Engineering Mechanics:
Statics
Chapter 6B: Applications of Friction in
Machines
Wedges
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Used to produce small position adjustments of a body
or to apply large forces
When sliding is impending, the resultant force on each
surface is inclined from the surface normal by an
amount equal to the friction angle.
The component of the resultant along the surface is
the friction force - direction to oppose the motion of
the wedge.
Force P required to lift up a large mass m can be
obtained from force equilibrium
R1 and R2 make an angle f to the surface normal
Wedges
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If P is removed and the wedge is sliding out, slippage must occur at
both upper and lower surfaces simultaneously
 Fig (a) – upper surface slips – R2 is of angle f to the inclined surface
 Fig (c) – lower surface slips – R1 is of angle f to the surface
Wedges
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Therefore, the wedge will be self-locking (not sliding out when P is
removed) if the wedge angle is in between the two cases or
a < 2f -- self-locking condition
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If the wedge is self-locking and is to be withdrawn, a pull P on the
wedge will be required.
Sample Problem 6/6
ms for both pairs of wedge surfaces = 0.30
ms between the block and the horizontal surface = 0.60
Determine the least P to move the block
Screws
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Used to fastening and for transmitting power or motion
For transmitting power, square thread is more efficient than the V-thread
o Consider a square-threaded jack under an axial load
W and a moment M,
• screw lead = L (advancement per revolution)
• mean radius = r
helix angle
-- To raise load
W = SRcos(a + f)
M = rsin(a + f) SR
M = Wr tan(a + f)
Screws
Conditions for unwinding
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If M is removed, the friction force changes direction.
The screw will be self-locking if a < f. An equivalent force P = M/r must
be applied to pull the thread down.
For a > f, the moment is required to prevent unwinding.
If a < f ,
M = Wr tan(f - a)
If a > f ,
M = Wr tan(a - f)
Sample Problem 6/7
150 mm
200 mm 250 mm
Screw has a square thread with mean diameter 25 mm
and a lead of 5 mm. ms in the threads = 0.2
Apply a 300-N pull at A. This produces a clamping force
of 5 kN between the jaws of the vise
(a) Find frictional moment MB
(b) Find force Q applied normal to the handle at A
required to loosen the vise
Journal Bearings
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give lateral support to a shaft (not axial)
As the shaft begins to turn in the direction shown, it will roll up the inner
surface of the bearing until it slips
Reaction R (caused by radial load L and torque M) made an angle f
to normal
SFy = 0;
SM = 0;
R = L
M = Lrf = Lr sin f
radius of the ‘friction circle’
For small m, (m = tan f ~ sin f)
M = m Lr
M = Applied moment to overcome friction
Journal Bearings
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Unwinding the cable from this spool
requires overcoming friction from the
supporting shaft
FBD of the shaft
Example
A torque M of 1510 N.m is applied to the 50-mm-diameter
shaft of the hoisting drum to raise the 500-kg load at
constant speed. The drum and shaft together have a mass
of 100 kg. Calculate the coefficient of friction m for the
bearing.
Sample Problem
The diameter of the bearing for the upper pulley is 20
mm and that for the lower puller is 12 mm. For m = 0.25
for both bearings, calculate T, T1 and T2 if the block is
being raised slowly.
Thrust Bearings; Disk Friction
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Friction between circular surfaces under distributed normal pressure
Pivot bearings, clutch plates, disk brakes
1. New surface
p = uniform = P/R2
M = mprdA
=
= 2/3 mPR
-- equivalent to moment due to friction force
mP acting at 2/3 R from the shaft center
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For ring friction disks (ex. collar bearing)
Ro3 - Ri3
2
M  mP 2
3
Ro - Ri2
Thrust Bearings; Disk Friction
2. After wearing-in period, further wear is constant over the surface
Wear depends on
- circumferential distance (proportional to r)
- pressure p
\ rp = K
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P = pdA =
= 2KR
M = mprdA =
= ½ mPR = ¾ times of the new plate
For rings, M = ½ mP(Ro + Ri)
Flexible Belts: cable, rope
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If friction is neglected, T1 = T2
If friction is consider, M (CW) -- T2 > T1
Frictional moment to resist rotation
Equilibrium
SFt = 0
mdN = dT
1
dN = Tdq
2
SFn = 0
Flexible Belts: cable, rope
Substitute 2 into 1
T2/T1 = emb
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b = total angle of belt contact (-- in radians!)
If a rope were wrapped around a drum n times:
b = 2n radians
Can also be used for a noncircular contact where the
total angle of contact is b
Sample Problem 6/9
Let m between the cable and the fixed drum be 0.30
(a) For a = 0, determine the maximum and minimum
values which P may have in order not to raise or
lower the load
(b) for P = 500 N, determine the minimum value of a
before the load begins to slip
Sample Problem
Calculate the horizontal force P required to raise the 100-kg
load. The coefficient of friction between the rope and the
fixed bars is 0.40