Chapter 8: Friction, Exercises of Engineering

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Chapter 8: Friction

School of Mechanical Engineering

hjchoi@cau.ac.kr

Contents

Introduction Laws of Dry Friction. Coefficients of Friction. Angles of Friction Problems Involving Dry Friction Sample Problem 8. Sample Problem 8. Wedges Square-Threaded Screws Sample Problem 8. Journal Bearings. Axle Friction. Thrust Bearings. Disk Friction. Wheel Friction. Rolling Resistance. Sample Problem 8. Belt Friction. Sample Problem 8. School of Mechanical Introduction Laws of Dry Friction. Coefficients of Friction. Angles of Friction Problems Involving Dry Friction Sample Problem 8. Sample Problem 8. Wedges Square-Threaded Screws Sample Problem 8. Journal Bearings. Axle Friction. Thrust Bearings. Disk Friction. Wheel Friction. Rolling Resistance. Sample Problem 8. Belt Friction. Sample Problem 8.

The Laws of Dry Friction. Coefficients of

Friction

• Block of weight W placed on horizontal surface. Forces acting on block are its weight and reaction of surface N.
• Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force. School of Mechanical
• Small horizontal force P applied to block. For block to remain stationary, in equilibrium, a horizontal component F of the surface reaction is required. F is a static-friction force.
• As P increases, the static-friction force F increases as well until it reaches a maximum value Fm.

Fm = m s N

• Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk. Fk = m k N

The Laws of Dry Friction. Coefficients of

Friction

• Maximum static-friction force:

Fm = m s N

• Kinetic-friction force: k s Fk kN m m m @ 0. 75

School of Mechanical

• Maximum static-friction force and kinetic- friction force are: - proportional to normal force - dependent on type and condition of contact surfaces - independent of contact area

Angles of Friction

• It is sometimes convenient to replace normal force N and friction force F by their resultant R : School of Mechanical
• No friction • No motion • Motion impending s s m s s N

N

N

F

f m

m

f

tan tan

• Motion k k k k k N

N

N

F

f m

m

f

tan tan

Angles of Friction

• Consider block of weight W resting on board with

variable inclination angle q.

School of Mechanical

• No friction • No motion • Motion impending - Motion

Sample Problem 8.

SOLUTION:

• Determine values of friction force and normal reaction force from plane required to maintain equilibrium.
• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide. School of Mechanical A 100 N force acts as shown on a 300 N block placed on an inclined plane. The coefficients of friction between the block

and plane are m s = 0.25 and m k = 0.20.

Determine whether the block is in equilibrium and find the value of the friction force.

• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide.
• If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.

Sample Problem 8.

SOLUTION:

• Determine values of friction force and normal reaction force from plane required to maintain equilibrium. å Fx^ =^0 :^100 N- (^300 N)^0 5 3 - F = F =- 80 N å F^ y =^0 : -^ (^300 N)^0 5 4 N = School of Mechanical å F^ y =^0 : -^ (^300 N)^0 5 4 N = N = 240 N
• Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide. Fm = m s N Fm = 0.25 240 N ( )=60 N The block will slide down the plane.

Sample Problem 8.

SOLUTION:

• When W is placed at minimum x , the bracket is about to slip and friction forces in upper and lower collars are at maximum value.
• Apply conditions for static equilibrium to find minimum x. School of Mechanical The moveable bracket shown may be placed at any height on the 3-cm diameter pipe. If the coefficient of friction between the pipe and bracket is 0.25, determine the minimum distance x at which the load can be supported. Neglect the weight of the bracket.
• Apply conditions for static equilibrium to find minimum x.

Sample Problem 8.

SOLUTION:

• When W is placed at minimum x , the bracket is about to slip and friction forces in upper and lower collars are at maximum value. B s B B A s A A F N N

F N N

m

m

• Apply conditions for static equilibrium to find minimum x. School of Mechanical
• Apply conditions for static equilibrium to find minimum x. å Fx^ =^0 : N^ B -^ NA =^0 N^ B = NA å Fy =^0 : N W

N N W

F F W

A A B A B =

N A = NB = 2 W

å M^ B =^0 : (^ )^ (^ )^ (^ ) ( ) ( ) 6 ( 2 ) 0. 75 ( 2 ) ( 1. 5 ) 0

6 cm 3 cm 1. 5 cm 0

• • • =

W W W x N N W x N F W x A A A A x = 12 cm

Square-Threaded Screws

• Square-threaded screws frequently used in jacks, presses, etc. Analysis similar to block on inclined plane. Recall friction force does not depend on area of contact.
• Thread of base has been “unwrapped” and shown as straight line. Slope is 2p r horizontally and lead L vertically.
• Moment of force Q is equal to moment of force P. (^) Q = Pa r School of Mechanical
• Impending motion upwards. Solve for Q.
• Self-locking, solve for Q to lower load. f (^) s > q , • Non-locking, solve for Q to hold load. f (^) s > q ,

Sample Problem 8.

A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of mean diameter equal to 10 mm with a pitch of 2 mm. The coefficient of friction

between threads is m s = 0.30.

If a maximum torque of 40 N*m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, and (b) the torque required to loosen the clamp.

SOLUTION

• Calculate lead angle and pitch angle.
• Using block and plane analogy with impending motion up the plane, calculate the clamping force with a force triangle.
• With impending motion down the plane, calculate the force and torque required to loosen the clamp. School of Mechanical A clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of mean diameter equal to 10 mm with a pitch of 2 mm. The coefficient of friction

between threads is m s = 0.30.

If a maximum torque of 40 N*m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, and (b) the torque required to loosen the clamp.

• With impending motion down the plane, calculate the force and torque required to loosen the clamp.

Sample Problem 8.

• With impending motion down the plane, calculate the force and torque required to loosen the clamp.

tan ( - ) = Q =( 17. 97 kN) tan 9. 4 °

W

Q

f s q Q = 2. 975 kN

( 2. 975 10 N)( 5 10 m)

2. 975 kN 5 mm 3 - 3 = ´ ´ Torque = Qr = School of Mechanical

( 2. 975 10 N)( 5 10 m)

2. 975 kN 5 mm 3 - 3 = ´ ´ Torque = Qr = Torque = 14. 87 N× m

Journal Bearings. Axle Friction

• Journal bearings provide lateral support to rotating shafts. Thrust bearings provide axial support
• Frictional resistance of fully lubricated bearings depends on clearances, speed and lubricant viscosity. Partially lubricated axles and bearings can be assumed to be in direct contact along a straight line.
• Forces acting on bearing are weight W of wheels and shaft, couple M to maintain motion, and reaction R of the bearing. School of Mechanical
• Forces acting on bearing are weight W of wheels and shaft, couple M to maintain motion, and reaction R of the bearing.
• Reaction is vertical and equal in magnitude to W.
• Reaction line of action does not pass through shaft center O ; R is located to the right of O , resulting in a moment that is balanced by M.
• Physically, contact point is displaced as axle “climbs” in bearing.