Trigonometry - solving force into components and determining magnitude of components

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Homework Statement


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.

Homework Equations


asinθ
bcosθ

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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Answers and Replies

  • #2
SammyS
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Homework Statement


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.

Homework Equations


asinθ
bcosθ

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
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
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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.
Sorry, tried using the information you have given and can't seem to solve it still
 
  • #4
gneill
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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
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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
 

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