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Fri Nov 8 You may copy this down AFTER the test, if you’d like!

Fri Nov 8 You may copy this down AFTER the test, if you’d like!. Projectiles unit test/quiz MAKE SURE YOUR CALCULATOR IS IN DEGREES! AFTER test: Turn in anything without a grade-mark (…???)

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Fri Nov 8 You may copy this down AFTER the test, if you’d like!

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  1. Fri Nov 8 You may copy this down AFTER the test, if you’d like! • Projectiles unit test/quiz • MAKE SURE YOUR CALCULATOR IS IN DEGREES! • AFTER test: Turn in anything without a grade-mark (…???) • Asst:a) GUIDED Notes pp 88-107 & 126-132; this means you have a 2-sided worksheet to fill out as you read the text…. • Sure, you could always copy someone else’s Guided Notes, and no, we’d probably never know, but if you do so you would be missing an IMPORTANT LEARNING OPPORTUNITY about our next unit & all of the concepts we will be covering!

  2. 2014 Possible additional asst:SKIP 2013 find a website with a video clip and/or a demo for EACH of Newton’s 3 laws (in English!); email me the URLS of those sites; more points the better the site AND more points the fewer students who find that particular site 2014 – minus 2 days because of staff development! Go back & look at 2012 for better agendas when not so rushed!

  3. Tues Nov 12(no school Monday, Veteran's Day) • Binder due today??? • Lecture Notes – Newton’s 3 laws • Asst: (a) Wksheet – front side only Backside assigned tomorrow; note on back part C cross-out “action-reaction pairs” leaving only “Equal but opposite forces” (b) Consider starting back-side (c) PRINT OFF STUDY GUIDE! – We will do this for you in 2013 Unit Test = Thursday November 21st

  4. Newton’s 1st Law: “A body in motion tends to stay in motion” (unless a force acts on it) • Demo 1: sliding book • … versus throwing an object in outer space • Demo 2: tomato on knife • Demo 3: brick with person on “Human Dynamics Cart” • Video clip: tablecloth trick(start at 2:40) • Video clip: eggs, pizza pan, … • Demo 4: card & marble • Demo 5: Ballistics car - if didn’t do before • Demo 6: wheelie-chair (draw pictures in notes – see next slide)

  5. This arrow is twice as big because he/she has more mass!) Wheelie chair:(A “system” in motion must keep that same total motion) + = zero movement ? + =

  6. F = -16N a = -20 m/s2 Newton’s 2nd Law: “F=ma” • Units: [N] = [kg] ·[m/s2] • so 1 Newton = 1 kg x 1 m/s2 • Ex 1: If a=2.0 m/s2, m=10. kg, F=? N • Vernier LabPro w/ accel & force, mass = ? kg • Ex 2: If a 3000.-kg car starts from rest and accelerates to 100. km/hr in 10.0 seconds, what is the force the engine exerts on the car? (3 steps!) • F = 8340 N

  7. specific example of Newton’s 2nd Law: mass versus weight The amount of matter that an object has kilograms (kg) m The force of gravity on an object Newtons (N) Fg

  8. MORE about: mass versus weight • Formula that connects those two things (7th pt on test) • if F = ma, then • Fg = mg (“g” is the “absolute value of the accel of gravity of a planet/moon”) • I mass 65 kg, calculate my weight on .... • Earth • The moon (g = 1.62 m/s2) • Jupiter (g = 2.53*9.8 = 24.794 m/s2) • outerspace • 8th point on test question: how can I lose weight withOUT losing mass? Note – For HW, use values given on HW! (Values for g’s on ditto are listed as something different.)

  9. Newton’s 3rd Law: “Equal and opposite forces” • Demo 1: Brick and skateboard: Person pushes brick left, so brick pushes person _________. • Demo 2: Rolling chairs: Small person pushes big person. But… • Demo 3: Rolling chair against wall: Person pushes wall. But… • Demo 3: Everybody jump up at the same time. Do you feel the Earth move? (1010 people on earth, each mass 102 kg; Earth 1024 kg) • Question: Why do YOU move? You “convince” Earth to push you …You push down on the Earth, so Earth pushes up on you! • Question: How does a person climb a rope in gym class? • Example N’s 3rd Law w/ math: F = F (equal & opposite forces) so: F1 = F2 m1a1 = m2a2 60 kg · 2.0 m/s2 = 80 kg · ???

