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- 1. INTRODUCTION TO TECHNOLOGY B FRICTION AND WEDGES
- 2. FRICTION A block of weight W is placed on a horizontal plane surface. The force acting on the block are its Weight and the reaction force of the table. Since the weight force has no horizontal component, the reaction of the surface also has no horizontal component. The reaction is therefore, normal to the surface and is represented by the normal force N
- 3. FRICTION Suppose now that a horizontal force P is applied to the block. If P is small the block will not move, some other horizontal force therefore must exist which balances P . this other force is known as the STATIC FRICTION FORCE .
- 4. FRICTION If the force P is increased, the friction force F also increases, continuing to oppose P , until its magnitude reaches a certain maximum value Fm . If P is increased the friction force cannot balance it any more and the block starts to move. As soon as the block starts to move the value of F drops to a lower value of Fk and is known as the kinetic force of friction
- 5. FRICTION Fm of the static friction force is proportional to the normal component of N of the reaction of the surface and so we have: Fmax = μ s N Fkinetic = μ k N Where μ s and μ k are called the coefficient of static and kinetic friction respectively.
- 6. ANGLES OF FRICTION Tan Φ s = μ s Tan Φ k = μ k If the applied force P has a horizontal component Px which tens to move the block, the force R will have a horizontal component F and thus will form an angle Φ with the normal to the surface. This value is called the angle of static friction and is denoted Φ s . Then from the geometry Tan Φ s = Fm / N
- 7. FRICTIONAL FORCES ON A INCLNED PLANE
- 8. SAMPLE PROBLEM <ul><li>A sand sled of mass 10kg is used on a sand dune having a lope of 15 °. If the coefficient of friction is 0.35, what force must the boy exert on the tow rope to drag the sand sled up the dune? </li></ul>
- 9. SAMPLE PROBLEM FBD P – W sin 15 ° - Fr = 0 Since Fr = μ N = 0.35 mg cos 15° = 0.35 (10)(9.8) cos 15° = 33.1N P – 9.8 sin15° - 33.1 = 0 P = 58.5N
- 10. Sample Problem A support block is acted upon the by two forces as shown. Knowing that coefficients of friction between the block and the incline plane are μ s = 0.35 and μ k = 0.25, determine the force P required to (a) to start the block moving up the inclined plane (b) to keep it moving up (c) to prevent it from sliding down. Solution For each part we draw the FBD of the block and a force triangle including the 800N vertical force, the horizontal force P and the force R exerted on the block by the inclined plane. The direction of R must be determined in each separate case.
- 11. Sample Problem A support block is acted upon the by two forces as shown. Knowing that coefficients of friction between the block and the incline plane are μ s = 0.35 and μ k = 0.25, determine the force P required to (a) to start the block moving up the inclined plane FBD Tan Φ s = μ s = 0.35 Φ s = 19.29° 25° + 19.29° = 44.29° Force P to start block moving up P = (800N) tan 44.29 ° = 780N
- 12. Sample Problem A support block is acted upon the by two forces as shown. Knowing that coefficients of friction between the block and the incline plane are μ s = 0.35 and μ k = 0.25, determine the force P required to (b) to keep it moving up FBD Tan Φ k = μ k = 0.25 Φ k = 14.04° 25° + 14.04° = 39.04° Force P to keep block moving up P = (800N) tan 39.04 ° = 649N
- 13. Sample Problem A support block is acted upon the by two forces as shown. Knowing that coefficients of friction between the block and the incline plane are μ s = 0.35 and μ k = 0.25, determine the force P required to (c) to prevent it from sliding down. FBD Φ s = 19.29° 25° - 19.29° = 5.71° Force P to prevent block sliding down P = (800N) tan 5.71 ° = 80.0N
- 14. The Wedge <ul><li>A wedge is one of the simplest and most useful machine. A wedge is used to produce small adjustments in the position of the body or to apply large forces. Wedges largely depend on friction to function. </li></ul>
- 15. Simple application example A wedge used to position or lift a large mass m The FBD showing the force triangle for the mass and the wedge.
- 16. Sample Problem <ul><li>A heavy block of mass 500kg is to be lifted vertically by the system of two wedges as shown. If the coefficient of the static friction for the all the surfaces is 0.3, determine the force P which will cause the system to be on the point of movement. </li></ul>
- 17. Sample Problem Forces on the block Triangle of forces for the block <ul><li>For the block: </li></ul><ul><li>= R2 sin 56.6 ° sin106.7° </li></ul>R2 = 5667N
- 18. Sample Problem Forces on the top wedge. Force triangle for the wedge P = 5667 Sin 39.4 ° sin 67.3° P = 3899N
- 19. Tutorial Questions A Piece of flat steel is clamped in the jaws of an engineers vice with the clamping force of 600N. If the coefficient of friction ( μ ) = 0.3, what minimum pull on the steel would just cause it to move? The horizontal position of the 500kg rectangular block of concrete is adjusted by the 5 ° wedge under the action of the force P. is the coefficient of static friction for both pairs of wedge surfaces is 0.30 and if the coefficient of static friction between the block and the horizontal surface is 0.60, determine the least force P required to move the block.
- 20. Tutorial Questions If the coefficient of the friction between the steel wedge an the moist fibers of the newly cut stump is 0.20, determine the maximum angle which the wedge may have and not pop out of the wood after being driven by the sledge. Determine the range of values which the mass mo may have so that the 100kg block neither start moving up the plane nor slip down. μ s = 0.30
- 21. Tutorial Questions A 100N force acts on a 300N block placed on an inclined plane. Determine whether the block is equilibrium and find the value of the friction force. μ s = 0.25 and μ k = 0.20

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