to the left. The directions of the friction forces switch since the
impending motion of the wedge is now to the right. Switching the signs of P, FA, and
FC, the solution to the new equations becomes
(d)
P  1.70 lb , FC  5 lb , FA  0 lb ,
N A  25 .9 lb, N C  25 lb,
N B  6.70 lb
Notice that | FA | <A NA, so there is no sliding between the wedge and the wheel.
However, the force P is negative so the wedge is not self-locking.
Fig. 8a
4 – 8 In a furniture manufacturing process, boards are clamped together with a
clamping force of 100 lb. The wood is clamped by an automated wrench that turns a
square-threaded screw that has a lead of 1/8 in, a radius of 3/8 in, and a coefficient of
static friction of 0.5. The wrench needs to be programmed to exert a moment of Ma
when unclamping the boards and a moment of Mr when clamping the boards.
Determine the moments Ma and Mr.
Solution: This type of system was already analyzed in the sub-section on squarethreaded screws. The unclamping moment is an assisting moment and the clamping
moment is a resisting moment. First, the lead angle needs to be calculated. From Eq. (4
– 11),
17
1/ 8
 0.053052 ,
2 (3 / 8)
which yields and angle of 3.04○. (You don’t need to know the angle for the calculations
that follow.) From Eq. (4 – 13) and (a) the unclamping force is
tan 
(a)
Ma 
(b)

 s  tan 
RP
1   s tan 
0.5  0.053052 3 / 8
100  1.36 lb  ft,
1  0.5  0.053052 12
and from Eq. (4 – 15) and (a) the clamping force is
Mr 
(c)

 s  tan 
RP
1   s tan 
0.5  0.053052 3 / 8
100  1.78 lb  ft.
1  0.5  0.053052 12
Notice that the difference between the unclamping force and the clamping force is
about 25%.
Fig. 9a
Fig. 9b
4 – 9 A 48 in long, 8 in wide, 550 lb precision table top in a manufacturing plant needs
to be leveled to within 0.1○. The table top is leveled by turning a nut in a squarethreaded screw that is built into the table’s left leg, as shown. The screw has a lead of
1/8 in, a radius of 3/8 in, and a coefficient of static friction of 0.4. The nut is turned by
exerting a force F on the end of an 8 in long wrench. Currently the table is tilting
down to the right by 1.5○. Find the force F needed to turn the nut and the number of
turns (revolutions) to level the table. Also assume that a 1200 lb machine is resting on
the table, as shown.
Solution: The system is best divided into two sub-systems; one is the table top and the
other is the leg sub-system. The table top is a planar rigid body and the leg subsystem
consists of several parts, but the important part is the square-threaded screw, which was
analyzed in the sub-section on square-threaded screws. The free body diagram of the
table top is shown in Fig. 9b. Summing forces in the y direction and summing
moments about point A,
0  PA  1200  550  PB ,
(a)
0  2(550 )  3(1200 )  4 PB .
Set-up – The system needed to be
analyzed is the table top.
Equations (a) are a pair of equations expressed in terms of the unknowns PA and PB.
Solving for the unknowns, PA = 575 lb and PB = 1175 lb. The force acting on the end of
the square-threaded screw is therefore 575 lb. From Eq. (4 – 11), the lead angle of the
screw is
Transition – The free-body-diagram
of the leg was not needed because it
was analyzed earlier in the text.
(b)
Equation – The leg reactions were
first found. Then it was observed that
the required moment is assisting. Eq.
(4 – 13) was used.
tan 
1/ 8
 0.053052 ,
2 (3 / 8)
which yields and angle of 3.04○. Since the table is being lowered at point A, the applied
moment is an assisting moment. From Eq. (4 – 13), and (b)
18
(c)
Ma 
0.4  0.053052 3 / 8
575  6.1047 lb  ft,
1  0.4  0.053052 12
Answer – A small angle assumption was
made to calculate the amount y that the
left leg needed to be lowered. This
wasn’t necessary, but the approximation
was a little simpler.
The force exerted on the end of the wrench to turn the nut is then
Knowledge – The wrench was turned
about 10 times with a force of 9.16 lb.
Since the tilt angle is small, the left end of the table needs to be lowered about y =
(48/12)(1.5·/180) = 0.1047 ft. But the lead of the screw is 1/8 in = 0.010417 ft.
Therefore, the nut needed to be turned n = 0.1047/0.010417 =10.1 times.
(d)
F
Mr
6.1047

 9.1571  9.16 lb.
8 / 12 0.6667
4 – 9B Develop formulas, similar to Eq. (4 – 13) and (4 – 15), for the assisting
moments and the resisting moments associated with V-shaped screws.
Fig. 9Ba
Solution: The screw shown in Fig. 9Ba is supported by the triangular shaped side
walls in the thread of the screw. Let’s look at the normal force dN0 acting on the side
walls. As shown in Fig. 9Bb, the resultant of the normal force produced by the side
walls is
N0 
(a)
N
cos  / 2
Referring to Eq. (4 – 12), the friction force F is now related to the normal force N0 by F
= sN0, so from (a)
Fig. 9Bb
F  eff N ,
(b)
where eff is the effective coefficient of static friction, given by
s
.
cos  / 2
(c)
 eff 
Using the trigonometric identity
1
 1  tan 2 a , (c) can be rewritten as
cos a
 eff   s 1  tan 2  / 2 ,
(d)
h
cos . Following the same steps that led from (4 –
2a
12) to (4 – 13), the assisting moment for a V-shaped thread becomes
where, from Fig. 9Bb, tan  / 2 
Ma 
(e)
 eff  tan 
1   eff tan 
RP,
in which eff is given by (d). Notice that Eq. (e) is the same as Eq. (4 – 13) except that
the coefficient of static friction s has been replaced with the effective coefficient of
19
static friction eff. Similarly, Eq. (4 – 15) can be used for V-shaped screws if the
coefficient of static friction used in that equation is the effective coefficient of static
friction, which is related to the coefficient of static friction by (d). In this way, Eqs. (4
– 13) and (4 – 15) can be used for V-shaped screws. Also,
T2  T1e
(d)
s w
sin  / 2
.
