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

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SUMMARY

The discussion focuses on calculating the angle between coupled forces given a moment of 600 kN-m and a force magnitude of 100 N. The forces are defined at specific coordinates, with one force acting in the positive x and y directions and the other in the negative x and y directions. The solution involves using the moment equation M = rxF and setting up a system of equations to solve for the components of the forces. The quadratic equation derived from the magnitude formula indicates that the discriminant is negative, suggesting no real solutions exist for the given parameters.

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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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