  10. SKIP some lines so we can finish the lecture notes!!!!Wed – November 13 • Finish lecture notes on N’s 3rd Law • Questions from HW? • Experiment – Newton's 2nd Law/Forces • Fix catapult if must (2 rubberbands + 3 paperclips) • Calibrate catapult in Newtons – USE PENCIL ONLY!!!!!! • Get a penny “launcher” baggie (2 pennies attached to launcher always + 4 additional pennies) • Record & collect BOTH data sets – 2 data tables on piece of paper • Erase calibration marks; detach all but 2 pennies; put stuff away • start HW… • Asst: (a) Newton’s wksht back side (both sides due tomorrow! • Asst (b) Email question #’s, there will be no time in class to answer! 2012: HAND-write first 2 paragraphs of Introduction of Experiment Write-up (see back of lab handout!) Unit Test = Thursday November 21st

  11. Newton's 1st Law on the test: • What is it? • Examples/demos/etc Newton's 2nd Law on the test: • What is it? • Pennies experiment (18 points – see study guide) • chart above (6 points) • Fg = mg (the “formula that connects mass & weight” =1pt) • How can I lose weight withOUT losing mass (1 point for any example) Newton's 3rd Law on the test: • What is it? • “Locomotion” (moving, jumping, climbing, etc) of any “animal” • F1 = F2 (equal & opposite forces) • so: m1a1 = m2a2 • 60 kg · 2.0 m/s2 = 80 kg · ??? (solve for a2) • (This was like the last 3 or 4 problems on Newton’s wksht!)

  12. F=ma (“Pennies”) experiment = 18 points – see below ... • In a nutshell, what did you do? (Hint – nutshell’s can not contain more than 4 words, and are what 5 years olds like to hear. NOT what you looked for! It’s the procedure, not the purpose!)

  13. Experiment Write-up so far(NOT doing write-up 2013 due to lack of time) • Introduction • Paragraph 1: start off with a defn of a force, examples, units, etc. Then say (in your own words!) something like: “In the 1600’s Isaac Newton developed three laws about forces”. • Paragraphs 2, 3, and 4: Three big paragraphs, one each about Newton’s 3 Laws and their implications / ramifications. (Use your textbook, book notes, lecture notes, and the “Newton’s Laws of Motion” worksheet!) • Purpose • The purpose of this experiment was to prove Netwon’s second law, F = ma. • Materials • Vertical, uncapitalized list of: manila file folder, 3 paper clips, 2 rubber bands, masking tape, spring scale, pennies, metersticks

  14. Experiment “Results” this is shorthand, obviously; re-word it yourself IN COMPLETE PROPER-GRAMMAR SENTENCES; see email !!!! • As the force was increased, d up. Since d proportional to a, a also up. This showsthat … a direct relationship. • As the mass was increased, d down. Since d proportional to a, a also down. This showsthat … an indirect relationship.

  15. Thurs Nov 14 • QUICKLY log in, go to Shared, Out folder • Find F=ma 2013 excel file & fill in data. Email home to print. • Quickly Discuss Test that is next THURSDAY!!!!… • See back of Study Guide handed out yesterday. • Questions 1-3 = N’s 3 Laws • Questions 4-9 = 1 of each of 6 types of FVD we are learning the next 3 class days • You can NOT memorize the equations! You must UNDERSTAND where they come from. • If you need to do MORE than the number that are assigned, do so! • Lecture Notes – FVD type 1 (elevators) • Asst (a): FVD 1 #1-11 (#7 we will grade like a quiz tomorrow; do it without looking at notes and/or previous work!) (b): 2013: Hand-write R&C (see email) on printed Data sheet (c): 2013: Check email about Newton’s wrksht questions you or others may have had! 2012: HAND-write Next 2 paragraphs of intro in write-up Unit Test = Thursday November 21st

  16. FT: Force of Tension Fg : Force of Gravity or “Weight” FT Fg Force Vector Diagram #1: An elevator is ascending or descending and you want to know what is the tension in the rope.

