Determining magnitude and direction of vectors

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Homework Help Overview

The discussion revolves around determining the magnitude and direction of a vector force (F1) such that the resultant force is vertically upward with a magnitude of 800N. The problem involves vector components and their relationships.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss using vector components to solve for F1, suggesting the need to analyze horizontal and vertical components separately. There is mention of confusion regarding how to obtain values for the vectors involved.

Discussion Status

Some participants have provided guidance on breaking down the problem into components, and one participant indicates they have successfully solved the problem with assistance. However, there is no explicit consensus on the approach, as multiple interpretations of the problem are being explored.

Contextual Notes

The original poster expresses uncertainty about obtaining values for the vectors, indicating potential missing information or assumptions that may need clarification.

savva
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Homework Statement


Determine the magnitude and direction θ of F1 so that the resultant force is directed vertically upward and has a magnitude of 800N.

Homework Equations


F1=FR-F2-F3


The Attempt at a Solution


I have attached the question with diagram and my attempt at solving the question. I know that by using the equation above, F1=FR-F2-F3, however I am slightly confused as to obtaining the values for these vectors and how to approach doing that.
 

Attachments

  • Chp 2 - Q2.21 Problem.jpg
    Chp 2 - Q2.21 Problem.jpg
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  • Chp 2 - Q2.21 Problem WORKING.jpg
    Chp 2 - Q2.21 Problem WORKING.jpg
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Solve for F1x using the horizontal components, solve for F1y using vertical components, then take F1x and F1y to solve for F1 and θ
 
Thanks for your help Joffran, I managed to solve the problem with your help, I will add the working I used to solve it tomorrow
 
\SigmaFx=0 ----> 400cos30+F1x-(4/5)(600)=0 ----> F1x=133.6N
\SigmaFy=800 ----> 400sin30+F1y+(3/5)(600)=800 ----> F1y=240N

F1= (F1x^2+F2y^2)^.5 = (133.6^2+240^2)^.5=274.7N

tanθ=F1x/F1y=133.6/240

θ=tan^-1(133.6/240)=29.1
 

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