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FRICTION (Sections 8.1 - 8.2)

FRICTION (Sections 8.1 - 8.2). Section’s Objective : Students will be able to: a) Understand the characteristics of dry friction. b) Draw a FBD including friction. c) Solve problems involving friction . In-Class Activities : Check homework, if any Reading quiz Applications

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FRICTION (Sections 8.1 - 8.2)

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  1. FRICTION (Sections 8.1 - 8.2) Section’s Objective: Students will be able to: a) Understand the characteristics of dry friction. b) Draw a FBD including friction. c) Solve problems involving friction. • In-Class Activities: • Check homework, if any • Reading quiz • Applications • Characteristics of dry friction • Problems involving dry friction • Concept quiz • Group problem solving • Attention quiz

  2. READING QUIZ • 1. A friction force always acts _____ to the contact surface. • A) normal B) at 45° • C) parallel D) at the angle of static friction • 2. If a block is stationary, then the friction force acting on it is ________ . • A) s N B) = sN • C)  sN D) = k N

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

  4. APPLICATIONS (continued) Consider pushing a box as shown here. How can you determine if it will slide, tilt, or stay in static equilibrium? What physical factors affect the answer to this question?

  5. CHARACTERISTICS OF DRY FRICTION (Section 8.1) Friction is defined as a force of resistance acting on a body which prevents or retards 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. For the body shown in the figure to be in equilibrium, the following must be true: F = P, N = W, and Wx = Ph.

  6. CHARACTERISTICS OF FRICTION (continued) To study the characteristics of the friction force F, let us assume that tipping does not occur (i.e., “h” is small or “a” is large). Then we gradually increase the magnitude of the force P. Typically, experiments show that the friction force F varies with P, as shown in the left figure above.

  7. FRICTION CHARACERISTICS (continued) The maximum friction force is 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 materials in contact. Once the block begins to move, the frictional force typically drops and is given by Fk = k N. The value ofk (coefficient of kinetic friction) is less than s.

  8. 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 A block with weight w is placed on an inclined plane. The plane is slowly tilted until the block just begins to slip. Using these two equations, we get s = (W sin s) / (W cos s) = tan sThis simple experiment allows us to find the S between two materials in contact.

  9. PROCEDURE FOR ANALYSIS (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.

  10. IMPENDING TIPPING versus SLIPPING For a given W and h, how can we determine if the block will slide first or tip first? In this case, we have four unknowns (F, N, x, and P) and only three EofE. Hence, we have to make an assumption to give us another equation. Then we can solve for the unknowns using the three EofE. Finally, we need to check if our assumption was correct.

  11. 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

  12. EXAMPLE Given: A uniform ladder weighs 100 N. The vertical wall is smooth (no friction). The floor is rough and s = 0.8. Find: The minimum force P needed to move ( tip or slide) the ladder. Plan: a) Draw a FBD. b) Determine the unknowns. c) Make any necessary friction assumptions. d) Apply EofE (and friction equations, if appropriate ) to solve for the unknowns. e) Check assumptions, if required.

  13. NB A FBD of the ladder P 100 N 1.2 m FA NA 0.9 m 0.9 m •   FY = NA – 100 = 0 ; so NA = 100 N +  MA = 100 ( 0.9 ) – P( 1.2 ) = 0 ; so P = 75 N +   FX = 75 – FA = 0 ; so FA = 75 N EXAMPLE (continued) 1.2 m There are four unknowns: NA, FA, NB, and P. Let us assume that the ladder will tip first. Hence, NB = 0

  14. NB A FBD of the ladder P 100 N 1.2 m FA NA 0.9 m 0.9 m EXAMPLE (continued) Now check the assumption. Fmax = s NA = 0.8 * 100 N = 80 N Is FA = 75 N  Fmax = 80 N? Yes, hence our assumption of tipping is correct.

  15. P(A) P(B) 100 N P(C) 2. A ladder is positioned as shown. Please indicate the direction of the friction force on the ladder at B. A)  B)  C) D) B A CONCEPT QUIZ 1. A 100 N box with wide base is pulled by a force P and s = 0.4. Which force orientation requires the least force to begin sliding? A) A B) B C) C D) Can not be determined

  16. GROUP PROBLEM SOLVING Given: Drum weight = 500 N, s = 0.5, a = 0.75 m and b = 1m. Find: The smallest magnitude of P that will cause impending motion (tipping or slipping) of the drum. Plan: a) Draw a FBD of the drum. b) Determine the unknowns. c) Make friction assumptions, as necessary. d) Apply EofE (and friction eqn. as appropriate) to solve for the unknowns. e) Check assumptions, as required.

  17. P 5 0.375 m 0.375 m 3 4 500 N 1 m 0 F X N GROUP PROBLEM SOLVING (continued) A FBD of the drum: There are four unknowns: P, N, F and x. First, let’s assume the drum slips. Then the friction equation is F = s N = 0.5 N.

  18. P 5 0.375 m 0.375 m 3 4 500 N 1 m 0 F X N P = 500 N and N = 800 N +  MO = (3 /5) 500 (0.375) – (4 / 5) 500 (1) + 800 (x) = 0 Check: x = 0.359  0.375 so OK! Drum slips as assumed at P = 500 N GROUP PROBLEM SOLVING (continued) A FBD of the drum: +   FX = (4 / 5) P – 0.5 N = 0 +   FY = N – (3 / 5) P – 500 = 0 These two equations give:

  19.  S = 0.3 2 N 2. The ladder AB is postioned as shown. What is the direction of the friction force on the ladder at B. A) B) C)  D)  B A ATTENTION QUIZ 1. A 10 N block is in equilibrium. What is the magnitude of the friction force between this block and the surface? A) 0 N B) 1 N C) 2 N D) 3 N

  20. CENTER OF GRAVITY AND CENTROID (Chapter 9) Section’s Objective : Students will: a) Understand the concepts of center of gravity, center of mass, and centroid. b) Be able to determine the location of these points for a system of particles or a body. • In-Class Activities: • Check homework, if any • Reading quiz • Applications • Center of gravity, etc. • Determine their location • Concept quiz • Group problem solving • Attention quiz

  21. READING QUIZ • 1. The _________ is the point defining the geometric center of an object . • A) center of gravity B) center of mass • C) centroid D) none of the above 2. To study problems concerned with the motion of matter under the influence of forces, i.e., dynamics, it is necessary to locate a point called ________. A) center of gravity B) center of mass C) centroid D) none of the above

  22. APPLICATIONS 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 are acting. How can we determine these weights and their locations?

