Presentation on theme: "1 ME 302 DYNAMICS OF MACHINERY FRICTION FORCE Dr. Sadettin KAPUCU © 2007 Sadettin Kapucu."— Presentation transcript:

Gaziantep University 2Friction When two bodies are in contact with each other, with or without relative motion between, then fiction forces come into account. This forces result in a loss of energy which is dissipated in the form of heat. Figures are taken from “Dynamics, Meriam and Kraige”

Gaziantep University 3 Friction types Dry friction : Dry friction is encountered when the unlubricated surface of two solids are in contact under a condition of sliding or tendency to slide. This type of friction is also called Coulomb’s friction. This friction force will always be in a direction to appose motion or the tendency toward motion of the body on which it acts. m P m P mg N FfFf R  FfFf P Static friction Kinetic friction

Gaziantep University 4 Friction types Fluid Friction: Fluid friction is developed when adjacent layers in a fluid are moving at different velocities. It may be either viscous or turbulent. Bearing surface Shaft lubrication V V=0 Bearing Shaft surface Lubrication layers Damping coefficient

Gaziantep University 5Example The Withworth quick return mechanism is acted upon by a 10 Nm external torque acting on crank BC in CW direction. Calculate the amount of torque required on crank AB at the instant when crank AB starts rotating CCW, If the coefficient of friction between the mating surfaces of the prismatic joints is 0,3. AB = AC = 10 cm.

Gaziantep University 6Example By definition friction force always opposes the motion, in order to put the friction force in proper direction we need velocity analysis. The velocity of the crank pin B as a point on AB link is easily found, so that B will be used as a reference point for determining the relative velocity of link 3 and link 4. The relative velocity equation may now be written; is normal to AB link and direction, dictated by problem definition. is normal to BC link. is along BC link.

Gaziantep University 7Example 3 4 x y

8 Example 2 An external torque of 100 N-m is acting on link 2 of the mechanism shown in CCW direction. Calculate the magnitude and direction of the external force required on link 4 at the instant when the link just starts moving rightwards if the coefficient of friction at the sliding joint at B is 0.3. AB = AC = 10 cm.

Gaziantep University 9 Example 2 AB = AC = 10 cm. is normal to AB link. is along BC link. direction dictated by problem definition is along AB link.

Gaziantep University 10 Example 2 3 2 x y +