Embed Size (px)
Citation preview
Ch 5 lecture.notebook
1
January 08, 2015
kok. 307:11 PM
kok. 307:11 PM
Chapter Five: Forces in Two Dimensions
Vector review:
vectors have magnitude and direction
add vectors tail to tip, any order
draw a vector from tail of first vector to the tip of the last vector
= resultant
Ch 5 lecture.notebook
2
January 08, 2015
kok. 307:11 PM
Vectors at right angle to each other : use pythagorean theorem
Bob walks 3 miles west then turns and walks 4 miles north, how far is Bob from his original starting point?
R2 = A2 + B2
kok. 307:11 PM
Any triangle: can find resultant with Law of Cosines
R2 = A2 + B2 2AB cos
Bob walks 10 miles due east, then turns 25 degrees to the north and walks 4 more miles. How far is Bob from his original starting point?
Ch 5 lecture.notebook
3
January 08, 2015
kok. 307:11 PM
Practice drawing vectors on a coordinate plane
15 degrees north of west 40 degrees east of north30 degrees west of south
kok. 307:11 PM
Find the magnitude of the sum of two forces, one 20.0 N and the other 7.0 N when the angle between them is 30.0 degrees.
Ch 5 lecture.notebook
4
January 08, 2015
kok. 307:11 PM
Components of vectors
+ x
y
+ x
+ y
x
+ y
x
y
SOHCAHTOA
sin = opposite
hypotenuse
cos = adjacent
hypotenuse
tan = opposite
adjacent
kok. 307:11 PM
7.0 m
35 degrees north of east
vector resolution: breaking a vector into its x and y components
xcomponent : parallel to xaxis
ycomponent : parallel to yaxisR=
Rx
Ry
in this case, Ry = side opposite the angle
so use sin 0 = opposite
hypotenuse
Rx = side adjacent to the angle
use cos 0 = adjacent
hypotenuse
Ch 5 lecture.notebook
5
January 08, 2015
kok. 307:11 PM
find the x and y components of the vector
kok. 307:11 PM
50 degres west of north
23.5 N
Find the x and y components
Ch 5 lecture.notebook
6
January 08, 2015
kok. 307:11 PM
43 degrees west of south
Find the x and y components
14.5 m/s
kok. 307:11 PM
73.7 degrees south of east
173 m
Ch 5 lecture.notebook
7
January 08, 2015
kok. 307:11 PM
Algebraic Addition of Vectors
travel 5.0 m east, then 7.5 m north
kok. 307:11 PM
but what about more than 2 vectors?
5.0 m east, 7.4 m north, 1.5 m east, 2.2 m south
Draw resultant
find x and y components of the right triangle of the resultant
Ch 5 lecture.notebook
8
January 08, 2015
Dec 31:38 PM
What about more complicated vectors?
5.0 m Northeast, then 3.0 m due east, then 7.0 m 15 north of west
Find x and y components of each, THEN add all x's to = Rx, and add all y's to = Ry
But.........also need a direction.
tan 0 = opposite
adjacent
Dec 31:38 PM
A pack of four arctic wolves are exerting four different forces on the carcass of a 500 kg dead polar bear. Wolf 1 is pulling with a force of 20. N due North. Wolf 2 is pulling with 45 N, 30 north of west. Wolf 3 is pulling with 25 N due West, and Wolf four is pulling with 35 N, 10 west of south.
Ch 5 lecture.notebook
9
January 08, 2015
Dec 31:38 PM
5.2 Friction
force that opposes motion
requires contact between 2 surfaces
2 types:
kinetic at least one surface is in motion
static no motion between the surfaces
Depends on:
materials the surfaces are made of
normal force
Dec 31:38 PM
graph on p 127
as Fn , kinetic friction force also
(lines are sloping upward)
sandpaper has steeper slope than polished table
slope = coefficient of kinetic friction
F f, kinetic = normal force
F f, static =
coefficient of static friction
Ch 5 lecture.notebook
10
January 08, 2015
Dec 31:38 PM
if an object is sitting on the floor and not moving, draw all of the forces in both directions acting upon it
What if someone pushed on the object to the right, but it still didn't move? Draw all of the forces acting upon it.
Dec 31:38 PM
Now someone pushes it with enough force that it travels at a constant velocity to the right. Draw all of the forces acting upon it.
Ch 5 lecture.notebook
11
January 08, 2015
Dec 31:38 PM
someone pushes on the box so that it is accelerating to the right. Draw all of the forces acting upon it.
