concepts about friction

Engineering Mechanics Engineering Mechanics

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Friction

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Chapter OutlineChapter Outline

Theory of Dry Friction Applications Computational Mechanics

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9.1 Theory of Dry Friction9.1 Theory of Dry Friction

To examine the nature of friction forces: Place a book on a table & push it with small

horizontal force:

If the force you exert is sufficiently small, the book does not move

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Free-body diagram:

The force W is the book’s weight & N is the total normal force exerted by the table on the surface of the book that is in contact with the table

The force F is the horizontal force you apply & f is the friction force exerted by the table

Because the book is in equilibrium, f = F

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As long as the book remains in equilibrium, when you increase the force you apply to the book, the friction force must increase correspondingly

When the force you apply becomes too large, the book moves (slips on the table)

After reaching some maximum value, the friction force can no longer maintain the book in equilibrium

Notice that the force you must apply to keep the book moving on the table is smaller than the force required to cause it to slip

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If the surfaces of the table & the book are magnified sufficiently, they will appear rough:

Friction forces arise in part from the interactions of the roughness or asperities of the contacting surfaces

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Suppose that we idealize the asperities of the book & table as the mating 2-D “saw-tooth” profiles in Fig. a

As the horizontal force F increases, the book will remain stationary until the force is sufficiently large to cause the book to slide upward as shown in Fig. b

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The normal force Ci exerted on the ith saw-tooth asperity of the book:

Notice that in this simple model we assume the contacting surfaces on the asperities to be smooth

Denote the sum of the normal forces exerted on the asperities of the book by the table by

ii

CC

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Equilibrium equations:

Eliminating C from these equations, we obtain the force necessary to cause the book to slip on the table:

The force necessary to cause the book to slip is proportional to the force pressing the saw-tooth surfaces together (the book’s weight)

0cos

0sin

WCF

CFF

y

x

WF tan

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Also, the angle is a measure of the roughness of the saw-tooth surfaces:

As 0, the surfaces become smooth & the force necessary to cause the book to slip approaches zero

As increases, the roughness increases & the force necessary to cause the book to slip increases

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Coefficients of Friction: The theory of dry friction or Coulomb friction

predicts the maximum friction forces that can be exerted by dry, contacting surfaces that are stationary relative to each other

It also predicts the friction forces exerted by the surfaces when they are in relative motion or sliding

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The Static Coefficient: The magnitude of the maximum friction force

that can be exerted between 2 plane, dry surfaces in contact that are not in motion relative to 1 another is:

(9.1)

where N is the normal component of the contact force between the surfaces & S is a constant called the coefficient of static friction


Nf S

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The value of S is assumed to depend only on the materials of the contacting surfaces & the conditions (smoothness & degree of contamination by other materials) of the surfaces

Typical values of S for various materials are shown in Table 9.1

The relatively large range of values for each pair of materials reflects the sensitivity of S to the conditions of the surfaces

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Table 9.1:

MaterialsCoefficient of

Station Friction, S

Metal on metal 0.150.20

Masonry on masonry

0.600.70

Wood on wood 0.250.50

Metal on masonry 0.300.70

Metal on wood 0.200.60

Rubber on concrete 0.500.90

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Returning to the example of the book on the table:

If we know the coefficient of static friction between the book & the table, Eq. (9.1) tells us the largest friction force that the table can exert on the book:

F = f = SN Also, from the free-body diagram: N = W, so

the largest force that will not cause the book to slip is F = SW

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Eq. (9.1) determines the magnitude of the maximum friction force but not its direction

The friction force is a maximum & Eq. (9.1) is applicable when 2 surfaces are on the verge of slipping relative to each other slip is impending & the friction forces resist the impending motion

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E.g. the lower surface is fixed & slip of the upper surface toward the right is impending:

The friction force on the upper surface resists its impending motion

The friction force on the lower surface is in the opposite direction

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The Kinetic Coefficient: According to the theory of dry friction, the

magnitude of the friction force between 2 plane dry contacting surfaces that are in motion (sliding) relative to each other is:

(9.2)

where N is the normal force between the surfaces & k is the coefficient of kinetic friction

Nf k

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The value of k is assumed to depend only on the compositions of the surfaces & their conditions

For a given pair of surfaces, its value is generally smaller than that of S

Once you have caused the book to begin sliding on the table, the friction force:

f = kN = kW

Therefore, the force you must exert to keep the book in uniform motion is: F = f = kW

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When 2 surfaces are sliding relative to each other, the friction forces resist the relative motion:

