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Maths Question - algebra / forces?
I think this is pretty easy, but I'm having a moment, and it's been over 14 years since I've done this stuff.
It's to do with vectors, and stuff... so the hypotenuse would be the net force which I want to break down into x,y components.
If I've got a triangle, say a 3,4,5 for easy maths: These are measurements in meters.
I'm actually calculating forces, and have calculated force of 20 for the 5m side of the triangle (hypotenuse).
I'm trying to figure out how to calculate the forces for other sides of the triangle.
Effectively I think I'm trying to enlarge a 3,4,5 triangle.... and having been through simultaneous equasions, I think I've REALLY overcomplicated something that should be simple.
Bear in mind, that these figures are just figures I'm using for testing, I need the method to convert x,y forces into z, and vice-versa.
Many thanks!
i've sorted it already (thanks for the replies anyway), I was just mixing up forces and distances in my head... so I was trying to equate two seperate things.
2 Answers
- jsardi56Lv 71 decade agoFavorite Answer
You're right. In this case you're working with similar triangles, and you can set up a simple proportion:
Short leg:
a/3 = 20/5
5a = 60
a = 12 units of force
Long leg:
b/4 = 20/5
5b = 80
b = 16 units of force
- 1 decade ago
just use pythagoras. a^2 +b^2=c^2 so if your triangle is 3, 4, 5 the 5= force of 20 thenn each meter represents a force of 4 so side a is 12 side b is 16. it is all proportional.
