Friction
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Transcript of Friction
Friction
Relative velocity
Cause of dry frictionContact between two surfaces.
Hence first task in a friction problem is correct identification of contact surfaces
Identify the surface, the normal and the tangential vectors.
Also important is to get an idea of probable direction of relative motion
The contact force acts along the normal.Gravity is the most common cause of normal force.
Friction acts along the tangent plane opposite to the direction of relative motion
Normal
Friction problems are essentially equilibrium problems with one f the
forces being functions of another
Friction
Relative velocity
Normal
N
Fr=f(N,V)N
Fr=f(N,V)
The correct way of writing the dry friction force
r
V ˆF N N VV
µ µ= − = −
N=Normal force vector
V=Relative velocity vector of the bodyµ= coefficient of dry friction or Coulomb friction
Problem 1
Knowing that the coefficient of friction between the 13.5 kg block and the incline is µs = 0.25, determine
a) the smallest value of P required to maintain the block in equilibrium,
b) the corresponding value of β.
Problem 1
x
y
( )
N cos60 mg f sin60 P sin 0N sin60 f cos60 P cos 0
f Nmg sin60 fP
sin sin60 cos cos60N cos60 mg N sin60 P sin 0
mg P sinNcos60 sin60
N sin60 N cos60 P cos 0mg P sin sin60 cos60 P cos 0
cos60 sin60mg P
ββ
µ
β βµ β
βµ
µ ββ µ β
µ
− + + =− + + =
=−
⇒ =+
− + + =−
⇒ =−
− + + =−
⇒ − − + =−
⇒ − −( )( ) ( )( ) ( ) ( )
( ) ( )( )
( ) ( )( )
sin sin60 cos60 P cos cos60 sin60 0
mg sin60 cos60 P sin sin60 cos60 P cos cos60 sin60 0
P sin60 sin cos60 sin cos60 cos sin60 cos mg sin60 cos60
sin60 cos60P mg
cos 60 sin 60
cos 60 sin 61 mgP
β µ β µ
µ β µ β µ
β µ β β µ β µ
µβ µ β
β µ
− + − =
⇒ − − + − + − =
⇒ − + − = −
−⇒ =
− − −
− −⇒ =
( )( )
( ) ( )
[ ]
o
0sin60 cos60
d 1 d0 cos 60 sin 60 0d P d
d sin60 sin cos60 sin cos60 cos sin60 cos 0dsin60 cos cos60 cos cos60 sin sin60 sin 0
sin60 cos60tan 2.614 69cos60 sin60
sinP mg
βµ
β µ ββ β
β µ β β µ ββ
β µ β β µ β
µβ βµ
−−
= ⇒ − − − =
⇒ − + − =
⇒ − − + =
−⇒ = = ⇒ = −
=( )
( ) ( )( )( ) ( )
60 cos60 0.866 0.125mg 0.72mg
cos 60 sin 60 cos 9 sin 9µ
β µ β µ− −
= =− − − − − −
Problem 2
Knowing that P = 110 N, determine the range of values value of θfor which equilibrium of the 8 kg block is maintained.
x
y
( )
k
P cos N 0P sin mg f 0f N
N P cosP sin mg N 0
P sin mg P cos 0P sin cos mg
mgPsin cos
Hence downward movement will not start before
mgPsin cos
θθ
µθ
θ µθ µ θθ µ θ
θ µ θ
θ µ θ
− + =− + =
=⇒ =
− + =⇒ − + =
⇒ + =
⇒ =+
=+
x
y
Problem 2
mg mg
( )
s
P cos N 0P sin mg f 0f N
N P cosP sin mg N 0
P sin mg P cos 0P sin cos mg
mgPsin cos
Hence upward movement will not start before
mgPsin cos
θθ
µθ
θ µθ µ θθ µ θ
θ µ θ
θ µ θ
− + =− − =
=⇒ =
− − =⇒ − − =
⇒ − =
⇒ =−
=−
Problem 3
The coefficients of friction are µs = 0.40 and µk = 0.30 between all the surfaces of contact. Determine the force P for which motion of the 27 kg block is impending if cable
a) is attached as shown,
b) is removed
Problem 3
( )
( )
1
1 1
1 1 1
1
1 2
2 1 2
2 1 2 2 1 2
1 2
1 1 1 2
1 2
T N 0N m g 0
T N ,N m gT m g
T N N P 0N N m g 0N N m g 0 N m m gT N N P 0
m g m g m m g P 0P 3 m g m g
µ
µµ
µ µ
µ µ
µ µ µ
µ µ
− =
− =
⇒ = =
⇒ =
+ + − =
− − =
− − = ⇒ = +
+ + − =
⇒ + + + − =
⇒ = +
T
m1gN1
f1
v
T
N1 f1
m2gN2
f2
Problem 3
( )
( )
1 1
1 1
1 1
1 1
1 2
2 1 2
2 1 2 2 1 2
1 2
1 1 2
1 2
T N m aNow T 0 N m aN m g 0
N m gN N P 0
N N m g 0N N m g 0 N m m g
N N P 0m g m m g P 0
P 2 m g m g
µµ
µ µ
µ µ
µ µ
µ µ
− =
= ⇒ − =
− =
⇒ =
+ − =
− − =
− − = ⇒ = +
+ − =
⇒ + + − =
⇒ = +
f1
0
m1gN1
f1
v
0
N1 f1
m2gN2
f2
Additional Problems
The 8 kg block A and the 16 kg block B are at rest on an incline as shown. Knowing that the coefficient of static friction is 0.25 between all surfaces of contact, determine the value of θ for which motion is impending.
