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Fri Nov 8 You may copy this down AFTER the test, if you’d like! 1.Projectiles unit test/quiz MAKE SURE YOUR CALCULATOR IS IN DEGREES! 2.AFTER test: Turn in anything without a grade- mark (…???) 3.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!

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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!

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Tues Nov 12 (no school Monday, Veteran's Day) 1.Binder due today??? 2.Lecture Notes – Newton’s 3 laws 3.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

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Newton’s 1 st 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)tablecloth trick Video clip: eggs, pizza pan, …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)

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

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Newton’s 2 nd Law: “F=ma” Units: [N] = [kg] ·[m/s 2 ] so 1 Newton = 1 kg x 1 m/s 2 Ex 1: If a=2.0 m/s 2, 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 F = -16N a = -20 m/s 2

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

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MORE about: mass versus weight Formula that connects those two things (7 th pt on test) if F = ma, then F g = 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/s 2 ) –Jupiter (g = 2.53*9.8 = 24.794 m/s 2 ) –outerspace 8 th 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.)

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Newton’s 3 rd 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? (10 10 people on earth, each mass 10 2 kg; Earth 10 24 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: F 1 = F 2 m 1 a 1 = m 2 a 2 60 kg · 2.0 m/s 2 = 80 kg · ???

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SKIP some lines so we can finish the lecture notes!!!! Wed – November 13 1.Finish lecture notes on N’s 3 rd Law 2.Questions from HW? 3.Experiment – Newton's 2 nd Law/Forces a.Fix catapult if must (2 rubberbands + 3 paperclips) b.Calibrate catapult in Newtons – USE PENCIL ONLY!!!!!! c.Get a penny “launcher” baggie (2 pennies attached to launcher always + 4 additional pennies) d.Record & collect BOTH data sets – 2 data tables on piece of paper e.Erase calibration marks; detach all but 2 pennies; put stuff away f.start HW… 4.Asst: (a) Newton’s wksht back side (both sides due tomorrow! 5.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

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

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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!) mini-experiment 1 or 2 variable held constant variable YOU changed (independent) AND how did you CHANGE IT? variable changed as a result (dependent)** RELATIONSHIP between indep & dep variable (direct or indirect) **(Hint – think about the axes on the graphs, although we’ll say accel instead of distance!)...We didn’t actually measure accel, but distance – which is proportional to distance. Again, please answer this question in terms of acceleration, even though that’s not what we actually measured.

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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

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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 shows that … a direct relationship. As the mass was increased, d down. Since d proportional to a, a also down. This shows that … an indirect relationship.

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Thurs Nov 14 1.QUICKLY log in, go to Shared, Out folder Find F=ma 2013 excel file & fill in data. Email home to print. 2.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! 3.Lecture Notes – FVD type 1 (elevators) 4.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

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FgFg FTFT F T : Force of Tension F g : Force of Gravity or “Weight” Force Vector Diagram #1: An elevator is ascending or descending and you want to know what is the tension in the rope.

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

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When the acceleration is upwards use the following equation… F y = F T – F g = ma F y is the “sum of the forces” in the y direction F T is the force of tension in Newtons F g 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/s 2 FTFT FgFg a

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But, when the acceleration is downwards use the following equation… F y = F g – F T = ma FTFT FgFg a

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Other things you need to remember: To find the mass of the object when given the weight in Newtons, divide by 9.8 m/s 2. To find the weight of the object when given the mass in kg’s, multiply by 9.8 m/s 2.

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Now let’s see an example… Example 1: There is an elevator that is going upwards and slowing down at a rate of 3 m/s 2. It masses 200 kg’s. What is the tension in the cable from which it is hanging? F g = 200 x 9.8 = 1960N a = 3m/s 2 ** ** (use the acceleration table to decide on the direction of this arrow) F T = ?

