Presentation on theme: "WEDGES AND FRICTIONAL FORCES ON FLAT BELTS"— Presentation transcript:
1WEDGES AND FRICTIONAL FORCES ON FLAT BELTS Today’s Objectives:Students will be able to:a) Determine the forces on a wedge.b) Determine the tensions in a belt.In-Class Activities:Check Homework, if anyReading QuizApplicationsAnalysis of a WedgeAnalysis of a BeltConcept QuizGroup Problem SolvingAttention QuizTitle: font size – 24, color – green, centered and boldSubheadings: font – 24, underlined, color - white and boldSlide Text: font – 24-22, color – whiteIn Today’s Objectives, the list is indicated by lower cased alphabets enclosed in right parenthesis.In-class Activities is a bulleted list of green bullets with single space between bullet and text.Use blue color for emphasis on the specified lecture topics.On this first slide, use only the forward action buttonStatics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
21. A wedge allows a ______ force P to lift a _________ weight W. READING QUIZ1. A wedge allows a ______ force P to lifta _________ weight W.A) (large, large) B) (small, small)C) (small, large) D) (large, small)W2. Considering friction forces and the indicated motion of the belt, how are belt tensions T1 and T2 related?A) T1 > T B) T1 = T2C) T1 < T D) T1 = T2 eAnswers:1. C2. CAll quizzes (Reading, Attention, Concept):The questions are numbered.One space is left between the numeral and question text. Do not use automatic bullets as this is not compatible with ’97 version of PowerPointAnswer choices are listed by upper case alphabets A, B, C, D, enclosed by right parenthesis. Single space is Left between the Alphabet and the Answer choice. The choices have hanging indentation aligned with the starting of the question text.For multiple fill up the blanks,answers are enclosed in parenthesis. (see #1 above)Figures/Pictures are positioned on the right side of the slide.Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
3How can we determine the force required to pull the wedge out? APPLICATIONSWedges are used to adjust the elevation or provide stability for heavy objects such as this large steel vessel.How can we determine the force required to pull the wedge out?For other lecture slides like these, the picture is positioned on the left side of the slide.The font size can vary from , White color, regular style. Sometimes important text is highlighted by blue color and underlined.Position the forward and backward action buttons to the lower right corner of the slide. Size it proportionally.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?Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
4APPLICATIONS (continued) Belt drives are commonly used for transmitting power from one shaft to another.For slides that are continuation from previous topic, include ‘continued’ as part of slide heading, enclose it in parenthesis.Have a font size of 24, color – white, regular style.How can we decide that the belts will function properly, i.e., without slipping or breaking?Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
5APPLICATIONS (continued) In the design of a band brake, it is essential to analyze the frictional forces acting on the band (which acts like a belt).How can we determine the tensions in the cable pulling on the band?How are these tensions, the applied force P and the torque M, related?Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
6ANALYSIS OF A WEDGEA wedge is a simple machine in which a small force P is used to lift a large weight W.WTo 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.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.As noted earlier, important text is given a blue color and underlined.Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
7ANALYSIS OF A WEDGE (continued) Next, a FBD of the object on top of the wedge is drawn. Please note that: a) at the contacting surfaces between the wedge and the object the forces are equal in magnitude and opposite in direction to those on the wedge; and, b) all other forces acting on the object should be shown.To determine the unknowns, we must apply EofE, Fx = 0 and Fy = 0, to the wedge and the object as well as the impending motion frictional equation, F = S N.Now of the two FBDs, which one should we start analyzing first? We should start analyzing the FBD in which the number of unknowns are less than or equal to the number of equations.
8ANALYSIS OF A WEDGE (continued) 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, then the wedge is self-locking, i.e., it will not come out on its own.WHowever, if the value of P is negative, or zero, then the wedge will come out on its own unless a force is applied to keep the wedge in place. This can happen if the coefficient of friction is small or the wedge angle is large.
9Wedges or or Wedges - simple machines used to raise heavy loads. Force required to lift block is significantly less than block weight.Friction prevents wedge from sliding out.Want to find minimum force P to raise block.Block as free-bodyorWedge as free-bodyor
10Square-Threaded Screws Square-threaded screws frequently used in jacks, presses, etc. Analysis similar to block on inclined plane. Recall friction force does not depend on area of contact.Thread of base has been “unwrapped” and shown as straight line. Slope is 2pr horizontally and lead L vertically.Moment of force Q is equal to moment of force P.Impending motion upwards. Solve for Q.Self-locking, solve for Q to lower load.Non-locking, solve for Q to hold load.
