Transcript Chapter 3: Two-Dimensional Motion and Vectors

Chapter 3–2:
Vector Operations
Physics
Coach Kelsoe
Pages 86–94
Objectives
• Identify appropriate coordinate systems for
solving problems with vectors.
• Apply the Pythagorean theorem and tangent
function to calculate the magnitude and
direction of a resultant vector.
• Resolve vectors into components using the
sine and cosine functions.
• Add vectors that are not perpendicular.
Coordinate Systems in 2-D
• In Chapter 2 we discussed the motion of the gecko
up the tree by a positive or negative sign. In
Chapter 3 we’re going to describe this motion using
an arrow pointing in a given direction.
• The most versatile system for diagramming the
motion of an object is using the x- and y-axes at the
same time.
• The addition of another axis not only helps describe
motion in two dimensions but also simplifies analysis
of motion in one dimension.
Coordinate Systems in 2-D
• Two methods can be used to describe the
motion of an object .
– In the first approach, the coordinate system can
be turned so that the object moves along the yaxis. The problem with this method is that the
system must be turned again if the motion
changes.
– In the second approach, we keep our axes fixed
so that the positive y-axis points north and the
positive x-axis points east.
Coordinate Systems in Two
Dimensions
• One method for
diagramming the motion of
an object employs vectors
and the use of the x- and yaxes.
• Axes are often designated
using fixed directions.
• In the figure shown here,
the positive y-axis points
north and the positive xaxis points east.
Determining Resultant
Magnitude and Direction
• In the previous section, we learned how to
find resultant vectors graphically. The
problem with this method is that it is very time
consuming and the accuracy depends on how
well the diagram is drawn.
• A more efficient way to find resultant vectors
is to use the Pythagorean Theorem.
• Imagine climbing a pyramid in Egypt. You
know the height and width of the pyramid, but
you want to know the distance covered in a
climb from the bottom to the top.
Pyramid Physics
• The magnitude of the
vertical displacement is
the height of the pyramid.
• The magnitude of the
horizontal displacement is
the distance from one
edge of the pyramid to the
middle, or half the width.
• These two vectors are
perpendicular.
Δy
d
Δx
Pythagorean Theorem
• The Pythagorean theorem
states that for any right
triangle, the square of the
hypotenuse equals the
sum of the squares of the
other two sides, or legs.
• So in this illustration, we
see that
d2 = Δx2 + Δy2
Δy
d
Δx
Don’t Forget Direction!!!
• Once you find the
displacement, it is
important to know the
direction.
• We can use the inverse
tangent function to find
the angle, which denotes
the direction.
• We can use sine or
cosine functions, but our
tangent values are given.
Δy
d
θ
Δx
Determining Resultant
Magnitude and Direction
• The Tangent Function
– Use the tangent function to find the direction of
the resultant vector.
– For any right triangle, the tangent of an angle is
defined as the ratio of the opposite and adjacent
legs with respect to a specified acute angle of a
right triangle.
Opposite leg
Tan θ =
Adjacent leg
Sample Problem
• Finding Resultant Magnitude and Direction
An archaeologist climbs the Great Pyramid in
Giza, Egypt. The pyramid’s height is 136 m
and its width is 2.30 x 102 m. What is the
magnitude and the direction of the
displacement of the archaeologist after she
has climbed from the bottom of the pyramid to
the top?
Sample Problem
• Define
– Given:
• Δy = 136 m
• Δx = ½(width) = 115 m
– Unknown:
• d=?
• θ=?
– Diagram:
• Choose the archaeologist’s starting position as the
origin of the coordinate system, as shown above.
Sample Problem
• Plan
– Choose an equation or situation:
• The Pythagorean theorem can be used to find the
magnitude of the archaeologist’s displacement. The
direction of the displacement can be found by using the
inverse tangent function.
d2 = Δx2 + Δy2
Δy
tan θ = Δx
– Rearrange the equations to isolate the unknowns:
d = √Δx2 + Δy2
Δy
θ = tan-1 Δx
Sample Problem
• Calculate
d = √Δx + Δy
d = √1152 + 1362
d = 178 m
• Evaluate
Δy
θ = tan-1 Δx
136
θ = tan-1 115
θ = 49.8°
• Because d is the hypotenuse, the archaeologist’s
displacement should be less than the sum of the height
and half of the width. The angle is expected to be more
than 45° because the height is greater than half width.
