WEDGES AND FRICTIONAL FORCES ON FLAT BELTS Today’s Objectives: Students will be able to: a) Determine the forces on a wedge. Title: font size – 24, color – green, centered and bold Subheadings: font – 24, underlined, color - white and bold Slide Text: font – 24-22, color – white In 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 button Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5

Answers: 1. C READING QUIZ 1. A wedge allows a ______ force P to lift a _________ weight W. A) (large, large) B) (small, small) C) (small, large) D) (large, small) W Answers: 1. C All 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 PowerPoint Answer 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

How can we determine the force required to pull the wedge out? APPLICATIONS Wedges 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? 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? For other lecture slides like these, the picture is positioned on the left side of the slide. The font size can vary from 24-22 , 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. Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5

ANALYSIS OF A WEDGE A wedge is a simple machine in which a small force P is used to lift a large weight W. 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. 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

ANALYSIS 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.

ANALYSIS 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. W However, 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.

Find: The force P needed to lift the load. EXAMPLE Given: 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. Plan: 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. 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 underlined Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5

EXAMPLE (continued) 10º N2 The FBDs of wedges A and B are shown in the figures. Applying the EofE to wedge B, we get A P +  FX = N2 sin 10 – N3 = 0 +  FY = N2 cos 10 – 100 – 0.3 N3 = 0 Solving the above two equations, we get N2 = 107.2 lb and N3 = 18.6 lb F1= 0.3N1 N1 N2 10º B F3= 0.3N3 N3 100 lb Known 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 Point Symbols 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. Applying the E-of-E to the wedge A, we get +  FY = N1 – 107.2 cos 10 = 0; N1 = 105.6 lb +  FX = P – 107.2 sin 10 – 0.3 N1 = 0; P = 50.3 lb Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5

Answers: 1. C CONCEPT QUIZ 1. Determine the direction of the friction force on object B at the contact point between A and B if P > 0. A)  B)  C) D) Answers: 1. C Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5

GROUP PROBLEM SOLVING 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 (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.

GROUP PROBLEM SOLVING (continued) Case a: +  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) 20º 30 lb 0.6 N P N Case b:  +  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 Case b has the lowest P and will occur first. A Block D weighing 5.81 lb will cause the block B to slide over the block A.

Answers: 1.A 2.B ATTENTION QUIZ 1. 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 W Answers: 1.A 2.B Statics:The Next Generation (2nd Ed.) Mehta, Danielson, & Berg Lecture Notes for Sections 8.3,8.5