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08. Friction
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.01 Friction
Chapter Objectives
β’ To introduce the concept of dry friction and show how to
analyze the equilibrium of rigid bodies subjected to this force
β’ To present specific applications of frictional force analysis on
wedges, screws, belts, and bearings
β’ To investigate the concept of rolling resistance
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.02 Friction
Β§1.Characteristics of Dry Friction
- Definition
β’ Friction: force that resists the movement of two contacting
surfaces that slide relative to one another
β’ Friction force always acts tangent to the surface at the points
of contact and is directed so as to oppose the possible or
existing motion between the surfaces
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.03 Friction
The heat generated
by the abrasive
action of friction
can be noticed
when using this
grinder to sharpen
a metal blade
Regardless of the weight
of the rake or shovel that
is suspended, the device
has been designed so
that the small roller holds
the handle in equilibrium
due to frictional forces
that develop at the points
of contact, π΄, π΅, πΆ
Β§1.Characteristics of Dry Friction
- Types of friction
β’ Fluid friction exists when the contacting surfaces are
separated by a film of fluid (i.e.,gas or liquid)
Fluid friction is studied in fluid mechanics
β’ Dry friction or Coulomb friction (C.A. Coulomb, 1781) occurs
between the contacting surfaces of bodies in absence of a
lubricating fluid
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.04 Friction
Β§1.Characteristics of Dry Friction
- Theory of dry friction
β’ Development of frictional force πΉ
+ Consider a pulling force π applied to a block of uniform
weight, π, resting on a rough horizontal surface
+ Under the effect of π and π, two reacting forces are developed
normal force perpendicular to the rough surface: π = βππ
frictional force parallel to the rough surface: πΉ = β πΉπHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.05 Friction
Β§1.Characteristics of Dry Friction
β’ Equilibrium π+ π + π π = 0
π π₯ β π β = 0
β’ Impending function
+ πΉ increases with increase in applied force π till πΉ reaches to a
maximum value, called limiting frictional force πΉπ πΉπ = ππ π ππ : the coefficient of static friction
+ Angle of static friction
ππ = π‘ππβ1 πΉπ /π = π‘ππβ1ππ HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.06 Friction
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Β§1.Characteristics of Dry Friction
β’ Motion
+ When π > | πΉπ |, πΉπ gets significantly reduced, body starts moving
+ The frictional force which acts between the rough surface
and the body under motion is called kinetic frictional force πΉππΉπ = πππ ππ: the coefficient of kinetic friction
+ Angle of kinetic friction
ππ = π‘ππβ1 πΉπ/π = π‘ππβ1ππ
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.07 Friction
Β§1.Characteristics of Dry Friction
+ ππ depends on the contact materials. Some typical values
metal on ice ππ = 0.03 Γ· 0.05
wood on wood ππ = 0.30 Γ· 0.70
leather on wood ππ = 0.20 Γ· 0.50
leather on metal ππ = 0.30 Γ· 0.60
aluminum on aluminum ππ = 1.10 Γ· 1.70
β’ The variation of the frictional force πΉ versus the applied load π
+ πΉ is a static frictional force if
equilibrium is maintained
+ πΉ is a limiting static frictional force
when it reaches a maximum value
needed to maintain equilibrium
+ πΉ is termed a kinetic frictional force when sliding occurs at
the contacting surfaceHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.08 Friction
Β§1.Characteristics of Dry Friction
- Cause of friction
β’ Friction is mainly caused by the surface roughness of the
objects in contact to each other. In general applies: the
rougher the surface, the higher the friction
β’ If both surfaces become ultra-smooth, friction from molecular
attraction comes into play, often becoming greater than the
mechanical friction
β’ There is especially the case with soft materials, like rubber
and other soft synthetics: soft materials will deform when
under pressure, material deformation is also increasing the
friction
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.09 Friction
Β§1.Characteristics of Dry Friction
- Characteristics of dry friction (Coulombβs dry friction law)
β’ The frictional force acts tangent to the contacting surfaces in
a direction opposed to the motion or tendency for motion of
one surface relative to another and is proportional to the
normal force π
+ slipping at the surface of contact is about to occur πΉπ = ββ ππ₯
ππ |π|
+ slipping at the surface of contact is occurring πΉπ = ββ ππ₯
ππ|π|
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.10 Friction
Β§1.Characteristics of Dry Friction
β’ The coefficient of friction depend on
+ both friction partners (material composition, surface roughness)
+ the surface conditions (cleanliness, humidity)
+ time of contact
β’ The coefficient of friction does not depend on
+ the contact surface area
+ normal contact pressure
+ the relative velocity between two contact surfaces
β’ The maximum static frictional force is generally greater than
the kinetic frictional force for any two surfaces of contact.
