University Tun Hussein Onn Malaysia (UTHM), Mechanics of Machines BDA2033 Lecture #06 By, Rosli Asmawi University Tun Hussein Onn Malaysia (UTHM), Faculty of Mechanical and Manufacturing, Department of Mechanics

CHAPTERS : FRICTION (PART I) Today’s Objectives: Students will be able to: 1. understand the characteristics of dry friction 2. solve problems involving dry friction Learning topics: • Introduction • Application dry friction angles of friction

INTRODUCTION Friction is defined as a force of resistance acting on a body which prevents or retards slipping of the body relative to a second body.

APPLICATION In designing a brake system for a bicycle, car, or any other vehicle, it is important to understand the frictional forces involved. For an applied force on the brake pads, how can we determine the magnitude and direction of the resulting friction force?

THEORY OF DRY 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. As P increases, the static-friction force F increases as well until it reaches a maximum value Fm. Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk.

THEORY OF DRY FRICTION (Cont.’d) Maximum static-friction force: Kinetic-friction force: 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

THEORY OF DRY FRICTION (Cont.’d) Four situations can occur when a rigid body is in contact with a horizontal surface: No friction, (Px = 0) No motion, (Px < Fm) Motion impending, (Px = Fm) Motion, (Px > Fm)

ANGLE OF FRICTION Consider block of weight W resting on board with variable inclination angle q. The value of θ when slip is impending is called the angle of static friction θs , and its value when the surfaces are sliding relative to each other is called the angle of kinetic friction θk. we can express the angles of static and kinetic friction in terms of the coefficients of friction. tan θs = μs , tan θk = μk In some situations it is more convenient to express the reaction in terms of its magnitude R and angle of friction θ between the reaction and the normal to surface. Motion impending No motion Motion No friction

PROBLEM INVOLVING DRY FRICTION Coefficient of static friction is known Motion is impending Determine magnitude or direction of one of the applied forces All applied forces known Motion is impending Determine value of coefficient of static friction. All applied forces known Coefficient of static friction is known Determine whether body will remain at rest or slide

EXAMPLE 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. If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force. A 450 N force acts as shown on a 1350 N block placed on an inclined plane. The coefficients of friction between the block and plane are ms = 0.25 and mk = 0.20. Determine whether the block is in equilibrium and find the value of the friction force.

SOLUTION 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. The block will slide down the plane.

SOLUTION (Cont.’d) If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.

IN CLASS TUTORIAL The arrangement in figure above exerts a horizontal force on the stationary 180 N crate. The coefficient of static friction between the crate and the ramp is μs = 0.4. If the rope exerts a 90N force on the crate, what is the friction force exerted on the crate by the ramp? [ans : f = -23.0 N] What is the largest force the rope can exert on the crate without causing it to slide up the ramp? [ans : T = 161 N]

IN CLASS TUTORIAL

End of Lecture