E.g. the lower surface is fixed & the upper surface is moving to the right

The friction force on the upper surface acts in the direction opposite to its motion

The friction force on the lower surface is in the opposite direction

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Angles of Friction: Expressing the reaction exerted

on a surface due to its contact with another surface in terms of its components parallel & perpendicular to the surface, the friction force f & normal force N

Expressing the reaction in terms of its magnitude R & the angle of friction between the reaction & the normal to the surface

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The forces f & N are related to R & by:

(9.3)

(9.4)

The value of when slip is impending is called the angle of static friction S & its value when the surfaces are sliding relative to each other is called the angle of kinetic friction k

cos

sin

RN

Rf

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By using Eqs. (9.1)—(9.4), we can express the angles of static & kinetic friction in terms of the coefficients of friction:

(9.5)

(9.6)kk

SS

tan

tan

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Summary: If slip is impending, the magnitude of the

friction is given by Eq. (9.1) & the angle of friction by Eq. (9.5)

If the surfaces are sliding relative to each other, the magnitude of the friction force is given by Eq. (9.2) & the angle of friction by Eq. (9.6)

Otherwise, the friction force must be determined from the equilibrium equations

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The sequence of decisions in evaluating the friction force & angle of friction is summarized in the Fig 9.8:

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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce

The arrangement in Fig. 9.9 exerts a horizontal force on the stationary 180-N crate. The coefficient of static friction between the crate & the ramp is S = 0.4.

(a) If the rope exerts a 90-N force on the crate, what is the friction force exerted on the crate by the ramp?(b) What is the largest force the rope can exert on the crate without causing it to slide up the ramp?

Fig. 9.9

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StrategyStrategy(a) We can follow the logic in Fig. 9.8 to decide how

to evaluate the friction force. The crate is not sliding on the ramp & we don’t know whether slip is impending, so we must determine the friction force by using the equilibrium equations.

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StrategyStrategy(b) We want to determine the value of the force exerted by the rope that causes the crate to be on the verge of slipping up the ramp. When slip is impending, the magnitude of the friction force

is f = SN & the friction force opposes the impending slip. We can use the equilibrium equations to determine the force exerted by the rope.

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SolutionSolution(a) Draw the free-body diagram of the crate, showing the force T exerted by the rope, the weight W of the crate & the normal force N & friction force f exerted by the ramp:

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SolutionSolutionWe can choose the direction of f arbitrarily & our solution will indicate the actual direction of the friction force.

By aligning the coordinate system with the ramp as shown, we obtain the equilibrium equation:

Σ Fx = f + T cos 20° W cos 20° = 0

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SolutionSolutionSolving for the friction force, we obtain: f = T cos 20° + W sin 20°

= (90 N) cos 20° + (180 N) sin 20° = 23.0 N

The minus sign indicates that the direction of the friction force on the crate is down the ramp.

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SolutionSolution(b) In this case the friction force f = SN & it opposes the impending slip.

To simplify our solution for T, we align the coordinate system as shown:

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SolutionSolutionThe equilibrium equations:

Σ Fx = T N sin 20° SN cos 20° = 0

Σ Fy = N cos 20° SN sin 20° W= 0

Solving the 2nd equilibrium equation for N, we obtain:

N 22420sin4020cos

N 180

20sin20cos S

.

WN

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SolutionSolutionThen, from the 1st equilibrium equation, the force T is: T = N (sin 20° + S cos 20°) = 0

= (224 N) (sin 20° + 0.4 cos 20°) = 161 N

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Critical ThinkingCritical Thinking When you use the equilibrium equations to

determine a friction force, often you will not know its direction beforehand: Depending on the value of the force T exerted

on the crate by the rope, the friction force exerted on the crate by the ramp can point either up or down the ramp

In drawing the free-body diagram of the crate in (a), we arbitrarily assumed that the friction force pointed up the ramp

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Critical ThinkingCritical Thinking The negative value obtained from the

equilibrium equations, f = 23.0 N, tells us that the force is in the opposite direction, down the ramp

In contrast, when you use the equation f = SN, the friction force must point in the correct direction on the free-body diagram

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Critical ThinkingCritical Thinking In drawing the free-body diagram in (b), we

wanted to determine the largest force T that would not cause the crate to slide up the ramp, so we assumed that the slip of the crate up the ramp was impending

This told us that the friction force, resisting the impending slip, pointed down the ramp

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Example 9.2 A Friction Problem in 3 Example 9.2 A Friction Problem in 3 DimensionsDimensions

The 80-kg climber at A in Fig. 9.12 is supported on an icy slope by friends. The tensions in the ropes AB & AC are 130 N & 220 N respectively, the y axis is vertical & the unit vector e = −0.182i + 0.818j + 0.545k is perpendicular to the ground where the climber stands. What minimum coefficient of static friction between the climber’s shoes & the ground is necessary to prevent him from slipping?

