* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Download Chapter 4B. Friction and Equilibrium
Survey
Document related concepts
Virtual work wikipedia , lookup
Coriolis force wikipedia , lookup
N-body problem wikipedia , lookup
Jerk (physics) wikipedia , lookup
Classical mechanics wikipedia , lookup
Fictitious force wikipedia , lookup
Brownian motion wikipedia , lookup
Frictional contact mechanics wikipedia , lookup
Rigid body dynamics wikipedia , lookup
Centrifugal force wikipedia , lookup
Newton's theorem of revolving orbits wikipedia , lookup
Work (physics) wikipedia , lookup
Equations of motion wikipedia , lookup
History of fluid mechanics wikipedia , lookup
Seismometer wikipedia , lookup
Classical central-force problem wikipedia , lookup
Newton's laws of motion wikipedia , lookup
Transcript
Chapter 4B. Friction and Equilibrium AA PowerPoint PowerPoint Presentation Presentation by by Paul Paul E. E. Tippens, Tippens, Professor Professor of of Physics Physics Southern Southern Polytechnic Polytechnic State State University University © 2007 Equilibrium: Until motion begins, all forces on the mower are balanced. Friction in wheel bearings and on the ground oppose the lateral motion. Objectives: After completing this module, you should be able to: •• Define Define and and calculate calculate the the coefficients coefficients of of kinetic kinetic and and static static friction, friction, and and give give the the relationship relationship of of friction friction to to the the normal normal force. force. •• Apply Apply the the concepts concepts of of static static and and kinetic kinetic friction friction to to problems problems involving involving constant constant motion motion or or impending impending motion. motion. Friction Forces When two surfaces are in contact, friction forces oppose relative motion or impending motion. P Friction forces are parallel to the surfaces in contact and oppose motion or impending motion. Static Friction: No relative motion. Kinetic Friction: Relative motion. motion Friction and the Normal Force 12 N n 8N n 4N n2N 4N 6N The force required to overcome static or kinetic friction is proportional to the normal force, n. ffss == ssn n ffkk == kkn n Friction forces are independent of area. 4N 4N If the total mass pulled is constant, the same force (4 N) is required to overcome friction even with twice the area of contact. For this to be true, it is essential that ALL other variables be rigidly controlled. Friction forces are independent of temperature, provided no chemical or structural variations occur. 4N 4N Heat can sometimes cause surfaces to become deformed or sticky. In such cases, temperature can be a factor. Friction forces are independent of speed. 5 m/s 2 N 20 m/s 2 N The force of kinetic friction is the same at 5 m/s as it is for 20 m/s. Again, we must assume that there are no chemical or mechanical changes due to speed. The Static Friction Force When an attempt is made to move an object on a surface, static friction slowly increases to a MAXIMUM value. value fs n P W f s s n In this module, when we use the following equation, we refer only to the maximum value of static friction and simply write: ffss == ssn n Constant or Impending Motion For motion that is impending and for motion at constant speed, the resultant force is zero and F = 0. (Equilibrium) fs P Rest P – fs = 0 fk P Constant Speed P – fk = 0 Here the weight and normal forces are balanced and do not affect motion. Friction and Acceleration When P is greater than the maximum fs the resultant force produces acceleration. fk a P Constant Speed This case will be discussed in a later chapter. fk = kn Note that the kinetic friction force remains constant even as the velocity increases. EXAMPLE 1: If k = 0.3 and s = 0.5, what horizontal pull P is required to just start a 250-N block moving? 1. Draw sketch and freebody diagram as shown. n fs P + W 2. List givens and label what is to be found: k = 0.3; s = 0.5; W = 250 N Find: P = ? to just start 3. Recognize for impending motion: P – fs = 0 EXAMPLE 1(Cont.): s = 0.5, W = 250 N. Find P to overcome fs (max). Static friction applies. n P fs + For this case: P – fs = 0 4. To find P we need to know fs , which is: fs = sn 250 N 5. To find n: Fy = 0 W = 250 N n=? n–W=0 n = 250 N (Continued) EXAMPLE 1(Cont.): s = 0.5, W = 250 N. Find P to overcome fs (max). Now we know n = 250 N. 6. Next we find fs from: fs = sn = 0.5 (250 N) 7. For this case: case P – fs = 0 P = fs = 0.5 (250 N) PP == 125 125 NN n fs P + 250 N s = 0.5 This force (125 N) is needed to just start motion. Next we consider P needed for constant speed. EXAMPLE 1(Cont.): If k = 0.3 and s = 0.5, what horizontal pull P is required to move with constant speed? (Overcoming kinetic friction) Fy = may = 0 k = 0.3 fk n n-W=0 P n=W Now: fk = kn = kW + mg P = (0.3)(250 N) Fx = 0; P - fk = 0 P = fk = kW PP == 75.0 75.0 NN The Normal Force and Weight The normal force is NOT