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Use the following equation: F y = F g – F T = ma where: F g = 200 kg x 9.8 m/s 2 = 1960 N F T = ? m = 200 kg a = 3 m/s 2 F y = 1960 – F T = 200 x 3 – F T = 600 – 1960 F T = 1360 N Plug what you know into the correct equation and solve for the object’s force of tension: NOW START YOUR HW! 8 points total on test

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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

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Fri Nov 15 Mistake on email about R&C; will resend today 1.Grade “quiz” #7 in red pen; put score on top ( /8) in red; turn in HW; turn in (& Guided Notes too!) 2.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. 3.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. 4.Asst (a): FVD 2 #4-11 (#11 we will grade like a quiz tomorrow) 5.Asst (b): FVD 4 #3-10 (#7 we will grade like a quiz tomorrow) 6.LAST CALL: email questions on Newton’s Laws wrksht! 7.Asst (b) 2012: HAND-write Write-up: all of Introduction should be done; do Purpose & Materials Unit Test = Thursday November 21st

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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/s 2. Find the tension in the rope: F T = ? F g = 87 x 9.8 = _____ N a=0.531 m/s 2 F y = 87*9.8 – F T = 87 x 0.531 F T = ____ N

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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. F TL F TR FgFg F TL sin F TL cos F TR sin F TR cos So = 90 - 38 = ____ , and = 15 . First sum the forces in the x direction: F x = F TR cos 15 – F TL cos 52 = 0 Then sum the forces in the y direction: F y = F TR sin 15 + F TL sin 52 – 2727 = 0

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F x = F TR cos 15 – F TL cos 52 = 0 F y = F TR sin 15 + F TL sin 52 – 2727 = 0 Use a matrix to solve: F TR F TL x eq: cos15 -cos52 y eq: sin15 sin52 [ ] [ ] constants on other side of equation: 0 2727 Since F TR was the first variable in the matrix, then F TR _____ N and F TL _____ N (since F TL was the second variable in the matrix.). LECTURE NOTES - FVD type #2 continued….

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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. 25 F T = 150 NFNFN F g = 421.4 N FfFf First sum the forces in the y direction: F y = F T sin 25 + F N – F g = 0 Solve for F N Then sum the forces in the x direction: F x = F T cos 25 – 0.100F N = (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 F f = * F N. The is pronounced “mu”, and is called the “coefficient of friction”.

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Mon Nov 18 1.Grade type2 HW “quiz” #11 in red pen; put score on top ( /19) in red & turn in HW 2.Grade type4 HW “quiz” #7 in red pen; put score on top ( /16) in red & turn in HW 3.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? 4.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? 5.Asst (a): FVD 3 #5-11 (go over #10 tomorrow like a quiz) 6.Asst (b): FVD 5 #3-9 (go over #8 tomorrow like a quiz) 7.Asst (c, review): FVD’s 1,2,4 #14 8.Asst (b) 2012: HAND-write Write-up - Procedure Unit Test = Thursday November 21st

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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 F TL F TR FgFg F TL sin 25 F TL cos 25 F TR sin 41 F TR cos 41 F x = F TR cos 41 – F TL cos 25 = 0 F y = F TR sin 41 + F TL sin 25 – 84.9*9.8 = 0 F TR F TL x eq: cos41 -cos25 y eq: sin41 sin25 [ ] [ ] constants on other side of equation: 0 832 5 pts FVD; 4 pts + 5 pts for equations; 3 pts for matrices F TR = _____ N (with THREE sig figs!) F TL = _____ N (with THREE sig figs!) 2 pts for answers written correctly!

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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. 20 F T = 350 NFNFN F g = 735 N F f =. 4 F N F y = 350sin 20 + F N – 735 = 0 F N = ____N F x = 350cos 20 – 0.400F N = (735 / 9.8) a a = ____ m/s 2 5 pts FVD 5 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel

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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? FTFT F gL FTFT F gR a?? 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.

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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? First sum the forces in the y direction on the left: F yL = F T – F gL = ma or: F yL = F T – 441 = 45a Then sum the forces in the y direction on the right: F yR = F gR – F T = ma or: F yR = 539 – F T = 55a Use a matrix to solve …. FTFT F gL FTFT F gR a a You don’t need to draw the pulley! = 539 N = 441 N F T a right eq: -1 -55 left eq: 1 -45 [ ] [ ] constants on other side of equation: -539 441 Since F T was the first variable in the matrix, then F T ____ N and a ____ m/s 2 (since a was the second variable in the matrix).

