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I think this is a very important point, but why my teacher didn't clarify a lot?

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- Thread starter primarygun
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- #1

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I think this is a very important point, but why my teacher didn't clarify a lot?

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arildno

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No, we don't.primarygun said:

I think this is a very important point, but why my teacher didn't clarify a lot?

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.

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Doc Al

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When you do a free body diagram, you are isolating a particular object and showing all the forces acting on it. These forces are certainlyprimarygun said:When we do free body diagram, we usually consider the vector as scalar.

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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)

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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.

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jtbell

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primarygun said:

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.

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Daniel.

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