However, if one of the bodies is moving with a very low
velocity over the surface of another, πΉπ becomes
approximately equal to πΉπ , i.e., ππ β ππ
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.11 Friction
Β§2.Problems Involving Dry Friction
- Types of friction problems
β’ No apparent impending motion
+ The number of unknowns = The number of available
equilibrium equations
+ Determine the frictional forces from the equilibrium equations
+ Check the numerical value of πΉ
If πΉ β€ ππ π, the body will not remain in equilibrium
If πΉ > ππ π, slipping will occur
+ Example
The bars will remain
in equilibrium if
πΉπ΄ β€ 0.3ππ΄
πΉπΆ β€ 0.5ππΆ
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.12 Friction
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Β§2.Problems Involving Dry Friction
β’ Impending motion at all points of contact
+ The number of unknowns = The total number of available
equilibrium equations + The total number of available
frictional equations, πΉ = ππ
+ When motion is impending at the points of contact, then
πΉπ = ππ π whereas if the body is slipping, then πΉπ = πππ
+ Example
Five unknowns: ππ΄, πΉπ΄, ππ΅, πΉπ΅, π
Five available equations
βπΉπ₯ = 0
βπΉπ¦ = 0
βππ΅ = 0
πΉπ΄ = ππ΄ππ΄
πΉπ΅ = ππ΅ππ΅HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.13 Friction
Β§2.Problems Involving Dry Friction
β’ Impending motion at some points of contact
+ The number of unknowns < The total number of available
equilibrium equations + The total number of available
frictional equations or conditional equations for tipping
+ Several possibilities for motion or impending motion will
exist and the problem will involve a determination of the
kind of motion which actually occurs
+ Example: Determine force π needed to cause movement
Seven unknowns
ππ΄, πΉπ΄, ππΆ, πΉπΆ, π΅π₯, π΅π¦, π
Available equations
Six equilibrium equations
One of two possible
static frictional equations
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.14 Friction
Β§2.Problems Involving Dry Friction
Seven unknowns
ππ΄, πΉπ΄, ππΆ, πΉπΆ, π΅π₯, π΅π¦, π
Available equations
Six equilibrium equations
One of two possible static frictional equations
This means that as π increases it will either cause
slipping at π΄, no slipping at πΆ: πΉπ΄ = 0.3ππ΄, πΉπΆ β€ 0.5ππΆ
slipping occurs at πΆ, no slipping at π΄: πΉπΆ = 0.5ππΆ, πΉπ΄ β€ 0.3ππ΄The actual situation can be determined by calculating π for
each case and then choosing the case for which π is
smaller
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.15 Friction
Β§2.Problems Involving Dry Friction
- Impending tipping versus slipping
As π increases the crate will either be on the verge of slipping
on the surface (πΉ = ππ π) or if the surface is very rough (large
ππ ) then the resultant normal force π will shift to the corner
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.16 Friction
Β§2.Problems Involving Dry Friction
β’ How can we determine if the block will
slide or tip first?
