Solve Resultant Forces Homework Problem 1.23 | Attached Figure & Variables

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Homework Help Overview

The discussion revolves around a problem involving resultant forces, specifically focusing on vector resolution into components. The original poster is attempting to solve a problem related to forces and their resultant values, referencing a figure and given variables.

Discussion Character

  • Exploratory, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • The original poster attempts to apply the Pythagorean theorem to the components of forces but expresses difficulty in proceeding. Some participants suggest resolving the vectors into their x and y components, questioning the correctness of their equations.

Discussion Status

Participants are actively engaging with the problem, offering guidance on how to approach the resolution of vector components. There is a recognition of potential corrections needed in the equations presented, indicating a collaborative effort to clarify the setup.

Contextual Notes

There is mention of a figure and specific angles associated with the vectors, which may influence the interpretation of the problem. The original poster has indicated a prior problem was solved, suggesting a continuity in the learning process.

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Homework Statement


See Attachments for Figure and Given Variables

Homework Equations


R^2 = Fx^2 + Fy^2
@=tan^-1 (Fy/Fx)

The Attempt at a Solution


I've already solved problem 1.22
I just need help at Problem 1.23
in problem 1.23
given the resultant as 4m/s, to the right
I tried solving it
4^2 = Fx^2 + Fy^2
But i can't seem to proceed from here.
I need help
 

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Try resolving the two vectors into components in the x and y directions. The sum of the x components has to equal 4, and the sum of the y components has to equal zero.
 
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sum of x = V1cos60 + v2cos30 = 4?
and
sum of y = v1sin60+v2sin30 = 0?
Corrent me if I am wrong.
 
MrMechanic said:
sum of x = V1cos60 + v2cos30 = 4?
and
sum of y = v1sin60+v2sin30 = 0?
Corrent me if I am wrong.
The 2nd eqn. should have a minus sign. See the figure.

Otherwise, great job.

Chet
 
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Likes   Reactions: 1 person

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