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