ch.08 friction

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08. Friction

HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien

Engineering Mechanics – Statics 8.01 Friction

Chapter Objectives

• To introduce the concept of dry friction and show how to

analyze the equilibrium of rigid bodies subjected to this force

• To present specific applications of frictional force analysis on

wedges, screws, belts, and bearings

• To investigate the concept of rolling resistance



§1.Characteristics of Dry Friction

- Definition

• Friction: force that resists the movement of two contacting

surfaces that slide relative to one another

• 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



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, 𝐴, 𝐵, 𝐶


- Types of friction

• 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

• Dry friction or Coulomb friction (C.A. Coulomb, 1781) occurs

between the contacting surfaces of bodies in absence of a

lubricating fluid




- Theory of dry friction

• Development of frictional force 𝐹

+ Consider a pulling force 𝑃 applied to a block of uniform

weight, 𝑊, resting on a rough horizontal surface

+ Under the effect of 𝑃 and 𝑊, two reacting forces are developed

normal force perpendicular to the rough surface: 𝑁 = ∑𝑁𝑖

frictional force parallel to the rough surface: 𝐹 = ∑ 𝐹𝑖HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



• Equilibrium 𝑊+ 𝑃 + 𝑅𝑠 = 0

𝑊 𝑥 − 𝑃 ℎ = 0

• Impending function

+ 𝐹 increases with increase in applied force 𝑃 till 𝐹 reaches to a

maximum value, called limiting frictional force 𝐹𝑠𝐹𝑠 = 𝜇𝑠𝑁 𝜇𝑠: the coefficient of static friction

+ Angle of static friction

𝜙𝑠 = 𝑡𝑎𝑛−1 𝐹𝑠/𝑁 = 𝑡𝑎𝑛−1𝜇𝑠HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien


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• Motion

+ When 𝑃 > | 𝐹𝑠|, 𝐹𝑠 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 𝐹𝑘𝐹𝑘 = 𝜇𝑘𝑁 𝜇𝑘: the coefficient of kinetic friction

+ Angle of kinetic friction

𝜙𝑘 = 𝑡𝑎𝑛−1 𝐹𝑘/𝑁 = 𝑡𝑎𝑛−1𝜇𝑘




+ 𝜇𝑠 depends on the contact materials. Some typical values

metal on ice 𝜇𝑠 = 0.03 ÷ 0.05

wood on wood 𝜇𝑠 = 0.30 ÷ 0.70

leather on wood 𝜇𝑠 = 0.20 ÷ 0.50

leather on metal 𝜇𝑠 = 0.30 ÷ 0.60

aluminum on aluminum 𝜇𝑠 = 1.10 ÷ 1.70

• The variation of the frictional force 𝐹 versus the applied load 𝑃

+ 𝐹 is a static frictional force if

equilibrium is maintained

+ 𝐹 is a limiting static frictional force

when it reaches a maximum value

needed to maintain equilibrium

+ 𝐹 is termed a kinetic frictional force when sliding occurs at

the contacting surfaceHCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- Cause of friction

• 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

• If both surfaces become ultra-smooth, friction from molecular

attraction comes into play, often becoming greater than the

mechanical friction

• 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




- Characteristics of dry friction (Coulomb’s dry friction law)

• 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 𝑁

+ slipping at the surface of contact is about to occur 𝐹𝑠 = ↑↓ 𝑂𝑥

𝜇𝑠|𝑁|

+ slipping at the surface of contact is occurring 𝐹𝑘 = ↑↓ 𝑂𝑥

𝜇𝑘|𝑁|




• The coefficient of friction depend on

+ both friction partners (material composition, surface roughness)

+ the surface conditions (cleanliness, humidity)

+ time of contact

• The coefficient of friction does not depend on

+ the contact surface area

+ normal contact pressure

+ the relative velocity between two contact surfaces

• 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, 𝐹𝑘 becomes

approximately equal to 𝐹𝑠, i.e., 𝜇𝑠 ≈ 𝜇𝑘



§2.Problems Involving Dry Friction

- Types of friction problems

• 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 𝐹

If 𝐹 ≤ 𝜇𝑠𝑁, the body will not remain in equilibrium

If 𝐹 > 𝜇𝑠𝑁, slipping will occur

+ Example

The bars will remain

in equilibrium if

𝐹𝐴 ≤ 0.3𝑁𝐴

𝐹𝐶 ≤ 0.5𝑁𝐶



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• 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, 𝐹 = 𝜇𝑁

