Algebra vs physics

Question

Before you read this, this is no homework request and I do not expect a solution since I also have written down a solution for the problem (see pictures) . I am not asking for solving the equation but for the following:

Can it be that the algebraic solution is different to the physical solution - So like in this example that we paste a "logical minus"?

I have calculated a force and looked up the solution. But I have a contradiction between algebraic and physicaI solution, I really got stucked on.

This is the force diagramme

This is my solution. The last line is the logical conclusion, the prelast is the algebraic solution. Here you can see that this last minus is just added for logical reasons by myself. Sorry I have forgotten to mark all forces as vectors!

And this is the "logical minus" in the author's solution which seems also just a logical but an algebraic conclusion.

So am I correct that this negative sign is added due to logical but algebraic reasons? Since when I solve this equation algebraicly I would just end up at

$|F_{S,1}|=0.5\cdot|F_G|$

Answer 1

You are right: there is no minus sign "popping up" from nowhere.

$\sin 30 = \cos 60 = \dfrac 1 2$

$\sin 60 = \cos 30 = \dfrac {\sqrt 3} {2}$.

Thus:

$|F_{S,1}| \sin 30 + |F_{S,1}| \dfrac { \cos 30 } { \cos 60 } \sin 60 = |F_{S,1}| \dfrac 1 2 + |F_{S,1}| \dfrac { \dfrac {\sqrt 3} {2} } { \dfrac 1 2 } \dfrac {\sqrt 3} {2} = |F_{S,1}| \dfrac 1 2 + |F_{S,1}| \dfrac { \dfrac 3 4 } { \dfrac 1 2 } = |F_{S,1}| \dfrac 1 2 + |F_{S,1}| \dfrac 3 2= 2 |F_{S,1}|.$

Thus:

$|FG|=2 |F_{S,1}|.$

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Algebra and physics agree: the magnitude of a force is a non-negative number by definition and thus it is "modelled" with the absolute value function, which is always non-negative.