Notice by comparing (d) with (4 – 26) that the sin /2 term, which is less than 1,
increases the exponent, which in turn increases the effectiveness of the friction. One
could say that the static coefficient of dry friction becomes effectively  seff 
s
.
sin  / 2
Fig. 10a
4 – 10 A heavy man-hole cover needs to be slid across a floor and put in position.
Doing this requires two people. It has already been learned that each person needs to
exert 60 lb of force to slide the man-hole cover straight. How much force does each
person need to exert to rotate it (without translation)?
Fig. 10b
Solution: This problem requires two situations to be analyzed; translating a cover and
rotating a cover. The free body diagram of the translating cover is shown in Fig. 10a.
In Fig. 10a, the friction force is related to the normal force by the kinetic model of dry
friction, so
Set-up – Two problems needed to
be looked at.
Transition – The free-bodydiagram of the translating system
was drawn. The rotating system was
analyzed earlier in this section.
Equation/Answer – The general
solutions to the translating problem
was found and Eq. (4 – 19) was
used.
Answer – The two solutions needed
to be compared to find the answer.
Knowledge – It’s 33% easier to
rotate a cover than to translate it.
Interestingly, we solved this
problem even though we didn’t have
enough information to find k or W.
2FT   k W
(a)
where FT = 60 lb is the force to translate the cover and W is the weight of the cover.
The free body diagram of the rotating cover was already analyzed in the sub-section on
disks. From Eq. (4 – 19),
M
(b)
2
 k RW .
3
where R is the radius of the cover. The applied moment M in Eq. (b) is related to the
force FR each person exerts to rotate the cover by M = 2RFR, so from (a) and (b)
2 RFR 
(c)

2
2
 k RW  R k W .
3
3
2
R2 FT 
3
Dividing both sides by 2R,
FR 
(d)
2 FT 2(60 )

 40 lb.
3
3
We see that it’s 33% easier to rotate the cover than to translate it. Would that be true
for covers having other shapes? What shapes do you think would be the most difficult
to rotate?
Fig. 11a
20
4 – 11 A cross-beam supports a steel tension rod that is being pulled on by a P = 250 N
load. The tension rod is connected to the beam by a collar bearing having an inner
radius of R1 = 1 cm and an outer radius of R2 = 2 cm. The coefficient of static friction
between the bearing and the beam is s =0.3 and the coefficient of kinetic friction
between the bearing and the beam is k =0.2. Find the minimum moment needed to
begin twisting the rod and the minimum moment needed to sustain the twisting.
Set-up – This type of problem was
set-up in the section on collar
bearings.
Transition – Not necessary
Equation/Answer – Equation (4 – 18)
was used.
Answer – The equations were solved.
Knowledge – The beginning moment
is 1.94 N·m and the sustaining
moment is 1.30 N·m, which is 33%
less than the beginning moment.
Fig. 12a
Fig. 12b
Solution: Assume that the rod is rigid and that, by twisting the rod, the collar bearing
twists on the supporting beam. This type of problem was analyzed in the section on
collar bearings. From Eq. (4 – 18), the moment required to begin the twisting is
(a)
2  2 3  13
M  0.3(250 )  2 2
3  2 1
The moment required to sustain the twisting is
(b)
2  2 3  13  1
M  0.2(250 )  2 2 
 1.29627  1.30 N  m.
3  2  1  100
4 – 12 An axle is supported on each end by journal bearings. Find the moment M
needed to turn the axle while it’s being subjected to a P = 200 lb load, as shown. The
radius of the axle is R = 3 in and the coefficient of static friction between the bearing
and the shaft is s =0.25. Neglect the weight of the axle.
Solution: This type of problem was treated in the sub-section on journal bearings. To
find the moment M, the loads PA and PB at bearings A and B first need to be found.
They’re found by looking at the equilibrium of the shaft. The free body diagram of the
shaft is shown in Fig. 12b. Summing forces in the vertical direction and summing
moments about point A
0  PA  PB  200 , 0  3(200 )  5PB .
(a)
Set-up – The shaft is a planar rigid
body.
Solving (a), PA = 80 lb and PB = 120 lb. The moment needed to turn the shaft
overcomes the friction at both bearings. Using Eq. (4 – 22), the total moment that
needs to be applied is
Transition – The free body diagram
of the shaft was drawn.
Equation
–
Translational
equilibrium in the y direction, and
rotational equilibrium about the x
and z axes were enforced. Equation
(4 – 22) was also used.
Answer
two linear
/Answer –– The
Equation
(4 – algebraic
18) was
equations
were
first
solved
and then
used.
the formula for the minimum
moment was used.
Knowledge – The minimum
moment is 12.1 lb·ft. We also
discovered that the reactions at A
and B actually didn’t need to be
calculated.
 1

 100  1.9444  1.94 N  m.

M  MA  MB 
(b)


s
1   s2
s
1
 s2
R( PA  PB ) 
0.25
1  0.25 2
RPA 
s
1   s2
s
1   s2
RPB
RP
(3 / 12 )200  12 .127  12 .1 lb  ft.
Notice in (b) that the different reactions at A and B actually didn’t need to be
calculated, so the development of (a) wasn’t actually necessary. In (b) we found that P
could have been used in Eq. (4 – 22) to find M.
21
Fig. 13a
4 – 13 A sailor temporarily secures a boat by first wrapping a rope around a capstan
and then pulling on the other end of the rope to prevent it from slipping. Assuming that
the boat would exert no more than 1000 lb of force on the rope, and that the sailor
doesn’t want to pull on the rope with more than 50 lb of force, how many times around
the capstan must the sailor wrap the rope? The coefficient of static friction between the
rope and the capstan is 0.2.
Solution: This type of problem was treated in the sub-section on friction wrappings.
From Eq. (4 – 26)
Knowledge – The number of turns
around the capstan was under 3 even
though the coefficient of restitution was
only 0.2 and the tension was reduced by
a factor of 1000/50 = 200. Capstans are
very effective!
