Download Chap. 8 Friction

Chap. 8 Friction
Chapter Outline
Characteristics of Dry Friction
Problems Involving Dry Friction
Wedges
Frictional Forces on Screws
Frictional Forces on Flat Belts
Frictional Forces on Collar Bearings,
Pivot Bearings, and Disks
Frictional Forces on Journal Bearings
Rolling Resistance
8-1-2
APPLICATIONS
In designing a brake system for a
bicycle, car, or any other vehicle, it
is important to understand the
frictional forces involved.
For an applied force on the brake
pads, how can we determine the
magnitude and direction of the
resulting friction force?
8-1-3
APPLICATIONS (continued)
Consider pushing a box as
shown here. How can you
determine if it will slide, tilt,
or stay in static equilibrium?
What physical factors affect
the answer to this question?
8-1-4
CHARACTERISTICS OF DRY FRICTION (Section 8.1)
Friction is defined as a force of
resistance acting on a body which
prevents or retards slipping of the
body relative to a second body.
Experiments show that frictional
forces act tangent (parallel) to the
contacting surface in a direction
opposing the relative motion or
tendency for motion.
For the body shown in the figure to
be in equilibrium, the following must
be true: F = P, N = W, and Wx = Ph.
8-1-5
CHARACTERISTICS OF FRICTION (continued)
To study the characteristics of the friction force F, let us
assume that tipping does not occur (i.e., “h” is small or “a”
is large). Then we gradually increase the magnitude of the
force P. Typically, experiments show that the friction force
F varies with P, as shown in the left figure above.
8-1-6
FRICTION CHARACERISTICS (continued)
The maximum friction force is attained just before the
block begins to move (a situation that is called
“impending motion”). The value of the force is found
using Fs = μs N, where μs is called the coefficient of
static friction. The value of μs depends on the
materials in contact.
8-1-7
Once the block begins to move, the frictional force
typically drops and is given by Fk = μk N. The value of
μk (coefficient of kinetic friction) is less than μs .
• μk are typically 25% smaller than s
Contact Materials
Coefficient of Static Friction μs
Metal on ice
0.03 – 0.05
Wood on wood
0.30 – 0.70
Leather on wood
0.20 – 0.50
Leather on metal
0.30 – 0.60
Aluminum on aluminum
1.10 – 1.70
8-1-8
DETERMING μs EXPERIMENTALLY
A block with weight w is placed on
an inclined plane. The plane is
slowly tilted until the block just
begins to slip.
The inclination, θs, is noted. Analysis
of the block just before it begins to
move gives (using Fs = μs N):
+ ∑ Fy = N – W cos θs = 0
+ ∑ FX = μS N – W sin θs = 0
Using these two equations, we get μs
= (W sin θs ) / (W cos θs ) = tan θs
This simple experiment allows us to
find the μS between two materials in
contact.
8-1-9
PROCEDURE FOR ANALYSIS (Section 8.2)
Steps for solving equilibrium problems involving dry
friction:
1. Draw the necessary free body diagrams. Make sure
that you show the friction force in the correct direction
(it always opposes the motion or impending motion).
2. Determine the number of unknowns. Do not assume
F = μS N unless the impending motion condition is
given.
3. Apply the equations of equilibrium and appropriate
frictional equations to solve for the unknowns.
8-1-10
IMPENDING TIPPING versus SLIPPING
For a given W and h, how
can we determine if the
block will slide first or tip first?
In this case, we have four
unknowns (F, N, x, and P)
and only three EofE.
Hence, we have to make an
assumption to give us another
equation. Then we can solve
for the unknowns using the
three EofE. Finally, we need
to check if our assumption
was correct.