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

  24. 4N CONCEPT OF CG and CM 3m 1m The center of gravity (G) is a point which locates the resultant weight of a system of particles or body.   • A B G 1 N 3 N 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. For the figure above, try taking moments about A and B. Also, note that the sum of moments due to the individual particle’s weights about point G is equal to zero. Similarly, the center of mass is a point which locates the resultant mass of a system of particles or body. Generally, its location is the same as that of G.

  25. If an object has an axis of symmetry, then the centroid of object lies on that axis. In some cases, the centroid is not located on the object. CONCEPT OF CENTROID The centroid C is a point which defines the geometric center of an object. The centroid coincides with the center of mass or the center of gravity only if the material of the body is homogenous (density or specific weight is constant throughout the body).

  26. Summing the moments about the y-axis, we get x WR = x1W1 + x2W2 + ……….. + xnWn where x1 represents x coordinate of W1, etc.. ~ ~ ~ ~ CG / CM FOR A SYSTEM OF PARTICLES (Section 9.1) Consider a system of n particles as shown in the figure. The net or the resultant weight is given as WR = W. Similarly, we can sum moments about the x and z-axes to find the coordinates of G. By replacing the W with a M in these equations, the coordinates of the center of mass can be found.

  27. CG / CM / CENTROID OF A BODY (Section 9.2) A rigid body can be considered as made up of an infinite number of particles. Hence, using the same principles as in the previous slide, we get the coordinates of G by simply replacing the discrete summation sign (  ) by the continuous summation sign (  ) and W by dW. Similarly, the coordinates of the center of mass and the centroid of volume, area, or length can be obtained by replacing W by m, V, A, or L, respectively.

  28. ~ ~ 3. Determine coordinates (x , y ) of the centroid of the rectangular element in terms of the general point (x,y). STEPS FOR DETERMING AREA CENTROID 1. Choose an appropriate differential element dA at a general point (x,y). Hint: Generally, if y is easily expressed in terms of x (e.g., y = x2 + 1), use a vertical rectangular element. If the converse is true, then use a horizontal rectangular element. 2. Express dA in terms of the differentiating element dx (or dy). 4. Express all the variables and integral limits in the formula using either x or y depending on whether the differential element is in terms of dx or dy, respectively, and integrate. Note: Similar steps are used for determining CG, CM, etc.. These steps will become clearer by doing a few examples.

  29. EXAMPLE (continued) ~ 4. x = ( A x dA ) / ( A dA ) 3 0  x ( 9 – x2) d x [ 9 (x2)/2– (x4) / 4] 3 0  x ( 9 – x2) d x [ 9 x– (x3) / 3 ] 3 = ( 9 ( 9 ) / 2 – 81 / 4 ) / ( 9 ( 3 ) – ( 27 / 3 ) ) = 1.13 m 0 = = 3 0 ~ 3 A y dA ½ 0  ( 9 – x2) ( 9 – x2) dx A dA 0  ( 9 – x2) d x = 3.60 m y = = 3

  30. CONCEPT QUIZ 1. The steel plate with known weight and non-uniform thickness and density is supported as shown. Of the three parameters (CG, CM, and centroid), which one is needed for determining the support reactions? Are all three parameters located at the same point? A) (center of gravity, no)B) (center of gravity, yes)C) (centroid, yes)D) (centroid, no) 2. When determining the centroid of the area above, which type of differential area element requires the least computational work? A) Vertical B) Horizontal C) Polar D) Any one of the above.

  31. Given: The area as shown. Find: The x of the centroid. Plan: Follow the steps. Solution 1. Choose dA as a horizontal rectangular strip. (x1,,y) (x2,y) • 2. dA = ( x2– x1) dy • = ((2 – y) – y2) dy • 3. x = ( x1 + x2) / 2 • = 0.5 (( 2 – y) + y2 ) ~ GROUP PROBLEM SOLVING

  32. ~ 4. x = ( A x dA ) / ( A dA ) 1 A dA = 0 ( 2 – y – y2) dy [ 2 y – y2 / 2 – y3 / 3] 1 = 1.167 m2 0 ~ 1 A x dA = 0 0.5 ( 2 – y + y2 ) ( 2 – y – y2 ) dy = 0.5 0 ( 4 – 4 y + y2 – y4 ) dy = 0.5 [ 4 y – 4 y2 / 2 + y3 / 3 – y5 / 5 ] 1 = 1.067 m3 1 0 PROBLEM SOLVING (continued) x = 1.067 / 1.167 = 0.914 m

  33. 2. If a vertical rectangular strip is chosen, then what are the values of x and y ? A) (x , y) B) (x / 2 , y / 2) C) (x , 0) D) (x , y / 2) ATTENTION QUIZ 1. If a vertical rectangular strip is chosen as the differential element, then all the variables, including the integral limit, should be in terms of _____ . A) x B) y C) z D) Any of the above.

  34. End of the Lecture Let Learning Continue

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