  17. Speeding Up Slowing Down Going Up Acceleration Acceleration Going Down Acceleration Acceleration First you will need to find the direction of the acceleration.To do so use the chart below:

  18. FT a Fg When the acceleration is upwards use the following equation… • Fy = FT – Fg = ma åFy is the “sum of the forces” in the y direction FT is the force of tension in Newtons Fg is the weight or force of gravity in Newtons m is the mass of the object in kilograms a is the acceleration of the object in m/s2

  19. FT a Fg But, when the acceleration is downwards use the following equation… • Fy = Fg – FT = ma

  20. Other things you need to remember: To find the mass of the object when given the weight in Newtons, divide by9.8 m/s2. To find the weight of the object when given the mass in kg’s, multiply by9.8 m/s2.

  21. Now let’s see an example… FT = ? Example 1: There is an elevator that is going upwards and slowing down at a rate of 3 m/s2. It masses 200 kg’s. What is the tension in the cable from which it is hanging? a = 3m/s2 ** Fg= 200 x 9.8 = 1960N ** (use the acceleration table to decide on the direction of this arrow)

  22. Plug what you know into the correct equation and solve for the object’s force of tension: Use the following equation: åFy = Fg – FT = ma where: Fg = 200 kg x 9.8 m/s2 = 1960 N FT = ? m = 200 kg a = 3 m/s2 8 points total on test åFy= 1960 – FT = 200 x 3 – FT = 600 – 1960 FT = 1360 N NOW START YOUR HW!

  23. Do #1 of HW together • Get 2 & 3 done (at least) before you go • Email home Experiment data from excel if you didn’t yet

  24. Fri Nov 15 Mistake on email about R&C; will resend today • Grade “quiz” #7 in red pen; put score on top ( /8) in red; turn in HW; turn in (& Guided Notes too!) • Lecture notes FVD type#2 - Signs • A sign is hung by two ropes. The left rope makes an angle of 38.0 degrees from the vertical. The right rope makes an angle of 15 degrees from the horizontal. The sign weighs 2727 Newtons. Find the tensions in the two ropes. • Lecture notes FVD type#4 - Friction on flat surfaces • A 421.4-Newton suitcase is pulled along the floor with a force of 150. Newtons at an angle of 25.0 degrees from the horizontal. If the coefficient of friction between the floor and the suitcase is equal to 0.100, find the acceleration of the suitcase. • Asst (a): FVD 2 #4-11 (#11 we will grade like a quiz tomorrow) • Asst (b): FVD 4 #3-10 (#7 we will grade like a quiz tomorrow) • LAST CALL: email questions on Newton’s Laws wrksht! • Asst (b) 2012: HAND-write Write-up: all of Introduction should be done; do Purpose & Materials Unit Test = Thursday November 21st

  25. FT = ? a=0.531 m/s2 Fg= 87 x 9.8 = _____ N HW “QUIZ” FVD 1 #7:An elevator with a mass of 87.0 kg is going up & slowing down with an acceleration of 0.531 m/s2. Find the tension in the rope: åFy= 87*9.8 – FT = 87 x 0.531 FT = ____ N

  26. FTL FTR FTL sin  FTR sin    FTL cos  FTR cos  Fg LECTURE NOTES - FVD type #2: A sign is hung by two ropes. The left rope makes an angle of 38.0 degrees from the vertical. The right rope makes an angle of 15 degrees from the horizontal. The sign weighs 2727 Newtons. Find the tensions in the two ropes. So  = 90- 38 = ____ , and  = 15. First sum the forces in the x direction: • Fx = FTRcos 15 – FTLcos 52 = 0 Then sum the forces in the y direction: • Fy = FTRsin 15 + FTLsin 52 – 2727 = 0

  27. constants on other side of equation: 0 2727 FTR FTL x eq: cos15 -cos52 y eq: sin15 sin52 -1 ] ] [ [ LECTURE NOTES - FVD type #2 continued…. • Fx = FTRcos 15 – FTLcos 52 = 0 • Fy = FTRsin 15 + FTLsin 52 – 2727 = 0 Use a matrix to solve: Since FTR was the first variable in the matrix, then FTR _____ N and FTL  _____ N (since FTL was the second variable in the matrix.).

