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- 1. Engineering Mechanics: Statics 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. . 8 - 1 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. NF sm µ= • Further increase in P causes the block to begin to move as F drops to a smaller kinetic-friction force Fk. NF kk µ=
- 2. Engineering Mechanics: Statics The Laws of Dry Friction. Coefficients of Friction • Four situations can occur when a rigid body is in contact with a horizontal surface: . 8 - 2 • No friction, (Px = 0) • No motion, (Px < Fm) • Motion impending, (Px = Fm) • Motion, (Px > Fm)
- 3. Engineering Mechanics: Statics Angles of Friction • It is sometimes convenient to replace normal force N and friction force F by their resultant R: . 8 - 3 • No friction • Motion impending• No motion ss sm s N N N F µφ µ φ = == tan tan • Motion kk kk k N N N F µφ µ φ = == tan tan
- 4. Engineering Mechanics: Statics Angles of Friction • Consider block of weight W resting on board with variable inclination angle θ. . 8 - 4 • No friction • No motion • Motion impending • Motion
- 5. Engineering Mechanics: Statics Problems Involving Dry Friction . 8 - 5 • All applied forces known • Coefficient of static friction is known • Determine whether body will remain at rest or slide • All applied forces known • Motion is impending • Determine value of coefficient of static friction. • Coefficient of static friction is known • Motion is impending • Determine magnitude or direction of one of the applied forces
- 6. Engineering Mechanics: Statics Sample Problem 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. . 8 - 6 A 100 lb force acts as shown on a 300 lb block placed on an inclined plane. The coefficients of friction between the block and plane are µs = 0.25 and µk = 0.20. Determine whether the block is in equilibrium and find the value of the friction force. greater, block will not slide. • If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force.
- 7. Engineering Mechanics: Statics Sample Problem SOLUTION: • Determine values of friction force and normal reaction force from plane required to maintain equilibrium. :0=∑ xF ( ) 0lb300-lb100 5 3 =− F lb80−=F :0=∑F ( ) 0lb300- 4 =N . 8 - 7 :0=∑ yF ( ) 0lb300- 5 4 =N lb240=N • Calculate maximum friction force and compare with friction force required for equilibrium. If it is greater, block will not slide. ( ) lb48lb24025.0 === msm FNF µ The block will slide down the plane.
- 8. Engineering Mechanics: Statics Sample Problem • If maximum friction force is less than friction force required for equilibrium, block will slide. Calculate kinetic-friction force. ( )lb24020.0= == NFF kkactual µ lb48=actualF . 8 - 8

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