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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? +y +x First sum the forces in the “y” direction: F y = F N – F g cos 65 = 0 (note F g = 200*9.8) Solve for F N Then sum the forces in the “x” direction: F x = F g sin 65 – 0.500F N = 200 a Solve for a 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! a 65 FNFN FgFg FfFf = 200*9.8 N =.5 F N

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Tues Nov 19 1.Catch-up day on FVD types 1-5!! 1.Grade type3 HW “quiz” #10 in red pen; put score on top ( /18) in red & turn in HW 2.Grade type5 HW “quiz” #8 in red pen; put score on top ( /15) in red & turn in HW 2.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) 3.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!!!! 4.Asst B: Do #15 & #16 from FVD’s 1-5 Unit Test = Thursday November 21st

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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): F yL = F T – 2205 = 225a F yR = 2401 – F T = 245a Use a matrix to solve …. (3 pts) FTFT F gL FTFT F gR a a = 2401 N = 2205 N F T a left eq: 1 -225 right eq: -1 -245 [ ] [ ] constants on other side of equation: 2205 -2401 F T = ____ N and a = ____ m/s 2 2 pts for answers!

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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? F y = F N – 2156cos 40 = 0 F N = 1652 N F x = 2156sin 40 – 0.10F N = 220 a a = 5.55 m/s 2 5 pts FVD (includes 1 pt for EITHER drawing accel or weird axes) You don’t have to draw the incline. 4 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel 40 FNFN FgFg FfFf +y +x a = 2156 N =.1F N

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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? FNFN F f =.5F N F gL = 157*9.8 N FTFT FTFT F gR = 999.6 N Hanging Mass Left mass on table: F yL = F N – 157*9.8 = 0 so F N = 1538.6N 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:

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Table: F yT = F N – 157*9.8 = 0 so F N = 1538.6N Left mass on table: F xL = – (.5)(1538.6) + F T = 157a Right hanging mass: F yR = 999.6 – F T = (999.6/9.8)a Matrices: F T a # F xL F yR.5*1538.6 999.6 1 -157 -1 999.6/9.8 Answers to the Problem: F T = 909 N a =.889 m/s 2 Lecture Notes FVD #6 continued: EQUATIONS:

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Wed November 20 (collab) 1.Go over Review Sheet in red pen; put score on top ( / 99 ) in red 2.Talk about important Guided Notes 3.Look at Study Guide 4.Bill Nye “Friction” ?? 5.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 1.Forces Test Test is LONG! Come early & be VERY prepared! 2.Asst: Guided Notes worksheet on Momentum, pages 228-245 COPY DOWN BOTH DAYS! Test is LONG! Come early & be VERY prepared! Remember our rules about test attendance!

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“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?

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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)

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1.Forces Test Test is LONG! Come early & be VERY prepared! 2.Asst: Guided Notes on Momentum, pages 228-245 COPY DOWN BOTH DAYS! Test is LONG! Come early & be VERY prepared! Remember our rules about test attendance! Thurs 11/ 21

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Things used in 2012 and before….

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Review QUIZ You have exactly 10 minutes! 1.An elevator is going down but speeding up at a rate of 9.7999 m/s 2. The elevator masses 10,000. kg. Find the tension in the cable. (8 pts) 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. (16 pts) ANSWERS ON NEXT SLIDE (Get out a red pen!!!)

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F T = ? F g = 98,000 N a=9.7999 m/s 2 1.An elevator is going down but speeding up at a rate of 9.7999 m/s 2. The elevator masses 10,000. kg. Find the tension in the cable. F y = 98000 – F T = 10,000 x 9.7999 F T = 1.0000 N Why is it so small? What’s almost happening? 3 pts FVD 1 pt for answer 4 pts equation

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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. F TL F TR F g = 1176 N 50 4 pts for x equation: F x = F TR – F TL cos 50 = 0 OR F x = F TR cos 0 – F TL cos 50 = 0 4 pts for y equation: F y = F TL sin 50 – 1176 = 0 OR F y = F TR sin 0 + F TL sin 50 – 1176 = 0 4 pts FVD F TL = 1540 N (3 sig figs!) F TR = 987 N (3 sig figs!) 2 pts for answers written correctly! You actually don’t need a matrix to solve this, but you can use one. (+2 pts for math work or matrix) F TR F TL x eq: 1 -cos50 y eq: 0 sin50 [ ] [ ] constants on other side of equation: 0 1176 You can have cos 0 and sin 0 in that F TR column too

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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 F T, but need left/right subscripts on F g 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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HW Quiz FVD 1 #14: (8 pts) An elevator with a mass of 15.0 kg is going down & speeding up with an acceleration of 7.03 m/s 2. Find the tension in the rope: F y = 15*9.8 – F T = 15 x 7.03 F T = 41.6 N 3 pts FVD 4 pts equation F T = ? F g = 15*9.8 N a=7.03 m/s 2 1 pt for answer written correctly! (label, units, sig figs)