Toppling
The magnitude of the force P is slowly increased. Does the homogeneous box of mass m slip or tip first? State the value of P which would cause each occurrence.
Slip or topple?
mg
N1
f1
N2
f2
( )( )
( )( )
1 2
1 2
2
1 2 1 2
1 2
2
1 2
P cos 30 N N 0P sin30 N N mg 0
P cos 30 d mg d N 2d 0P cos 30 N N N N
P sin30 N N mgP cos 30 mg 2 N
P sin30 N N mg P cos 30 mg
P sin30 P cos 30 mg P sin30 cos 30 mgmgP
sin30 cos 300.5mgP
µ µ
µ
µ µ µ
µ
µ µ µ µ
µ µ µ µµ
µ
− − =
+ + − =
− × − × + × =
= + = +
= − + +
+ =
= − + + = − +
+ = ⇒ + =
⇒ =+
⇒ = 0.448mg0.25 0.866
=+
Slip or topple?
mg
N1
f1
N2
f2
( )
2
2
2
2
P cos 30 f 0P sin30 N mg 0
P cos 30 d P sin30 2d mg d 0
P cos 30 fP sin30 N mgP cos 30 2P sin30 mg 0
P cos 30 2 sin30 mgmgP
cos 30 2 sin30mgP 0.536mg
0.866 1
− =
+ − =
− × − × + × =
=
= − +
+ − =
⇒ + =
⇒ =+
⇒ = =+
Additional Problems
Additional Problems
Additional Problems
Wedge
mg
f=µN
N
P
θ
( )
( )
P f cos N sin 0f sin N cos mg 0f N
mgN sin N cos mg 0 Nsin cos
P N cos N sin 0cos sinP cos sin N mgsin cos
P tanmg tan 1P tan tan tan
mg tan tan 1
θ θθ θ
µ
µ θ θµ θ θ
µ θ θµ θ θµ θ θµ θ θ
µ θµ θ
φ θ φ θφ θ
− + =+ − =
=
+ − = ⇒ =+
− + =−
⇒ = − =+
−⇒ =
+−
⇒ = = −+
A screw thread is a wedge
A screw thread is a wedge
M=Pa
W
Q
Nf=µN
Q=Pa/r
= equivalent force
A screw thread is a wedge
W
Q
Nf=µN
( )
Q f cos N sin 0f sin N cos W 0
f NWN sin N cos W 0 N
sin cosQ N cos N sin 0
cos sinQ cos sin N Wsin cos
Q tanW 1 tan
θ θθ θ
µ
µ θ θµ θ θ
µ θ θµ θ θµ θ θµ θ θ
µ θµ θ
− − =− + − =
=
− + − = ⇒ =− +
− − =+
⇒ = + =− +
+⇒ =
−
θ
Pa tanWr 1 tan
1 Pif tanW
µ θµ θ
θµ
+=
−
= ⇒ = ∞
Screw locks
A screw thread is a wedge
W
P
Nf=µNθ
1 PatanWr
1 Wr 0tan Pa
θµ
µθ
= ⇒ = ∞
= = =
Screw locks
The
There is a critical value of friction coefficient beyond which the thread
does not move irrespective of the force applied.
This happens when a screw is not maintained properly. Because of dirt
and rust µ becomes more than critical.
A screw thread is a wedge
W
Nf=µN
A screw thread is a wedge
W
Nf
For no movementf cos N sin 0f sin N cos W 0
f sinN cos
sin tancos
θ θθ θ
θθ
θµ θθ
− =+ − =
⇒ =
⇒ = =
θ
Self locking
Therefore after raising the load if we let go of the screw the load will not cause the screw to unscrew by itself.
Terminologies
Lead L npwheren=no. of parallely running threads = starts
Ltan =2 r
θπ
= =
Pitch (p)
2πr
Lea
d (L
)
Turnbuckle
T1 T2
( )2 1
M tanT T r 1 tan
µ θµ θ
+=
− −
Used to apply tension.
The sleeve is rotated to pull the threads together.
