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!
Transcript
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!
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!
Tues Nov 12(no school Monday, Veteran's Day)
1. Binder due today???2. Lecture Notes – Newton’s 3 laws3. Asst: (a) Wksheet – front side onlyBackside 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
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
• 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.)
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 · ???
SKIP some lines so we can finish the lecture notes!!!!
Wed – November 13
1. Finish lecture notes on N’s 3rd Law2. Questions from HW?3. Experiment – Newton's 2nd 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
papere. Erase calibration marks; detach all but 2 pennies; put stuff awayf. 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
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!)
symbol units definition
mass
weight
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.
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
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.
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-upUnit Test = Thursday November 21st
Fg
FT
FT: Force of Tension
Fg : 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.
First you will need to find the direction of the acceleration. To do so use the chart below:
Acceleration Acceleration
Acceleration Acceleration
Going Down
Going Up
Slowing DownSpeeding Up
When the acceleration is upwards use the following
equation…
Fy = FT – Fg = ma
Fy 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
FT
Fg
a
But, when the acceleration is downwards use the following
equation…
Fy = Fg – FT = ma FT
Fg
a
Other things you need to remember:
To find the mass of the object when given the weight in Newtons, divide by
9.8 m/s2.
To find the weight of the object when given the mass in kg’s, multiply by 9.8
m/s2.
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/s2. It masses 200 kg’s. What is the tension in the cable from which
it is hanging?
Fg= 200 x 9.8 = 1960N
a = 3m/s2
**
** (use the acceleration table to decide on the direction of this arrow)
FT = ?
Use the following equation:
Fy = Fg – FT = ma
where:
Fg = 200 kg x 9.8 m/s2 = 1960 N
FT = ?
m = 200 kg
a = 3 m/s2Fy= 1960 – FT = 200 x 3
– FT = 600 – 1960
FT = 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
• 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
Fri Nov 15
Mistake on email about R&C; will resend today1. 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 & MaterialsUnit Test = Thursday November 21st
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:
FT = ?
Fg= 87 x 9.8 = _____ N
a=0.531 m/s2
Fy= 87*9.8 – FT = 87 x 0.531
FT = ____ N
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.
FTL FTR
Fg
FTL sin
FTL cos
FTR sin
FTR cos
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
Fx = FTRcos 15 – FTLcos 52 = 0
Fy = FTRsin 15 + FTLsin 52 – 2727 = 0
Use a matrix to solve:
FTR FTL
x eq: cos15 -cos52y eq: sin15 sin52[ ]
-1
[ ]constants on other side
of equation:
02727
Since FTR was the first variable in the matrix, then FTR _____ N and FTL _____ N (since FTL was the second variable in the matrix.).
LECTURE NOTES - FVD type #2 continued….
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
FT = 150 NFN
Fg = 421.4 N
FfFirst 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”.
Mon Nov 181. Grade type2 HW “quiz” #11 in red pen; put score on top (
/19) in red & turn in HW2. Grade type4 HW “quiz” #7 in red pen; put score on top ( /16) in
red & turn in HW3. 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
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
FTL FTR
Fg
FTL sin 25
FTL cos 25
FTR sin 41
FTR cos 41
Fx = FTRcos 41 – FTLcos 25 = 0
Fy = FTRsin 41 + FTLsin 25 – 84.9*9.8 = 0
FTR FTL
x eq: cos41 -cos25y eq: sin41 sin25[ ]
-1
[ ]constants on other side
of equation:
0832
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!
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
FT = 350 NFN
Fg = 735 N
Ff = .4 FN Fy = 350sin 20 + FN – 735 = 0
FN = ____N
Fx = 350cos 20 – 0.400FN = (735 / 9.8) a
a = ____ m/s2
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
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?
FT
FgL
FT
FgR
a??
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.
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:FyL = FT – FgL = ma
or: FyL = FT – 441 = 45a
Then sum the forces in the y direction on the right:
FyR = FgR – FT = maor:
FyR = 539 – FT = 55a
Use a matrix to solve ….
FT
FgL
FT
FgR
a
a
You don’t need to draw the pulley!
= 539 N
= 441 N
FT a
right eq: -1 -55 left eq: 1 -45[ ]
-1
[ ]constants on other side
of equation:
-539441
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).
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:
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
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
65
FN
Fg
F f
= 200*9.8 N
=.5 FN
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
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 = 225aFyR = 2401 – FT = 245a
Use a matrix to solve …. (3 pts)
FT
FgL
FT
FgR
a
a
= 2401 N
= 2205 N
FT a
left eq: 1 -225 right eq: -1 -245[ ]
-1
[ ]constants on other side
of equation:
2205-2401
FT = ____ N and a = ____ m/s2
2 pts for answers!
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?
Fy = FN – 2156cos 40 = 0
FN = 1652 N
Fx = 2156sin 40 – 0.10FN = 220 a
a = 5.55 m/s2
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
40
FN
Fg
F f
+y
+x
a
= 2156 N
= .1FN
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
Ff = .5FN
FgL = 157*9.8 N
FTFT
FgR = 999.6 N
Hanging Mass
Left mass on table: FyL = FN – 157*9.8 = 0
so FN = 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:
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
Matrices:FT a #
FxL
FyR
.5*1538.6
999.6
1 -157
-1 999.6/9.8
-1Answers to the
Problem:
FT = 909 N
a = .889 m/s2
Lecture Notes FVD #6 continued: EQUATIONS:
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!
“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?
GO OVER STUDY GUIDE TOGETHERTest is LONG! Come very early & be VERY prepared!
Be able to do all 6 FVD’s in about 35 minutes total!!