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Quiz FVD 2 #14: (19 pts) A 6697-N sign is suspended by two ropes. The left rope makes an angle of 4.00 degrees from the vertical, while the right makes an angle of 50.0 degrees from the horizontal. Find the tension in the two ropes F TL = 6.20 x 10 3 N F TR = 673 N F TL F TR F g = 6697N 86 50 F x = F TR cos 50 – F TL cos 86 = 0 F y = F TR sin 50 + F TL sin 86 – 6697 = 0 F TR F TL x eq: cos50 -cos86 y eq: sin50 sin86 [ ] [ ] constants on other side of equation: 0 6697 5 pts FVD; 4 pts + 5 pts for equations; 3 pts for matrices 2 pts for answers written correctly! (labels, units, sig figs)

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Quiz FVD 3 #14: (18 pts) Two objects are hanging from ropes, that are strung around a pulley mounted in the ceiling. The left mass weighs 1470 N, while the right masses 170 kg. Find the accel of the objects, and the tension in the rope. F T = 1600 N a = 0.61 m/s 2 5 pts FVD Remember you don’t have to draw the pulley or extra string!) 4 pts each equation (total 8 pts): F yL = F T – 1470 = 150a F yR = 1666 – F T = 170a Use a matrix to solve …. (3 pts) FTFT F gL FTFT F gR a a = 170*9.8 N = 1470 N F T a left eq: 1 -150 right eq: -1 -170 [ ] [ ] constants on other side of equation: 1470 -1666 2 pts for answers written correctly! (labels, units, sig figs)

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Quiz FVD 4 #14: (16 pts) My mother has one of those suitcases with a neat little handle you can pull behind you. She pulls at an angle of 20.0 degrees with the horizontal and with about a 400. N pull. Her suitcases are usually packed, with about 189 kg of stuff in it. The suitcase has rollers, so it has very little friction, only about 0.200 with the floor. What is the acceleration of her suitcase? 20 F T = 400 NFNFN F g = 189*9.8 N F f =.2 F N F y = 400sin 20 + F N – 189*9.8 = 0 F N = 1715 N F x = 400cos 20 – 0.200F N = 189 a a = 0.174 m/s 2 5 pts FVD 5 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel

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Quiz FVD 5 #14: (15 pts) A 1078-Newton box is on a plane inclined at 20.0 ° with the horizontal. The coefficient of friction between the box and the inclined plane is 0.300 What is the box’s acceleration? F y = F N – 1078cos 20 = 0 F N = 1013 N F x = 1078sin 20 – 0.30F N = 110 a a =.589 m/s 2 5 pts FVD (includes 1 pt for EITHER drawing accel or weird axes) You don’t have to draw the incline. 4 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel 20 FNFN FgFg FfFf +y +x a = 1078 N =.3F N

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Quiz FVD 1 #20: (8 pts) An elevator with a mass of 96.0 kg is going up & speeding up with an acceleration of 9.80 m/s 2. Find the tension in the rope: F y = F T – 96*9.8 = 96 x 9.8 F T = 1880 N 4 pts equation 3 pts FVD F T = ? F g = 96*9.8 N a=9.8 m/s 2 1 pt for answer written correctly! (label, units, sig figs)

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Quiz FVD 2 #20: (19 pts) A 337-kg sign is suspended by two ropes. The left rope makes an angle of 12.0 degrees from the vertical, while the right makes an angle of 13.0 degrees from the horizontal. Find the tension in the two ropes F TL = 3220 N F TR = 687 N F TL F TR F g = 337*9.8 N 78 13 F x = F TR cos 13 – F TL cos 78 = 0 F y = F TR sin 13 + F TL sin 78 – 3302.6 = 0 F TR F TL x eq: cos13 -cos78 y eq: sin13 sin78 [ ] [ ] constants on other side of equation: 0 3303 5 pts FVD; 4 pts + 5 pts for equations; 3 pts for matrices 2 pts for answers written correctly! (labels, units, sig figs)