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Fig. 9.12

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StrategyStrategy We know the forces exerted on the climber by the

2 ropes & by his weight so we can use equilibrium to determine the force R exerted on him by the ground. The components of R normal & parallel to the ground are the normal & friction forces exerted on him by the ground. By calculating them, we can obtain the minimum necessary coefficient of static friction.

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SolutionSolutionDraw the free-body diagram of the climber showing the forces TAB & TAC exerted by the ropes, the force R exerted by the ground & his weight:

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SolutionSolutionThe sum of the forces equals zero:

R + TAB + TAC – mgj = 0

By expressing TAB & TAC in terms of their components, we can solve this equation for the components of R. The force TAB is:

kji

kji

kjiTT

5.1137.564.28

873.0436.0218.0 N 130

400232

400232

222

ABAB

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SolutionSolutionAnd the force TAC is:

Substituting these expressions into the equilibrium equation & solving for R, we obtain:

R = –48.2i + 651.5j + 305.0k (N)

kji

kji

kjiTT

5.1916.766.76

870.0348.0348.0 N 220

410235

410235

222

ACAC

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SolutionSolutionThe normal force on the climber is the component of R perpendicular to the surface, which is the component of R parallel to the unit vector e:N = (e · R)e = [(−0.182)(−48.2 N) + (0.818)(651.5 N) + (0.545)(305.0 N)]e = −129i + 579j + 386k (N)

The magnitude of the normal force is:N = |N| = 708 N.

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SolutionSolutionThe friction force on the climber is the magnitude of the component of R parallel to the surface:

f = |R – N| = 135 N

The minimum coefficient of friction necessary to prevent the climber from slipping is therefore:

191.0N 708

N 135S

N

f

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Critical ThinkingCritical Thinking Notice the role of the unit vector e in this example:

By using equilibrium, we were able to determine the total force R exerted on the climber by the ground

This force consists of components normal & parallel to the ground (the normal & friction forces respectively)

The unit vector e specified the orientation of the ground on which the climber stood & allowed us to determine the normal & friction forces

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9.2 Applications9.2 Applications

Belt Friction: If a rope is wrapped around a fixed post, a

large force T2 exerted on 1 end can be supported by a relatively small force T1 applied to the other end:

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Consider a rope wrapped through an angle around a fixed cylinder:

Assume that the tension T1 is known

The objective is to determine the largest force T2 that can be applied to the other end of the rope without causing the rope to slip

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Draw the free-body diagram of an element of the rope whose boundaries are at angles & + ∆ from the point where the rope comes into contact with the cylinder:

The force T is the tension in the rope at the position defined by the angle

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The tension in the rope varies with position, because it increases from T1 at = 0 to T2 at =

Therefore, we write the tension in the rope at the position + ∆ as T + ∆T

The force ∆N is the normal force exerted on the element by the cylinder

Assume that the friction force is equal to its maximum possible value S∆N, where S is the coefficient of static friction between the rope & the cylinder

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The equilibrium equations in the directions tangential to & normal to the centerline of the rope are:

(9.16)

Eliminating ∆N, we can write the resulting equilibrium equation as:

02

sin2

sin

02

cos2

cos

normal

Stangential

TΔα

TTNF

TTTNF

0

2/2/sin

2 sin

2 cos SS

TT

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Evaluating the limit of this equation as ∆ 0 & observing that:

We obtain:

This differential equation governs the variation of the tension in the rope

0S TddT

1

2/2/sin

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9.2 Applications9.2 Applications Separating the variables yields:

We can now integrate to determine the tension T2 in terms of the tension T1 & the angle :

Thus, we obtain the largest force T2 that can be applied without causing the rope to slip when the force on the other end is T1:

(9.17)

0

2

1 d

TdT

S

T

T

S12 eTT

dTdT

S

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The angle in this equation must be expressed in radians

Replacing S by the coefficient of kinetic friction k gives the force T2 required to cause the rope to slide at a constant rate

Eq. (9.17) explains why a large force can be supported by a relatively small force when a rope is wrapped around a fixed support:

The force required to cause the rope to slip increases exponentially as a function of the angle through which the rope is wrapped

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Suppose S = 0.3: When the rope is wrapped 1 complete

turn around the post ( = 2), the ratio T2/T1 = 6.59

When the rope is warped 4 complete turns around the post ( = 8), the ratio T2/T1 = 1880

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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders

The 100-N crate in Fig. 9.29 is suspended from a rope that passes over 2 fixed cylinders. The coefficient of static friction is 0.2 between the rope & the left cylinder & 0.4 between the rope & the right cylinder. What is the smallest force the woman can exert & support the crate?

