Vector Resolution
Objectives
1. Define sine, cosine and tangent
2. Resolve vectors into horizontal and vertical
components
General Information
• Resultants are the combination of ANY two
vectors
• Vectors that do not form right triangles can be
broken down into the components that form a
right triangle
– Horizontal and vertical components can be
combined by addition
– Those two types of components form right
triangles
Advice
•
•
•
•
Always draw a diagram
Include the reference angle in your diagram
Sketch in the horizontal and vertical components
Make sure you accurately represent the exact
direction of the components
– The two components should combine head-to-tail
– Be sure the components are drawn as arrows
SOH CAH TOA
• Sine = opposite ÷ hypotenuse
• Cosine = adjacent ÷ hypotenuse
• Tangent = opposite ÷ adjacent
So What Now?
• Multiply the magnitude of the hypotenuse by
the sine of the angle for the magnitude of the
opposite side.
• Multiply the magnitude of the hypotenuse by
the cosine of the angle for the magnitude of the
adjacent side.
Common Pitfalls
• Vector resolution does not use the inverse
functions if you already know the angle.
• Use your diagram to figure out which
components are the opposite and adjacent
sides
• You can use the inverse of sine or cosine to
calculate the angle
1. What are the vertical and horizontal
components of a velocity vector with a
magnitude of 75.3 m/s at 27° E of S?
Vector Diagram
Draw in Components
Horizontal Component
• In this case the horizontal component is the
opposite side
• Use sine
• Sin 27 x 75.3 = 0.454 x 75.3 = 34.2
34.2 m/s east
Vertical Component
• In THIS case the vertical component is the
adjacent side.
• Use cosine
• cos 27 x 75.3 = 0.891 x 75.3 = 67.1
67.1 m/s south
2. What are the horizontal
and vertical components of
a 55 m/s vector at 215º?
Angle Calculation
•
•
•
•
215º is not suitable for the calculator
Need to convert it to an angle less than 90º
The vector is going southwest
Could convert it one of two ways
• 270 – 215 = 55º S of W
• 215 – 180 = 35º W of S
Sketch
Add In
Components
Adjacent Side
• In THIS case, the vertical component is the
adjacent side of the triangle.
• Since you know the angle and the hypotenuse,
use the cosine
• Cos 35 x 55 = 0.819 x 55 = 45
45 m/s south
Opposite Side
• In THIS case, the horizontal component is the
opposite side of the triangle.
• Since you know the angle and the hypotenuse,
use the sine
• Sin 35 x 55 = 0.574 x 55 = 31.5
31.5 m/s West
3. 365 m/s at an angle
15º to the horizon
Information
• Since no specific direction is given, the best
you can say is that the two components go up
and across
• This is the common description for artillery
shots
Sketch
Horizontal Component
• In THIS case the horizontal component is the
adjacent side.
• Use Cosine
• cos 15 x 365 = 0.966 x 365 = 353
353 m/s across
Vertical Component
• In THIS case the vertical component is the
opposite side.
• Use sine
• sin 15 x 365 = 0.259 x 365 = 94.5
94.5 m/s up
4. A pilot wishes to fly his plane to an airport
north of his current location. The plane has a
speed of 290 m/s.
A. If he is confronted by a wind blowing
east at 50 m/s, in what direction will
he need to head in order to reach the
desired destination?
Questions About the Question
• If the pilot heads due north, where will he end
up?
– East of where he wants to be
• So in what general direction must he head?
– Northwest
• Draw the sketch to reflect this.
Sketch
Angle Calculation
•
•
•
•
•
•
•
The question only asks for direction.
Only need to find the angle
What information is known about the triangle?
Which function do you use?
Sin = O ÷ H = 50 ÷ 290 = 0.172
To get the angle, use the inverse
Sin-1 0.172 = 9.93º
9.93º W of N
B. How long will it take him to reach the
airport if it is 800 km away?
• 800 km = 800,000 m
• The airport is due north of where the plane is.
• The velocity vector and displacement need to be in
the same direction. (north)
• You need to solve for the vector heading due north.
• There are three ways to do this
– Use Pythagorean Theorem
– Use cosine
– Use tangent
The Three Methods
• Using Pythagorean Theorem
b=
c −a =
2
2
290 − 50 =
2
2
84100 − 2500 =
81600 = 286
• Using cosine
=
a (=
h) cos θ (290) cos
=
9.93 286
• Using tangent
Opposite
50
50
a
=
=
= = 286
tanθ
tan 9.93 0.175
Time Calculation
• The airport is 800,000 m away.
• The plane is moving north at 286 m/s
• t = d ÷ v = 800000 ÷ 286 = 2797 seconds
Or about 46 minutes and 37 seconds