8-1-11
IMPENDING TIPPING versus SLIPPING (continued)
Assume: Slipping occurs
Known: F = μs N
Solve:
x, P, and N
Check:
0 ≤ x ≤ b/2
Or
Assume: Tipping occurs
Known: x = b/2
Solve:
P, N, and F
Check:
F ≤ μs N
8-1-12
Characteristics of Dry Friction
Angle of Friction
靜摩擦角
FS
)
N
−1 μ S N
= tan (
)
N
= tan −1 μ S
φs = tan −1 (
動摩擦角
Fk
)
N
μk N
φk = tan −1 (
= tan −1 (
N
= tan −1 μ k
)
ΦS ≥ ΦK
8-1-13
Characteristics of Dry Friction
Angle of Repose 安息角
N = W cos θ S
FS = W sin θ S
= μS N
= μ S (W cos θ S )
∴θ S = tan −1 μ S
安息角 = 靜摩擦角
8-1-14
Problems Involving Dry Friction
Equilibrium
Impending Motion at All Points
8-1-15
Problems Involving Dry Friction
Tipping or Impending Motion at Some Points
，
Ph
Ph = wx x =
w
Impending
Motion
0 ≤ x ≤ b/2
(F = μ S N )
Tipping
x=b/2
(F ≤ μ S N )
h=
一、p增大，x也會增大而往外移
二、h增大，P、F形成的couple增大、x增大 當
wx
P
x=
b
2
h = hmax
8-1-16
Example 8.1
The uniform crate has a mass of 20kg. If a
force P = 80N is applied on to the crate,
determine if it remains in equilibrium. The
coefficient of static friction is = 0.3.
8-1-17
Solution
Resultant normal force NC act a distance x
from the crate’s center line in order to
counteract the tipping effect caused by P.
3 unknowns to be determined by 3
equations of equilibrium.
8-1-18
Solution
+ → ∑ Fx = 0;
80 cos 30o N − F = 0
+ ↑ ∑ Fy = 0;
− 80 sin 30o N + N C − 196.2 N = 0
∑ M O = 0;
80 sin 30o N (0.4m) − 80 cos 30o N (0.2m) + N C ( x) = 0
F = 69.3 N , N C = 236 N
x = −0.00908m = −9.08mm
Solving
8-1-19
Solution
Since x is negative, the resultant force acts
(slightly) to the left of the crate’s center line.
No tipping will occur since x ≤ 0.4m
Max frictional force which can be
developed at the surface of contact
Fmax = sNC = 0.3(236N) = 70.8N
8-1-20
p. 401, 8-8
The uniform pole has a weight of 150 N and a length of 7.8 m.
Determine the maximum distance d it can be placed from the smooth
wall and not slip. The coefficient of static friction between the floor and
the pole is μs = 0.3.
8-1-21
p. 403, 8-15
The spool has a mass of 200 kg and rests against the wall and on the floor.
If the coefficient of static friction at B is (μs)B = 0.3, the coefficient of kinetic
friction is (μK)B = 0.2, and the wall is smooth, determine the friction force
developed at B when the vertical force applied to the cable is P = 800 N.
8-1-22
p. 405, 8-28
Determine the minimum force P needed to push the two 75-kg cylinders
up the incline. The force acts parallel to the plane and the coefficients of
static friction of the contacting surfaces are μA = 0.3, μB = 0.25, , and μC =
0.4. Each cylinder has a radius of 150 mm.
8-1-23
8-1-24
8-1-25
p. 408, 8-45
The 45-kg disk rests on the surface for which the coefficient of
static friction is μA = 0.2. Determine the largest couple moment M
that can be applied to the bar without causing motion.
8-1-26
8-1-27
p. 410, 8-60
If θ =15°, determine the minimum coefficient of static friction between the
collars A and B and the rod required for the system to remain in
equilibrium, regardless of the weight of cylinder D. Links AC and BC
have negligible weight and are connected together at C by a pin.
8-1-28
8-1-29
Wedges
(楔)
抬重物省力的工具
若P=0，則摩擦力會使Block(A)靜止稱為self-locking
8-2-1
Screws
„
An inclined plane or wedge wrapped around a
cylinder lead l, lead angle θ
M = P⋅L = s⋅r
8-2-2
Screws
上移
M PL
=
r
r
W , S , R三力平衡
止滑
S=
自動鬆開
必須加外力才不會下滑
使轉開
鎖住，必須加力才可鬆開
8-2-3
p.418, Problem 8-72
The wedge blocks are used to hold the specimen in a tension testing machine.