β’ In this case, we have four unknowns
(πΉ,π,π₯ and π) and only three equations
of equilibrium
β’ We have to make an assumption to
give us another equation (the friction
equation!). Then we can solve for the
unknowns
β’ Finally, we need to check if our
assumption was correct
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.17 Friction
Β§2.Problems Involving Dry Friction
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.18 Friction
Assumption: Slipping occurs
Known: πΉ = ππ πSolve for: π₯, π, and πCheck: 0 β€ π₯ β€ π/2
Assumption: Tipping occurs
Known: π₯ = π/2Solve for: π₯, π, and πCheck: πΉ β€ ππ π
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Β§2.Problems Involving Dry Friction
- Equilibrium versus frictional equations
When the frictional equation πΉ = ππ πis used in the solution of a problem, πΉ must always be shown acting with
its correct sense on the free-body
diagram
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.19 Friction
The applied vertical force π on this roll must be
large enough to overcome the resistance of
friction at the contacting surfaces π΄ and π΅ in order
to cause rotation
Β§2.Problems Involving Dry Friction
- Example 8.1 The uniform crate has a mass of 20ππ. If a
force π = 80π is applied to the crate, determine if it remains in
equilibrium. The coefficient of static friction is ππ = 0.3
Solution
Free-body diagram
Equations of equilibrium
+ββπΉπ₯ = 0: β80πππ 300 β πΉ = 0
+ β βπΉπ¦ = 0: β80π ππ300+ππΆ β196.2 = 0
+βΊβππ = 0: 80π ππ300 Γ0.4+
β80πππ 300 Γ0.2+ππΆπ₯ = 0
βΉ πΉ = 69.3π, ππΆ = 236π, π₯ = β9.08ππ
π₯ < 0.4π βΉ no tipping will occur
πΉπππ₯ = ππ ππΆ = 0.3 Γ 236 = 70.8π > πΉ βΉ no slip will occur
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.20 Friction
Β§2.Problems Involving Dry Friction
- Example 8.2 It is observed that when the bed of the dump
truck is raised to an angle of π = 250 the
vending machines will begin to slide off
the bed. Determine the static coefficient
of friction between a vending machine
and the surface of the truckbed
Solution
Idealized model of a vending
machine resting on the
truckbed
Free-body diagram
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.21 Friction
Β§2.Problems Involving Dry Friction
Equations of equilibrium
+ββπΉπ₯ = 0: ππ ππ250 β πΉ = 0
+ β βπΉπ¦ = 0: π βππππ 250 = 0
+βΊβππ = 0: βππ ππ250 Γ0.75+
ππππ 250 Γ π₯ = 0
Since slipping impends at π = 250
πΉ = πΉπ = ππ π
or
ππ ππ250 = ππ ππππ 250
βΉ ππ = π‘ππ250 = 0.466
Note: we can show that the
vending machine will slip
before it can tip as observed
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.22 Friction
Β§2.Problems Involving Dry Friction
- Example 8.3 The uniform 10ππ ladder rests against the
smooth wall at π΅, and the end π΄ rests on the rough horizontal
plane for which the coefficient of static friction is ππ = 0.3.
