Transcript PPTX
• • • • MedAdvance Internship Program MedAdvance is an internship program based out of the Scientific Education and Research Institute in Thornton, CO. As a member you may lead and participate in Anatomy/Suture labs and Shadow Doctors. In addition to these opportunities, the Scientific Education and Research Institute is a medical device training facility that has doctors visit from all around the world to learn about new medical technology. This gives interns the opportunity to, not only meet health professionals from around the nation, but also to become familiar with the latest medical technology!!! MedAdvance is having an Open House on Thursday September 29th at 6-8:30pm at the Scientific Education and Research Institute. The address is 9005 Grant st. Thornton, CO 80229. If you have any more questions about the internship or would like to RSVP for the Open House, please feel free to contact [email protected]. Announcements • CAPA Set #5 due Friday at 10 pm • This week in Section Assignment #3: Forces, Newton’s Laws, Free-Body Diagrams Print out and bring the assignment to your section… • Complete reading all of Chapter 4 Step 1: Draw a free-body diagram Note that “Normal Force” is always Perpendicular (i.e. normal) to the surface. Fg = mg In this case, it is not in the opposite direction to the gravitational force. Step 2: Choose a coordinate system Fg = mg y x We could choose our “usual” axes. However, we know that we will have motion in both the x and y directions. If we pick these rotated axes, we know that the acceleration along y must be zero. Step 3: Write down the equations Σ Fx = m ax , Σ Fy = m ay Fg = mg Problem: Fg doesn’t point in a coordinate direction. Must break it into its x- and y-components. Room Frequency BA Clicker Question F(gravity)y F(gravity)x 90 – q a 90 |F(gravity)| = mg What is the angle a labeled above? A) a = 90 degrees B) a = q C) a = 90 – q D) Cannot be determined Step 3: Write down the equations Σ Fx = m ax , Σ Fy = m ay N Fx = mg sin(q) Fy = N – mg cos(q) Fg = mg We also know that ay = 0, there is no motion in that direction. Fy =may=0= N – mg cos(q) Step 4: Solve the Equations Fx ma x mg sin(q ) a x g sin(q ) Fy may N mg cos(q ) N a y g cos(q ) 0 m N mg cos(q ) As expected, maximum acceleration if straight down (q=90 degrees), ax = g. Recall x-y definition Room Frequency BA Clicker Question A mass m accelerates down a frictionless inclined plane. +y -mg cos θ θ +x θ - Which statement is true? A) N < mg 0 = Fy = N – mg cos θ B) N > mg N = mg cos θ < mg C) N = mg Clicker Question Room Frequency BA The driver of a car parked on a hill releases the brake and puts the car in neutral Let N = normal force F g = force of gravity Which of the following is true? a = acceleration of the car. A) A a, E N, B F g B) E N, D a,C F g C) E N, D Fg ,C a D) A a, B Fg , E N Room Frequency BA Clicker Question Two boxes with masses M and m are glued together. Force F is applied to Box 1. Force fc is the “contact force” that Box 1 exerts on Box 2. M fc F A)F = fc m Box 1 +x Box 2 * Frictionless surface B) F > fc C) F < fc D)Indeterminate from information given Draw the free-body diagram for each box! Box 1 F Box 2 fc m M m +x F M fc fc m What other forces are acting on either box? Gravitational force down on both boxes, and Normal force up on both boxes. However, since there is no vertical motion we sometimes do not bother to write them down. What about Newton’s Third Law? F M fc fc m Apply Σ Fx = ma. (ignore vertical forces, no vertical motion) F-fc = MaM fc = mam Since the boxes are fixed (glued) together aM = am aM=(F-fc)/M = am =fc/m Fm f c m f c M f c ( M m) Fm m fc F M m fc F Room Frequency BA Clicker Question Box 1 F M m fc F M m Box 2 fc m +x Assume Box 1 has a mass of 2 kg and Box 2 has a mass of 1 kg. If a force of 1 N is applied to Box 1, what is the contact force between the boxes? A) B) C) D) E) 1N ½N 1/3 N 2N 3N 1kg 1 m fc F 1N N M m 2kg 1kg 3 Room Frequency BA Clicker Question F Box 1 F sinθ θ F cosθ M Previous equation: Box 2 fc m m fc F M m m +x Assume a force F is applied to Box 1 at an angle θ from the horizontal. What will be the equation for the contact force fc? mM A) fc F tan q m m B) fc F sin q M m m C) fc F cosq M m m(M M ) D) fc F cosq The horizontal component of F produces horizontal motion.