+ When motion is impending at the points of contact, then

𝐹𝑠 = 𝜇𝑠𝑁 whereas if the body is slipping, then 𝐹𝑘 = 𝜇𝑘𝑁

+ Example

Five unknowns: 𝑁𝐴, 𝐹𝐴, 𝑁𝐵, 𝐹𝐵, 𝜃

Five available equations

∑𝐹𝑥 = 0

∑𝐹𝑦 = 0

∑𝑀𝐵 = 0

𝐹𝐴 = 𝜇𝐴𝑁𝐴

𝐹𝐵 = 𝜇𝐵𝑁𝐵HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



• 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 𝑃 needed to cause movement

Seven unknowns

𝑁𝐴, 𝐹𝐴, 𝑁𝐶, 𝐹𝐶, 𝐵𝑥, 𝐵𝑦, 𝑃

Available equations

Six equilibrium equations

One of two possible

static frictional equations




Seven unknowns

𝑁𝐴, 𝐹𝐴, 𝑁𝐶, 𝐹𝐶, 𝐵𝑥, 𝐵𝑦, 𝑃

Available equations

Six equilibrium equations

One of two possible static frictional equations

This means that as 𝑃 increases it will either cause

slipping at 𝐴, no slipping at 𝐶: 𝐹𝐴 = 0.3𝑁𝐴, 𝐹𝐶 ≤ 0.5𝑁𝐶

slipping occurs at 𝐶, no slipping at 𝐴: 𝐹𝐶 = 0.5𝑁𝐶, 𝐹𝐴 ≤ 0.3𝑁𝐴The actual situation can be determined by calculating 𝑃 for

each case and then choosing the case for which 𝑃 is

smaller




- Impending tipping versus slipping

As 𝑃 increases the crate will either be on the verge of slipping

on the surface (𝐹 = 𝜇𝑠𝑁) or if the surface is very rough (large

𝜇𝑠) then the resultant normal force 𝑁 will shift to the corner




• How can we determine if the block will

slide or tip first?

• In this case, we have four unknowns

(𝐹,𝑁,𝑥 and 𝑃) and only three equations

of equilibrium

• We have to make an assumption to

give us another equation (the friction

equation!). Then we can solve for the

unknowns

• Finally, we need to check if our

assumption was correct






Assumption: Slipping occurs

Known: 𝐹 = 𝜇𝑠𝑁Solve for: 𝑥, 𝑃, and 𝑁Check: 0 ≤ 𝑥 ≤ 𝑏/2

Assumption: Tipping occurs

Known: 𝑥 = 𝑏/2Solve for: 𝑥, 𝑃, and 𝑁Check: 𝐹 ≤ 𝜇𝑠𝑁

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- Equilibrium versus frictional equations

When the frictional equation 𝐹 = 𝜇𝑠𝑁is used in the solution of a problem, 𝐹 must always be shown acting with

its correct sense on the free-body

diagram



The applied vertical force 𝑃 on this roll must be

large enough to overcome the resistance of

friction at the contacting surfaces 𝐴 and 𝐵 in order

to cause rotation


- Example 8.1 The uniform crate has a mass of 20𝑘𝑔. If a

force 𝑃 = 80𝑁 is applied to the crate, determine if it remains in

equilibrium. The coefficient of static friction is 𝜇𝑠 = 0.3

Solution

Free-body diagram

Equations of equilibrium

+→∑𝐹𝑥 = 0: −80𝑐𝑜𝑠300 − 𝐹 = 0

+ ↑ ∑𝐹𝑦 = 0: −80𝑠𝑖𝑛300+𝑁𝐶 −196.2 = 0

+↺∑𝑀𝑂 = 0: 80𝑠𝑖𝑛300 ×0.4+

−80𝑐𝑜𝑠300 ×0.2+𝑁𝐶𝑥 = 0

⟹ 𝐹 = 69.3𝑁, 𝑁𝐶 = 236𝑁, 𝑥 = −9.08𝑚𝑚

𝑥 < 0.4𝑚 ⟹ no tipping will occur

𝐹𝑚𝑎𝑥 = 𝜇𝑠𝑁𝐶 = 0.3 × 236 = 70.8𝑁 > 𝐹 ⟹ no slip will occur




- Example 8.2 It is observed that when the bed of the dump

truck is raised to an angle of 𝜃 = 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




Equations of equilibrium

+→∑𝐹𝑥 = 0: 𝑊𝑠𝑖𝑛250 − 𝐹 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑁 −𝑊𝑐𝑜𝑠250 = 0