always equal to the weight. The following are examples: n m P Here the normal force is less than weight due to upward component of P. 300 W P n W Here the normal force is equal to only the component of weight perpendicular to the plane. Review of Free-body Diagrams: For For Friction Friction Problems: Problems: •• Read Read problem; problem; draw draw and and label label sketch. sketch. •• Construct Construct force force diagram diagram for for each each object, object, vectors vectors at at origin origin of of x,y x,y axes. axes. Choose Choose xx or or yy axis axis along along motion motion or or impending impending motion. motion. •• Dot Dot in in rectangles rectangles and and label label xx and and yy compo components nents opposite opposite and and adjacent adjacent to to angles. angles. •• Label Label all all components; components; choose choose positive positive direction. direction. For Friction in Equilibrium: •• Read, Read, draw draw and and label label problem. problem. •• Draw -body diagram Draw free free-body diagram for for each each body. body. •• Choose -axis along Choose xx or or yy-axis along motion motion or or impending impending motion motion and and choose choose direction direction of of motion motion as as positive. positive. •• Identify Identify the the normal normal force force and and write write one one of of following: following: ffs ==snn or or ffkk==kknn s s •• For For equilibrium, equilibrium, we we write write for for each each axis: axis: F Fx ==00 x F Fy ==00 y •• Solve Solve for for unknown unknown quantities. quantities. Example 2. A force of 60 N drags a 300-N block by a rope at an angle of 400 above the horizontal surface. If uk = 0.2, what force P will produce constant speed? W = 300 N fk n P=? 400 m W The force P is to be replaced by its components Px and Py. 1. Draw and label a sketch of the problem. 2. Draw free-body diagram. P sin 400 Py n fk W P 400 Py Px P cos 400 + Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2. 3. Find components of P: Px = P cos 400 = 0.766P Py = P sin 400 = 0.643P Px = 0.766P; Py = 0.643P P sin 400 n fk P 400 P cos 400 mg + Note: Vertical forces are balanced, and for constant speed, horizontal forces are balanced. F x 0 F y 0 Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2. Px = 0.766P Py = 0.643P 4. Apply Equilibrium conditions to vertical axis. F Fyy == 00 0.643P n fk 300 N P 400 0.766P + n + 0.643P – 300 N= 0 [Py and n are up (+)] n = 300 N – 0.643P; Solve for n in terms of P nn == 300 300 NN –– 0.643 0.643PP Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2. nn == 300 300 NN –– 0.643 0.643PP 5. Apply Fx = 0 to con- stant horizontal motion. F Fxx == 0.766 0.766PP –– ffkk = = 00 0.643P n fk 300 N P 400 0.766P + fk = k n = (0.2)(300 N - 0.643P) fk = (0.2)(300 N - 0.643P) = 60 N – 0.129P 0.766P – fk = 0; 0.766 0.766PP –– (60 (60 NN –– 0.129 0.129PP)) = = 00 Example 2 (Cont.). P = ?; W = 300 N; uk = 0.2. 0.643P n P 400 fk 0.766P 300 N + 0.766 0.766PP –– (60 (60 NN –– 0.129 0.129PP )=0 )=0 6. Solve for unknown P. 0.766P – 60 N + 0.129P =0 0.766P + 0.129P = 60 N 0.766P + 0.129P = 60 N 0.895P = 60 N P = 67.0 N If P = 67 N, the block will be dragged at a constant speed. P = 67.0 N Example 3: What push P up the incline is needed to move a 230-N block up the incline at constant speed if k = 0.3? P Step 1: Draw free-body including forces, angles and components. y W sin 600 n fk P 230 N 600 x W cos 600 600 W =230 N Step 2: Fy = 0 n – W cos 600 = 0 n = (230 N) cos 600 nn == 115 115 NN Example 3 (Cont.): Find P to give move up the incline (W = 230 N). n y W sin 600 x P fk W cos 600 600 W n = 115 N 600 W = 230 N Step 3. Apply Fx= 0 P - fk - W sin 600 = 0 fk = kn = 0.2(115 N) fk = 23 N, P = ? P - 23 N - (230 N)sin 600 = 0 P - 23 N - 199 N= 0 PP == 222 222 NN Summary: Important Points to Consider When Solving Friction Problems. • The maximum force of static friction is the force required to just start motion. fs n P W f s s n Equilibrium exists at that instant: Fx 0; Fy 0 Summary: Important Points (Cont.) • The force of kinetic friction is that force required to maintain constant motion. n fk P W f k k n • Equilibrium exists if speed is constant, but fk does not get larger as the speed is increased. Fx 0; Fy 0 Summary: Important Points (Cont.) • Choose an x or y-axis along the direction of motion or impending motion. k = 0.3 fk n P + W The F will be zero along the x-axis and along the y-axis. In this figure, we have: Fx 0; Fy 0 Summary: Important Points (Cont.) • Remember the normal force n is not always equal to the weight of an object. n m P It is necessary to draw the free-body diagram and sum forces to solve 300 W P n W for the correct Fx 0; n value. Fy 0 Summary Static Friction: No relative motion. Kinetic Friction: Relative motion. ffss ≤≤ ssn n ffkk == kkn n Procedure for solution of equilibrium problems is the same for each case: Fx 0 Fy 0 CONCLUSION: Chapter 4B Friction and Equilibrium