  28. FN FT = 150 N 25 Ff Fg = 421.4 N LECTURE NOTES: FVD type#4: A 421.4-Newton suitcase is pulled along the floor with a force of 150. Newtons at an angle of 25.0 degrees from the horizontal. If the coefficient of friction between the floor and the suitcase is equal to 0.100, find the acceleration of the suitcase. First sum the forces in the y direction: • Fy = FTsin 25 + FN– Fg = 0 Solve for FN Then sum the forces in the x direction: • Fx = FTcos 25 – 0.100FN= (421.4 / 9.8) a Solve for a Note new forces of friction and “normal”… Normal is an old math word that means “perpendicular”. It is always perpendicular to the surface. It is the support force that a surface provides, and is only there when the object is on the ground. Friction depends on the normal force. It has the equation Ff =  * FN . The  is pronounced “mu”, and is called the “coefficient of friction”.

  29. Mon Nov 18 • Grade type2 HW “quiz” #11 in red pen; put score on top ( /19) in red & turn in HW • Grade type4 HW “quiz” #7 in red pen; put score on top ( /16) in red & turn in HW • Lecture notes FVD type #3 – pulleys in the ceiling • Two masses are being hung by a rope that is strung over a pulley hanging from the ceiling. The left object weighs 441 N, and the right one masses 55 kg. The heavier object is allowed to fall down. What is the acceleration of the objects, and the tension in the rope? • Lecture notes FVD type #5 - inclined planes • A 200.-gram box is on a plane inclined at 65.0° with a 0.500 coefficient of friction. What is the normal force applied by the inclined plane and the box’s acceleration? • Asst (a): FVD 3 #5-11 (go over #10 tomorrow like a quiz) • Asst (b): FVD 5 #3-9 (go over #8 tomorrow like a quiz) • Asst (c, review): FVD’s 1,2,4 #14 • Asst (b) 2012: HAND-write Write-up - Procedure Unit Test = Thursday November 21st

  30. FTL FTR FTL sin 25 FTR sin 41   FTL cos 25 FTR cos 41 Fg constants on other side of equation: 0 832 FTR FTL x eq: cos41 -cos25 y eq: sin41 sin25 -1 ] ] [ [ HW “QUIZ” FVD 2 #11:An 84.9-kg sign is suspended by two ropes. The left rope makes an angle of 65 from the vertical, while the right makes an angle of 41 from the horizontal. Find the tension in the two ropes Fx = FTRcos 41 – FTLcos 25 = 0 Fy = FTRsin 41 + FTLsin 25 – 84.9*9.8 = 0 5 pts FVD; 4 pts + 5 pts for equations; 3 pts for matrices FTR = _____ N (with THREE sig figs!) FTL = _____ N (with THREE sig figs!) 2 pts for answers written correctly!

  31. FN FT = 350 N 20 Ff = .4 FN Fg = 735 N HW “QUIZ” FVD 4 #7:A 735-Newton suitcase is pulled along the floor with a force of 350. Newtons at an angle of 20.0 degrees from the horizontal. If the coefficient of friction between the floor and the suitcase is equal to 0.400, find the acceleration of the suitcase. 5 pts FVD • Fy = 350sin 20 + FN– 735 = 0 FN = ____N • Fx = 350cos 20 – 0.400FN= (735 / 9.8) a a = ____ m/s2 5 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel

  32. LECTURE NOTES: FVD type#3: Two masses are being hung by a rope that is strung over a pulley hanging from the ceiling. The left object weighs 441 N, and the right one masses 55 kg. The heavier object is allowed to fall down. What is the acceleration of the objects, and the tension in the rope? You don’t need to draw the pulley! The givens for this problem are the masses or weights of the objects. You must determine which object masses more (or weighs more) to determine the two directions of acceleration. FT a?? FT FgR a?? FgL

  33. constants on other side of equation: -539 441 FT a right eq: -1 -55 left eq: 1 -45 -1 ] [ ] [ LECTURE NOTES: FVD type#3: Two masses are being hung by a rope that is strung over a pulley hanging from the ceiling. The left object weighs 441 N, and the right one masses 55 kg. The heavier object is allowed to fall down. What is the acceleration of the objects, and the tension in the rope? You don’t need to draw the pulley! First sum the forces in the y direction on the left: • FyL = FT – FgL = ma or: • FyL = FT – 441 =45a Then sum the forces in the y direction on the right: FyR = FgR– FT= ma or: • FyR = 539 – FT= 55a Use a matrix to solve …. FT a FT FgR = 539 N a FgL = 441 N Since FT was the first variable in the matrix, then FT ____ N and a  ____ m/s2 (since a was the second variable in the matrix).