Resolving Vectors
• In the previous example, the horizontal and
vertical parts that add up to give the actual
displacement are called components.
• The x component is parallel to the x-axis and
the y component is parallel to the y-axis.
• You can often describe an object’s motion
more conveniently by breaking a single vector
into two components, a process we call
resolving the vector.
Resolving Vectors Into
Components
• Consider an airplane flying at 95 km/hr.
– The hypotenuse (vplane) is the resultant vector
that describes the airplane’s total velocity.
– The adjacent leg represents the x component
(vx), which describes the airplane’s horizontal
speed.
– The opposite leg represents the y component
(vy), which describes the airplane’s vertical
speed.
Resolving Vectors Into
Components
• The sine and cosine functions can be used to
find the components of a vector.
• The sine and cosine functions are defined in
terms of the lengths of the sides of right
triangles.
Sine of angle θ =
opposite
hypotenuse
Cosine of angle θ =
adjacent
hypotenuse
Adding Vectors That Are
Not Perpendicular
• Suppose that a plane travels first 5 km at an angle
of 35°, then climbs at 10 for 22 km, as shown below.
How can you find the total displacement?
• Because the original displacement vectors do not
form a right triangle, you can not directly apply the
tangent function or the Pythagorean Theorem.
d2
d1
Adding Vectors That Are
Not Perpendicular
• You can find the magnitude and the direction
of the resultant by resolving each of the
plane’s displacement vectors into its x and y
components.
• Then the components along each axis can be
added together.
d2
d1
Δx1
Δx2
Δy2
Δy1
Adding Non-Perpendicular
Vectors
• The goal of resolving vectors is to allow us to
have components that are aligned along the
x- and y-axis.
• Once the components are set up on these
axes, our math becomes simple– we add our
x components together and y components
together to find our new x and y components.
• With our new-found x and y components, we
can find our new resultant vector.
Sample Problem
• Adding vectors algebraically
A hiker walks 27.0 km from her base camp at
35° south of east. The next day, she walks
41.0 km in a direction of 65° north of east and
discovers a forest ranger’s tower. Find the
magnitude and direction of her resultant
displacement.
Sample Problem Solution
• Select a coordinate system, then sketch and
label each vector.
– Given
d1 = 27.0 km
d2 = 41.0 km
θ1 = -35°
θ2 = 65°
• Tip: θ1 is negative because
clockwise movement from
the positive x-axis is negative.
– Unknown
d=?
θ=?
Sample Problem Solution
• Find the x and y components of
both vectors.
– Separate your displacements.
– For Day 1:
Δx1 = d1cos θ1 = (27.0 km)(cos -35°) =
22 km
Δy1 = d1sin θ1 = (27.0 km)(sin -35°) =
-15 km
– For Day 2:
Δx2 = d2cos θ2 = (41.0 km)(cos 65°) =
17 km
Δy2 = d2sin θ2 = (41.0 km)(sin 65°) =
37 km
Sample Problem Solution
• Find the x and y components of the total
displacement.
Δxtotal = Δx1+ Δx2 = 22 km + 17 km = 39 km
Δytotal = Δy1+ Δy2 = -15 km + 37 km = 22 km
• Use the Pythagorean theorem to find the
magnitude of the resultant vector.
d2 = (Δxtotal)2 + (Δytotal)2
d = √(39 km)2 + (22 km)2
d = 45 km
Sample Problem Solution
• Use a suitable trigonometric function to find
the angle.
Tan θ = opposite/adjacent
Tan θ = Δy/Δx = 22 km/39 km
Tan θ = 0.564
θ = 29° north of east
Vocabulary
• Components of a vector
Important Formulas
•
•
•
•
c2 = a2 + b2
sin θ = opposite/hypotenuse
cos θ = adjacent/hypotenuse
tan θ = opposite/adjacent