#### Transcript PPTX

Clicker Question Room Frequency BA A glider (Yoda) is gliding along an air track at constant speed. There is no friction and no air resistance. What can you say about the net force (total force) on the glider? A) The net force is zero. B) The net force is non-zero and is in the direction of motion. C) The net force is non-zero and is in the direction opposite the motion. D) The net force is non-zero and is perpendicular to the motion. Velocity is constant, thus acceleration is zero. F = ma = 0. So, net force is zero. 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… • Reading in Chapter 4.1-4.6 (Forces and Force diagrams) • Exam #1 grades posted on CULearn Solutions for version #1 posted on CULearn More details at “Exam Information” link Newton’s Three Laws Newton’s Laws (as stated) hold in an “Inertial Reference Frame” (non-rotating / non-accelerating): I. “Law of Inertia”: Every object continues in its state of rest, or uniform motion (constant velocity in a straight line), as long as no net force acts on it. II. Acceleration is proportional to force and inversely proportional to mass. Famous F net ma equation. III.Whenever one object exerts a force on a second object, the second exerts an equal force in the opposite direction on the first. F AB F BA Newton’s Second Law Acceleration is proportional to force and inversely proportional to mass. F net ma Knowing the forces acting on an object of mass (m), we can calculate the accelerations. Then we can use these accelerations to calculate the kinematics (motion). Note that the direction of the acceleration is the same as the direction of the net force vector. The unit of force in the SI system is the Newton (N). Note that the pound is a unit of force, not of mass, and can therefore be equated to Newtons but not to kilograms. 1 Newton ~ 0.22 pounds. Force is a vector, so Newton’s 2nd Law is a vector equation: F ma Fx max Fy may All Kinds of Forces Pulling / Pushing Friction Gravity Air Resistance / Drag Springs Lots of others too (electric, magnetic, etc.) Clicker Question Room Frequency BA Forces add like vectors, not like scalars. Consider two forces, F1 and F 2, both of magnitude 1 N. One points to the right and other straight up. What’s the magnitude and direction of the net force? A) B) C) D) E) 2 N, straight up. √2 N, straight up. 0 N. √2 N, up and right. -√2 N, left. F2 Fnet F1 F 2 √2 N 1N 1N F1 Newton’s Third Law Whenever one object exerts a force on a second object, the second object exerts an equal and opposite direction force on the first. F AB F BA Room Frequency BA Clicker Question A moving van collides with a sports car in a high-speed head-on collision. Crash! M F truck F car m During the impact, the truck exerts a force with magnitude Ftruck on the car and the car exerts a force with magnitude Fcar on the truck. Which of the following statements about these forces is true: A) The magnitude of the force exerted by the truck on the car is the same size as the magnitude of the force exerted by the car on the truck: Ftruck = Fcar B) Ftruck > Fcar C) Ftruck < Fcar Guaranteed by Newton’s Third Law Room Frequency BA Clicker Question A moving van collides with a sports car in a high-speed head-on collision. Crash! M F truck F car m During the collision, the imposed forces cause the truck and the car to undergo accelerations with magnitudes atruck and acar. What is the relationship between atruck and acar? A) atruck > acar B) acar > atruck Newton’s Second Law: F = ma C) atruck = acar D) Indeterminate from information given. Free-Body Diagram Draw all forces for each object (truck and car). Consider only horizontal direction. m F truck CAR F car Ftruck = m acar Force of truck acting on the car Fcar = M atruck M TRUCK Force of car acting on the truck | Ftruck || Fcar | By Newton’s Third Law (and with opposite directions) ma car Matruck acar M atruck m Demonstration with Skateboard Consider an Everyday Occurrence Assume you are in a tie with your sumo-wrestler friend. Draw the free-body diagrams (i.e. forces) acting on you. F (tension in rope) F (pulling on the rope)? F (friction with floor) Consider an Everyday Occurrence Draw the free-body diagrams (i.e. forces) acting on you. F (tension in rope) F (Normal) F (friction with floor) F (gravity) Any other forces acting on you? Room Frequency BA Clicker Question What is the direction of the friction force acting on the sumo wrestler’s feet? A) Left B) Down C) Right D) Up F (tension in rope) F (Normal) F (friction with floor) F (gravity) We often draw free-body diagrams schematically on you) F ( tension pulling F ( tension pulling on Sumo) m M N = “normal force” N F friction F friction Fgr mg Fgr M g = force of gravity Be very careful to indicate on which object the force is acting (not who does the acting). Point-Like Objects We often treat objects as point-like with all forces acting on that point. Reasonable for rigid (non-deformable) objects and ignoring rotations. Since you are not moving…. No acceleration in x or y directions N F friction m F ( tension) F mg y x Label Coordinate Axes Net force in x must be zero. F (friction) F (tension) 0 Net force in y must be zero. F (gravity) F ( Normal) 0 Slip-n-Slide Part I – going down the slide Physics approximation…. 1) You are a small rectangular box (that we will treat as a point-like object anyway). 2) The slope of the slide is constant (angle = q) How large a net force will you feel down the slope? What is the acceleration down the slope? How fast will you be going after traveling down the slope for some time t? 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.