+↺∑𝑀𝑂 = 0: −𝑊𝑠𝑖𝑛250 ×0.75+

𝑊𝑐𝑜𝑠250 × 𝑥 = 0

Since slipping impends at 𝜃 = 250

𝐹 = 𝐹𝑠 = 𝜇𝑠𝑁

or

𝑊𝑠𝑖𝑛250 = 𝜇𝑠𝑊𝑐𝑜𝑠250

⟹ 𝜇𝑠 = 𝑡𝑎𝑛250 = 0.466

Note: we can show that the

vending machine will slip

before it can tip as observed




- Example 8.3 The uniform 10𝑘𝑔 ladder rests against the

smooth wall at 𝐵, and the end 𝐴 rests on the rough horizontal

plane for which the coefficient of static friction is 𝜇𝑠 = 0.3.

Determine the angle of inclination 𝜃 of the ladder and the

normal reaction at 𝐵 if the ladder is on the verge of slipping

Solution

Free-body diagram




Equations of equilibrium and friction

+→∑𝐹𝑥 = 0: 𝐹𝐴 − 𝑁𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐴 − 10 × 9.81 = 0

+↺∑𝑀𝐴 = 0: 𝐹𝐴 × 4𝑠𝑖𝑛𝜃 − 10× 9.81× 2𝑐𝑜𝑠𝜃 = 0

Since the ladder is on the verge of slipping

𝐹𝐴 = 𝜇𝑠𝑁𝐴 = 0.3𝑁𝐴

Solving the above equations, we obtain

𝑁𝐴 = 98.1𝑁

𝑁𝐵 = 29.4𝑁

θ = 59.00



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- Example 8.4 Beam 𝐴𝐵 is subjected to a uniform load of

200𝑁/𝑚 and is supported at 𝐵 by post 𝐵𝐶. The coefficients of

static friction at 𝐵 and 𝐶 are 𝜇𝐵 = 0.2 and 𝜇𝐶 = 0.5. Determine

the force 𝑃 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


+→∑𝐹𝑥 = 0: 𝑃 − 𝐹𝐵 − 𝐹𝐶 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐶 − 400 = 0

+↺∑𝑀𝐴 = 0: −𝑃×0.25+ 𝐹𝐵 ×1 = 0




+→∑𝐹𝑥 = 0: 𝑃 − 𝐹𝐵 − 𝐹𝐶 = 0+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐶 − 400 = 0

+↺∑𝑀𝐴 = 0: −𝑃×0.25+ 𝐹𝐵 ×1 = 0

Post slips at 𝐵 and rotates about 𝐶

𝐹𝐶 ≤ 𝜇𝐶𝑁𝐶,𝐹𝐵 = 𝜇𝐵𝑁𝐵 = 0.2×400= 80𝑁

⟹ 𝑃 = 320𝑁, 𝐹𝐶 = 240𝑁, 𝑁𝐶 = 400𝑁

Since 𝐹𝐶 =240> 𝜇𝐶𝑁𝐶 =0.5×400 =200𝑁,

slipping at 𝐶 occurs ⟹ other case of

movement must be investigated

Post slips at 𝐵 and rotates about 𝐶

𝐹𝐵 ≤ 𝜇𝐵𝑁𝐵, 𝐹𝐶 = 𝜇𝐶𝑁𝐶 = 0.5𝑁𝐶

⟹ 𝑃=267𝑁,𝐹𝐶 =200𝑁,𝑁𝐶 =400𝑁,𝐹𝐵 =66.7𝑁

This case occurs first since it requires

a smaller value for 𝑃HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



- Example 8.5 Blocks 𝐴 and 𝐵 have a mass of 3𝑘𝑔 and 9𝑘𝑔,

respectively, and are connected to the

weightless links. Determine the largest

vertical force 𝑃 that can be applied at the

pin 𝐶 without causing any movement.

The coefficient of static friction between

the blocks and the contacting surfaces is

𝜇𝑠 = 0.3

Solution

Free-body diagram


Pin 𝐶

+→∑𝐹𝑥 = 0: 𝐹𝐴𝐶𝑐𝑜𝑠300 − 𝑃 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐴𝐶𝑠𝑖𝑛300 − 𝐹𝐵𝐶 = 0




Pin 𝐶

+→∑𝐹𝑥 = 0: 𝐹𝐴𝐶𝑐𝑜𝑠300 − 𝑃 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝐹𝐴𝐶𝑠𝑖𝑛300 − 𝐹𝐵𝐶 = 0

