# Homework Help: Angle between coupled forces

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1. Sep 30, 2015

### whitejac

1. The problem statement, all variables and given/known data
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

2. Relevant equations
M = rxF
M = D

3. The attempt at a solution

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!

2. Sep 30, 2015

### haruspex

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.

3. Sep 30, 2015

### whitejac

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

4. Sep 30, 2015

### haruspex

why the 25 and the 16? If F has orthogonal components Fx and Fy, what is the magnitude of F?