Fig. 9.29

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StrategyStrategy She exerts the smallest possible force when slip of

the rope is impending on both cylinders. Because we know the weight of the crate, we can use Eq. (9.17) to determine the tension in the rope between the 2 cylinders & then use Eq. (9.17) again to determine the force she exerts.

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SolutionSolutionThe weight if the crate is W = 100 N. Let T be the tension in the rope between the 2 cylinders. The rope is wrapped around the left cylinder through an angle = /2 rad.

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SolutionSolutionThe tension necessary to prevent the rope from slipping on the left cylinder is related to W by:

Solving for T, we obtain:

N 0.73

N 100 2/2.02/2.0

eWeT

2/2.0S TeTeW

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SolutionSolutionThe rope is also wrapped around the right cylinder through an angle = /2 rad.The force F the woman must exert to prevent the rope from slipping on the right cylinder is related to T by:

The solution for F is:

N 0.39

N 3.07 2/4.02/4.0

eTeF

2/4.0S FeFeT

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Critical ThinkingCritical Thinking To determine the force that would need to be

exerted on the rope to cause the crate to begin moving upward: Assume that the slip of the rope is impending

on both cylinders but in the opposite direction to our analysis in this example

For the left cylinder, the tension T necessary for slip of the rope to be impending in the direction that would cause the crate to move upward is:

N 137N 001 2/2.02/2.0 eWeT

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Critical ThinkingCritical Thinking For the right cylinder, the force F necessary

for slip to be impending in the direction that would cause the crate to move upward is:

Although the young woman would be able to support the stationary crate, she might need to call for help to raise it.

N 257

N 371 2/4.02/4.0

eTeF

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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys

The pulleys in Fig. 9.30 turn at a constant rate. The large pulley is attached to a fixed support. The small pulley is supported by a smooth horizontal slot & is pulled to the right by the force F = 200 N. The coefficient of static friction between the pulleys & the belt is S = 0.8, the dimension b = 500 mm & the radii of the pulleys are RA = 200 mm & RB = 100 mm. What are the largest values of the couples MA & MB for which the belt will not slip?

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Fig. 9.30

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StrategyStrategy By drawing the free-body diagrams of the pulleys,

we can use the equilibrium equations to relate the tensions in the belt to MA & MB & obtain a relation between the tensions in the belt & the force F. When slip is slipping, the tensions are also related by Eq. (9.17). From these equations we can determine MA & MB.

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SolutionSolution

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SolutionSolutionFrom the free-body diagram of the large pulley, we obtain the equilibrium equation:

MA = RA (T2 T1) (1)

and from the free-body diagram of the small pulley, we obtain:

F = (T1 + T2) cos (2)

MB = RB (T2 T1) (3)

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SolutionSolutionThe belt is in contact with the small pulley through the angle 2:

From the dashed line parallel to the belt, we see that the angle satisfies the relation:

2.0mm 500

mm 100mm 200sin

bRR BA

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SolutionSolutionTherefore = 15° = 0.201 rad.

If we assume that the slip is impends between the small pulley & the belt, Eq. (9.17) states that:

We solve this equation together with Eq. (2) for the 2 tensions, obtaining T1 = 20.5 N & T2 = 183.6 N.

Then from Eqs. (1) & (3), the couples are: MA = 32.6 N-m & MB = 16.3 N-m

1

28.01

S12 95.8 TeTeTT

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SolutionSolutionIf we assume that slip impends between the large pulley & the belt, we obtain:

MA = 36.3 N-m & MB = 18.1 N-m,

so the belt slips on the small pulley at smaller values of the couples.

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Design IssuesDesign Issues Belts & pulleys are used to transfer power in

cars & many other types of machines, including printing presses, farming equipment & industrial robots: Because 2 pulleys of different diameters

connected by belt are subjected to different torques & have different rates of rotation, they can be used as a mechanical “transformer” to alter torque or rotation rate

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Design IssuesDesign Issues In this example we assumed that the

belt was flat but “V-belts” that fit into matching grooves in the pulleys are often used in applications This configuration keeps the belt

in place on the pulleys & also decreases the tendency of the belt to slip

Suppose that a V-belt is wrapped through an angle around a pulley

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Design IssuesDesign Issues If the tension T1 is known, what is the largest

tension T2 that can be applied to other end of the belt without causing it to slip?