Block 𝐴

+→∑𝐹𝑥 = 0: 𝐹𝐴 − 𝐹𝐴𝐶𝑠𝑖𝑛300 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐴 − 𝐹𝐴𝐶𝑐𝑜𝑠300

−3× 9.81 = 0

Block 𝐵

+→∑𝐹𝑥 = 0: 𝐹𝐵𝐶 − 𝐹𝐵 = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐵 − 9 × 9.81 = 0

⟹ 𝐹𝐴𝐶 = 1.155𝑃, 𝐹𝐵𝐶 = 0.5774𝑃

𝐹𝐴 = 0.5774𝑃, 𝑁𝐴 = 𝑃 + 29.43

𝐹𝐵 = 0.5774𝑃, 𝑁𝐵 = 88.29𝑁HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien



𝐹𝐴𝐶 = 1.155𝑃, 𝐹𝐵𝐶 = 0.5774𝑃

𝐹𝐴 = 0.5774𝑃, 𝑁𝐴 = 𝑃 + 29.43

𝐹𝐵 = 0.5774𝑃, 𝑁𝐵 = 88.29𝑁

Movement of the system may be caused by

the initial slipping of either block𝐴 or block𝐵

Assume that block 𝐴 slips first

𝐹𝐴 = 𝜇𝑠𝑁𝐴 = 0.3𝑁𝐴

⟹ 0.5774𝑃 = 0.3(𝑃 + 29.43) ⟹ 𝑃 = 31.8𝑁

𝐹𝐵 = 0.5774 × 31.8 = 18.4𝑁

Since the maximum static frictional force at 𝐵

𝐹𝐵𝑚𝑎𝑥 = 𝜇𝑠𝑁𝐵 = 0.3 × 88.29 = 26.5𝑁 > 𝐹𝐵

block 𝐵 will not slip. Thus, the above assumption is correct

Note: If the inequality were not satisfied, we would have to

assume slipping of block 𝐵 and then solve for 𝑃HCM City Univ. of Technology, Faculty of Mechanical Engineering Nguyen Tan Tien


Fundamental Problems

- F.8.1 If 𝑃 = 200𝑁, determine the friction developed between

the 50𝑘𝑔 crate and the ground. The coefficient of static friction

between the crate and the ground is 𝜇𝑠 = 0.3



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- F.8.2 Determine the minimum force 𝑃 to prevent the 30𝑘𝑔 rod

