1. Friction
Cause of dry friction
Contact 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
Relative velocity

2. Friction
Normal
Relative velocity
Friction problems are
essentially equilibrium
problems with one f the
forces being functions of
another
N
Fr=f(N,V)
Fr=f(N,V)
N

3. The correct way of writing the dry friction force
V
−µ N
Fr = =ˆ −µ N V
V
N=Normal force vector
V=Relative velocity vector of the body
µ= coefficient of dry friction or Coulomb
friction

4. 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.

5. Problem 1 y
N cos60 − mg + f sin60 + P sin β = 0
− N sin60 + f cos 60 + P cos β = 0
f = µN x
mg sin60 − f
⇒P=
sin β sin60 + cos β cos 60
N cos60 − mg + µ N sin60 + P sin β = 0
mg − P sin β
⇒N=
cos 60 − µ sin60
− N sin60 + µ N cos 60 + P cos β = 0
mg − P sin β
⇒− ( sin60 − µ cos 60 ) + P cos β =0
cos60 − µ sin60
⇒ − ( mg − P sin β )( sin60 − µ cos 60 ) + P cos β ( cos 60 − µ sin60 ) =
0
⇒ − mg ( sin60 − µ cos 60 ) + P sin β ( sin60 − µ cos 60 ) + P cos β ( cos 60 − µ sin60 ) =
0
⇒ P ( sin60 sin β − µ cos 60 sin β + cos 60 cos β − µ sin60 cos β ) = ( sin60 − µ cos 60 )
mg
( sin60 − µ cos 60 )
⇒P=
mg
cos ( 60 − β ) − µ sin ( 60 − β )
1 cos ( 60 − β ) − µ sin ( 60 − β )
⇒ =mg
P ( sin60 − µ cos 60 )
d 1 d
  =⇒0  cos ( 60 − β ) − µ sin ( 60 − β )  =
dβ P dβ   0
d
⇒ [ sin60 sin β − µ cos 60 sin β + cos 60 cos β − µ sin60 cos β ] =
0
dβ
⇒ sin60 cos β − µ cos 60 cos β − cos 60 sin β + µ sin60 sin β = 0
 sin60 − µ cos 60 
β
⇒ tan =  β
= 2.614 ⇒ = 69 o

 cos 60 − µ sin60 
( sin 60 − µ cos 60 ) ( 0.866 − 0.125 )
P = mg = mg = 0.72mg
cos ( 60 − β ) − µ sin ( 60 − β ) cos ( −9 ) − µ sin ( −9 )

6. Problem 2
Knowing that P = 110 N, determine the range of values value of θ
for which equilibrium of the 8 kg block is maintained.

7. Problem 2
y y
x x
mg mg
− P cos θ + N =0
− P cos θ + N =0
P sinθ − mg − f = 0
P sinθ − mg + f = 0
f = µN
f = µN
⇒N= θ P cos
⇒N= θ P cos
P sinθ − mg − µ N = 0
P sinθ − mg + µ N = 0
⇒ P sinθ − mg − µ P cos θ =
0
⇒ P sinθ − mg + µ P cos θ =
0
⇒ P ( sinθ − µ cos θ ) =
mg
⇒ P ( sinθ + µ cos θ ) =
mg
mg
mg ⇒P=
⇒P= sinθ − µ cos θ
sinθ + µ cos θ
Hence
Hence
upward movement will not start before
downward movement will not start before
mg
mg P=
P= sinθ − µ s cos θ
sinθ + µ k cos θ

8. 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

9. Problem 3
v
T N1
f1
m1g f1
N1 T
m2g f2
N2
T − µ N1 =
0
N 1 − m1 g =
0
= µ N= m1 g
⇒T 1
, N1
µm
⇒ T = 1g
T + µ N1 + µ N2 − P =
0
N 2 − N 1 − m2 g =
0
N 2 − N 1 − m2 g = ⇒ N 2 =m1 + m2 ) g
0 (
T + µ N1 + µ N2 − P =
0
⇒ µ m1 g + µ m1 g + µ ( m1 + m 2 ) g − P =
0
⇒ P 3 µ m1 g + µ m 2 g
=

10. Problem 3
v
0 N1
f1
m1g f1
N1 0
m2g f2
N2
T − µ N1 =
m1 a
Now T = 0 ⇒ − µ N 1 = m1 a
N 1 − m1 g =
0
⇒ N1 =
m1 g
µ N1 + µ N2 − P =
0
N 2 − N 1 − m2 g =
0
N 2 − N 1 − m2 g = ⇒ N 2 =m1 + m2 ) g
0 (
µ N1 + µ N2 − P =
0
⇒ µ m1 g + µ ( m1 + m 2 ) g − P =
0
⇒ P 2 µ m1 g + µ m 2 g
=

11. 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.

