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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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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
9.1 Theory of Dry Friction9.1 Theory of Dry Friction
Nf S
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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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9.1 Theory of Dry Friction9.1 Theory of Dry Friction
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.1 Determining the Friction Example 9.1 Determining the Friction ForceForce
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
Fig. 9.12
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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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Example 9.4 A Friction Problem in 3 Example 9.4 A Friction Problem in 3 DimensionsDimensions
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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9.2 Applications9.2 Applications
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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9.2 Applications9.2 Applications
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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9.2 Applications9.2 Applications
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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9.2 Applications9.2 Applications
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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9.2 Applications9.2 Applications
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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9.2 Applications9.2 Applications
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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9.2 Applications9.2 Applications
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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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders
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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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders
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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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders
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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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders
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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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders
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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Example 9.9 Rope Wrapped Around 2 Example 9.9 Rope Wrapped Around 2 CylindersCylinders
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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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
Fig. 9.30
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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
SolutionSolution
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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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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Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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 Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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
72
Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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
73
Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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:
74
Design Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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 Example 9.10 Belts & PulleysDesign Example 9.10 Belts & Pulleys
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
76
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
77
Computational Example 9.11Computational Example 9.11
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.
78
Computational Example 9.11Computational Example 9.11
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
79
Computational Example 9.11Computational Example 9.11
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
80
Computational Example 9.11Computational Example 9.11
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
81
Computational Example 9.11Computational Example 9.11
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
82
Computational Example 9.11Computational Example 9.11
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:
83
Computational Example 9.11Computational Example 9.11
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
84
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 :
85
Chapter SummaryChapter Summary
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
86
Chapter SummaryChapter Summary
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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Chapter SummaryChapter Summary
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
88
Chapter SummaryChapter Summary
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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Chapter SummaryChapter Summary
Thrust Bearings & Clutches: The couple required to rotate the shaft at a
constant rate is:
(9.13)
22
33k
cos32
io
io
rr
rrFM
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Chapter SummaryChapter Summary
Belt Friction: The force T2 required for impending slip in the
direction of T2 is:
(9.17)
where is in radians
S12 eTT