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.

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

f = µN tan Ø = f / N tan Ø = µN / N tan Ø = µ

N

Ø

R

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 W= 500kN down the incline (c) If P=400 kN, P what is the amount and direction of the 45○ friction force?

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 f k = 0.20 while at block B they are fs = 0.40 and f k = 0.30. Compute the 200kN magnitude 300kN and direction A of the friction B 3 θ force acting 4 θ on each block.

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. P α

300lb

A

. B 30○

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 4’’ force P parallel to the 8’’ incline maybe applied so that the block will P slide up the incline h without tipping over. 3 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.

A B θ

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 P A is the tension in the chord B attached to A? θ=30

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 P and the plane is 0.45. 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 P 28 kg block. 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 282 kN M start if rotating clockwise. 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. B

A

30○

60○

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. B

P

A

30○

60○

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

P 400 kN

8

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. 200 kN

400 kN

P 30○

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 x of friction at all contact surface is 15○ . 60○

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 300 kN required to start the M cylinder rotating counterclockwise.

3m

75○

15○

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. P A

B

45○

30○

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. W

P

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. P A The weight of A is 100 lb and the angle of friction 30○ for all surfaces in contact is 15°. B

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. 60○

A

P

10○

B

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

200-lb

30○

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.

200 kg

P 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 45○ B plane to block A if B moves upward. P (c) Compute the force A P required to start block A to move 20○ to the right.

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 F force exerted by the horizontal plane to C block B if C is moving upward. (b) Compute A the normal force exerted by the wall to B 10○ block A if C is moving 12○ upward. (c) Compute the force F required to start C moving upward.

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 T1 = 150 kN his leg and the rope. Neglect his pull on the rope. 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 50lb at the other end of the rope?

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 A 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 200 lb without rotating the pulley P or slipping the belt on the pulley P.

3 ft

2 ft P

B

W

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 8m

T1

C

P

T2 16m

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