An improved screw jack
θ
θ
W
W
T T
T
2T cos θ
T
An improved screw jack
φ
φ
W
W=2T cos φ M=Pa
Pa cos sinWr sin cos
M cos sin2Tr cos sin cos
µ θ θµ θ θ
µ θ θφ µ θ θ
+=
− ++
⇒ =− +
Worm gear
R
MG
GG
G
G
MM WR W
RPa tanWr 1 tan
M tanM 1 tanrR
MR tanM r 1 tan
µ θµ θ
µ θµ θ
µ θµ θ
= ⇒ =
+=
−+
⇒ =−
+⇒ =
−
M0/R
NM f
Belt drives
( )
( )
( )
( )
x
y
s
s
F 0 T cos F T T cos 02 2
F 0 T sin N T T sin 02 2
F N
T cos F T T cos 02 2
T cos F T cos T cos 02 2 2
F T cos 0 T cos F N2 2
T sin N T T sin 02 2
T sin N T sin2 2
∆θ ∆θ∆ ∆
∆θ ∆θ∆ ∆
∆ µ ∆∆θ ∆θ∆ ∆
∆θ ∆θ ∆θ∆ ∆
∆θ ∆θ∆ ∆ ∆ ∆ µ ∆
∆θ ∆θ∆ ∆
∆θ ∆θ∆ ∆
= ⇒ − − + + =
= ⇒ − − − + =
=
− − + + =
⇒ − − + + =
⇒ − + = ⇒ = =
− − − + =
⇒ − + − −
∑
∑
T sin 02
2T sin N T sin 02 2
2T sin N 0 2T sin N2 2
∆θ
∆θ ∆θ∆ ∆
∆θ ∆θ∆ ∆
=
⇒ − + − =
⇒ − + = ⇒ =
Belt drives
s
s
s
s0 , T 0 0 , T 0
s
T cos N2
2T sin N2
T cos 2T sin2 2
sin1 T 2cos2T 2
sin1 T 2Lim cos Lim2T 2
1 dT 12T d 2
dT dsT
∆θ ∆ ∆θ ∆
∆θ∆ µ ∆
∆θ ∆
∆θ ∆θ∆ µ
∆θ∆ ∆θ µ∆θ ∆θ
∆θ∆ ∆θ µ∆θ ∆θ
µθ
µ θ
→ → → →
=
=
⇒ =
⇒ =
=
⇒ =
⇒ =
Belt drives
2
1
2
1
T
s 0T
2 1 s
2s
1
T
sT 0
dT dT
lnTlnT lnT
TlnT
dT dsT
T2 seT1
β
β
µ θ
µ β
µ β
µ θ
µ θ
µ β
−
=
=
=
=
∫ ∫⇒
⇒⇒
⇒
⇒
=
=
Belt drives
Belt is just about to slide to the right
22 1
1
2 1 1
T s se T e TTTorque required to drive the pulley
sT T e 1 T
µ β µ β
µ β
= ⇒ =
− = −
T2 seT1
µ β⇒ =
Belt drives : Important points
Angle β must be expressed in radians Smallest µβ determines which pulley slips
first Larger tension occurs at that end of the
belt where relative motion is about to begin or is already moving
T2 is used to denote the larger tension A freely rotating pulley implies no friction For a rotating pulley where slipping is about to start friction is µs since relative velocity between belt and pulley is zero.
Once slipping starts friction coefficient is dynamic or kinetic i.e. µk
If pulley does not rotate at all then rope has to slide and not slip, hence friction is µk
Belt drives
θ
θ
θ
θ
A B
A B
Belt drives
θ
θ
( )2
1
Texp 2 ???
Tµ π θ = +
T2
T1
T2
T1( )1
2
Texp 2 ???
Tµ π θ = −
A
B
Which expression is correct???? Is T2>T1
Or T1>T2.One pulley must slip.
Friction force is larger for the larger pulley since
angle of wrap is larger. Hence smaller pulley
slips and determines the tension
Always check for µβ value for each pulley in a system. The one
with the smallest µβ value will
determine the tensions.
Band brakes
Band brakes
T1 T3
T2
T4
( )
2 1 1
2 1
3 4 4
4 3
0 3 4 2 1
x 1 3 1 3
y 1 3 B 1
75 75T T exp T exp 0.25180 180
T 1.39T135 135T T exp T exp 0.25180 180
T 0.55T
M T T T T RConsider the pin B
F 0 T cos 45 T cos 45 T T
F 0 T cos 45 T cos 45 T 2T cos 45
µ π π
µ π π
= =
=
= =
=
= − − + −
= ⇒ = ⇒ =
= ⇒ + = ⇒
∑∑
( )
B 1 B
2 1 3 1 4 1
2 1 3 4 B
0
B B
T 2T T
T 1.39T ,T T ,T 0.55TThus the largest tension is T 5.6 T T 4.03,T 2.22,T 5.7
M 5.6 2.22 R 3.38 0.16 0.54KNm 540NmTaking moments about D50T 250P P 0.2T 1.14KN
= ⇒ =
∴ = = =
= ⇒ = = = =
∴ = − = × = =
= ⇒ = =
B
T1 T3
TB