Determine the angle of inclination π of the ladder and the
normal reaction at π΅ if the ladder is on the verge of slipping
Solution
Free-body diagram
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.23 Friction
Β§2.Problems Involving Dry Friction
Equations of equilibrium and friction
+ββπΉπ₯ = 0: πΉπ΄ β ππ΅ = 0
+ β βπΉπ¦ = 0: ππ΄ β 10 Γ 9.81 = 0
+βΊβππ΄ = 0: πΉπ΄ Γ 4π πππ β 10Γ 9.81Γ 2πππ π = 0
Since the ladder is on the verge of slipping
πΉπ΄ = ππ ππ΄ = 0.3ππ΄
Solving the above equations, we obtain
ππ΄ = 98.1π
ππ΅ = 29.4π
ΞΈ = 59.00
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.24 Friction
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Β§2.Problems Involving Dry Friction
- Example 8.4 Beam π΄π΅ is subjected to a uniform load of
200π/π and is supported at π΅ by post π΅πΆ. The coefficients of
static friction at π΅ and πΆ are ππ΅ = 0.2 and ππΆ = 0.5. Determine
the force π needed to pull the post out
from under the beam. Neglect the
weight of the members and the
thickness of the beam
Solution
Free-body diagrams
Equations of equilibrium and friction
+ββπΉπ₯ = 0: π β πΉπ΅ β πΉπΆ = 0
+ β βπΉπ¦ = 0: ππΆ β 400 = 0
+βΊβππ΄ = 0: βπΓ0.25+ πΉπ΅ Γ1 = 0
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.25 Friction
Β§2.Problems Involving Dry Friction
+ββπΉπ₯ = 0: π β πΉπ΅ β πΉπΆ = 0+ β βπΉπ¦ = 0: ππΆ β 400 = 0
+βΊβππ΄ = 0: βπΓ0.25+ πΉπ΅ Γ1 = 0
Post slips at π΅ and rotates about πΆ
πΉπΆ β€ ππΆππΆ,πΉπ΅ = ππ΅ππ΅ = 0.2Γ400= 80π
βΉ π = 320π, πΉπΆ = 240π, ππΆ = 400π
Since πΉπΆ =240> ππΆππΆ =0.5Γ400 =200π,
slipping at πΆ occurs βΉ other case of
movement must be investigated
Post slips at π΅ and rotates about πΆ
πΉπ΅ β€ ππ΅ππ΅, πΉπΆ = ππΆππΆ = 0.5ππΆ
βΉ π=267π,πΉπΆ =200π,ππΆ =400π,πΉπ΅ =66.7π
This case occurs first since it requires
a smaller value for πHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.26 Friction
Β§2.Problems Involving Dry Friction
- Example 8.5 Blocks π΄ and π΅ have a mass of 3ππ and 9ππ,
respectively, and are connected to the
weightless links. Determine the largest
vertical force π that can be applied at the
pin πΆ without causing any movement.
The coefficient of static friction between
the blocks and the contacting surfaces is
ππ = 0.3
Solution
Free-body diagram
Equations of equilibrium and friction
Pin πΆ
+ββπΉπ₯ = 0: πΉπ΄πΆπππ 300 β π = 0
+ β βπΉπ¦ = 0: πΉπ΄πΆπ ππ300 β πΉπ΅πΆ = 0
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.27 Friction
Β§2.Problems Involving Dry Friction
Pin πΆ
+ββπΉπ₯ = 0: πΉπ΄πΆπππ 300 β π = 0
+ β βπΉπ¦ = 0: πΉπ΄πΆπ ππ300 β πΉπ΅πΆ = 0
Block π΄
+ββπΉπ₯ = 0: πΉπ΄ β πΉπ΄πΆπ ππ300 = 0
+ β βπΉπ¦ = 0: ππ΄ β πΉπ΄πΆπππ 300
β3Γ 9.81 = 0
Block π΅
+ββπΉπ₯ = 0: πΉπ΅πΆ β πΉπ΅ = 0
+ β βπΉπ¦ = 0: ππ΅ β 9 Γ 9.81 = 0
βΉ πΉπ΄πΆ = 1.155π, πΉπ΅πΆ = 0.5774π
πΉπ΄ = 0.5774π, ππ΄ = π + 29.43
πΉπ΅ = 0.5774π, ππ΅ = 88.29πHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.28 Friction
Β§2.Problems Involving Dry Friction
πΉπ΄πΆ = 1.155π, πΉπ΅πΆ = 0.5774π
πΉπ΄ = 0.5774π, ππ΄ = π + 29.43
πΉπ΅ = 0.5774π, ππ΅ = 88.29π
Movement of the system may be caused by
the initial slipping of either blockπ΄ or blockπ΅
Assume that block π΄ slips first
πΉπ΄ = ππ ππ΄ = 0.3ππ΄
βΉ 0.5774π = 0.3(π + 29.43) βΉ π = 31.8π
πΉπ΅ = 0.5774 Γ 31.8 = 18.4π
Since the maximum static frictional force at π΅
πΉπ΅πππ₯ = ππ ππ΅ = 0.3 Γ 88.29 = 26.5π > πΉπ΅
block π΅ will not slip. Thus, the above assumption is correct
Note: If the inequality were not satisfied, we would have to
assume slipping of block π΅ and then solve for πHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.29 Friction
Fundamental Problems
- F.8.1 If π = 200π, determine the friction developed between
