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Doing free body diagram

by primarygun
Tags: body, diagram, free
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primarygun
#1
Feb10-05, 04:41 AM
P: 233
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?
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arildno
#2
Feb10-05, 07:50 AM
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Quote Quote by primarygun
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?
No, we don't.
What you might be thinking of, is that we often write the vector as a MAGNITUDE (a non-negative scalar quantity) multiplied with a DIRECTION (a unit vector)
The direction is, of course, "readily" seen from the diagram.
Doc Al
#3
Feb10-05, 07:51 AM
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Quote Quote by primarygun
When we do free body diagram, we usually consider the vector as scalar.
When you do a free body diagram, you are isolating a particular object and showing all the forces acting on it. These forces are certainly vectors, usually depicted as arrows in the diagram. What makes you think you are treating them as scalars?

primarygun
#4
Feb11-05, 02:07 AM
P: 233
Doing free body diagram

Note: Forces in x-axis are 5N and -5N. Find the acceleration.
My teacher would suggest us to
F=ma
5N-5N=ma
Hence, a=0.
The equation is the same,though, he never tells us 5N+(-5N)
dextercioby
#5
Feb11-05, 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.
jtbell
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Feb11-05, 07:28 AM
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Quote Quote by primarygun
Note: Forces in x-axis are 5N and -5N. Find the acceleration.
My teacher would suggest us to
F=ma
5N-5N=ma
Hence, a=0.
The equation is the same,though, he never tells us 5N+(-5N)
It would make things more explicit to write out all the steps like this:

[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 y-components 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.
dextercioby
#7
Feb11-05, 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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