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Doing free body diagram 
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#1
Feb1005, 04:41 AM

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When we do free body diagram, we usually consider the vector as scalar.
I think this is a very important point, but why my teacher didn't clarify a lot? 


#2
Feb1005, 07:50 AM

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PF Gold
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What you might be thinking of, is that we often write the vector as a MAGNITUDE (a nonnegative scalar quantity) multiplied with a DIRECTION (a unit vector) The direction is, of course, "readily" seen from the diagram. 


#3
Feb1005, 07:51 AM

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#4
Feb1105, 02:07 AM

P: 233

Doing free body diagram
Note: Forces in xaxis are 5N and 5N. Find the acceleration.
My teacher would suggest us to F=ma 5N5N=ma Hence, a=0. The equation is the same,though, he never tells us 5N+(5N) 


#5
Feb1105, 02:11 AM

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Forces are vectors.Period.The laws of Newton must be written in vector form...ALWAYS.
In your example,what if one of the forces would act as to make an angle (different from 0 or pi) with the direction of the other force...? Daniel. 


#6
Feb1105, 07:28 AM

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[tex] \bold F_{net} = m \bold a [/tex] [tex] {\bold F}_1 + {\bold F}_2 = m \bold a [/tex] [tex] F_{1x} + F_{2x} = m a_x [/tex] [tex] (+5N) + (5N) = m a_x [/tex] [tex] 5N  5N = m a_x [/tex] [tex] 0 = m a_x [/tex] [tex] 0 = a_x [/tex] And of course if the ycomponents of the forces are zero, then [tex]a_y[/tex] is zero also, so [tex] \bold a [/tex] (the vector) equals zero. But nobody ever actually writes out all those steps, in practice. I might do it that way once, when teaching it, just to clarify things. 


#7
Feb1105, 11:02 AM

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The way the problem is actually posted,it says nothing about an accelereration (or simply a nonzero velocity) in the "y" direction,so even if you come up with the conclusion that a_{x}=0,you still wouldn't tell how that body's moving.
Daniel. 


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