the 50ππ crate and the ground. The coefficient of static friction
between the crate and the ground is ππ = 0.3
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.30 Friction
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Fundamental Problems
- F.8.2 Determine the minimum force π to prevent the 30ππ rod
π΄π΅ from sliding. The contact surface at π΅ is smooth, whereas
the coefficient of static friction between the rod and the wall at
π΄ is ππ = 0.2
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.31 Friction
Fundamental Problems
- F.8.3 Determine the maximum force π that can be applied
without causing the two 50ππ crates to move. The coefficient
of static friction between each crate and the ground is ππ =0.25
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.32 Friction
Fundamental Problems
- F.8.4 If the coefficient of static friction at contact points π΄ and
π΅ is ππ = 0.3, determine the maximum force π that can be
applied without causing the 100ππ spool to move
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.33 Friction
Fundamental Problems
- F.8.5 Determine the minimum force π that can be applied
without causing movement of the 250π crate which has a
center of gravity at πΊ. The coefficient of static friction at the
floor is ππ = 0.4
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.34 Friction
Β§3.Wedges
- A wedge is a simple machine in which a small force π is used
to lift a large weight π
- Wedges are used to adjust the elevation or provide stability for
heavy objects
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.35 Friction
Β§3.Wedges
- Analysis of a wedge
β’ Draw the free body diagram of the wedge
Note
+ the friction forces are always in the
direction opposite to the motion
+ the friction forces are along the contacting
surfaces
+ the normal forces are perpendicular to the
contacting surfaces
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.36 Friction
5/6/2013
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Β§3.Wedges
β’ Look at the object on top of the wedge
Note
+ At the contacting surfaces between the
wedge and the object the forces are equal
in magnitude and opposite in direction to
those on the wedge
+ All other forces acting on the object should
be shown
+ For the wedge and the object
βπΉπ₯ = 0
βπΉπ¦ = 0
For the impending motion frictional equation
πΉ = ππ π
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.37 Friction
Β§3.Wedges
β’ Start by analyzing the free body diagram in which the
number of unknowns are less than or equal to the number of
equations of equilibrium and frictional equations
β’ If the object is to be lowered,
then the wedge needs to be
pulled out
β’ If the value of the force πneeded to remove the wedge is
positive, then the wedge is self-
locking, i.e., it will not come out
on its own
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.38 Friction
Β§3.Wedges
- Example 8.6 The stone has a mass of 500ππ and is held in
the horizontal position using a wedge at π΅. If
ππ = 0.3 at the surfaces of contact, determine
the minimum force π needed to remove the
wedge. Assume that the stone does not slip at π΄
Solution
The free body diagrams
For the wedge
+ββπΉπ₯ = 0: ππ΅π ππ70β0.3ππ΅πππ 7
0β0.3ππΆ+π=0
+ β βπΉπ¦ = 0: ππΆ β 0.3ππ΅π ππ70 βππ΅πππ 7
0 = 0
For the stone
+βΊβππ΄ = 0: β4905Γ0.5+ππ΅πππ 70Γ1+0.3ππ΅π ππ7
0Γ1=0
βΉ ππ΅ = 2383.1π,ππΆ = 2452.5π, π = 1154.9π
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.39 Friction
Β§4.Frictional Forces on Screws
- Screws are used as fasteners or to transmit power or motion
from one machine part to another
- A screw is considered a cylinder
called a barrel or shaft, with the
