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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
• Using cosine • Using tangent
2 2 2 2290 50 84100 2500 81600 286b c a= − = − = − = =
( ) cos (290)cos9.93 286a h θ= = =
50 50 286tan 9.93 0.175
Oppositeatanθ
= = = =
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