Friction is the contact resistance exerted by one body when the second body moves or tends to move past the first body. Friction is a retarding force that always acts opposite to the motion or to the tendency to move.

Types of Friction

1. Dry Friction

2. Fluid Friction

3. Skin Friction

4. Internal Friction

Dry friction, also called Coulomb

friction, occurs when unlubricated

surfaces of two solids are in contact and

slide or tend to slide from each other. If

lubricant separates these two surfaces, the

friction created is called lubricated

friction. This section will deal only with

dry friction.

P W

N

f

R Ø

Elements of Dry Friction N = Total reaction perpendicular to the contact surface f = Friction force µ = Coefficient of friction R = Resultant of f and N Ø = angle of friction f = µN tan Ø = f / N tan Ø = µN / N tan Ø = µ

A block weighing W lb is placed upon a plane inclined at an angle of θ with the horizontal. What will happen if the angle of friction Ø is (a) greater than θ, (b) equal to θ, (c) less than θ

a) If Ø is greater than θ the block will not slide down instead it will retain its position because the frictional force is so much that it will hold the block.

b) If Ø is equal to θ the block will still not slide down because having θ equal to Ø the system will still be in equilibrium.

c) If Ø is less than θ, then slipping occurs because the frictional force is not enough to hold the block.

Example 1: The 500kN block shown is in contact with a 45○

incline. For which the coefficient of friction is 0.25. Compute the value of the horizontal force P necessary to:

(a) Just start the block up the incline

(b) Just prevent motion

down the incline

(c) If P=400 kN,

what is the amount

and direction of the

friction force? 45○

P

W= 500kN

Example 2: The blocks shown are connected by flexible,

inextensible cords passing over frictionless pulleys. At block A the coefficient of friction are fs = 0.30 and fk = 0.20 while at block B they are

fs = 0.40 and fk = 0.30.

Compute the

magnitude

and direction

of the friction

force acting

on each block.

B A

3

4 θ

θ

200kN 300kN

Example 3: Find the least value of P required to cause the

system of blocks shown in the figure to have impending motion to the left. The coefficient of friction is 0.2 under each block.

.

30○

300lb P α

A

B

Example 4: A homogeneous block of weight W rest upon the

inclined plane shown in the figure. If the coefficient of friction is 0.30, determine the greatest height h at which a

force P parallel to the

incline maybe applied

so that the block will

slide up the incline

without tipping over. 3

4

h

P

8’’

4’’

Example 5: Block A in the figure below weighs 120 N, block

B weighs 200 N and the cord is parallel to the incline. If the coefficient of friction for all surfaces in contact is 0.25, determine the angle θ of the incline at which motion of B impends.

θ

A

B

Example 6: Block A in the figure below weighs 120 N, block

B weighs 200 N and the cord is parallel to the incline. If the coefficient of friction for all surfaces in contact is 0.60 and θ= 30, what force P applied to B acting down and parallel to the incline will start

motion? What

is the tension in

the chord

attached to A? θ=30

A

B

P

Example 7: The three flat blocks are positioned on a 25○

inclined as shown and a force parallel to the inclined is applied to the middle block. The upper block is prevented from moving by a wire which attaches it to the fixed support. The coefficient of static friction

between 28 kg and 45 kg is

0.30, between 45kg and

36 kg is 0.40 and

between the 36 kg

and the plane is 0.45. P

25○

Example 7: (a) Which of the following gives the value of P

before any slipping takes place.

(b) Which of the following gives the max. value of the friction force between the 36 kg

and 45 kg block.

(c) Which of the following

gives the tension of the

cable supporting the

28 kg block. P

25○

Example 8: A homogenous cylinder weighing 282 kN rests

on a horizontal surface in contact with a block which weighs 150 kN. If µ= 0.40 of all contact surfaces, find the couple M acting on the cylinder which will

start if rotating

clockwise.

282 kN M

1.5m

Example 9: In the figure shown below, two blocks are

connected by a solid strut attached to each block with frictionless pins. If the coefficient of friction under each block is 0.25 & B weighs 270 kN. Find the minimum weight of A to prevent motion.

60○ 30○ A

B

Example 10: In the figure shown below, block A weighs 400

kN and block B is 200 kN. The coefficient of friction µ=0.30 for both A and B. The blocks are connected by a solid strut that is a uniform rod weighing 300 kN. Find the value of P to prevent motion.

60○ 30○ A

B

P

Example 11: A force of 400 kN is applied to the pulley shown

in figure below. The pulley is prevented from rotating by a force P applied to the end of the lever. If the coefficient of friction at the brake surface is 0.20, determine the value of P.

10

20

8

400 kN

P

16 32

Example 12: The two crates shown are pin-connected by the

horizontal strut AB. Find the minimum value of P parallel to the incline which will maintain equilibrium.

30○

P

400 kN 200 kN

Example 13: A ladder 20 ft long weighs 40 lb & its center of

gravity is 8ft long from the bottom. The ladder is placed against a vertical wall so that it makes an angle of 60○ with the ground. How far up the ladder can a 160 lb

man climb before the

ladder is on the verge

of slipping. The angle

of friction at all

contact surface is 15○ . 60○

x

Example 14: A homogeneous cylinder 3m in diameter and

weighing 300 kN is resting on two inclined planes as shown in the figure. If the angle of friction is 15○ for all contact surfaces, compute the magnitude of the couple

required to start the

cylinder rotating

counterclockwise.