11C-ClampA clamp is used to hold two pieces of wood together as shown. The clamp has a double square thread of mean diameter equal to 10 mm with a pitch of 2 mm. The coefficient of friction between threads is ms = 0.30.If a maximum torque of 40 N*m is applied in tightening the clamp, determine (a) the force exerted on the pieces of wood, and (b) the torque required to loosen the clamp.SOLUTIONCalculate lead angle and pitch angle.Using block and plane analogy with impending motion up the plane, calculate the clamping force with a force triangle.With impending motion down the plane, calculate the force and torque required to loosen the clamp.
12SOLUTIONCalculate lead angle and pitch angle. For the double threaded screw, the lead L is equal to twice the pitch.Using block and plane analogy with impending motion up the plane, calculate clamping force with force triangle.
13With impending motion down the plane, calculate the force and torque required to loosen the clamp.
14BELT ANALYSISBelts are used for transmitting power or applying brakes. Friction forces play an important role in determining the various tensions in the belt. The belt tension values are then used for analyzing or designing a belt drive or a brake system.
15BELT ANALYSIS (continued) Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to radians.If the belt slips or is just about to slip, then T2 must be larger than T1 and the friction forces. Hence, T2 must be greater than T1.Detailed analysis (please refer to your textbook) 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!!
16Find: The force P needed to lift the load. EXAMPLEGiven: The load weighs 100 lb and the S between surfaces AC and BD is 0.3. Smooth rollers are placed between wedges A and B. Assume the rollers and the wedges have negligible weights.Find: The force P needed to lift the load.In the Example slide, do not align the colons, but, rather the question text.Given, Find and Plan text are always in bold.‘Plan’ is always underlinedPlan:1. Draw a FBD of wedge A. Why do A first?2. Draw a FBD of wedge B.3. Apply the E-of-E to wedge B. Why do B first?4. Apply the E-of-E to wedge A.Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
17EXAMPLE (continued) 10º N2 The FBDs of wedges A and B are shown in the figures. Applying the EofE to wedge B, we getAP+ FX = N2 sin 10 – N3 = 0+ FY = N2 cos 10 – 100 – 0.3 N3 = 0Solving the above two equations, we getN2 = lb and N3 = lbF1= 0.3N1N1Known forces are indicated by blue color arrow, Unknown forces indicated by yellow color.The thickness of arrow lines are 2 1/4 pt.Bolden the Decimal PointSymbols used in the equations are from the Symbols library.do not use sentence case on ‘cos’ and ‘sin’, they are to be in lower case.N210ºBF3= 0.3N3N3100 lbApplying the E-of-E to the wedge A, we get+ FY = N1 – cos 10 = 0; N1 = lb+ FX = P – sin 10 – 0.3 N1 = 0; P = lbStatics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
18A) be lifted up. B) slide down. CONCEPT QUIZ1. Determine the direction of the friction force on object B at the contact point between A and B.A) B) C) D)Answers:1. C2. C The maximum force she can apply before she slides is 150 (0.6) = 90 lb. This will not be enough to lift the boy up as she needs to apply more than 100 lb to lift the boy up. Without knowing the coefficient of friction between the cliff and the rope, we can not say whether the boy will slide down or not.2. The boy (hanging) in the picture weighs 100 lb and the woman weighs 150 lb. The coefficient of static friction between her shoes and the ground is The boy will ______ ?A) be lifted up B) slide down.C) not be lifted up. D) not slide down.Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
19GROUP PROBLEM SOLVINGGiven: 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.
20GROUP PROBLEM SOLVING (continued) 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.
21GROUP PROBLEM SOLVING (continued) Case a:+ FY = N – 80 = 0N = 80 lb+ FX = 0.4 (80) – P = 0P = lbP30 lbB50 lbAF=0.4 NN
22GROUP PROBLEM SOLVING (continued) 20º30 lb0.6 NPNCase b: + Fy = N cos 20 N sin 20 – 30 = N = lb + Fx = – P ( 26.2 ) cos 20 – 26.2 sin 20 = P = lbCase b has the lowest P and will occur first. Next, using the frictional force analysis of belt, we getWD = P e = e 0.5 ( 0.5 ) = lbA Block D weighing 12.7 lb will cause the block B to slide over the block A.
23A) the wedge B) the block ATTENTION QUIZ1. 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 blockC) the horizontal ground D) the vertical wallWAnswers:1.A2.B2. In the analysis of frictional forces on a flat belt, T2 = T1 e . In this equation, equals ______ .A) angle of contact in degrees B) angle of contact in radiansC) coefficient of static friction D) coefficient of kinetic frictionStatics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5
24End of the Lecture Let Learning Continue In the last slide add the ‘home’ action buttonLet Learning ContinueStatics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5