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Quiz FVD 3 #20: (18 pts) Two objects are hanging from ropes, that are strung around a pulley mounted in the ceiling. The left mass is 235 kg, while the right is 196 N. Find the accel of the objects, and the tension in the rope. F T = 361 N a = 8.26 m/s 2 5 pts FVD Remember you don’t have to draw the pulley or extra string!) 4 pts each equation (total 8 pts): F yL = 2303 – F T = 235a F yR = F T – 196 = 20a Use a matrix to solve …. (3 pts) FTFT F gL FTFT F gR a a = 196 N = 235 *9.85 = 2303 N F T a left eq: -1 -235 right eq: 1 -20 [ ] [ ] constants on other side of equation: -2303 196 2 pts for answers written correctly! (labels, units, sig figs)

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Quiz FVD 4 #20: (16 pts) My mother has one of those suitcases with a neat little handle you can pull behind you. She pulls at an angle of 35.0 degrees with the horizontal and with about a 450. N pull. Her suitcases are usually packed, with about 578.2 N of stuff in it. The suitcase has rollers, so it has very little friction, only about 0.200 with the floor. What is the acceleration of her suitcase? 35 F T = 450 NFNFN F g = 578.2 N F f =.2F N F y = 450sin 35 + F N – 578.2 = 0 F N = 320 N F x = 450cos 35 – 0.200*320 = 59 a a = 5.16 m/s 2 (.0973 m/s 2 if you did #25!) 5 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel 5 pts FVD Remember you don’t have to draw the floor

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Quiz FVD 5 #20: (15 pts) A 90.0-kilogram box is on a plane inclined at 40.0 ° with the horizontal. The coefficient of friction between the box and the inclined plane is 0.400 What is the box’s acceleration? F y = F N – 882cos 40 = 0 F N = 676 N F x = 882sin 40 – 0.4F N = 90 a a = 3.30 m/s 2 4 pts for y equation 1 pt for solving for Normal 4 pts for x equation 1 pt for solving for accel 5 pts FVD Remember you don’t have to draw the incline 40 FNFN FgFg FfFf +y +x a = 90*9.8 N =.4 F N

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Quiz FVD 6 #20: (24 pts) An object is on top of a (flat, level) table. This object masses 182 kg and has a coefficient of friction of 0.300 with the table. There is a pulley on the edge of the table, with a rope over it, attached to a “hanging mass” of 335 kg. What will be the acceleration of the objects over the edge? FNFN F f = F N =. 3 F N F gT = 182*9.8 N FTFT FTFT F gH = 335*9.8 N Hanging Mass F yT = F N – 182*9.8 = 0 so F N = 1783.6N F xT = F T – (. 3)(1783.6) = 182 a F yH = 3283 – F T = 335 a F T = 1.50 x 10 3 N a = 5.32 m/s 2 F T a x T eq: 1 -182 y H eq: -1 -335 [ ] [ ] constants on other side of equation: H.3*1783.6 -3283 6 pts FVD (don’t need table, pulley, extra rope) 4 pts 1 pt 4 pts Matrices = 3 pts 2 pts for answers written correctly! (labels, units, sig figs)

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Practice Quiz (stamp when done; grade online at home) 1.Write each of Newton’s 3 laws, in order. 2.Give one application/use/etc of EACH of N’s 3 Laws, in order. 3.If you have no accel, does it mean there are no forces acting on you? 4.What does equilibrium mean, and how do you find the equilibrant? 5.There are two force vectors acting on an object: The first force is acting north and has a magnitude of 30.0 N. The second force is acting east and has a magnitude of 40.0 N. i) Find the total force on the object by using our old “head-to-tail” stuff. ii) Determine the force that will put this system in equilibrium. iii) Draw a force-vector diagram for this object in equilibrium.

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Quiz KEY (23 pts total) 1.1) Body in motion stays in motion; 2) F = ma; 3) Equal & opposite forces (3 pts) 2.1) When you turn right quickly in your car, your body appears to move left (it really just keeps going straight!); 2) with the same force, a bigger mass won’t accelerate as much; 3) I jump by pushing down on the ground, which then pushes up on me (6 pts) 3.NO, there could be forces that just cancel out! (2 pts) 4.Equilibrium means acceleration = 0 because the forces are balanced out. (Note: there could be constant movement/velocity, but the acceleration = zero!!)) (2 pts) 5.see at right (10 pts)

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Chapter 4 Dynamics: Newton’s Laws of Motion. Units of Chapter 4 Force Newton’s First Law of Motion Mass Newton’s Second Law of Motion Newton’s Third Law.

Chapter 4 Dynamics: Newton’s Laws of Motion. Units of Chapter 4 Force Newton’s First Law of Motion Mass Newton’s Second Law of Motion Newton’s Third Law.

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