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About Midterm Exam 1

About Midterm Exam 1. Lect. 8, Chapter 4: Newton’s Laws Today: The 3 rd Law; and applications. When and where Thurs Feb. 17 th 5:45-7:00 pm Rooms: See course webpage. Be sure report to your TA’s room Your TA will give a review during the discussion session next week. Format

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About Midterm Exam 1

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  1. About Midterm Exam 1 Lect. 8, Chapter 4: Newton’s LawsToday: The 3rd Law; and applications • When and where • Thurs Feb. 17th 5:45-7:00 pm • Rooms: See course webpage. Be sure report to your TA’s room • Your TA will give a review during the discussion session next week. • Format • Closed book, 20 multiple-choices questions (consult with practice exam) • 1page 8x11 formula sheet allowed, must be self prepared, no photo copying/download-printing of solutions, lecture slides, etc. • Bring a calculator (but no computer). Only basic calculation functionality can be used. Bring a 2B pencil for Scantron. • Fill in your ID and section # ! • Special requests: • Email me at than@hep.wisc.edu, w/ valid excuse for approval. • One alternative exam: 3:30pm – 4:45pm, Thurs Feb.17, in our 201 lab rm.

  2. Important technique:Free-Body Diagrams Free-body diagrams are diagrams of the forces on an object. First, isolate the object in question. Then, identify the individual forces on it. Forces acting on sled: 1. The gravitational force on the sled-rope 2. The contact force exerted by the ice on the runners. (Without friction, the contact force is directed normal to the ice.) 3. The contact force exerted by the dog on the rope. (Since the sled remains on the ice, the y-components of the force sum to zero.)

  3. Example: Hanging a Picture Free-body diagram for the picture Picture is not accelerating Forces on picture sum to zero.

  4. Question 2 N mg Consider a person standing in an elevator that is accelerating upward. The upward normal force N exerted by the elevator floor on the person is a) larger than b) identical to c) less than the downward weight W of the person. Person is accelerating upwards - net upwards force is non zero N – mg = ma

  5. Newton’s Third Law Ffingerbox Fboxfinger • For every action, there is an equal and opposite reaction. • Finger pushes on box • Ffingerbox = force exerted on box by finger • Box pushes on finger • Fboxfinger = force exerted on finger by box • Third Law: • Fboxfinger = -Ffingerbox

  6. Newton’s Third Law… Fw,m Fm,w Fm,f Ff,m • FA ,B = - FB ,A. is true for all types of forces Gravity Fm,f = -mg Whenever one body exerts a force on a second body, the first body experiences a force that is equal in magnitude and opposite in direction to the one it exerts – either stationary or in motion.

  7. System of Interest Newton’s second law, combined with third law, applies to every system of interest. f - All forces opposing the motion System 1: Acceleration of the professor and the cart System 2: Force the professor exerts on the cart

  8. The Horse Before the Cart Free-body diagram for the cart:

  9. Example: Forces on an object on an incline Coordinates: Most conveniently, Parallel & perpendicular (normal) Normal (no motion): Fn = Fgy = mg cosθ Parallel (moving): Fgx – Ffrict = mg sinθ – Ffrict = ma

  10. Summary: • Newton’s First Law: • Themotion (velocity)of an object does not change unless it is acted on by a net force • Newton’s Second Law: • Fnet = ma • Newton’s Third Law: For every action, there is an equal and opposite reaction. • Fa,b = -Fb,a • When considering action reaction pairs it’s important to identify which forces act on a given object. Physics 103, Fall 2009, U.Wisconsin

  11. Solving Problems • Identify force using Free Body Diagram • This is the most important step! • Set up axes and origin • x and y • Write Fnet=ma for each axis (components of forces) • Calculate acceleration components • Setup kinematic equations • Solve! • Strong suggestion: • work problem algebraically (using symbols) • plug in numbers only at the end

  12. Example: Pulley (Atwood Machine) T T a a W1 W2 What is the tension in the string (assume the pulley is frictionless)? A) T<mg B) T=mg C) mg<T<2mg D) T=2mg Moving together, acceleration a is same for both blocks. T With 2 free-body diagrams, set up 2 equations: T a a • T – m1g = m1a • m2g – T = m2a m2=2m m1=m • + (2)  a = (m2-m1)g/(m2+m1), • Thus, T = 2m1m2g/(m2+m1) • Note: • Sige of a matters; T is only a magnitude. • If not frictionless, then T not the same.

  13. Building a Space Station You are an astronaut constructing a space station, and you push on a box of mass m1 with force FA1 The box is in direct contact with a second box of mass m2 . (a) What is the acceleration of the boxes? • FA1 - F21 = m1a • F12 = m2a • a = FA1 /(m1 + m2), as expected (b) What is the magnitude of the force each box exerts on the other? FA1 = (m1 + m2) a F12 = m2a = F21 (c) Generalization: What if for n boxes ? FA1 = (m1 + m2+ … mn) a; F2 = (m2+ … mn) a; F3 = (m3+ … mn) a … … Fn = mn a

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