1000  50e
(a)
0.2 w
.
Solving for w in (a),  w  ln(1000 / 50) / 0.2  14.979 rad. The number of turns is then
n   w / 2  14.979 / 2  2.38.
4 – 14 Develop a formula, similar to Eq. (4 – 26), for the V-shaped belt shown.
Fig. 14a
Solution: The belt shown in Fig. 14a is supported by the side walls of a groove in a
disk. Let’s look at the normal force produced by the side walls on a slice of the disk.
As shown in Fig. 14b, the resultant of the normal forces produced by the side walls is
dN  2dN 0 sin  / 2
(a)
The friction force dF in Eq. (4 – 23) is related to dN0 by dF = 2sdN0, in which dN0 is
related to dN by (a). So, from (a),
Fig. 14b
dF  2 s dN 0 
(b)
 s dN
sin  / 2
Substituting (b) into (4 – 23), yields
1
1
dT 
 s d
T
sin  / 2
(c)
instead of Eq. (4 – 24). Following the same steps that led to (4 – 26) yields
T2  T1e
(d)
s w
sin  / 2
.
Notice by comparing (d) with (4 – 26) that the sin /2 term, which is less than 1,
increases the exponent, which in turn increases the effectiveness of the friction. One
could say that the static coefficient of dry friction becomes effectively  seff 
Fig. 15a
s
.
sin  / 2
4 – 15 A W = 3250 lb corvette sits on a level road with its brakes disengaged.
Determine the force Q needed to begin moving the car. The rolling resistance of the
wheels is a = 0.3 in and the wheel radii are R = 13 in.
22
Solution: This type of problem was treated in the sub-section on rolling resistance.
First, the equilibrium of the car is examined. The free body diagram of the car is shown
in Fig. 15b. Summing forces in the vertical direction and summing moments about
point A
0  N A  N B  3250 ,
(a)
0  (53 / 12 )3250  (105 / 12 ) N B  (36 / 12 )Q
Fig. 15b
The force needed to push the car overcomes the friction at each of the wheels. Using
Eq. (4 – 28), summing forces in the horizontal direction yields
(b)
Set-up – The car is a planar rigid body.
Transition – The free body diagram of
the car was drawn.
Q  Q A  QB  N A a / R  N B a / R
 ( N A  N B )a / R  3250(0.3/1 3)  75 lb.
Notice in (b) that NA and NB didn’t need to be individually calculated, so the second
equation in (a) wasn’t necessary. In (b) we saw that W was used in Eq. (4 – 28) to find
Q. In any event, from (a) and (b), NA = 1580 lb and NB = 1670 lb.
Equation – Translational equilibrium in
the x and y directions, and rotational
equilibrium about the z axis were
enforced. Equation (4 – 28) was also
used.
Answer
The (4two
/Answer – –Equation
– 18)translational
was used.
equilibrium equations were solved to
determine the minimum force Q.
Knowledge – The minimum force Q is
75 lb. We also observed that the
reactions at A and B didn’t really need to
be individually calculated.
23
Problems
4 – 1 (L)
A broom is pushed across the floor. If the coefficient of static friction between the broom and the floor is 0.4, find the
largest angle that the broom can be pushed.
Answer: = 68.2○.
4 – 2 (L)
A block is resting on an inclined surface. What is the smallest that the coefficient of static friction can be?
Answer: s= 1.
4 – 3 (L)
An industrial slide in an automated production process moves delicate parts from one station to the next. If the coefficient
of kinetic friction between the part and the slide is 0.4, find the slide angle that causes the part to slide at a constant
speed.
Answer: = 21.8○.
4 – 4 (L)
Find the smallest coefficient of static friction that allows the 180 lb person to hoist the 60 lb crate without slipping.
Answer: s = 0.340.
4 – 5 (L)
A 100 lb block on an inclined surface is being pulled on by force P, as shown. The coefficient of static friction between the
block and the surface is 0.6. What is the smallest force P1 needed to move the block upward? What is the smallest force P2
needed to prevent the block from sliding downward?
Answer: P1 = 110 lb, P2 = 18.3 lb.
4 – 6 (L)
Find the smallest force P that can move the 50 N block. The coefficient of static friction is 0.5,  = 20○ and = 30○.
Answer: P = 36.4 N.
4 – 7 (L)
A 100 lb piece of metal has been machined with triangular shaped grooves and a contact surface has been machined with
matching grooves. The groove angles are 35○, as shown. If the contact surface is smooth, find the minimum force P needed
to move the piece of steel. What is the equivalent value of the coefficient of static friction?
Answer: P = 70.0 lb, s= 0.700.
4 – 8 (L)
A 50 N block of wood resting on an inclined surface is acted on by a horizontal force of P = 300 N, as shown. What
minimum downward force Q1 prevents the block from sliding to the right? What minimum downward force Q2 causes the
block to slide to the left? Let s= 0.3 and = 60○.
Answer: Q1 = 268 N, Q2 = 20.8 N.
4 – 9 (L)
A 20 lb beam is pinned at one end and leaning against a block of wood on the other end. Find how much force P is needed
to remove the block. The coefficients of static friction are A = 0.2 and B = 0.3.
Answer: P = 5 lb.
4 – 10 (L)
A ladder is resting on a smooth wall, as shown. Find the smallest coefficient of static friction between the ladder and the
ground that’ll support the ladder.
Answer: s = 0.188.
24
4 – 11 (L)
A wire is being pulled with 40 lb of force using a wire stripper. Find the smallest gripping force P needed to prevent the
wire from slipping through the stripper’s teeth. The coefficient of static friction between the wire and the teeth is 0.8.
Answer: P = 12.5 lb.
4 – 12 (L)
Two 10 lb bars are being pressed down with force P, as shown. Find the minimum force needed to cause the bars to slip on
the floor
Answer: P = 20 lb.
4 – 13 (L)
A 30 lb post and a 50 lb post are resting on a concrete slab, as shown. Find the maximum force P that can be applied
without causing either post to slide.
Answer: P = 7.69 lb.
