1 / 83

CHARACTERISTICS OF DRY FRICTION & PROBLEMS INVOLVING DRY FRICTION

Today’s Objective : a ) Understand the characteristics of dry friction. b) Draw a FBD including friction. c) Solve problems involving friction. CHARACTERISTICS OF DRY FRICTION & PROBLEMS INVOLVING DRY FRICTION.

pascale-blanchard
Download Presentation

CHARACTERISTICS OF DRY FRICTION & PROBLEMS INVOLVING DRY FRICTION

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. Today’s Objective: a) Understand the characteristics of dry friction. b) Draw a FBD including friction. c) Solve problems involving friction. CHARACTERISTICS OF DRY FRICTION & PROBLEMS INVOLVING DRY FRICTION

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

  3. The rope is used to tow the refrigerator. In order to move the refrigerator, is it best to pull up as shown, pull horizontally, or pull downwards on the rope? APPLICATIONS (continued)

  4. CHARACTERISTICS OF DRY FRICTION (Section 8.1) Friction is defined as a force of resistance acting on a body which prevents or slows slipping of the body relative to a second body. Experiments show that frictional forces act tangent (parallel) to the contacting surface in a direction opposing the relative motion or tendency for motion.

  5. CHARACTERISTICS OF DRY FRICTION (Section 8.1) For the body shown in the figure to be in equilibrium, the following must be true: F = P, N = W, and W*x = P*h.

  6. CHARACTERISTICS OF DRY FRICTION (continued) • To study the characteristics of the friction force F • Assume that tipping does not occur • Gradually increase the magnitude of the force • Experiments show the friction F varies with P • Shown in the right figure above.

  7. CHARACTERISTICS OF DRY FRICTION (continued) • The maximum friction force • Attained just before the block begins to move (a situation that is called “impending motion”). • The value of the force is found using Fs = sN, where sis called the coefficient of static friction. • The value of sdepends on the two materials in contact.

  8. CHARACTERISTICS OF DRY FRICTION (continued) • Once the block begins to move • Frictional force typically drops and is given by Fk = k N. • The value ofk (coefficient of kinetic friction) is less than s.

  9. CHARACTERISTICS OF DRY FRICTION (continued) • Note that the friction force maybe less than the maximum friction force. • Cannot assume that because the object is not moving, the friction force is at its maximum of Fs = sN • Unless motion is impending

  10. When the block is on the verge of sliding • Normal force N and • frictional force Fs, create a resultant Rs • From the figure, • tan s = ( Fs / N ) = (s N / N ) = s DETERMING s EXPERIMENTALLY

  11. A block with weight w is placed on an inclined plane. The plane is slowly tilted until the block just begins to slip. The inclination, s, is noted. Analysis of the block just before it begins to move gives (using Fs = s N): +  Fy = N – W cos s = 0 +  FX = S N – W sin s = 0 DETERMING s EXPERIMENTALLY (continued)

  12. +  Fy = N – W cos s = 0 N = W cos s +  FX = S N – W sin s = 0 S N = W sin s S = W sin s / N DETERMING s EXPERIMENTALLY (continued) s = (W sin s) / (W cos s) = tan s This finds the S between two materials in contact.

  13. THINK ABOUT IT … • A friction force always acts _____ to the contact surface. • A) Normal B) At 45° • C) Parallel D) At the angle of static friction C

  14. PROBLEMS INVOLVING DRY FRICTION (Section 8.2) Steps for solving equilibrium problems involving dry friction: 1. Draw the necessary free body diagrams. Make sure that you show the friction force in the correct direction (it always opposes the motion or impending motion). 2. Determine the number of unknowns. Do not assume F = S N unless the impending motion condition is given. 3. Apply the equations of equilibrium and appropriate frictional equations to solve for the unknowns.

  15. For a given W and h of the box, how can we determine if the block will slide or tip first? In this case, we have four unknowns (F, N, x, and P) and only three E-of-E. IMPENDING TIPPING versus SLIPPING

  16. We have four unknowns (F, N, x, and P) and only three E-of-E. IMPENDING TIPPING versus SLIPPING • Make an assumption to give us another equation (the friction equation!). • Check if our assumption was correct.

  17. IMPENDING TIPPING versus SLIPPING (continued) Assume: Slipping occurs Known: F = s N Solve: x, P, and N Check: 0  x  b/2 Or Assume: Tipping occurs Known: x = b/2 Solve: P, N, and F Check: F s N

  18. Given: A uniform ladder weighs 30 lb. The vertical wall is smooth (no friction). The floor is rough and s = 0.2. Find: Whether it remains in this position when it is released. Plan: EXAMPLE a) Draw a FBD. b) Determine the unknowns. c) Make any necessary friction assumptions. d) Apply E-of-E (and friction equations, if appropriate ) to solve for the unknowns. e) Check assumptions, if required.

