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In previous articles we covered the topic of resolving vectors, mainly the headings: how to resolve a vector, definitions and a complete guide on how to go about splitting/resolving a vector into its perpendicular components and how to label them, in this article we will look at some examples on how to resolve vectors.

To read more on resolving vectors click here

Question # 1

Calculate the horizontal and vertical components of a 50N force which is acting 40 degrees to the horizontal.

vector
Vector


Answer: 


The question asked for the values of the horizontal and vertical components, so first you need to split the vector seen in the diagram above into its horizontal and vertical components:


resolved vector
Resolved Vector


Now that you have identified both components the next thing that needs to be done is to label the components. Look at the diagram above and find out which of the components is adjacent to the given angle and label it “F×cosθ”, where F is the force and θ is the given angle. Also find out which component is opposite to the given angle and label it “F×sinθ”. In this example the component adjacent to the given angle is the horizontal component so it is labeled 50×cos 40 degrees. The component opposite to the given angle is the vertical component so it is labeled 50×sin40 degrees.


Resolved Vector
Resolved Vector



Therefore your answer should be:

Horizontal component = 50×cos40
                                  = 38.30N

Vertical Component = 50×sin40
                               = 32.14N
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Projectile Motion
What is the projectile motion of a body?
- A body in projectile motion will move in two dimensions simultaneously. Bodies that are in projectile motion are resolved(divided) into two components. A vertical component(Vy) and an horizontal component(Vx). These components however are dealt with simultaneously.


Please refer to the diagram below showing projectile motion.

Projectile motion






Formulas associated with projectile motion include:

  • Vy = VsinA
  • Vx = VcosA
  • V = Sqrt [Vy^2 + Vx^2]
  • TanA = Vy / Vx => A = Tan^-1 A (Vy / Vx)
Where A is used as angle instead of theta in diagram

Example calculation involving projectile motion:
  1. A body is projected with a velocity of 200m/s at and angle of 30 °C to the horizontal.
  • Calculate the time taken to reach its maximum height.
  • Calculate its velocity after 16 seconds.

Answers:

  • Finding the time for maximum height.
    angle A = 30 °C
              V = 200m/s
    Acceleration(a) = -10 m/s2
    Vy = 0 m/s

    Therefore Uy = VsinA                                                [ Note: Uy represents initial velocity]
                      Uy = 200 x Sin 30
                      Uy = 100 m/s

                      
                      Vy = Uy = at
                      Vy - Uy = at
                            a
                       0 - 100 = t
                          - 10
                          10 sec = t

    Therefore Maximum Time (t) = 10 seconds

    • Finding the velocity after 16 seconds.

                      Vx = V cos A 
                      Vx = 200 x Cos 30
                      Vx = 173.2 m/s

                      Vy = Uy + at
                      Vy = 100 - 10 (16)
                      Vy = -60 m/s

    Therefore V =  √[Vy^2 + Vx^2]
                          =  √[(-60)^2 + (173.2)^2]
                          =  √33598
                          =  183.30 m/s

                    
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