4 – 14 (L)
It requires 200 lb·ft to loosen the pipe shown. If a wrench is used to loosen the pipe, first find how much the wrench needs
to be tightened (the normal forces N acting on the pipe). Then find how much force P is needed to turn the wrench. The
coefficient of static friction between the pipe and the wrench teeth is 6.0 and the diameter of the pipe is 1 in.
Answer: N =35.1 lb, P = 21.1 lb.
4 – 15 (L)
A 0.3 lb spool of textile yarn is pulled on with a force P, as shown. For what angle  will the spool remain in static
equilibrium? If the coefficient of static friction between the spool and the horizontal surface is 0.4, what is the largest force
P for which the spool will remain in static equilibrium?
Answer:  = 48.2○. P = 0.124 lb.
4 – 16 (L)
A 100 N disk is placed in a corner. Find the minimum moment M needed to turn the disk counter-clockwise. The radius of
the disk is 20 cm and the coefficients of static friction between the surfaces are 0.2.
Answer: M = 4.62 N·m.
4 – 17 (L)
The coefficient of static friction between a disk and a bar is A and the coefficient of static friction between a horizontal
surface and the disk is B. Find the smallest coefficients of static friction A and B that will support the bar.
Answer: A = 1, B = 0.667.
4 – 18 (L)
Find the smallest weight WA needed to cause the blocks to slide to the left. Let WB = 60 N, A = 0.4, B = 0.6, A = 30○ and
B = 60○.
Answer: WA = 456 lb.
4 – 19 (L)
A light mechanism is designed to slide a 50 lb block to the right. Find the minimum force P needed to slide the block. The
coefficient of static friction is 0.2 and  = 30○.
Answer: P = 13.1 lb.
4 – 20 (L)
Two structural members stay together by friction, as shown. Find the weight W needed to overcome the friction and cause
the horizontal bar to drop. The coefficient of static friction is s = 0.6,  = 60○, and the weights of the members are WAC =
15 lb and WBC = 30 lb.
Answer: W = 750 lb.
25
4 – 21 (M)
Find the shortest arm c on the clamp for which the clamp will bind.
a  sb
Answer: c 
.
2 s
4 – 22 (M)
A drawer is being pulled open. How far to the right (d) can the drawer be pulled without it getting stuck? Let s = 0.75.
Answer: d = 12 in.
4 – 23 (M)
A 20 lb wheel is secured on the left by a tension wire while being pulled on to the right by a force P. Find the minimum
force P needed to slide the wheel. The coefficient of static friction is s = 0.3.
Answer: P = 18 lb.
4 – 24 (M)
Find the smallest force P that can move block B. Let WA = 30 N, WB = 20 N, WC = 30 N, A = 0.5, B = 0.4, and C = 0.3.
Answer: P = 35 N.
4 – 25 (M)
An engineer is sanding a 5 lb block on a belt sander. If he’s pressing down on the block at a 30 ○ angle, what force P is he
applying to keep the block in equilibrium? The coefficient of kinetic friction between the block and the belt is 1.5.
Answer: P = 64.6 lb.
4 – 26 (M)
A heavy package needs to be dragged across the floor at a constant speed. The coefficient of static friction between the
package and the floor is 0.8. Find the angle that minimizes the force P needed to exert.
Answer: = 38.7○.
4 – 27 (M)
An engineer is sanding a 5 lb block on a belt sander. Find the minimum force that the engineer needs to apply to keep the
block in equilibrium. What is the angle of the force? The coefficient of kinetic friction between the block and the belt is
1.5.
Answer: P = 20.2 lb, = 21.8○.
4 – 28 (M)
A crane is pulling a 1000 lb shipping container with force of P at a 40○ angle, as shown. Find the maximum force for which
the container remains stationary. The coefficient of static friction between the container and the ground is s = 0.3.
Answer: P = 312 lb.
4 – 29 (M)
A 20 lb triangular block rests on a 25 lb rectangular block while being pressed on by a force P. Find the minimum force P
needed to initiate motion. The coefficient of static friction between the block and the horizontal surface is 0.25 and
between the two blocks is 0.2. The angle of the triangular block is  = 25○.
Answer: P = 57.4 lb.
4 – 30 (M)
Two 15 N blocks are arranged as shown. Find the minimum force P needed to slide the blocks. The coefficients of static
friction between the block and the horizontal surface and between the two blocks are 0.3. The vertical surface is smooth
and  = 30○.
Answer: P = 24.9 N.
4 – 31 (M)
An industrial drum sander in a furniture plant consists of a sanding disk, a motor and a uniform elevation bar. The disk and
the motor weigh a combined 20 lb and the elevation bar weighs 5 lb. Find the minimum moment needed to turn the disk
counter-clockwise. The coefficient of static friction between the disk and the wood is 1.5.
Answer: M = 3.84 lb·ft.
26
4 – 32 (M)
Three identical cylinders are stacked on top of each other, as shown. The coefficient of static friction between the cylinders
is A and the coefficient of static friction between the cylinders and the ground is B. Find the smallest coefficients of static
friction A and B for which the stack is stable.
Answer: A = 0.268, B = 0.0893.
4 – 33 (M)
A disk is resting in a triangular groove when a moment M is applied. The moment can either cause the disk to roll or cause
it to slip. Find the smallest coefficient of static friction for which the disk rolls. Let  = 30○.
Answer: s= 0.578.
.
4 – 34 (M)
Two identical blocks are secured, as shown. Find the largest angle  that supports the inner block. The coefficient of static
friction between the surfaces is s.
Answer:   tan 1 (3 s ).
4 – 35 (M)
Find the largest possible gripping angle if the coefficient of static friction between the nut and the pliers is s.
Answer:   2 tan 1  s .
4 – 36 (M)
A desert menu rests on a moderately smooth dinner table. Find a formula for the largest distance d that the menu can be
opened and still stand up. The coefficient of static friction between the menu and the table is s(s
2c s
Answer: d 
 2c s . 
1   s2
4 – 37 (M)
A simple bar can be used in mechanisms to prevent an object from sliding in one direction while allowing it to move freely
in the opposite direction. The coefficient of static friction is s. Find the largest distance d that prevents the object from
moving to the right.