Dec 31:38 PM
If you exert a 35.0 N force to the right on a 14.0 kg box of books so that it travels at a constant velocity of 0.75 m/s, what is the force of friction acting upon the box of books?
Ch 5 lecture.notebook
12
January 08, 2015
Dec 31:38 PM
What is the friction force on a 10.0 N sled that is moving at 3.0 m/s if the coefficient of static friction is 0.30 and the coefficient of kinetic friction is 0.12?
Dec 31:38 PM
What is the friction force when a refrigerator with a mass of 150.0 kg is being pushed at a constant velocity? 0.43 0.77
Ch 5 lecture.notebook
13
January 08, 2015
Dec 31:38 PM
If you use a horizontal force of 30.0 N to slide a 12.0 kg wooden crate across a floor at a constant velocity, what is the coefficient of kinetic friction between the crate and the floor?
Dec 31:38 PM
What is the minimum force with which you would need to push to get a dresser moved from its spot? The dresser has a mass of 450. kg and 0.62 0.89
Ch 5 lecture.notebook
14
January 08, 2015
Dec 31:38 PM
Unbalanced friction forcesIf there is an acceleration there is an Fnet
previous problems were BALANCED friction forces
= constant velocity
= no acceleration
= no Fnet
Dec 31:38 PM
Fg
Fn
Pushing on the box to the right with a force of 25.0 N and the friction force is 15.0 N.
What will be the acceleration of the box?
What is the coefficient of kinetic friction?
Ff Fpush14.0 kg
.
Ch 5 lecture.notebook
15
January 08, 2015
Dec 31:38 PM
A 32.0 kg fireman slides down a pole at constant velocity. What is the friction force of his hands on the pole?
If he loosens his grip and decreases the friction force to 310 N, what is his acceleration downward?
Dec 31:38 PM
At a wedding reception, a small, 25 kg boy is being drug by his heels across the dance floor by his cousin. If the coefficient of friction between they boy's pants and the floor is 0.15, what is the frictional force acting upon him as he slides?
If the cousin is pulling with 32 N of force eastward, what is the small boy's acceleration?
Ch 5 lecture.notebook
16
January 08, 2015
Dec 31:38 PM
Dec 31:38 PM
5.3 Force and motion in Two Dimensions
an object is in equilibrium when the net force on it is ZERO
ex:
Fg
Fair resistance object is NOT acceleration because the net force = zero
will be at rest OR at constant velocity
Ch 5 lecture.notebook
17
January 08, 2015
Dec 31:38 PM
IF there is a net force, then the object is accelerating
Fg
Fair resistance = 5 N
= 12N
Net force = 5N 12N
= 7N
the equilibrant is the force that will put an object in equilibrium
so.....if the net force is 7 N downward, the equilibrant is 7 N upward
because then all of the forces added together will equal zero
5N 12N + 7N = 0N
Dec 31:38 PM
Practice:Find the equilibrant for:
3 N north and 4 N east
12 N left plus 15 N right
20.0 N north, 5 N west, 15 N at 54 degrees north of east
Ch 5 lecture.notebook
18
January 08, 2015
Dec 31:38 PM
Motion along an inclined plane
motion of boxmo
tion of
box
you need to tip your coordinate system to match the slope of the hill
Dec 31:38 PM
Fg is straight down towards center of earth
Fgy
FgxFg
Fn = Fgy
Ch 5 lecture.notebook
19
January 08, 2015
Dec 31:38 PM
Dec 31:38 PM
Ch 5 lecture.notebook
20
January 08, 2015
Dec 31:38 PM
Dec 31:38 PM
Ch 5 lecture.notebook
21
January 08, 2015
Dec 31:38 PM
![Scanning curves and kinetics of the acetylcholine ...€¦ · 46 Since AcCh is a cation [(CH,),N+-CH,CH,OCOCH,] and the receptor protein at pH 7 is net anionic, the hysteresis may](https://img.pdfslide.us/doc/110x75/5e2ac614ffad924f260d9faa/scanning-curves-and-kinetics-of-the-acetylcholine-46-since-acch-is-a-cation.jpg)
![99mTcOAADT]-(CH -NEt : A Potential Small-Molecule Single ......[99mTcOAADT]-(CH 2) 2-NEt 2: A Potential Small-Molecule Single-Photon Emission Computed Tomography Probe for Imaging](https://img.pdfslide.us/doc/110x75/60d8da7e371b133b9a3433a6/99mtcoaadt-ch-net-a-potential-small-molecule-single-99mtcoaadt-ch.jpg)
