Understanding How to Solve Vector Magnitudes and Angles

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SUMMARY

This discussion focuses on solving vector magnitudes and angles, specifically using components in a statics context. The user describes a method for determining the x and y components of vectors AB and BE, utilizing trigonometric functions such as cosine and arctangent. The components are expressed as fractions of the magnitudes, specifically $\frac{3}{5}$ and $\frac{4}{5}$ of a given vector magnitude of -9.38. The conversation highlights the importance of setting the sum of components equal to zero for equilibrium in statics problems.

PREREQUISITES
  • Understanding of vector components in two dimensions
  • Familiarity with trigonometric functions, specifically cosine and arctangent
  • Knowledge of statics and equilibrium conditions
  • Ability to solve systems of equations
NEXT STEPS
  • Study vector decomposition techniques in physics
  • Learn how to apply trigonometric functions in vector analysis
  • Explore Free Body Diagrams and their applications in statics
  • Practice solving equilibrium problems involving multiple vectors
USEFUL FOR

This discussion is beneficial for physics students, engineers, and anyone involved in mechanics or statics, particularly those working with vector analysis and equilibrium conditions.

bergausstein
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I just don't understand others way solving this problem

to solve for the unknown magnitudes of the two vectors I have to sum up all of the components in x and y and set them equal zero and from there I'll get some systems of equation.

I saw a method where they let the x-component of AB and BE to be

$\displaystyle\frac{3}{5}\cdot -9.38$ and $\displaystyle\frac{3}{5}\cdot BE$ respectively.

and the y-components of AB and BE to be

$\displaystyle\frac{4}{5}\cdot -9.38$ and $\displaystyle\frac{4}{5}\cdot -BE$ respectively

Can you explain why they do that?

and how can we solve this using angle

I know that $\arctan(\frac{4}{3})=53.1\deg$ but I'm uncertain on how to plug it in my equation properly I also know that to get the x-component I have to multiply the magnitude of the vector I'm interested into by the $\cos(53.1)$ in the problem given. but this configuration makes me wonder if I'm doing it correctly. hope you can help me with this. thanks!

 

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I think some context would be good here. Is this a statics problem? I'm seeing forces on your diagram, which is typically what you find on a Free Body Diagram. If you have to set all the components equal to zero, I think you're going to have a hard time of it, as none of the vectors have any positive $y$-component.
 

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