Answer: d 
s
1   s2
L.
4 – 38 (M)
Find the largest angle that the fork can be rotated and still support the bar. The coefficient of static friction is s.
b  2c
Answer:   tan 1 ( s
).
b
4 – 39 (M)
A sphere of weight W is placed in a corner. Find the minimum moment M needed to turn the disk counter-clockwise if the
radius of the disk is R, the coefficient of static friction between the sphere and the horizontal surface is s and the vertical
surface is smooth.
Answer: M = sRW.
4 – 40 (M)
A disk of weight W is resting in a groove, as shown. Find the largest moment that maintains the disk in equilibrium.
Assume that stan
 s RW
.
Answer: M 
(1   s2 ) cos 
27
4 – 41 (M)
A 45 lb case is being subjected to the forces shown. How large can P2 be before the case slides on the ground? Let s = 0.4,
P1 = 10 lb,  = 50○, and  = 20.
Answer: P2 = 8.42 lb.
4 – 42 (M, C)
A spring pulls on a collar, as shown. The unstretched length of the spring is 4 cm and the coefficient of static friction
between the collar and the shaft is 0.3. Find the largest distance y that the collar can be placed and remain there.
Answer: y = 3.26 cm.
4 – 43 (M, C)
A spring pulls on a collar, as shown. The unstretched length of the spring is 4 cm and the coefficient of static friction
between the collar and the shaft is 0.3. Find the largest distance x that the collar can be placed and remain there.
Answer: x = 1.90 cm.
4 – 44 (H)
A collar of length L having an inner diameter of dC is designed to slide over a shaft having an outer diameter of dS. Find an
expression for the coefficient of friction at which the system shown “locks up.” You’ll see that the expression reveals that
sliding friction is overcome by minimizing the ratio between the clearance and the long sliding dimension. This ratio shows
up and is minimized in the many engineering systems.
Answer:  s 
L
dC

L

d d
S
 C




4 – 45 (H)
A 25 lb block rests next to a 30 lb wheel when the wheel is subjected to a moment M, as shown. The radius of the wheel is
2 in and the coefficient of static friction between the surfaces is 0.4. Find the minimum moment needed to cause the wheel
to move.
Answer: M = 2.41 lb·ft.
4 – 46 (H)
A 75 lb bar is resting on a 25 lb cylinder, as shown. The coefficients of static friction between all of the surfaces are s.
Find the smallest s for which the system remains at rest in the position shown. Let  = 60○.
Answer: s = 0.577.
4 – 47 (H)
A light rod with a 50 lb drum at its end is being hoisted up a smooth, inclined track using a 50 lb counterweight. At what
angle can the system jam? Let s = 0.8,  = 30○.
Answer:  = 37.5○.
4 – 48 (H, C)
An 8 in long mixing stick is resting on a 3 in wide shot glass. If the coefficient of static friction between the stick and the
glass is 0.2, find the smallest angle that supports the stick.
Answer:  =16.0○.
4 – 49 (H, C)
A pole is being lowered, as shown. At what angle does the pole slip?
Answer:  = 41.7○.
4 – 50 (H, C)
A 75 lb bar is resting on a 25 lb cylinder, as shown. Find the largest angle for which the system remains at rest. Let A =
0.4, B = 0.6 and C = 0.4.
Answer:  = 61.9○.
28
4 – 51 (M)
A light wedge is inserted between two 15 lb blocks, as shown. Find the minimum force P that would cause the blocks to
slide outward. Let s = 0.6 and  = 25○. Is the wedge self-locking?
Answer: P = 11.7 lb, no.
4 – 52 (M)
A light wedge is inserted between two 20 lb cylinders, as shown. Find the minimum force P that would cause the cylinders
to slide outward. Let A = 0.5, B = 0.8, and  = 30○. Is the wedge self-locking?
Answer: P = 20 lb, no.
4 – 53 (M)
A light wedge is inserted between a wall and a 100 lb block, as shown. Find the minimum force P that would cause the
block to slide. Let A = 0.6, B = 0.6, C = 0, and  = 20○. Is the wedge self-locking?
Answer: P = 285 lb, no.
4 – 54 (M)
A light wedge is inserted between a wall and a 100 lb block, as shown. Find the minimum force P that would cause the
block to slide. Let A = 0, B = 0.6, C = 0.6, and  = 20○. Is the wedge self-locking?
Answer: P = 285 lb, no.
4 – 55 (M)
A light wedge is inserted between a wall and a 100 lb block, as shown. Find the minimum force P that would cause the
block to slide. Let A = 0.6, B = 0.6, C = 0.6, and  = 20○. Is the wedge self-locking?
Answer: P = 141 lb, no.
4 – 56 (M)
A light wedge is between a wall and a 100 lb block that is acted on by a Q = 100 load, as shown. Find the minimum force P
needed to pull out the wedge. Let s = 0.6 and  = 20○.
Answer: P = 6.39 lb.
4 – 57 (M)
A light wedge is inserted between a smooth inclined surface and a 500 lb container, as shown. Find the minimum force P
that would cause the container to slide. Let s = 0.5 and  = 20○.
Answer: P = 45.3 lb.
4 – 58 (M)
A light wedge is tilting a 1000 lb crate against a wall, as shown. Find the minimum force P1 that would cause the crate to
slide up the wall. Then, find the minimum force P2 needed to remove the wedge. Let A = 0.5, B = 0, C = 0, and  = 10○.
Answer: P1 = 676 lb, P2 = 324 lb.
4 – 59 (M)
A light wedge is tilting a 1000 lb crate against a wall, as shown. Find the minimum force P1 that would cause the crate to
slide up the wall. Then, find the minimum force P2 needed to remove the wedge. Let A = 0, B = 0.5, C = 0.5, and  =
10○.
Answer: P1 = 1180 lb, P2 = 349 lb.
4 – 60 (M)
A light wedge is inserted between a wall and a cylinder, as shown. Find the minimum force P that would cause the cylinder
to slide. Let A = 0.4, B = 0.4, C = 0.7, and  = 30○.
