September 22nd, 2015, 12:34 PM  #1 
Senior Member Joined: Oct 2014 From: I don't know... 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. Thank you in advance for your help. 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: 2,591 Thanks: 1283 
your equation (1) should be ... $\displaystyle F_{net} = \sum F_x + w + F_y = F_x$ 
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: 2,591 Thanks: 1283 
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 ydirection, the net force is simply the xcomponent of $F_{block on runner}$, which is $F_x$, and the runner's acceleration will be strictly in the xdirection. 
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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