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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

force

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! enter image description here

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

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

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

J.Doe
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    What is $|FG|$ ? the "module" of the force ? If so, it is a positive number and thus $0.5$ (without minus) is correct. – Mauro ALLEGRANZA Dec 12 '17 at 14:13
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    The book appears to have that wrong. I disagree with your "logical conclusion" since $F_{s1}$ looks to be a vector, meaning there is no sense in saying it is equal to a scalar. – John Doe Dec 12 '17 at 14:16
  • @Mauro Allegranza Hopefully I do not make a mistake by translation, because I am no native speaker: |FG| is the magnitude of the gravitational force=m*g. So In my opinion it is positive. – J.Doe Dec 12 '17 at 14:19
  • @JohnDoe thank you, I totally forgot to mark the forces as vectors. My mistake, I added this Information to the question. – J.Doe Dec 12 '17 at 14:21
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    magnitude can't be negative so the minus sign is a mistake – Vasili Dec 12 '17 at 14:31
  • So where does this minus come from. Is it a mistake by the book like JohnDoe mentioned? – J.Doe Dec 12 '17 at 14:31

1 Answers1

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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:

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

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.

Mauro ALLEGRANZA
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