Answer: P = 55.5 lb.
4 – 61 (L)
Two plates are held together using a ½ in diameter bolt, as shown. The design calls for compressing the spacer 200 lb.
When tightening the bolt, how much torque should be applied? The lead of the bolt is 3/32 in and its coefficient of static
friction is s = 0.
29
Answer: Mr = 1.78 lb·ft.
4 – 62 (L)
A square threaded rod is attached to two gears, as shown. The left gear carries a 200 lb load and the right gear does not
carry any load. Find the minimum moment required to lower the rod. The radius of the thread is 3/8 in, its lead is 1/8 in and
its coefficient of static friction s = 0.5.
Answer: Mr = 3.55 lb·ft.
4 – 63 (L)
A wood-working table clamp is being used to clamp a piece of wood. Find the resisting moment needed to clamp down on
the wood with 250 lb of force. The radius of the thread is 3/8 in, its lead is 1/8 in and its coefficient of static friction s =
0.5. Also, let a = 4 in, and b = 6 in.
Answer: Mr = 11.1 lb·ft.
4 – 64 (M)
A 50 lb beam is supported by two square threaded screws, as shown. Each beam is cushioned using a rubber washer with a
spring constant of 250 lb/ft and the system is perfectly balanced, meaning that the beam is perfectly horizontal. Then a 70
lb load is placed on top of the beam, shift the mass center to point C. Find the moment needed to raise the left end and
balance the system. Also, predict how many turns will be needed to balance the system. The radius of the thread is 3/8 in,
its lead is 1/8 in and its coefficient of static friction s = 0.4. Also, let a =2 ft and b = 4 ft.
Answer: Mr = 1.16 lb·ft, n = 15.4 times.
4 – 65 (M)
A jack is subjected to a 2,200 lb automotive load, as shown. Find the moment needed to raise the jack. The radius of the
thread is 3/8 in, its lead is 1/8 in, its coefficient of static friction s = 0.3, and  = 30○.
Answer: Mr = 42.7 lb·ft.
4 – 66 (M)
A 150 lb concert stand supports a 250 lb audio mixer in the position shown. Find the moment needed to lower the mixer.
The radius of the thread is 3/8 in, its lead is 1/8 in and its coefficient of static friction s = 0.5.
Answer: Ma = 1.25 lb·ft.
4 – 67 (M)
A motorized platform supports a 250 lb load. Find the motor moment needed to raise the platform. The radii of the threads
are 3/8 in, their leads are 1/8 in and their coefficients of static friction are s = 0.4.
Answer: Mr = 3.46 lb·ft.
4 – 68 (M)
Truss ABC is being reinforced using the tension wire AD. How much torque is needed to tighten the tension wire in order
to reduce the compressive load in member AB to 240 lb? The radius of the thread is 3/8 in, its lead is 1/8 in and its
coefficient of static friction is s = 0.4.
Answer: Mr = 2.48 lb·ft.
4 – 69 (M)
The inclined platform weighs 200 lb. Find the minimum moment needed to raise the platform. The radius of the thread is
3/8 in, its lead is 1/8 in and its coefficient of static friction is s = 0.6.
Answer: Mr = 2.13 lb·ft.
4 – 70 (H)
A square threaded screw is being used to lower the 75 lb rafters in an experimental greenhouse roof (The total weight of
the roof with the rafters is 600 lb). First, find the moment needed to turn the screw when H/B = 2. Then, find the moment
needed to turn the screw when H/B = ½. The radius of the thread is 3/8 in, its lead is 1/8 in and its coefficient of static
friction is s = 0.4.
Answer: Mr = 7.13 lb·ft, Ma = 2.75 lb·ft,
30
4 – 71 (L)
A 300 lb load acts on a 1 ft radius column that has been placed directly over an 8 in radius hole, as shown. If the coefficient
of static friction between the column and the ground is s = 0.6, find the minimum moment M needed to turn the column.
Answer: M = 152 lb·ft.
4 – 72 (L)
A worker holds a 200 lb industrial sander in position at the end of a handle, as shown. Find the force F that the worker
exerts on the handle by each of his hands. The coefficient of kinetic friction between the sander and the floor k = 1.2.
Answer: F = 80 lb.
4 – 73 (L)
The base of a chair consists of column A fixed to the seat, and a tubular support B attached to the wheels. The inside radius
of the tubular support is r1 = 3/8 in and the outer radius is r2 = 1 in. If a 100 lb load acts on the chair, determine the
minimum moment M needed to swivel the chair. The coefficient of static friction between column A and tube B is s = 0.3.
Answer: M = 184 lb·ft.
4 – 74 (L)
A 250 lb load acts on a 150 lb, 6 in radius column, as shown. If the coefficient of static friction between the column and the
ground is s = 0.3, find the minimum moment M needed to turn the column.
Answer: M = 40 lb·ft.
4 – 75 (L)
A 25 lb force presses down on disk A which presses down on disk B. The top of disk B has a 2 in radius and a coefficient of
static friction of A = 0.4. The bottom of disk B has a 1 in inner radius, a 2 in outer radius and a coefficient of static friction
of B = 0.4. Find the minimum moment M needed to turn disk A. Which surface slips?
Answer: M = 1.11 lb·ft, Surface A slips.
4 – 76 (L)
Two geared cylinders are acted on by 100 lb and 80 lb loads, as shown. Find the minimum moment M needed to turn the
cylinders. Let s = 0.5, R = 3 in, RA = 5 in, and RB = 8 in.
Answer: M = 12.5 lb·ft.
4 – 77 (L)
A 100 lb load acts on a pin having an inner radius of 3 in and an outer radius of 4 in, as shown. The pin, in turn, presses
down on the bracket at two contact surfaces. Find the minimum moment M needed to turn the pin. Assume that the normal
forces acting each of the two contact surfaces are uniform although not necessarily equal to each other. Let s = 0.4.
Answer: M = 11.7 lb·ft.
