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Resolving vectors into components

  1. Nov 13, 2009 #1
    I have just looked at a problem in an A level physics book which asks you to work out the forward force acting on a boat. There are 7 N of force going straight ahead and a further 5 N of force going 30 degrees to the left. In order to work out the forward force of the boat the book says this:

    First you need to find the amount of the 5 N force that acts in the forward direction, using trigonometry:

    Part of 5 N force in forward direction = 5 cos 30° = 4.3 N

    Then this can be added to the 7 N force:

    4.3 + 7 = 11.3 N force in the forward direction.

    How do I work out that 5 cos 30° = 4.3 N?
     
  2. jcsd
  3. Nov 13, 2009 #2
    It stems from the trig identity

    [tex]\cos \theta = \frac{x}{r}[/tex]

    In this case, x represents your forward direction (draw yourself a sketch to see) and r represents the force vector. The rest is then simply algebraic manipulation of this identity and a final application of the trigonometric rules to find the value of x.

    Makes sense?
     
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