Determine the largest design angle θ of the wedges so that the specimen will
not slip regardless of the applied load. The coefficients of static friction are μ A
= 0.1 at A and μ B = 0.6 at B. Neglect the weight of the blocks.
8-2-4
8-2-5
p.420, Problem 8-84
The clamp provides pressure from several directions on the edges of the board.
If the square-threaded screw has a lead of 3 mm, mean radius of 10 mm, and
the coefficient of static friction is μ S = 0.4, determine the horizontal force
developed on the board at A and the vertical forces developed at B and C if a
torque of M = 1.5 N⋅m is applied to the handle to tighten it further. The blocks at
B and C are pin connected to the board.
8-2-6
8-2-7
Flat Belts
dθ
dθ
+ μdN − (T + dT ) cos
= 0...........(1)
2
2
dθ
dθ
ΣFy = 0 : dN − (T + dT ) sin
− T sin
= 0...............(2)
2
2
ΣFX = 0 : T cos
又dT ⋅ sin
dθ
dθ
dθ
dθ
→ 0, sin
→
, cos
→1
2
2
2
2
由(1) ⇒ μdN = dT
由(2) ⇒ dN = Tdθ
dT
= μdθ
相除 ⇒
T
β
T2 dT
= μ ∫ dθ
積分 ⇒ ∫
0
T1 T
T
⇒ ln 2 = μβ
T1
⇒ T2 = T1 ⋅ e
μβ
β以弳度 (radian)表示
8-2-8
p.424, 8-92.
The boat has a weight of 2500 N ( 250
kg) and is held in position off the side of
a ship by the spars at A and B. A man
having a weight of 650 N ( 65 kg) gets
in the boat, wraps a rope around an
overhead boom at C, and ties it to the
end of the boat as shown. If the boat is
disconnected from the spars, determine
the minimum number of half turns the
rope must make around the boom so
that the boat can be safely lowered into
the water at constant velocity. Also,
what is the normal force between the
boat and the man? The coefficient of
kinetic friction between the rope and
the boom is μs = 0.15. Hint: The
problem requires that the normal force
between the man’s feet and the boat be
as small as possible.
8-2-9
8-2-10
p.427, 8-108.
Determine the maximum number of 25-kg packages that can be
placed on the belt without causing the belt to slip at the drive wheel A
which is rotating with a constant angular velocity. Wheel B is free to
rotate. Also, find the corresponding torsional moment M that must be
supplied to wheel A. The conveyor belt is pre-tensioned with the
1500-N horizontal force. The coefficient of kinetic friction between the
belt and platform P is μK = 0.2, and the coefficient of static friction
between the belt and the rim of each wheel is μs = 0.35.
8-2-11
Bearings and Disks
p=
P
π ( R2 2 − R1 2 )
dF = μ S dN = μ S pdA =
μS P
π ( R2 − R1 )
2
2
dA
dA = rdrdθ
ΣM Z = 0 : M − ∫ A rdF = 0
∴M = ∫
R2 2π
∫
R1 0
r[
μS P
π ( R2 − R1 )
2
2
](rdrdθ ) =
μS P
π ( R2 − R1
2
2
)∫
R2
R1
2π
r dr ∫ dθ
2
0
⎛ R2 3 − R1 3 ⎞
2
⎟
∴ M = μ S P⎜⎜ 2
2 ⎟
3
⎝ R2 − R1 ⎠
For pivot bearing : R 2 = R,R1 = 0
2
⇒ M = μ S PR
3
8-2-12
Journal Bearings
r sin φk = rf
Σ M Z = 0 : − M + ( R sin φ k ) r = 0
⇒ M = Rr sin φ k
若 μ k 很小，則
μ k = tan φ k ≈ sin φ k ≈ φ k
⇒ M ≈ Rr μ k ≈ Rr f
8-2-13
Rolling Resistance
Wa＝P(r cosθ )
⇒ Wa ≈ Pr
Wa
P≈
r
a：滾動摩擦係數
1
8-2-14
p.443, 8-102.
8-2-15
p.446, 8-117.
8-2-16
8-2-17