(see next slide)
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
• Things used in 2012 and before….
Review QUIZYou have exactly 10 minutes!
1. 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)
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!!!)
FT = ?
Fg= 98,000 N
a=9.7999 m/s2
1. 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.
Fy= 98000 – FT = 10,000 x 9.7999
FT = 1.0000 N
Why is it so small?
What’s almost happening?
3 pts FVD
1 pt for answer
4 pts equation
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.
FTL
FTR
Fg = 1176 N
50
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
ORFy = FTRsin 0 + FTLsin 50 – 1176 = 0
4 pts FVD
FTL = 1540 N (3 sig figs!)
FTR = 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)
FTR FTL
x eq: 1 -cos50y eq: 0 sin50[ ]
-1
[ ]constants on other side
of equation:
01176
You can have cos 0 and sin 0 in that FTR column too
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!
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/s2.
Find the tension in the rope:
Fy= 15*9.8 – FT = 15 x 7.03
FT = 41.6 N
3 pts FVD
4 pts equationFT = ?
Fg= 15*9.8 N
a=7.03 m/s2
1 pt for answer writte
n
correctly! (label, units,
sig figs)
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
FTL= 6.20 x 103 N
FTR = 673 N
FTL FTR
Fg = 6697N
86
50
Fx = FTRcos 50 – FTLcos 86 = 0
Fy = FTRsin 50 + FTLsin 86 – 6697 = 0
FTR FTL
x eq: cos50 -cos86y eq: sin50 sin86[ ]
-1
[ ]constants on other side
of equation:
06697
5 pts FVD; 4 pts + 5 pts for
equations; 3 pts for matrices
2 pts for answers
written correctly!
(labels, units, sig figs)
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.
FT = 1600 N
a = 0.61 m/s2
5 pts FVD
Remember you
don’t have to draw
the pulley or extra
string!) 4 pts each equation (total 8 pts):
FyL = FT – 1470 = 150aFyR = 1666 – FT = 170a
Use a matrix to solve …. (3 pts)
FT
FgL
FT
FgR
a
a
= 170*9.8 N
= 1470 N
FT a
left eq: 1 -150 right eq: -1 -170[ ]
-1
[ ]constants on other side
of equation:
1470-1666
2 pts for answers
written correctly!
(labels, units, sig figs)
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
FT = 400 NFN
Fg = 189*9.8 N
Ff = .2 FN
Fy = 400sin 20 + FN – 189*9.8 = 0
FN = 1715 N
Fx = 400cos 20 – 0.200FN = 189 a
a = 0.174 m/s2
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
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?
Fy = FN – 1078cos 20 = 0
FN = 1013 N
Fx = 1078sin 20 – 0.30FN = 110 a
a = .589 m/s2
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
20
FN
Fg
F f
+y
+x
a
= 1078 N
= .3FN
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/s2.
Find the tension in the rope:
Fy= FT – 96*9.8 = 96 x 9.8
FT = 1880 N
4 pts equation
3 pts FVDFT = ?
Fg= 96*9.8 N
a=9.8 m/s2
1 pt for answer writte
n
correctly! (label, units,
sig figs)
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
FTL= 3220 N
FTR = 687 N
FTL FTR
Fg = 337*9.8 N
78
13
Fx = FTRcos 13 – FTLcos 78 = 0
Fy = FTRsin 13 + FTLsin 78 – 3302.6 = 0
FTR FTL
x eq: cos13 -cos78y eq: sin13 sin78[ ]
-1
[ ]constants on other side
of equation:
03303
5 pts FVD; 4 pts + 5 pts for
equations; 3 pts for matrices
2 pts for answers
written correctly!
(labels, units, sig figs)
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.
FT = 361 N
a = 8.26 m/s2
5 pts FVD
Remember you
don’t have to draw
the pulley or extra
string!) 4 pts each equation (total 8 pts):
FyL = 2303 – FT = 235aFyR = FT – 196 = 20a
Use a matrix to solve …. (3 pts)
FT
FgL
FT
FgR
a
a
= 196 N
= 235*9.85 = 2303 N FT a
left eq: -1 -235 right eq: 1 -20[ ]
-1
[ ]constants on other side
of equation:
-2303 196
2 pts for answers
written correctly!
(labels, units, sig figs)
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
FT = 450 NFN
Fg = 578.2 N
Ff
= .2FN
Fy = 450sin 35 + FN – 578.2 = 0
FN = 320 N
Fx = 450cos 35 – 0.200*320 = 59 a
a = 5.16 m/s2 (.0973 m/s2 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
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 0.400What is the box’s acceleration?
Fy = FN – 882cos 40 = 0
FN = 676 N
Fx = 882sin 40 – 0.4FN = 90 a
a = 3.30 m/s2
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
40
FN
Fg
F f
+y
+x
a
= 90*9.8 N
= .4 FN
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?FN
Ff = FN
= . 3 FN
FgT = 182*9.8 N
FTFT
FgH = 335*9.8 N
Hanging Mass
FyT = FN – 182*9.8 = 0
so FN = 1783.6N
FxT = FT – (. 3)(1783.6) = 182 a
FyH = 3283 – FT = 335 a
FT = 1.50 x 103 N
a = 5.32 m/s2
FT a
xT eq: 1 -182yH eq: -1 -335[ ]
-1
[ ]constants on other side
of equation: H.3*1783.6
-3283
6 pts FVD (don’t need
table, pulley, extra rope)4 pts
4 pts1 pt
4 pts
Matrices = 3 pts
2 pts for answers
written correctly!
(labels, units, sig figs)
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.
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)
