Transcript File - El Paso High School

VECTORS AND SCALARS
A scalar quantity has only magnitude and is completely specified
by a number and a unit. Examples are mass (a stone has a mass
of 2 kg), volume (1.5 L), and frequency (60 Hz). Scalar
quantities of the same kind are added by using ordinary
arithmetic.
A vector quantity has both magnitude and direction. Examples
are displacement (an airplane has flown 200 km to the
southwest), velocity (a car is moving at 60 km/h to the north),
and force (a person applies an upward force of 25 N to a
package). When vector quantities are added, their directions
must be taken into account.
A vector is represented by an arrowed line whose length is
proportional to the vector quantity and whose direction
indicates the direction of the vector quantity.
The resultant, or sum, of a number of vectors of a particular
type (force vectors, for example) is that single vector that would
have the same effect as all the original vectors taken together.
GRAPHICAL ADDITION OF VECTORS
PARALLELOGRAM METHOD
The resultant of two vectors acting at any angle may be
represented by the diagonal of a parallelogram.
The two vectors are drawn as the sides of the parallelogram
and the resultant is its diagonal.
The direction of the resultant is away from the origin of the two
vectors.
3.1 Find the magnitude and direction of the resultant force produced by a
vertically upward force of 40 N and a left horizontal force of 30 N.
A (40 N, up)
B (30 N, left)
Scale: 1 cm = 10 N
R
B
A
R (50 N, 127)
3.2 Using the graphical method, find the resultant of the following two
displacements:
A (2.0 m, 40)
Scale 1 cm = 1 m
B (4.0 m, 127)
B
R
A
R (4.5 m, 101)
POLYGON or TIP-TO-TAIL METHOD
This method for finding the resultant R of several vectors (A, B, C) consists
in beginning at any convenient point and drawing (to scale in the proper
directions) each vector arrow in turn.
They may be taken in any order of succession
A + B + C = C + A + B.
The tail of each arrow is positioned at the head of the preceding one.
The resultant is represented by an arrow
with its tail end at the starting point and
its head at the tip of the last vector added.
The size or magnitude of R is given by:
R = R
TIP-TO-TAIL METHOD
PARALLELOGRAM
METHOD
90
0
180
360
270
3.3 Add the following two forces by use of the tip-to-tail method:
A (30 N at 30)
Scale: 1 cm = 10 N
B (20 N at 140)
B
R
A
R (30 N, 72)
The resultant is drawn from tail end at the starting point
to the head of the last vector added.
3.4 Three ropes are tied to a stake. Find the resultant force.
A (20 N, 0)
B (30 N, 150)
C (40 N, 230)
Scale: 1 cm = 10 N
B
C
R
A
R (35 N, 206)
VECTOR COMPONENTS
A component of a vector is its
effective value in a given
direction.
For example, the x-component
of a displacement is the
displacement parallel to
the x-axis caused by the given
displacement.
A vector in two dimensions may
be resolved into two component
vectors acting along any two
mutually perpendicular
directions.
Rx = R cos 
Ry = R sin 
3.5 Find the components of the vector:
F (250 N, 235o)
Fx = F cos 
= 250 cos (235o)
= - 143.4 N
Fy = F sin 
= 250 sin (235o)
= - 204.7 N
F
x
Fy
VECTOR ADDITION: COMPONENT METHOD
To add two or more vectors A, B, C,… by the component method, follow
this procedure:
1. Resolve the initial vectors into components x and y.
2. Add the components in the x direction to give Rx and add the
components in the y direction to give Ry . That is, the magnitudes of Rx
and Ry are given by, respectively:
Rx = Ax + Bx + Cx…
Ry = Ay + By + Cy…
3. Calculate the magnitude and direction of the resultant R from its
components by using the Pythagorean theorem:
R 
and   tan  1 R y
Rx
Rx  Ry
2
2
3.6 Four coplanar forces act on a body at point O as shown in the figure.
Find their resultant graphically.
A (80 N, 0)
B (100 N, 45)
C (110 N, 150)
D (160 N, 200)
Scale: 1 cm = 20 N
D
C
B
R
A
R (119 N, 143)
3.7 Solve problem 3.6 by use of the component method.
A (80 N, 0)
C (110 N, 150)
B (100 N, 45)
D (160 N, 200)
x-component
80 cos 0
100 cos 45
110 cos 150
160 cos 200
y-component
80 sin 0
100 sin 45
110 sin 150
160 sin 200
Σx = -95 N
Σy = 71 N
R 
( 95 )  ( 71 )
2
2
= 118.6 N
  tan
1
 71 


