Resolving vectors into components

  • Thread starter Gringo123
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  • #1
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Main Question or Discussion Point

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?
 

Answers and Replies

  • #2
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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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