4 – 78 (L)
A 50 lb force acts on two large posts that are placed on top of each other, as shown. Find the minimum moment M needed
to turn the top post. Assume that the coefficients of static friction are s = 0.3 and that the radii of the post are R = 4 in.
Does the bottom post also turn?
Answer: M = 6 lb·ft, No.
4 – 79 (M)
A 90 lb beam rests on a pair of 2 in radius columns that are geared to each other, as shown. Find the minimum moment M
needed to turn the columns if the coefficients of static friction between the four contact surfaces are s = 0.2.
Answer: M = 3.33 lb·ft.
4 – 80 (M)
A brake system consists of a lever that pushes a disk into a wheel that’s geared to an axle. Find the lever force P needed to
stop the wheel from turning if a resisting moment of M = 2 lb·ft acts on the axle. Let k = 0.2.
Answer: P = 15 lb.
31
4 – 81 (M)
An applied force P acts on a disk that has a conically tapered contact surface. Derive an expression for the minimum
moment M needed to turn the disk if the coefficient of static friction is s. Express M in terms of the inner radius R1, the
outer radius R2, the coefficient of static friction s and the taper angle is  .
2 s P R23  R13
Answer: M 

.
3 sin  R22  R12
4 – 82 (M)
The Morse taper is used in machining to hold a work-piece. A standard Morse taper uses a taper of 5/8 in per foot. If a 20
lb load acts on the taper shown, find the minimum moment M needed to turn the taper, to remove it from its spindle. Use
the formula developed in Prob. 4.1 – 81 to solve this problem and let s = 0.3.
Answer: M = 3.55 lb·ft.
4 – 83 (M)
A 150 lb load acts on the bent shaft, as shown. Find the minimum moment M needed to turn the shaft and the location of
the point of contact A. Let s = 0.6 and let the radius of the shaft be R = 1 in.
Answer: M = 6.43 lb·ft,  = 31.0○.
4 – 84 (M)
A shaft is acted on by two loads, as shown. Find the minimum moment M needed to turn the shaft and the locations of the
points of contact A and B. Let s = 0.5 and let the radius of the shaft be R = 2 in.
Answer: M = 17.3 lb·ft,  = 124○,  = 103○.
4 – 85 (M)
A beam is acted on by a 45 load P, which would cause the beam to rotate clockwise if it were not for the friction in the
pinned joint. How much additional force P is needed to overcome the friction in the pinned joint and cause the shaft to
rotate clockwise? Let s = 0.8.
Answer: P = 0.309 lb.
4 – 86 (M)
A support column is being pulled to the left by a 150 lb tensile force. An unknown tensile reaction force T acts to the right
while the beam is supporting a 250 lb load, as shown. The column is held in place at its base by a slip connection that has a
coefficient of static friction of A = 0.4 on the bottom surface and B = 0.4 along the side surface. Find the minimum
moment needed to turn the column. The radius of the column is 4 in.
Answer: M = 53.2 lb·ft.
4 – 87 (H)
A light ring supports a heavy disk, as shown. The disk lines up with the vertical axis of the ring when a clockwise moment
M is first applied. The moment M causes the disk to roll up the ring and the ring to rotate clockwise, as well, to maintain
equilibrium. If the applied moment is the largest moment that can be applied without slip, find the angle  that the disk
rotates and the angle  that the ring rotates. Let s = 0.3. The inner radius of the ring is 1 in and the outer radius of the disk
is ¾ in.
Answer:  = 16.7○,  = 39.0○.
4 – 88 (L)
A weight of W = 50 N hangs from a cable that’s wrapped around a fixed disk. Find how much minimum force P is needed
to pull up the weight. Let s = 0.5.
Answer: P1 = 241 N.
4 – 89 (L)
A weight of W = 50 N hangs from a cable that’s wrapped around a fixed disk. Find how much minimum force P to prevent
the weight from moving downward. Let s = 0.5.
Answer: P = 10.4 N.
32
4 – 90 (L)
A thread is wound around a spool in three stages. First, the thread is placed in a hole in the spool. Secondly, the thread is
lightly wound around the spool n times until it “catches.” Third, the thread is wound around the spool more aggressively. If
the force, that keeps the end of the tread in the hole, needs to be lower than 0.2 lb, and if the force acting on the other end
of the thread during the third stage is held below 10 lb, find the number of times n the thread needs to be lightly wound
around the spool in stage two. Let s = 0.1.
Answer: n = 6.23 turns.
4 – 91 (L)
The force that it takes to take your belt off can be calculated by regarding your waste as a cylindrically shaped body with
an effective coefficient of static friction between the belt and the waste of s. If the force needed to take off your belt is 20
lb and a resistance force of about 1 lb acts on the belt at the other end, find the effective coefficient of static friction s.
Answer: s = 0.477.
4 – 92 (L)
Imagine that you happened to see your dry towel slide off your towel rack in the position shown. Find the coefficient of
static friction between the towel and the towel rack.
Answer: s = 0.350.
4 – 93 (L)
A clamp is tightened around a pipe with 40 lb of force. Then, a force P is applied to slide the clamp around the pipe. Find
the minimum force P needed to do that. The coefficient of static friction between the clamp and the pipe is s = 0.5.
Answer: P = 89.9 lb.
4 – 94 (L)
A belt is placed around a guide on the left and a motor disk on the right, as shown. The guide is then tightened until the
spring is stretched 0.5 in. Find the minimum moment that the motor needs to exert to cause the belt to slip around the
guide. Assume that there is no slip between the belt and the motor disk. Let s = 0.3, k = 100 lb/ft and R = 3 in.
Answer: M = 0.82 lb·ft.
4 – 95 (L)
A weight W is supported by a cable that’s wrapped around a round surface and attached to a spring, as shown. Assume that
the spring is stretched 0.1 ft. Find the maximum weight W supported by the cable in the position shown. Let s = 0.6 and k
= 200 lb/ft.
Answer: W = 132 lb.
4 – 96 (L)
A belt is placed around a guide on the left and a pulley on the right, as shown. The guide is first tightened until the spring is
stretched 0.1 m. Then an external force P is applied to the top of the belt. Find the minimum force P needed to slide the top
of the belt to the right. Let s = 0.4 and k = 250 N/m.