 95 
= 36.7
Since Σx = (-) and Σy = (+)
R is in the III Quadrant, therefore:
180 - 36.7= 143.3
R (118.6 N, 143.3)
3.8 The five coplanar forces shown in the figure act on an object. Find their
resultant with the component method.
A (19 N, 0)
B (15 N, 60)
C (16 N, 135)
D (11 N, 210)
E (22 N, 270)
x-component
19 cos 0
15 cos 60
16 cos 135
11 cos 210
22 cos 270
y-component
19 sin 0
15 sin 60
16 sin 135
11 sin 210
22 sin 270
Σx = 5.7 N
Σy = -3.2 N
R 
(5 .7 )  (  3 .2 )
  tan
2
1
 3 .2 


 5 .7 
2
= 6.5 N
= 29
Since Σx = (+) and Σy = (-)
R is in the IV Quadrant, therefore:
360 - 29= 331
R (6.5 N, 331)
3.9 A force of 100 N makes an angle  with the x-axis and has a
y-component of 30 N. Find both the x-component and the angle .
A (100 N,  )
Ay = 30 N
  sin
1
 30 


 100 
A

= 17.5
Ay
Ax = A cos 
= 100 cos 17.5
= 95 N
3.10 A child pulls on a rope attached to a sled with a force of 60 N. The
rope is at 40 with respect to the ground.
a. Calculate the effective value of the pull tending to move the sled along
the ground.
F = 60 N
θ = 40
Fx = F cos 
= 60 cos 40
= 46 N
b. Calculate the force tending to lift the sled vertically.
Fy = F sin 
= 60 sin 40
= 39 N
3.11 A plane is traveling eastward at a speed of 500 km/h. A 90 km/h wind
is blowing southward. What are the direction and speed of the plane
relative to the ground?
A (500 km/h, E)
B (90 km/h, S)
v 
( 500 )  ( 90 )
  tan
2
1
 90 


500


2
= 508 km/h
= 10
360 - 10 = 350
R (508 km/h, 350)
PROJECTILE MOTION
An object launched into space without motive power of its own
is called a projectile. If we neglect air resistance, the only force
acting on a projectile is its weight, which causes its path to
deviate from a straight line.
The projectile has a constant horizontal velocity and a vertical
velocity that changes uniformly under the influence of gravity.
THE MONKEY AND THE ZOOKEPER
The monkey spends most of its day
hanging from a branch of a tree.
The zookeeper feeds the monkey by
shooting bananas from a banana cannon
to the monkey in the tree.
The monkey usually drops from the tree
the moment that the banana leaves the
muzzle of the cannon.
The zookeeper is faced with the dilemma
of where to aim the banana cannon in
order to feed the monkey. If the monkey
lets go of the tree the moment that the
banana is fired, then where should he aim
the banana cannon?
http://www.fisica.uniud.it/~deangeli/a
pplets/Multimedia/ExplrSci/dswmedi
a/monkey.htm
HORIZONTAL PROJECTION
If an object is projected horizontally, its motion can best be
described by considering its horizontal and vertical motion
separately. In the figure we can see that the vertical velocity
and position increase with time as those of a free-falling body.
Note that the horizontal distance increases linearly with time,
indicating a constant horizontal velocity.
3.14 A stunt flier is moving at 15 m/s parallel to the flat ground 100 m
below. How large must the distance x from the plane from the target be if a
sack of flour released from the plane is to strike the target?
vx = 15 m/s
y = 100 m
vy = 0
t 
2y
g
y = ½ gt2

2 (100 )
9 .8
= 4.52 s
x = vx t
= 15(4.52)
= 67.8 m
3.15 A person standing on a cliff throws a stone with a horizontal velocity
of 15.0 m/s and the stone hits the ground 47 m from the base of the cliff.
How high is the cliff?
vx = 15 m/s
x = 47 m
vy = 0
t 
x
vx

47
15
= 3.13 s
y = ½ gt2
= ½ (9.8)(3.13)2
= 48 m
3.16 A cannonball is projected horizontally with an initial velocity of 120
m/s from the top of a cliff 250 m above a lake.
a. In what time will it strike the water at the foot of the cliff?
vx = 120 m/s
y = 250 m
vy = 0
t 
2y
g

2 ( 250 )
9 .8
= 7.14 s
b. What is the x-distance (range) from the foot of the cliff to the point of
impact in the lake?
x = vx t
= 120(7.14)
= 857 m
c. What are the horizontal and vertical components of its final velocity?
vx = 120 m/s
vy = voy + gt
= 9.8(7.14)
= 70 m/s
d. What is the final velocity at the point of impact and its direction?
vR 
  tan
vx  vy 
2
1
70
120
2
(120 )  ( 70 )
2
2
= 139 m/s
= 30.2 below horizontal
v (139 m/s, 30.2)
PROJECTILE MOTION AT AN ANGLE
The more general case of projectile motion occurs when the
projectile is fired at an angle.
Problem solution:
1. Upward direction is positive. Acceleration (g) is downward thus
negative.
2. Resolve the initial velocity vo into its x and y components:
and
v 0 y  v 0 sin 
v  v cos 
0x
0
3. The horizontal and vertical components of its position at any instant is
given by:
1
2
x  v0 x t
and
y  v 0 y t  2 gt
4. The horizontal and vertical components of its velocity at any instant are
given by:
v x  v0 x
v y  v 0 y  gt
and
5. The final position and velocity can then be obtained from their
components.
3.17 A ball is thrown from the top of one building toward a tall building 50
m away. The initial velocity of the ball is 20 m/s at 40. How far above or
below its original level will the ball strike the opposite wall?
x = 50 m
vo = 20 m/s, 40
g = - 9.8 m/s2
t 
x
vx