thread wrapped around it
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.40 Friction
Β§4.Frictional Forces on Screws
- Screws can be classified by the thread. E.g. square-threaded
screw, V-thread, β¦
- External / Internal thread
- Right / Left hand thread
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.41 Friction
Β§4.Frictional Forces on Screws
- If we unwind the thread by one revolution, the slope or lead
angle is given by
π = π‘ππβ1π
2ππ
π: the lead of the screw, is the distance advanced by turning
the screw one revolution
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.42 Friction
5/6/2013
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Β§4.Frictional Forces on Screws
- Upward Impending Motion
β’ Consider a square-threaded screw subject to impending
motion due to an applied torque π
β’ The free body diagram of the entire unraveled thread through
π: vertical force on the or the axial force on the shaft
π : reaction of the groove on the thread, π = πΉ + π
πΉ: frictional component, π: normal component
π/π: horizontal force
associated with the
couple moment π
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.43 Friction
Β§4.Frictional Forces on Screws
β’ The frictional component πΉ = ππ π
β’ The angle of static friction ππ = π‘ππβ1 πΉ/π = π‘ππβ1ππ
β’ By Equations of Equilibrium
+ββπΉπ₯ = 0: π/π β π π ππ(+π) = 0
+ β βπΉπ¦ = 0: π πππ ππ + π βπ = 0
βΉ π = πππ‘ππ(ππ + π)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.44 Friction
Β§4.Frictional Forces on Screws
- Self Locking Screw
β’ A screw is self-locking if it remains in place under any axial
load π when the moment π is removed
β’ In this case π acts on the other side of π
β’ If ππ = π, then π will act vertically to balance π,
and the screw will be on the verge of winding
downwards
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.45 Friction
Β§4.Frictional Forces on Screws
- Downward Impending Motion
β’ If a screw is self-locking, a couple πβ² must be applied in the
opposite direction to wind the screw downward
β’ ππ > π
β’ This causes a horizontal force in the reverse
direction that will push the thread downwards
β’ Using the previous procedure it can be shown
that πβ² = πππ‘ππ(π β ππ )
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.46 Friction
Β§4.Frictional Forces on Screws
β’ If the screw is not self-locking it is necessary to apply a
moment π" to prevent the screw from winding downwards
β’ ππ < π
β’ A horizontal force πβ²/π is required to push
against the thread to prevent it from sliding
downwards
β’ The magnitude of the moment required to
prevent this unwinding is π" = πππ‘ππ(ππ β π)
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.47 Friction
Β§4.Frictional Forces on Screws
- Example 8.7 The turnbuckle has
a square thread with a mean radius
of 5ππ and a lead of 2ππ. If the
coefficient of static friction between
the screw and the turnbuckle is
ππ = 0.25, determine the moment πthat must be applied to draw the
end screws closer together 2ππ
Since friction at two screws must be overcome, this requires
π = 2 Γ πππ‘ππ(ππ + π)
where π€ = 2000π, π = 5ππ
ππ = π‘ππβ1ππ = π‘ππβ1(0.25) = 14.040
π = π‘ππβ1(π/2ππ) = π‘ππβ1(2/(2π Γ 5)) = 3.640
βΉ π = 2 Γ 5 Γ 2000π‘ππ 14.040 + 3.640 = 6374.7πππ
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.48 Friction
Solution
5/6/2013
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Β§5.Frictional Forces on Flat Belts
Β§6.Frictional Forces on Collar Bearings, Pivot Bearings, and Disks
Β§7.Frictional Forces on Journal Bearings
Β§8.Rolling Resistances
HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien
Engineering Mechanics β Statics 8.49 Friction