What is the angle between coupled forces with a given moment and magnitude?

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Homework Help Overview

The problem involves determining the angle between two coupled forces given a specific moment and their magnitudes. The forces are defined in terms of their positions and directions, with a moment of 600 N-m specified.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the moment and the forces, attempting to set up equations based on the moment's definition. There are questions about the components of the forces and how they relate to the overall magnitude.

Discussion Status

Some participants have provided equations based on their interpretations of the problem, while others are questioning the assumptions made regarding the force components and their magnitudes. There is an ongoing exploration of the mathematical relationships without a clear consensus on the next steps.

Contextual Notes

Participants note the challenge of having one equation with two unknowns and the implications of the quadratic discriminant being negative, suggesting potential issues with the assumptions or setup of the problem.

whitejac
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Homework Statement


The moment of the couple is 600k (N-m). What is the angle A?

F = 100N located at (5,0)m and pointed in the positive x and positive y direction
-F = 100N located at (0,4)m and pointed in the negative x and negative y direction

Homework Equations


M = rxF
M = D

The Attempt at a Solution


[/B]
M = 600k
|F| = 100N
r = <5,-4>

rxF = a 3x3 determinant with
<i , j , k>
<5, -4, 0>
<Fx , Fy, 0>
=
5Fy + 4Fx +0 = 600k
but I've only got 1 equation and 2 unknowns now!
 
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I'm guessing that your Fx and Fy are the x and y components of one of the two given forces. Your other equation, then, is the given magnitude of that force.
 
haruspex said:
I'm guessing that your Fx and Fy are the x and y components of one of the two given forces. Your other equation, then, is the given magnitude of that force.

I tried that.
5Fy + 4Fx = 600
25Fy^2 + 16Fx^2 = 10,000 <- (Magnitude formula squared)
Simplifying:

Fx = 150 - 5/4Fy
25Fy^2 + 16(150- 4/5Fy)^2 = 10,000
This yields a quadratic whose discriminant is
6000^2 - 4*50*350000 < 0
 
whitejac said:
25Fy^2 + 16Fx^2 = 10,000 <- (Magnitude formula squared)
why the 25 and the 16? If F has orthogonal components Fx and Fy, what is the magnitude of F?
 

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