12. 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.

13. Slip or topple?
mg
f1 f2
N1 N2
P cos 30 − µ N 1 − µ N 2 =
0
P sin 30 + N 1 + N 2 − mg =
0
− P cos 30 × d − mg × d + µ N 2 × 2d =
0
P cos 30 = µ N 1 + µ N 2 = µ ( N 1 + N 2 )
P sin 30 = 1 + N 2 ) + mg
−( N
2µ N
P cos 30 + mg = 2
µ P sin 30 = N 1 + N 2 ) + µ mg = 30 + µ mg
−µ ( − P cos
µ P sin 30 + P cos 30 =mg ⇒ P ( µ sin 30 + cos 30 ) =mg
µ µ
µ mg
⇒P=
µ sin 30 + cos 30
0.5mg
⇒P= = 0.448mg
0.25 + 0.866

14. Slip or topple?
mg
f1 f2
N1 N2
P cos 30 − f 2 =
0
P sin 30 + N 2 − mg =
0
− P cos 30 × d − P sin 30 × 2d + mg × d =
0
P cos 30 = f 2
P sin 30 = 2 + mg
−N
P cos 30 + 2P sin 30 − mg =
0
⇒ P ( cos 30 + 2 sin 30 ) =
mg
mg
⇒P=
cos 30 + 2 sin 30
mg
=
⇒P = 0.536mg
0.866 + 1

15. Additional Problems

16. Additional Problems

17. Additional Problems

18. Wedge
f=µN
P
θ
N mg
P − f cos θ + N sinθ =0
f sinθ + N cos θ − mg = 0
f = µN
mg
µ N sinθ + N cos θ − mg = ⇒ N =
0
µ sinθ + cos θ
P − µ N cos θ + N sinθ = 0
µ cos θ − sinθ
⇒ P ( µ cos θ − sinθ= mg
= )N
µ sinθ + cos θ
P µ − tanθ
⇒ =
mg µ tanθ + 1
P tanφ − tanθ
⇒ = = tan (φ − θ )
mg tanφ tanθ + 1

19. A screw thread is a wedge

20. A screw thread is a wedge
M=Pa
Q=Pa/r
= equivalent force
W
Q
N
f=µN

21. A screw thread is a wedge
W
Q
N
θ f=µN
Q − f cos θ − N sinθ = 0
− f sinθ + N cos θ − W = 0
f = µN
W
− µ N sinθ + N cos θ − W = ⇒ N =
0
− µ sinθ + cos θ
Q − µ N cos θ − N sinθ = 0
µ cos θ + sinθ
⇒ Q ( µ cos θ + sinθ= W
= )N
− µ sinθ + cos θ
Q µ + tanθ
⇒ =
W 1 − µ tanθ Pa µ + tanθ
=
Wr 1 − µ tanθ
1 P
if tanθ =⇒ =
∞
µ W
Screw locks

22. A screw thread is a wedge
The
W
P
N
θ f=µN
1 Pa
tanθ = ⇒ =
∞ Screw locks
µ Wr
1 Wr
µ
= = = 0
tanθ Pa
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.

23. A screw thread is a wedge
W
N
f=µN

24. A screw thread is a wedge
W
N
θ f
For no movement
f cos θ − N sinθ = 0
f sinθ + N cos θ − W =
0
Self locking
f sinθ
⇒ =
N cos θ
sinθ
⇒= µ = tanθ
cos θ
Therefore after raising the load if we
let go of the screw the load will not
cause the screw to unscrew by itself.

25. Terminologies
Lead (L)
2πr
Pitch (p)
= =
Lead L np
where
n=no. of parallely running threads = starts
L
tanθ =
2π r

26. Turnbuckle
T1 T2
Used to apply tension.
The sleeve is rotated to pull the
threads together.
M µ + tanθ
=
( T2 − T1 ) r 1 − µ tanθ

27. An improved screw jack
W
θ
θ
W
T
T T
T
2T cos θ

28. An improved screw jack
W
φ
φ
M=Pa
W=2T cos φ
Pa µ cos θ + sinθ
=
Wr − µ sinθ + cos θ
M µ cos θ + sinθ
⇒ =
2Tr cos φ − µ sinθ + cos θ