Free-body diagram of an element of the belt whose boundaries are at angles & + ∆ from the point where the belt comes into contact with the pulley:

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Design IssuesDesign Issues The equilibrium equations in the directions

tangential to & normal to the centers of the belt are:

(4)

By the same steps leading from Eqs. (9.16) to Eq. (9.17), it can be shown that:

(5)

02

sin2

sin 2

sin2

02

cos2

cos2

normal

Stangential

TΔα

TTNF

TTTNF

2/sin/S12

eTT

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Design IssuesDesign Issues Thus, using a V-belt effectively increases the

coefficient of friction between the belt & the pulley by the factor 1/sin ( /2)

When it is essential that the belt not slip relative to the pulley, a belt with cogs & a matching pulley or a chain & sprocket wheel can be used

The chains & sprocket wheels in bicycles & motorcycles are examples

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Computational Example 9.11Computational Example 9.11

The mass of the block A in Fig. 3.33 is 20 kg & the coefficient of static friction between the block & the floor is S = 0.3. The spring constant k = 1 kN/m & the spring is unstretched. How far can the slider B be moved to the right without causing the block to slip?

Fig. 3.33

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StrategyStrategy Draw the free-body diagram of block A assuming

that the slider B is moved a distance x to the right & slip of block A is impending. Then by applying the equilibrium equations, we can obtain an equation for the distance x corresponding to impending slip.

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SolutionSolutionSuppose that moving the slider B a distance x to the right causing impending slip of the block. The resulting stretch of the spring is m. 11 2 x

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SolutionSolutionThe magnitude of the force exerted on the block by the spring is:

(1)

From the free-body diagram of the block, we obtain the equilibrium equations:

0 1

1

0 1

S2

SS2

mgNFx

F

NFx

xF

y

x

11 2S xk F

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SolutionSolutionSubstituting Eq. (1) into these 2 equations & then eliminating N, we can write the resulting equation in the form:

We must obtain the root of this function to determine the value of x corresponding to impending slip of the block.

0111 2S

2S xmgx xkxh

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SolutionSolutionFrom the graph of h(x), we estimate that h(x) = 0 at x = 0.43 m. By examining computed results near this value of x (see table), we see that h(x) = 0 & slip is impending, when x is approximately 0.4284 m.

x (m) h(x)

0.4281 0.1128

0.4282 0.0777

0.4283 0.0425

0.4284 0.0074

0.4285 0.0278

0.4286 0.0629

0.4287 0.0981

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Critical ThinkingCritical Thinking Many software packages are available that allow

you to obtain solutions to nonlinear algebraic equations such as the 1 we obtained in this example

Even when you have access to such software, it is a good idea to examine graphical results like those we have presented:

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Critical ThinkingCritical Thinking Nonlinear equations sometimes have multiple

roots & to insure that you have obtained all the solutions within the range of interest & you have identified the 1 you want

In addition, you can often gain insight by examining the behaviour of an equation over a range of its variables instead of obtaining just 1 solution

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Chapter SummaryChapter Summary

Dry Friction: The forces resulting from the contact of 2

plane surfaces can be expressed in terms of the normal force N & friction force f or the magnitude R & angle of friction :

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If slip is impending, the magnitude of the friction force is:

(9.1) and its direction opposes the impending slip.

The angle of friction equals the angle of static friction S = arctan (S)

If the surfaces are sliding, the magnitude of the friction force is:

(9.2) and its direction opposes the relative motion.

The angle of friction equals the angle of kinetic friction k = arctan (k)

Nf S

Nf k

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Threads: The slope of the thread is related

to its pitch p by:

(9.7)

The couple required for impending rotation & the axial motion opposite to the direction of F is:

M = rF tan (S + ) (9.9)

rp 2

tan

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The couple required for impending rotation & the axial motion of the shaft in the direction of F is:

M = rF tan (S ) (9.11)

When S < , the shaft will rotate & move in the direction of the force F with no couple applied

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Journal Bearings: The couple required for impending slip of the

circular shaft is:

M = rF sin S (9.12)

where F is the total load on the shaft

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Thrust Bearings & Clutches: The couple required to rotate the shaft at a

constant rate is:

(9.13)

22

33k

cos32

io

io

rr

rrFM

90


Belt Friction: The force T2 required for impending slip in the

direction of T2 is:

(9.17)

where is in radians

S12 eTT

concepts about friction

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theory of dry friction

theory of dry friction

dry surfaces

force w

ifthe force

maximum friction forces

smallhorizontal force