  19. There are three unknowns: NA, FA, NB. •   FY = NA – 30 = 0 ; so NA = 30 lb +  MA = 30 ( 5 ) – NB( 24 ) = 0 ; so NB = 6.25 lb +   FX = 6.25 – FA = 0 ; so FA = 6.25 lb EXAMPLE (continued) NB FBD of the ladder 12 ft 30 lb 12 ft FA NA 5 ft 5 ft

  20. Now check the friction force to see if the ladder slides or stays. Fmax = s NA = 0.2 * 30 lb = 6 lb Since FA = 6.25 lb  Ffriction max = 6 lb, the pole will not remain stationary. It will move. EXAMPLE (continued) NB FBD of the ladder 12 ft 30 lb 12 ft FA NA 5 ft 5 ft

  21. Given: Refrigerator weight = 160 lb, s = 0.25 Find: The smallest magnitude of P that will cause impending motion (tipping or slipping) of the refrigerator. Plan: GROUP PROBLEM SOLVING a) Draw a FBD of the refrigerator. b) Determine the unknowns. c) Make friction assumptions, as necessary. d) E-of-E (and friction equation as appropriate) to solve for unknowns. e) Check assumptions, as required.

  22. GROUP PROBLEM SOLVING (continued) Volunteer 1.5 ft 1.5 ft P 160 lb 4 ft 3 ft 0 F X N FBD of the refrigerator There are four unknowns: P, N, F and x. First, let’s assume the refrigerator slips. Then the friction equation is F = s N = 0.25 N.

  23. +  MO = -45(4) + 180 (x) = 0, x = 1 Check: x = 1.0  1.5 so OK! Refrigeratorslips as assumed at P = 45 lb GROUP PROBLEM SOLVING (continued) 1.5 ft 1.5 ft P P 180 lb 4 ft 3 ft +   FX = P – 0.25 N = 0 +   FY = N – 180 = 0 P = 45 lb and N = 180 lb Volunteer 0 F X N FBD of the refrigerator Volunteer Is our assumption correct?

  24. WEDGES &FRICTIONAL FORCES ON FLAT BELTS

  25. Today’s Objectives: a) Determine the forces on a wedge. b) Determine the tensions in a belt. WEDGES AND FRICTIONAL FORCES ON FLAT BELTS

  26. Wedges are used to adjust the elevation or provide stability for heavy objects such as this large steel pipe. APPLICATIONS How can we determine the force required to pull the wedge out? When there are no applied forces on the wedge, will it stay in place (i.e., be self-locking) or will it come out on its own? Under what physical conditions will it come out?

  27. Belt drives are commonly used for transmitting the torque developed by a motor to a wheel attached to a pump, fan or blower. APPLICATIONS (continued) How can we decide if the belts will function properly, i.e., without slipping or breaking?

  28. A wedge is a simple machine in which a small force P is used to lift a large weight W. ANALYSIS OF A WEDGE W To determine the force required to push the wedge in or out, it is necessary to draw FBDs of the wedge and the object on top of it. It is easier to start with a FBD of the wedge since you know the direction of its motion.

  29. ANALYSIS OF A WEDGE W Note that: a) the friction forces are always in the direction opposite to the motion, or impending motion, of the wedge;b) the friction forces are along the contacting surfaces; and, c) the normal forces are perpendicular to the contacting surfaces.

  30. A FBD of the object on top of the wedge is drawn. • a) at the contacting surfaces between the wedge and the object the forces areequal in magnitude and opposite in direction to those on the wedge; • b) all other forces acting on the object should be shown. ANALYSIS OF A WEDGE (continued)  Fy = 0, to the wedge and the object as well as the impending motion frictional equation, F = S N.

  31. ANALYSIS OF A WEDGE (continued) Start by analyzing the FBD in whichthe number of unknowns are less than or equal to the number of EofE and frictional equations.

  32. ANALYSIS OF A WEDGE (continued) W • NOTE: • If the object is to be lowered, then the wedge needs to be pulled out. • If the value of the force P needed to remove the wedge is positive • The wedge is self-locking, i.e., it will not come out on its own.

  33. When determining the force P needed to lift the block of weight W, it is easier to draw a FBD of ______ first. A) The wedge B) The block C) The horizontal ground D) The vertical wall THINK ABOUT IT… W A

  34. Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to  radians. BELT ANALYSIS If the belt slips or is just about to slip, then T2 must be larger than T1 and the motion resisting friction forces. Detailed analysis (textbook - 425) shows that T2 = T1 e   where  is the coefficient of static friction between the belt and the surface. Be sure to use radians when using this formula!!

