My Math Forum Doubt with a free body diagram problem

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September 22nd, 2015, 12:34 PM   #1
Senior Member

Joined: Oct 2014
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Posts: 220
Thanks: 26

Math Focus: Calculus
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.

Attached Images
 example.jpg (14.5 KB, 20 views)

Last edited by szz; September 22nd, 2015 at 12:37 PM.

 September 22nd, 2015, 02:19 PM #2 Math Team   Joined: Jul 2011 From: Texas Posts: 2,515 Thanks: 1238 your equation (1) should be ... $\displaystyle F_{net} = \sum F_x + w + F_y = F_x$ Thanks from szz
 September 22nd, 2015, 03:03 PM #3 Senior Member     Joined: Oct 2014 From: I don't know... Posts: 220 Thanks: 26 Math Focus: Calculus 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 ?
 September 22nd, 2015, 03:34 PM #4 Math Team   Joined: Jul 2011 From: Texas Posts: 2,515 Thanks: 1238 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. Thanks from szz
 September 22nd, 2015, 04:20 PM #5 Senior Member     Joined: Oct 2014 From: I don't know... Posts: 220 Thanks: 26 Math Focus: Calculus Thank you! Now it's more more & more clear !

 Tags body, diagram, doubt, free, problem

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