29. Worm gear
MG
M0/R
R N
M f
MG
M G = WR ⇒ W =
R
Pa µ + tanθ
=
Wr 1 − µ tanθ
M µ + tanθ
⇒ =
MG 1 − µ tanθ
r
R
MR µ + tanθ
⇒ =
M G r 1 − µ tanθ

30. Belt drives

31. Belt drives
∆θ ∆θ
∑F x
= 0 ⇒ −T cos − ∆ F + ( T + ∆T ) cos = 0
2 2
∆θ ∆θ
∑F y
= 0 ⇒ −T sin − ∆ N − ( T + ∆T ) sin = 0
2 2
∆ F = µs ∆ N
∆θ ∆θ
−T cos − ∆ F + ( T + ∆T ) cos =
0
2 2
∆θ ∆θ ∆θ
⇒ −T cos − ∆ F + T cos + ∆T cos =0
2 2 2
∆θ ∆θ
⇒ − ∆ F + ∆T cos = 0 ⇒ ∆T cos = ∆ F = µs ∆ N
2 2
∆θ ∆θ
−T sin − ∆ N − ( T + ∆T ) sin =
0
2 2
∆θ ∆θ ∆θ
⇒ −T sin + ∆ N − T sin − ∆T sin =0
2 2 2
∆θ ∆θ
⇒ −2T sin + ∆ N − ∆T sin =0
2 2
∆θ ∆θ
⇒ −2T sin + ∆ N = 0 ⇒ 2T sin = ∆N
2 2

32. Belt drives
∆θ
∆T cos = µs ∆ N
2
∆θ
2T sin = ∆N
2
∆θ ∆θ
⇒ ∆T cos 2T µ s sin
=
2 2
∆θ
sin
1 ∆T ∆θ 2
⇒ cos µs
=
2T ∆θ 2 ∆θ
∆θ
sin
1 ∆T ∆θ 2
Lim cos = Lim µ
∆θ → 0 ,∆T → 0 2T ∆θ 2 ∆θ →0 ,∆T →0 s ∆θ
1 dT µs
⇒ 1=
2T dθ 2
⇒
dT = µ dθ
T s

33. Belt drives
Belt is just about to
slide to the right
dT = µ dθ
T s
dT β
T2
µ β µ β
⇒ ∫ = ∫ µ s dθ T2
= e s ⇒T e s T
=
T T
1 0 2 1
T1
⇒  lnT  β
T2
  =  µ sθ 0
 
T1 Torque required to drive the pulley
⇒ lnT − lnT
2 1
= µs β  µ β 
T2 − T1  e
= s − 1 T
⇒ ln T = µ β
2 

 1

T 1
s
2 = e µsβ
T
⇒
T
1

34. Belt drives : Important points
2 = e µsβ
T
⇒
T
1
 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

35. Belt drives
A B
θ
θ
A B
θ
θ

36. Belt drives
T2
T2
= exp  µ ( π + 2θ )  ???
 
T1
A Always check
for µβ value for
each pulley in a
system. The one
with the
T1
smallest µβ
θ
value will
determine the
Which expression is tensions.
correct???? Is T2>T1
Or T1>T2.
One pulley must slip. T2
Friction force is larger
for the larger pulley since
angle of wrap is larger.
Hence smaller pulley
B
slips and determines the
tension
T1 θ
exp  µ ( π − 2θ )  ???
  T1
T2

37. Band brakes

38. Band brakes
TB T1 T3
B
T1 T3
 75   75 4 
Tπ
T2 T1 exp  µ π
= T1 exp  0.25 
 180   180  T2
T2 = 1.39T1
 135   135 
T3 T4 exp  µ π
= T4 exp  0.25 π
 180   180 
T4 = 0.55T3
M 0 =( T3 − T4 + T2 − T1 ) R
−
Consider the pin B
∑F x
=0 ⇒ T1 cos 45 =T3 cos 45 ⇒ T1 =T3
∑F y
= cos 45 + T3 cos 45 = 1 cos 45 = ⇒ 2T1 =
0 ⇒ T1 TB ⇒ 2T TB TB
∴ T2 1.39T1 ,T3 T1 ,T4 0.55T1
= = =
Thus the largest tension is T2 = 5.6 ⇒ T1 = T3 = 4.03,T4 = 2.22,TB = 5.7
∴ M0 = ( 5.6 − 2.22 ) R = 3.38 × 0.16 = 0.54 KNm = 540 Nm
Taking moments about D
50TB 250 P ⇒ = 0.2TB 1.14 KN
= P =