75○ 15○

3m

300 kN

M

Example 15: A plank 10 ft long is placed in a horizontal

position with its ends resting on two inclined plane as shown in the figure below. The angle of friction is 20○. Determine how close the load P can be placed to each end before slipping impends.

30○ 45○

P

A B

Example 16: The 10m ladder weighing 35 kg is resting on a

horizontal floor at A and on the wall at B making an angle of 60○ from the horizontal. The coefficient of friction between the ladder and the floor and between the ladder and the wall is 0.25

a) Determine the distance “x” to which the 72 kg man can climb on the ladder without causing the ladder to slip at its lower end A.

b) Determine the reaction at the wall at B.

c) Determine the reaction at the horizontal floor at A.

A WEDGE is a simple device

that is used for same purpose

as a lever that is to create

a mechanical advantage.

P

W

A wedge allows a small force P to lift a large weight W.

Wedges are also used

o adjust the elevation

or provide stability for

heavy objects such as

this large steel pipe.

Example 17: In the figure shown, determine the minimum

weight of block B that will keep it at rest while a force P starts blocks A up

the incline surface of B.

The weight of A is 100 lb

and the angle of friction

for all surfaces in contact

is 15°.

P A

B

30○

Example 18: In the figure shown below , determine the value

of P just sufficient to start the 10○ wedge under the 400-lb block. The angle of friction is 20○ for all contact surfaces.

10○

60○

P

B

A

Example 19: As shown in the figure, two blocks weighing

200-lb & resting on a horizontal surface are to be pushed apart by a 30○ wedge. The angle of friction is 15○ for all contact surfaces. What value of P is required to start movement of the blocks?

P

30○ 200-lb 200-lb

Example 20: Determine the force P required to start the

wedge shown in figure below. The angle of friction for all surfaces in contact is 15○.

P

2000

5000

75○

Example 21: From the given figure, the mass of the block is 200

kg. Coefficient of friction for all surfaces is 0.40.

(a) Determine the reaction between the wall and the block if the block is raised due to the force P.

(b) Determine the reaction between the wedge and the block if the block is raised due to the force P.

(c) Determine the horizontal force P to raise the 200 kg block.

P

200 kg

10○

Example 22: The masses of A and B are 42 kg and 50 kg

respectively. Between all contact surfaces µ=0.05.

(a) Compute the normal force exerted by block B to block A if B moves upward.

(b) Compute the normal force exerted by the inclined plane to block A if B moves upward. (c) Compute the force P required to start block A to move to the right.

P

20○

A

B 45○

Example 23: The masses of A, B and C are 8 kg and 12 kg and 80

kg respectively. Between all contact surfaces µ=0.40.

(a) Compute the normal

force exerted by the

horizontal plane to

block B if C is moving

upward. (b) Compute

the normal force

exerted by the wall to

block A if C is moving

upward. (c) Compute

the force F required to

start C moving upward.

C

A

B

F

10○

12○

Belt Friction The transmission of power by means of belt or rope

drives that depends upon the frictional resistance developed between the belt and the resisting surface with which it is in contact.

If the pulley is smooth, no driving torque is developed because no frictional resistance exists, and the tension all throughout the belt will be constant and will have the same value on both sides of the pulley.

If the surface of the pulley is rough, the tension in the belt will vary throughout the length of contact, the difference in the belt tensions being caused by the frictional resistance.

Belt Friction

T1 = tension in the tight side

T2 = tension in the slack side

T1 > T2

B = angle of contact expressed in radians

µ = Coefficient of friction

Example 24: An acrobat weighing 150 kN supports himself by

a wrapping a rope around one leg as shown. From his leg there hangs 5m of rope which weighs 0.40 kN/m. What is the minimum coefficient of friction between

his leg and the rope. Neglect

his pull on the rope.

T1 = 150 kN

T2

Example 25: A rope making 1 and ¼ turns around a stationary

horizontal drum is used to support a heavy weight. If the coefficient of friction is 0.4. What weight can be supported

by exerting a 50-lb force

at the other end of the

rope?

50lb

Example 26: The 180-lb farmer tries to restrain the cow from

escaping by wrapping the rope two turns around the trunk as shown. If the cow exerts a force of 250 lb on the rope, determine if the farmer can successfully restrain the

cow. The coefficient of

static friction between

the rope and the trunk is

µ= 0.15, and between

the farmer’s shoes and

the ground is µ=0.30

Example 27: In the figure shown below, a flexible belt

runs from A over the

compound pulley B and

back over P to a 200-lb

weight. The coefficient of

friction is 1/∏ between the

belt and the compound

pulley P. Find the maximum

weight W that can be supported

without rotating the pulley P or

slipping the belt on the pulley P.

3 ft

2 ft

200 lb

W

A

B

P

Example 28: A torque of 240 kN.m acts on the brake drum

shown in figure below. If the brake bond is in contact with the brake drum through 250○ and the coefficient of friction is 0.30, determine the force P at the end of the brake lever.

M

T1 T2 P

16m

8m

C

Example 29: In the figure shown below, the coefficient of

friction is 0.20 between the rope and fixed drum and between all surfaces in contact. Determine the minimum weight W to prevent down plane motion of the 1000 lb body.

W

3

4