Download Engineering Analysis: Determining Forces on Wedges and Belts and more Lecture notes Design Patterns in PDF only on Docsity! 1 Lecture 27 ENGR-1100 Introduction to Engineering Analysis WEDGES AND FRICTIONAL FORCES ON FLAT BELTS In-Class Activities: • Reading Quiz • Applications • Analysis of a Wedge • Analysis of a Belt • Concept Quiz • Group Problem Solving • Attention Quiz Today’s Objectives: Students will be able to: a) Determine the forces on a wedge. b) Determine tension in a belt. 2 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? Wedges are used to adjust the elevation or provide stability for heavy objects such as this large steel pipe. APPLICATIONS (continued) How can we decide if the belts will function properly, i.e., without slipping or breaking? Belt drives are commonly used for transmitting the torque developed by a motor to a wheel attached to a pump, fan or blower. 5 ANALYSIS OF A WEDGE (continued) 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, then the wedge is self-locking, i.e., it will not come out on its own. BELT ANALYSIS Detailed analysis (please refer to your textbook) shows that T2 = T1 e μ β where μ is the coefficient of static friction between the belt and the surface. Be sure to use radians when using this formula!! If the belt slips or is just about to slip, then T2 must be larger than T1 and the motion resisting friction forces. Hence, T2 must be greater than T1. Consider a flat belt passing over a fixed curved surface with the total angle of contact equal to β radians. 6 EXAMPLE 1. Draw a FBD of the crate. Why do the crate first? 2. Draw a FBD of the wedge. 3. Apply the E-of-E to the crate. 4. Apply the E-of-E to wedge. Given: The crate weighs 300 lb and μS at 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 (continued) →+ FX = NB – 0.3NC = 0 −+ 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 The FBDs of crate and wedge are shown in the figures. Applying the E-of-E to the crate, we get NB 300 lb FB=0.3NB FC=0.3NC NC 15º 15º FC=0.3NC NC FD=0.3ND ND P FBD of Crate FBD of Wedge 7 EXAMPLE (continued) NB = 82.6 lb 300 lb FB=0.3NB FC=0.3NC NC = 275 lb 15º 15º FC=0.3NC NC FD=0.3ND ND P FBD of Crate FBD of Wedge 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 READING QUIZ W 1. A wedge allows a ______ force P to lift a _________ weight W. A) (large, large) B) (small, large) C) (small, small) D) (large, small) 2. Considering friction forces and the indicated motion of the belt, how are belt tensions T1 and T2 related? A) T1 > T2 B) T1 = T2 C) T1 < T2 D) T1 = T2 eμ
