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September 22nd, 2015, 12:34 PM   #1
szz
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Doubt with a free body diagram problem

Hi,

I am strongly in doubt with an exemple of free body diagram and I hope you will make me understand.
This example is that one shown in the attachment:



For which I don't have any explanation. Just this representation.

Where we have:

$\displaystyle \begin{aligned}
& \boldsymbol w = (0,-w_y)\\
& \boldsymbol F_y = (0, F_y)\\
& \boldsymbol F_x = (0, F_x)\\
& \boldsymbol F_\mathrm{block\ on\ runner} = \boldsymbol F
\end{aligned}$

And from the representation I would formulate that:

$\displaystyle \boldsymbol F = \sum \boldsymbol F_x + \boldsymbol w + \boldsymbol F_y \qquad(1) $

But if:

$\displaystyle \boldsymbol w + \boldsymbol F_y = 0 \qquad(2) $

How can be that:

$\displaystyle \boldsymbol F \neq \boldsymbol F_x \qquad(3) $

and:

$\displaystyle \boldsymbol F = \boldsymbol F_x + \boldsymbol F_y \qquad(4) $

as the representation implies ? Should not be (at least this is what would formulate):

$\displaystyle \boldsymbol F = \boldsymbol F_x \qquad(5) $

due to equation (2) ?

I think I am missing some particular important concept.
But I don't understand the example.

Thank you in advance for your help.
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File Type: jpg example.jpg (14.5 KB, 21 views)

Last edited by szz; September 22nd, 2015 at 12:37 PM.
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September 22nd, 2015, 02:19 PM   #2
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your equation (1) should be ...

$\displaystyle F_{net} = \sum F_x + w + F_y = F_x$
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September 22nd, 2015, 03:03 PM   #3
szz
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OK, thank You, that's what I was thinking for 3 days, which is what I deduced in equation (5).

So why is represented $\displaystyle \boldsymbol F_\mathrm {block\ on\ runner}$ ?

Just to confuse the ideas ?
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September 22nd, 2015, 03:34 PM   #4
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Note that the only two forces acting on the runner are $w$ and $F_{block on runner}$

$\displaystyle F_{net} = \sum w + F_{block on runner}$

$F_y$ and $F_x$ are the components of $F_{block on runner}$ ...

$\displaystyle F_{net} = \sum w + F_y + F_x$

since $w$ and $F_y$ create an equilibrium condition in the y-direction, the net force is simply the x-component of $F_{block on runner}$, which is $F_x$, and the runner's acceleration will be strictly in the x-direction.
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September 22nd, 2015, 04:20 PM   #5
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Thank you! Now it's more more & more clear !
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