50
15 . 3
vox = 20 cos 40
= 15.3 m/s
voy = 20 sin 40
= 12.9 m/s
= 3.27 s
y = voy t + ½ gt2
= 12.9(3.27) + ½ (-9.8)(3.27)2
= - 10.2 m or 10.2 m below its original level
3.18 An artillery shell is fired with an initial velocity of 100 m/s at an angle
of 30 above the horizontal. Find:
a. Its position and velocity after 8 s
vo = 100 m/s, 40
t=8s
g = - 9.8 m/s2
x = vox t
= 86.6(8)
= 692.8 m
y = voy t + ½ gt2
= 50(8) + ½ (-9.8)(8)2
= 86.4 m
vox = 100 cos 30
= 86.6 m/s
voy = 100 sin 30
= 50 m/s
vx = vox = 86.6 m/s
vy = voy + gt
= 50 + (-9.8)(8)
= - 28.4 m/s
b. The time required to reach its maximum height
At top vy = 0
t 
 v oy
g
vy = voy + gt

 50
 9 .8
= 5.1 s
c. The horizontal range R
Total time T = 2t
= 2(5.1)
= 10.2 s
x = vox t
= 86.6(10.2)
= 883.7 m
3.19 A baseball is thrown with an initial velocity of 120 m/s at an angle of
40above the horizontal. How far from the throwing point will the baseball
attain its original level?
vox = 120 cos 40
= 91.9 m/s
voy = 120 sin 40
= 77.1 m/s
vo = 120 m/s, 40
g = - 9.8 m/s2
At top vy = 0
t 
 v oy
g

 77 . 1
 9 .8
= 7.9 s
x = vox (2t)
= 91.9(2)(7.9)
= 1452 m
3.20 a. Find the range of a gun which fires a shell with muzzle velocity v at
an angle . What is the maximum range possible?
At top vy = 0
vy = voy + gt
= vo sin θ - gt
 2 v o sin  
x = vxt  v o cos  

g


2
x 
2vo
g
(sin  cos  )
t 
v o sin 
g
Total time = 2t
sin θcos θ= ½ sin 2θ
2
x 
2vo
(sin  cos  )
g
2
x 
vo
sin 2
g
Maximum range is 45 since 2θ = 90
b. Find the angle of elevation  of a gun that fires a shell with muzzle
velocity of 120 m/s and hits a target on the same level but 1300 m distant.
vo = 120 m/s
x = 1300 m
2
x 
vo
sin 2
g
sin 2 
gx
v
2
o

9 . 8 (1300 )
(120 )
2
= 0.885
sin-1(2θ)= 62
θ = 31
RELATIVE VELOCITY
Velocity measurements differ in different frames of
reference. If the frame of reference is denoted with subscripts
(vab is the velocity of a with respect to b), then the velocity of an
object with respect to a different frame of reference can be
found by adding the known velocities.
vab = vac + vcb
3.12 The world’s fastest current is in Slingsby Channel, Canada, where the
speed of the water reaches 30.0 km/h. Suppose a motorboat crosses the
channel perpendicular to the bank at a speed of 18.0 km/h relative to the
bank. Find the velocity of the motorboat relative to the water.
vwb = 30.0 km/h
vmb = 18.0 km/h
vmw =
v mw 
  tan
vwb
vmw
v mb  v wb
2
1
2
30
18

= 59
vmb
(18 )  ( 30 )
2
2
= 35 km/h
180 - 59 = 121
= R (508 km/h, 350)
3.13 A polar bear swims 2.60 m/s south relative to the water. The bear is
swimming against a current that moves 0.78 m/s at an angle of 40.0° north
of west, relative to Earth. How long will it take the polar bear to reach the
shore, which is 5.50 km to the south?
N
vbe
vbc = 2.60 m/s
vbc
vce = 0.78 m/s, 40° N of W
vce
W
x = 5.5 km, S
vbe = vbc + vce
y- comp
vbc
vce
2.6 sin 270°
0.78 sin 140°
Σy = -2.10 m/s
t 
x
v

 5500
2 . 10
= 2619 s
= 43.6 min