𝐴𝐵 from sliding. The contact surface at 𝐵 is smooth, whereas

the coefficient of static friction between the rod and the wall at

𝐴 is 𝜇𝑠 = 0.2




- F.8.3 Determine the maximum force 𝑃 that can be applied

without causing the two 50𝑘𝑔 crates to move. The coefficient

of static friction between each crate and the ground is 𝜇𝑠 =0.25




- F.8.4 If the coefficient of static friction at contact points 𝐴 and

𝐵 is 𝜇𝑠 = 0.3, determine the maximum force 𝑃 that can be

applied without causing the 100𝑘𝑔 spool to move




- F.8.5 Determine the minimum force 𝑃 that can be applied

without causing movement of the 250𝑁 crate which has a

center of gravity at 𝐺. The coefficient of static friction at the

floor is 𝜇𝑠 = 0.4



§3.Wedges

- A wedge is a simple machine in which a small force 𝑃 is used

to lift a large weight 𝑊

- Wedges are used to adjust the elevation or provide stability for

heavy objects



§3.Wedges

- Analysis of a wedge

• 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



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§3.Wedges

• 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

∑𝐹𝑥 = 0

∑𝐹𝑦 = 0

For the impending motion frictional equation

𝐹 = 𝜇𝑠𝑁



§3.Wedges

• 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

• If the object is to be lowered,

then the wedge needs to be

pulled out

• If the value of the force 𝑃needed to remove the wedge is

positive, then the wedge is self-

locking, i.e., it will not come out

on its own



§3.Wedges

- Example 8.6 The stone has a mass of 500𝑘𝑔 and is held in

the horizontal position using a wedge at 𝐵. If

𝜇𝑠 = 0.3 at the surfaces of contact, determine

the minimum force 𝑃 needed to remove the

wedge. Assume that the stone does not slip at 𝐴

Solution

The free body diagrams

For the wedge

+→∑𝐹𝑥 = 0: 𝑁𝐵𝑠𝑖𝑛70−0.3𝑁𝐵𝑐𝑜𝑠7

0−0.3𝑁𝐶+𝑃=0

+ ↑ ∑𝐹𝑦 = 0: 𝑁𝐶 − 0.3𝑁𝐵𝑠𝑖𝑛70 −𝑁𝐵𝑐𝑜𝑠7

0 = 0

For the stone

+↺∑𝑀𝐴 = 0: −4905×0.5+𝑁𝐵𝑐𝑜𝑠70×1+0.3𝑁𝐵𝑠𝑖𝑛7

0×1=0

⟹ 𝑁𝐵 = 2383.1𝑁,𝑁𝐶 = 2452.5𝑁, 𝑃 = 1154.9𝑁



§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




- Screws can be classified by the thread. E.g. square-threaded

screw, V-thread, …

- External / Internal thread

- Right / Left hand thread




- If we unwind the thread by one revolution, the slope or lead

angle is given by

𝜃 = 𝑡𝑎𝑛−1𝑙

2𝜋𝑟

𝑙: the lead of the screw, is the distance advanced by turning

the screw one revolution



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- Upward Impending Motion

• Consider a square-threaded screw subject to impending

motion due to an applied torque 𝑀

• The free body diagram of the entire unraveled thread through

𝑊: vertical force on the or the axial force on the shaft

𝑅: reaction of the groove on the thread, 𝑅 = 𝐹 + 𝑁

𝐹: frictional component, 𝑁: normal component

𝑀/𝑟: horizontal force

associated with the

couple moment 𝑀




• The frictional component 𝐹 = 𝜇𝑠𝑁

• The angle of static friction 𝜙𝑠 = 𝑡𝑎𝑛−1 𝐹/𝑁 = 𝑡𝑎𝑛−1𝜇𝑠

• By Equations of Equilibrium

+→∑𝐹𝑥 = 0: 𝑀/𝑟 − 𝑅𝑠𝑖𝑛(+𝜃) = 0

+ ↑ ∑𝐹𝑦 = 0: 𝑅𝑐𝑜𝑠 𝜙𝑠 + 𝜃 −𝑊 = 0

⟹ 𝑀 = 𝑟𝑊𝑡𝑎𝑛(𝜙𝑠 + 𝜃)




- Self Locking Screw

• A screw is self-locking if it remains in place under any axial

load 𝑊 when the moment 𝑀 is removed

• In this case 𝑅 acts on the other side of 𝑁

• If 𝜙𝑠 = 𝜃, then 𝑅 will act vertically to balance 𝑊,

and the screw will be on the verge of winding

downwards




- Downward Impending Motion

• If a screw is self-locking, a couple 𝑀′ must be applied in the

opposite direction to wind the screw downward

• 𝜙𝑠 > 𝜃

• This causes a horizontal force in the reverse

direction that will push the thread downwards

• Using the previous procedure it can be shown

that 𝑀′ = 𝑟𝑊𝑡𝑎𝑛(𝜃 − 𝜙𝑠)




• If the screw is not self-locking it is necessary to apply a

moment 𝑀" to prevent the screw from winding downwards

• 𝜙𝑠 < 𝜃

• A horizontal force 𝑀′/𝑟 is required to push

against the thread to prevent it from sliding

downwards

• The magnitude of the moment required to

prevent this unwinding is 𝑀" = 𝑟𝑊𝑡𝑎𝑛(𝜙𝑠 − 𝜃)




- Example 8.7 The turnbuckle has

a square thread with a mean radius

of 5𝑚𝑚 and a lead of 2𝑚𝑚. If the

coefficient of static friction between

the screw and the turnbuckle is

𝜇𝑠 = 0.25, determine the moment 𝑀that must be applied to draw the

end screws closer together 2𝑘𝑁

Since friction at two screws must be overcome, this requires

𝑀 = 2 × 𝑟𝑊𝑡𝑎𝑛(𝜙𝑠 + 𝜃)

where 𝑤 = 2000𝑁, 𝑟 = 5𝑚𝑚

𝜙𝑠 = 𝑡𝑎𝑛−1𝜇𝑠 = 𝑡𝑎𝑛−1(0.25) = 14.040

𝜃 = 𝑡𝑎𝑛−1(𝑙/2𝜋𝑟) = 𝑡𝑎𝑛−1(2/(2𝜋 × 5)) = 3.640

⟹ 𝑀 = 2 × 5 × 2000𝑡𝑎𝑛 14.040 + 3.640 = 6374.7𝑁𝑚𝑚



Solution

5/6/2013

9

§5.Frictional Forces on Flat Belts

§6.Frictional Forces on Collar Bearings, Pivot Bearings, and Disks

§7.Frictional Forces on Journal Bearings

§8.Rolling Resistances



ch.08 friction

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