Objectives

1. Objectives • To introduce the concept of dry friction and show how to analyze the equilibrium of rigid bodies subjected to this force. • To present specific applications of frictional force analysis on wedges, screws, belts, and bearings. Course Outcome (CO) CO3: Ability to determine friction and properties of sections

2. Friction : • or retards slipping of the body relative to a second body or surface which it is in contact • Acts tangent to the surfaces at points of contact with other body • Opposing possible or existing motion of the body relative to points of contact • Two types of friction – Fluid and Coulomb Friction

3. Fluid friction exist when the contacting surface are separated by a film of fluid (gas or liquid) Depends on velocity of the fluid and its ability to resist shear force Coulomb friction, also known as dry friction, occurs between contacting surfaces of bodies in the absence of a lubricating fluid

4. Theory of Dry Friction Example P is applied at a height h from the surface Moment equilibrium about point O is satisfied if W x = Ph or x = Ph/W The block is on the verge on tipping if N acts at the right corner of the block, x = a/2

5. Impending Motion In cases where h is small or surfaces of contact are rather “slippery”, the frictional force F may not be great enough to balance P and consequently, the block will tend to slip before it can tip As P is slowly increased, F correspondingly increase until it attains a certain maximum value F, called the limiting static frictional force

6. When this value is reached, any further increase in P will cause deformations and fractures at the points of surface contact and consequently, the block will begin to move Limiting static frictional force Fs is directly proportional to the resultant normal force N Fs = μsN

7. Typical Values of μs

8. Motion If the magnitude of P acting on the block is increased so that it is greater than Fs, the frictional force at the contacting surfaces drops slightly to a smaller value Fs, called kinetic frictional force The block will not be held in equilibrium (P > Fs) but slide with increasing speed

10. Example 1: The uniform crate has a mass of 20kg. If a force P = 80N is applied on to the crate, determine if it remains in equilibrium. The coefficient of static friction is μ = 0.3.

11. Solution Resultant normal force NC act a distance x from the crate’s center line in order to counteract the tipping effect caused by P 3 unknowns to be determined by 3 equations of equilibrium

13. Since x is negative, the resultant force acts (slightly) to the left of the crate’s center line No tipping will occur since x ≤ 0.4m Maximum frictional force which can be developed at the surface of contact Fmax = μsNC = 0.3(236N) = 70.8N Since F = 69.3N < 70.8N, the crate will not slip thou it is close to doing so

14. Example 2: • It is observed that when the bed of the dump truck is raised to an angle of θ = 25° the vending machines will begin to slide off the bed. Determine the static coefficient of friction between a vending machine and the surface of the truckbed.

16. The uniform 1O-kg ladder in Fig. rests against the smooth wall at the end A rests on the rough horizontal plane for which the coefficient of static friction is µr = 0.3. Determine the angle of inclination of θ of the ladder and the normal reaction at B if the ladder is verge of slipping. • Example 3:

17. Example 4: Beam AB is subjected to a uniform load of 200N/m and is supported at B by post BC. If the coefficients of static friction at B and C are μB = 0.2 and μC = 0.5, determine the force P needed to pull the post out from under the beam. Neglect the weight of the members and the thickness of the post.

18. Solution FBD of beam AB and the post Apply ∑MA = 0, NB = 400N 4 unknowns 3 equilibrium equations and 1 frictional equation applied at either B or C

19. Solution Post slips only at B

20. Solving Post slips only at C Solving Choose second case as it requires a smaller value of P

21. Wedges • A simple machine used to transform an applied force into much larger forces, directed at approximately right angles to the applied force • Used to give small displacements or adjustments to heavy load • Consider the wedge used to lift a block of weight W by applying a force P to the wedge

22. FBD of the block and the wedge Exclude the weight of the wedge since it is small compared to weight of the block

23. Example 3 The uniform stone has a mass of 500kg and is held in place in the horizontal position using a wedge at B. if the coefficient of static friction μs = 0.3, at the surfaces of contact, determine the minimum force P needed to remove the wedge. Is the wedge self-locking? Assume that the stone does not slip at A.

24. Solution • Minimum force P requires F = μs NA at the surfaces of contact with the wedge • FBD of the stone and the wedge • On the wedge, friction force opposes the motion and on the stone at A, FA ≤ μsNA, slipping does not occur

25. 5 unknowns, 3 equilibrium equations for the stone and 2 for the wedge

26. Since P is positive, the wedge must be pulled out If P is zero, the wedge would remain in place (self-locking) and the frictional forces developed at B and C would satisfy FB < μsNB FC < μsNC

27. The End. Thank you for your attention.