# Trigonometry - solving force into components and determining magnitude of components

1. Mar 4, 2012

### savva

1. The problem statement, all variables and given/known data
Resolve the force F1 into components acting along u and v axes and determine the magnitudes of the components. Refer to first attachment for full problem with diagram.

2. Relevant equations
asinθ
bcosθ

3. The attempt at a solution
Please refer to attachments for attempt at solution, drawings of diagram involved. Could not get the correct answer solving the way in which I attempted, can anybody give me a hand solving it, answers in the book were:
f1u=205N, f2v=160N

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2. Mar 4, 2012

### SammyS

Staff Emeritus
Re: Trigonometry - solving force into components and determining magnitude of compone

The axes are not orthogonal.

The angle you have labeled as 60° is 40°. 30° + 40° = 70° .

(F1)v is a vector that's parallel to the v axis.

(F1)u is a vector that's parallel to the u axis.

(F1)v + (F1)u = F1

Superimpose an xy axis system onto the uv system, if that helps you.

3. Mar 4, 2012

### savva

Re: Trigonometry - solving force into components and determining magnitude of compone

Sorry, tried using the information you have given and can't seem to solve it still

4. Mar 4, 2012

### Staff: Mentor

Re: Trigonometry - solving force into components and determining magnitude of compone

Start with force vector F1. Your last drawing has the required angles labelled. What are the expressions for the u component of F1 and the v component of F1? Remember that u and v are not orthogonal (not at 90° to each other), so while the projection on one axis may be Fcos(θ), the projection on the other axis won't be Fsin(θ). Choose appropriate angles and use the cosine.

5. Mar 13, 2012

### savva

Re: Trigonometry - solving force into components and determining magnitude of compone

I managed to solve it guys, thanks for your help.
I used the parallelogram law to split up the vector into components and used sin law to find relevant information, so:

300/sin110=v/sin30
v=sin^-1(300sin30/sin110) = 160N

300/sin110=u/sin40 ---> u=sin^-1(300sin40/sin110) = 205N

Cheers