  34. FN +y =.5 FN Ff +x a 65 65 Fg = 200*9.8 N Lecture Notes FVD type#5: A 200.-kg box is on a plane inclined at 65.0° with a 0.500 coefficient of friction. What is the normal force applied by the inclined plane and the box’s acceleration? Remember how we've been saying this unit that our direction of acceleration dictates our axes? Well this problem takes that fact to the extreme! First sum the forces in the “y” direction: • Fy = FN–Fgcos 65 = 0 (note Fg = 200*9.8) Solve for FN Then sum the forces in the “x” direction: • Fx = Fgsin 65 – 0.500FN= 200 a Solve for a

  35. Tues Nov 19 • Catch-up day on FVD types 1-5!! • Grade type3 HW “quiz” #10 in red pen; put score on top ( /18) in red & turn in HW • Grade type5 HW “quiz” #8 in red pen; put score on top ( /15) in red & turn in HW • VERY BRIEF INTRODUCTION into FVD 6 … • An object is on top of a (flat, level) table. This object masses 157 kg and has a coefficient of friction of 0.500 with the table. There is a pulley on the edge of the table, with a rope over it, attached to a “hanging mass” of 999.6 Newtons. What will be the acceleration of the objects over the edge? • This problem will be extra credit on Thursday’s test; you will have to do #4 on the Review sheet + 2 others for HW to be able to get the XC points on the test. (due day of the test) • Asst A: Review worksheet 1-6, skip xc4 *****IT IS VERY IMPORTANT YOU DO THESE WITHOUT LOOKING AT ANY NOTES!!!! We will grade this like the test tomorrow in class!!!! • Asst B: Do #15 & #16 from FVD’s 1-5 Unit Test = Thursday November 21st

  36. constants on other side of equation: 2205 -2401 FT a left eq: 1 -225 right eq: -1 -245 -1 ] [ ] [ HW “QUIZ” FVD 3 #10:Two masses are being hung by a rope that is strung over a pulley hanging from the ceiling. The left object weighs 2205 N, and the right one masses 245 kg. The heavier object is allowed to fall down. What is the acceleration of the objects, and the tension in the rope? 5 pts FVD Remember you don’t have to draw the pulley or extra string!) 4 pts each equation (total 8 pts): • FyL = FT – 2205 =225a • FyR = 2401 – FT= 245a Use a matrix to solve …. (3 pts) FT a FT FgR = 2401 N a FgL = 2205 N FT= ____ N and a = ____ m/s2 2 pts for answers!

  37. FN +y = .1FN Ff +x a 40 40 Fg = 2156 N Quiz FVD 5 #8:(15 pts)A 2156-Newton box is on a plane inclined at 40.0° with the horizontal. The coefficient of friction between the box and the inclined plane is 0.100 What is the box’s acceleration? 5 pts FVD (includes 1 pt for EITHER drawing accel or weird axes) You don’t have to draw the incline. • Fy = FN–2156cos 40 = 0 FN = 1652 N • Fx = 2156sin 40 – 0.10FN= 220 a a = 5.55 m/s2 4 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel

  38. Lecture Notes FVD #6: An object is on top of a (flat, level) table. This object masses 157 kg and has a coefficient of friction of 0.500 with the table. There is a pulley on the edge of the table, with a rope over it, attached to a “hanging mass” of 999.6 Newtons. What will be the acceleration of the objects over the edge? FN FT FT Ff = .5FN • Note, this is the first problem on the HW packet, FVD type 6. • Do left (on table top) mass in vertical direction first to get normal force: Hanging Mass FgL = 157*9.8 N FgR = 999.6 N Left mass on table: FyL = FN – 157*9.8 = 0 so FN = 1538.6N

  39. Matrices: FT a # -1 FxL FyR 1-157 -1999.6/9.8 .5*1538.6 999.6 Lecture Notes FVD #6 continued: EQUATIONS: Table: FyT = FN – 157*9.8 = 0 so FN = 1538.6N Left mass on table: FxL = – (.5)(1538.6) + FT = 157a Right hanging mass: FyR = 999.6 – FT = (999.6/9.8)a Answers to the Problem: FT = 909 N a = .889 m/s2

