Transcript Applications of Vectors

APPLICATIONS
BASIC RESULTANT PROBLEMS
First let’s recall some basic facts from geometry about
parallelograms
opposite sides are
These 2 angles
add to 180°
These 2 angles
add to 180°
These 2 angles
add to 180°
These 2 angles
add to 180°
parallel and equal
alternate interior angles are congruent (equal sizes)
Let’s take these forces as
vectors and make a
parallelogram and use its
properties.
60°
280 lb
Two draft horses are pulling on a tree stump with forces of
250 pounds and 280 pounds as shown. If the angle
between the forces is 60°, then what is the magnitude of
the resultant force? What is the angle between the
resultant and the 280 pound force?
You can make a parallelogram and the
diagonal will be the resultant force
showing which direction and how
strong the resultant force will be.
v
60°
280 lb
Two draft horses are pulling on a tree stump with forces of
250 pounds and 280 pounds as shown. If the angle
between the forces is 60°, then what is the magnitude of
the resultant force? What is the angle between the
resultant and the 280 pound force?
Using geometry, can you figure out any of the angles or
sides of the lower triangle formed with the vector v as
one side?
Opposite sides are equal and the two angles add to 180°
  28.1
How can we
then find  as
shown?
v
60°
120°

280 lb
sin  sin120

250
v
Looking at the lower triangle we have side, angle, side so we
can use the Law of Cosines to find the magnitude (length) of v.
v  280  250  2  280  250  cos120
2
2
2
v  459.2
y
250cos 60, 250sin 60
We could also solve this
problem by making a
coordinate system with
the forces as vectors with
initial point at the origin.
v
60°

280 lb
280, 0
x
We then put the vectors in
component form and add them.
280,0  250cos 60, 250sin 60
v
 280  250cos 60   250sin 60
2
2
 459.2
 250sin 60 
  tan 
  28.1
 280  250 cos 60 
1
PROBLEMS
Think of pushing or pulling something up a ramp. Our model
will assume you have a well-oiled dolly and we will neglect
friction. What other forces are there in this problem?
Gravity does
is acting
NOTon
change.
the object
The
so thedue
force
weight
to gravity
of the stays
object is a
force. Gravity
vertical
and of pulls
the same
down so the
force vectorso
magnitude
for
the
the
gravity
weightvector
of
the object
remains
the
is same.
always vertical.
Can you see what
happens to the resultant
force as the ramp gets
steeper?
The force required to move
the object would
is in a direction
need to be
parallel to the ramp.
greater.
If we make the ramp
steeper,
even
steeper
will either
what of
would
theseto
have
forces
change?
change?
Let’s look at the resultant force in each case.
60°
250
30° lb
Workers at the zoo must move a
250-pound giant tortoise to his new
home. Find the amount of force
required to pull him up a ramp
leading into a truck if the angle of
elevation of the ramp is 30°.
First look at this triangle and find the
missing angle.
Now look at this triangle. We can easily find
the other angle and the hypotenuse since it
is part of a parallelogram and parallel to the
250 lb. side.
Okay---can you see how to use trig to find the magnitude of the force
vector to pull the turtle up the ramp?
opp
F
sin 30 

hyp 250
F  250sin30  125 lb.
We can represent the speed and direction of a plane in
still air as a vector. We’d need to add to that, the speed
and direction of the wind. The resultant vector would be
the speed and direction the plane would actually travel.
to northeast
45°
from southwest
In these problems, they will tell you the direction the
wind is coming FROM---NOT the direction it is blowing.
For example: If a wind is blowing from the southwest, it
is blowing towards the northeast at a 45° angle.
bearing
measured
clockwise
from north
In this case,
270° plus the
drift angle.
In air navigation, the bearing is a nonnegative angle
smaller than 360° measured in a clockwise direction
from due north.
An airplane heads west at 350 miles per hour in a 50 mph northwest
wind. Find the ground speed and bearing of the plane.
Let’s draw a picture on coordinate axes.
v  350,0
You could draw the parallelogram and use a
triangle and trig to find the resultant vector
whose magnitude is the groundspeed and
use the angle to determine the bearing, or
you could put these 2 vectors in component
form and add. Let’s do the second way this
time.
This is the
y
northwest
quadrant so
wind would
blow towards
southeast.
315°
v
50
350
u
c
x
u  50cos315,50sin 315
c = v + u  350,0  50cos315,50sin 315
50sin 315

1 
ground speed = c = 316.6 mph
tan 
  6.4
 350  50 cos 315 
An airplane heads west at 350 miles per hour in a 50 mph northwest
wind. Find the ground speed and bearing of the plane.
Remember the bearing is
measured clockwise from north
y
bearing = 270° – 6.4° = 263.6°
v  350,0 v
350
c
u
50
x
5.2°
u  50cos315,50sin 315
c = v + u  350,0  50cos315,50sin 315
50sin 315

1 
ground speed = c = 316.6 mph
tan 
  6.4
 350  50 cos 315 