  35. Given: The crate weighs 300 lb and Sat all contacting surfaces is 0.3. Assume the wedges have negligible weight. Find: The smallest force P needed to pull out the wedge. Plan: EXAMPLE Draw a FBD of the crate. Draw a FBD of the wedge. 3. Apply the E-of-E to the crate. 4. Apply the E-of-E to wedge.

  36. EXAMPLE (continued) 300 lb FB=0.3NB NB FC=0.3NC NC NC FC=0.3NC FBD of Crate FBD of Wedge • +  FX = NB– 0.3NC = 0, NB= 0.3NC • +  FY = NC– 300 + 0.3 NB = 0 Solving the above two equations, we get NB= 82.57 lb= 82.6 lb, NC= 275.3 lb= 275 lb P 15º FD=0.3ND 15º ND

  37. EXAMPLE (continued) 300 lb FB=0.3NB NB = 82.6 lb FC=0.3NC NC NC = 275 lb FC=0.3NC FBD of Wedge FBD of Crate Applying the E-of-E to the wedge, we get +  FY = ND cos 15 + 0.3 ND sin 15 – 275.2= 0; ND= 263.7 lb= 264 lb +  FX = 0.3(263.7) + 0.3(263.7)cos 15 – 0.3(263.7)cos 15 – P = 0; P = 90.7 lb P 15º FD=0.3ND 15º ND

  38. Given: Blocks A and B weigh 50 lb and 30 lb, respectively. Find: The smallest weight of cylinder D which will cause the loss of static equilibrium. GROUP PROBLEM SOLVING

  39. Plan: 1. Consider two cases: a) both blocks slide together, and, b) block B slides over the block A. 2. For each case, draw a FBD of the block(s). 3. For each case, apply the EofE to find the force needed to cause sliding. 4. Choose the smaller P value from the two cases. 5. Use belt friction theory to find the weight of block D. GROUP PROBLEM SOLVING (Plan)

  40. Case a (both blocks sliding together): • +  FY = N – 80 = 0 N = 80 lb • +  FX = 0.4 (80) – P = 0 P = 32 lb P 30 lb B 50 lb A F=0.4 N N GROUP PROBLEM SOLVING (continued) Volunteer!

  41. 30 lb Case b (block B slides over A): P 0.6 N 20º  +  Fy = N cos 20 + 0.6 N sin 20 – 30 = 0 N = 26.20 lb  +  Fx = – P + 0.6 ( 26.2 ) cos 20 – 26.2 sin 20 = 0 P = 5.812 lb N GROUP PROBLEM SOLVING (continued) Volunteer!

  42. 30 lb Case b (block B slides over A): P 0.6 N 20º N GROUP PROBLEM SOLVING (continued) Case b has the lowest P (case a was 32 lb) and thus will occur first. Next, using a frictional force analysis of belt, we get WD = P e   = 5.812 e 0.5 ( 0.5  ) = 12.7 lb A Block D weighing 12.7 lb will cause the block B to slide over the block A.

  43. Let’s take a 10 Minute Break

  44. Chapter 9 Center of Gravity and Centroid Dr. N. Norman

  45. Today’s Objective : a) Understand the concepts of center of gravity, center of mass, and centroid. b) Be able todetermine the location of these points for a body. CENTER OF GRAVITY, CENTER OF MASS AND CENTROID OF A BODY Dr. N. Norman

  46. To design the structure for supporting a water tank, we will need to know the weights of the tank and water as well as the locations where the resultant forces representing these distributed loads act. APPLICATIONS How can we determine these weights and their locations? Dr. N. Norman

  47. One concern about a sport utility vehicle (SUV) is that it might tip over while taking a sharp turn. APPLICATIONS (continued) One of the important factors in determining its stability is the SUV’s center of mass. Should it be higher or lower to make a SUV more stable? How do you determine the location of the SUV’s center of mass? Dr. N. Norman

  48. CONCEPT OF CENTER OF GRAVITY (CG) A body is composed of an infinite number of particles, and so if the body is located within a gravitational field, then each of these particles will have a weight dW. Dr. N. Norman

  49. CONCEPT OF CENTER OF GRAVITY (CG) The center of gravity (CG) is a point, often shown as G, which locates the resultant weight of a system of particles or a solid body. From the definition of a resultant force, the sum of moments due to individual particle weight about any point is the same as the moment due to the resultant weight located at G. Also, note that the sum of moments due to the individual particle’s weights about point G is equal to zero. Dr. N. Norman

More Related