  40. COPY DOWN BOTH DAYS! Test is LONG! Come early & be VERY prepared! Remember our rules about test attendance! Wed November 20 (collab) • Go over Review Sheet in red pen; put score on top ( / 99 ) in red • Talk about important Guided Notes • Look at Study Guide • Bill Nye “Friction” ?? • Asst: (a) Do #19 & #20 from FVD’s 1-5 (b) STUDY N’s 3 Laws from STUDY GUIDE! (c) XC work for FVD type 6 due tomorrow too Thurs 11/21 • Forces Test • Test is LONG! Come early & be VERY prepared! • Asst: Guided Notes worksheet on Momentum, pages 228-245

  41. “Guided Notes” of import • #10-13 actually refer to the F=ma Experiment (“penny lab”) relationships: • #10/11: bigger force, MORE accel  direct! • #12/13: bigger mass, LESS accel  indirect! • Recall….. • Highlight the bolded words in … • #5, 9, 14, 18, 19, 22, 24, 28, and 33. • These are all on your test! Questions?

  42. GO OVER STUDY GUIDE TOGETHER Test is LONG! Come very early & be VERY prepared! Be able to do all 6 FVD’s in about 35 minutes total!! (see next slide)

  43. Thurs 11/ 21 COPY DOWN BOTH DAYS! Test is LONG! Come early & be VERY prepared! Remember our rules about test attendance! • Forces Test • Test is LONG! Come early & be VERY prepared! • Asst: Guided Notes on Momentum, pages 228-245

  44. Things used in 2012 and before….

  45. Review QUIZYou have exactly 10 minutes! • An elevator is going down but speeding up at a rate of 9.7999 m/s2. The elevator masses 10,000. kg. Find the tension in the cable. (8 pts) • A 120.0 kg sign is suspended by 2 ropes. The left rope makes an angle of 50.0 degrees with the horizontal, and the right rope makes an angle of zero degrees with the horizontal. Find the tensions in the two ropes. (16 pts) ANSWERS ON NEXT SLIDE (Get out a red pen!!!)

  46. FT = ? a=9.7999 m/s2 Fg= 98,000 N 3 pts FVD • An elevator is going down but speeding up at a rate of 9.7999 m/s2. The elevator masses 10,000. kg. Find the tension in the cable. 4 pts equation åFy= 98000 – FT = 10,000 x 9.7999 1 pt for answer FT = 1.0000 N Why is it so small? What’s almost happening?

  47. constants on other side of equation: 0 1176 FTR FTL x eq: 1 -cos50 y eq: 0 sin50 -1 ] ] [ [ You can have cos 0 and sin 0 in that FTR column too FTL 4 pts for x equation: Fx = FTR – FTLcos 50 = 0 OR Fx = FTRcos 0 – FTLcos 50 = 0 4 pts for y equation: Fy = FTLsin 50– 1176 = 0 OR Fy = FTRsin 0 + FTLsin 50– 1176 = 0 4 pts FVD 2. A 120.0 kg sign is suspended by 2 ropes. The left rope makes an angle of 50.0 degrees with the horizontal, and the right rope makes an angle of zero degrees with the horizontal. Find the tensions in the two ropes. FTR 50 Fg = 1176 N You actually don’t need a matrix to solve this, but you can use one. (+2 pts for math work or matrix) FTL = 1540 N (3 sig figs!) FTR = 987 N (3 sig figs!) 2 pts for answers written correctly!

  48. 3. A pulley is attached to the ceiling. There are 2 masses attached to either end of a rope strung around the pulley. The one on the left masses 245 kg, and the one on the right weighs 490. N. Find the tension in the rope & the acceleration of the objects. ***NOT GIVEN TODAY! THINGS TO REMEMBER about pulleys in ceilings: • No subscript on FT, but need left/right subscripts on Fg • Compare weight to weight to see which way it will accelerate; draw in acceleration vectors off to sides • Because they have different directions of accelerations, the equations will be “opposite” of each other • Need additional left/right subscript on Fy equations • Make your equations in “matrix form” (variables on left side; constants on right) and solve!

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