Answer: P = 31.4 N.
4 – 97 (L)
A simple tension adjustment mechanism consists of a bar placed against a belt. The tension P needed to move the belt
increases when the bar is lowered. If the belt is acted on by a W = 50 lb weight on the left, how far down does the bar need
to be lowered (d) to cause the belt to slide to the right with a minimum of P = 100 lb. The radius of the head on the bar is
0.5 in and s = 0.5.
Answer: d = 2.82 in.
4 – 98 (M)
A tie will slip off a hanger if it’s not properly balanced. Find the smallest height difference h beyond which the tie slips off
the hanger in the clockwise direction. Let s = 0.1.
Answer: h = 13.1 in.
33
4 – 99 (M)
A force P pulls on a 200 lb crate, as shown. When the force is large enough, the crate will cease to be in equilibrium
because either: a) the crate slides to the left, b) it tips over clockwise, or c) it tips over counter-clockwise. Find the
minimum force P that causes the crate to move. Does the crate slide, tip over clockwise, or tip over counter-clockwise? Let
A = 0.4 andB = 0.8.
Answer: P = 339 lb, It slides.
4 – 100 (M)
A force P pulls on a 200 lb crate, as shown. When the force is large enough, the crate will cease to be in equilibrium
because either: a) the crate slides to the left, b) it tips over clockwise, or c) it tips over counter-clockwise. Find the
minimum force P that causes the crate to move. Does the crate slide, tip over clockwise, or tip over counter-clockwise? Let
A = 0.4 and B = 0.8.
Answer: P = 447 lb, It tips over clockwise.
4 – 101 (M)
A force P pulls on a 200 lb crate, as shown. When the force is large enough, the crate will cease to be in equilibrium
because either: a) the crate slides to the left, b) it tips over clockwise, or c) it tips over counter-clockwise. Find the
minimum force P that causes the crate to move. Does the crate slide, tip over clockwise, or tip over counter-clockwise? Let
A = 0.4 and B = 0.8.
Answer: P = 246 lb, It tips over counter-clockwise.
4 – 102 (M)
The tension on the left is 50 lb and the tension applied on the right is the largest tension for which the cable doesn’t slip to
the right. Find the reactions at connection A. Let s = 0.4 and  = 40○.
Answer: M = 15.6 lb·ft, Px = -28.7 lb, Py = 88.3 lb.
4 – 103 (M)
A height adjustment mechanism consists of a light bar placed between smooth bearings acted on by a load P. Find the
minimum load P needed to initiate upward motion of the weight W. Let s = 0.4 and  = 25○.
Answer: P = 169 lb.
4 – 104 (H)
A heavy belt can slide around a cylinder of radius R, as shown, making contact with the sides of the cylinder and the
bottom ledge of the cylinder. The cross section of the cylinder lies in the horizontal plane. Assume that the coefficient of
static friction of both surfaces is s and that the weight per unit length of the belt is w. First, show that the tension in the
dT
belt is governed by the differential equation
  s T   s wR . Next, assume that the tension in the cable is zero at  = 0,
d
and show that the maximum tension in the belt for which the belt remains in static equilibrium is T  wR(e  s  1) .
4 – 105 (M)
Referring to Problem 4 – 104, find the tension T needed to slide a 2 lb belt wound once around a 3 in radius cylinder,
letting s = 0.5.
Answer: T = 6.64 lb.
4 – 106 (H)
A heavy belt can slide around a cylinder of radius R, as shown. The cross section of the cylinder lies in the vertical plane.
Assume that the coefficient of static friction is s and that the weight per unit length of the belt is w. First, show that the
dT
tension in the belt is governed by the differential equation
  s T  wR( s cos  sin  ) . Next, show that the
d
maximum
tension
in
the
belt
for
which
the
belt
remains
in
static
equilibrium
is
Rw
T  Ae s 
[cos  (1   s2 )  sin  (2 s )], where A is a constant determined by evaluating T at an angle for which the
1   s2
tension is known.
34
4 – 107 (M)
Using the formulas developed in Problem 4 – 107, find the largest distance d for which the belt does not slip off the disk.
Let R = 0.1 ft, w = 10 lb/ft, and s = 0.8.
Answer: d = 0.245 in.
4 – 108 (M)
Find the largest angle for which the disk does not slip out from under the rope. Let WC = WD = 10 lb,A = 0.75 and B =
0.9.
Answer: = 15.6○.
4 – 109 (M)
The radii of the wheels in a 20,000 lb semi are 1.5 ft. If it takes 450 lb to move the semi on level ground, estimate the
rolling resistance a between the tires and the road.
Answer: a = 0.405 in.
4 – 110 (M)
A 100 N force presses down on a bar, as shown. Find the minimum force Q needed to cause the wheel to roll. Assume that
the wheel radius is R = 0.2 m, the rolling resistance on the bottom surface is aA = 1 mm, and that the rolling resistance on
the top surface is aB = 2 mm.
Answer: Q = 1.5 N.
4 – 111 (M)
The weight W of an object is hoisted with the help of a counter-balance weight W + W. When the hoist is locked, find
how large the % difference E = 100W/W can be without the weight slipping up the hoist. Let ,  = 0.1.
Answer: E = 36.9%.
4 – 112 (M)
Find the minimum force needed to move the 0.4 lb roller. The rolling resistance is a = 0.1 in and the radius of the roller is
1 in.
Answer: P = 0.247 lb.
4 – 113 (M)
Find the minimum force needed to move the 2 lb roller. The rolling resistance is a = 0.5 in and the radius of the roller is 2
in.
Answer: P = 6.89 lb.
4 – 114 (M)
In the figure shown the rolling resistance Q of a wheel of radius R is compared with the friction force Q acting on a sliding
block. Using the figure shown, determine the relationship between the coefficient of static friction and effective rolling
resistance that produce the same force Q.
Answer: eff = a/R.
4.1-115 to 4.1-120
Applied Problems
35