All language subtitles for LS1U5B2

So let's think about this direction
a little bit more.

I'm going to redraw this just to
make a little less junk on it.

Let's see, this was x.

This was y.

And our problem is, between 3 and 1,
we have a unit vector like that.

That's the direction of our
force, r-hat 3 to 1.

And we need to know that direction,
because we need to break it into x and

y-coordinates.

This one does not just naturally
rely on an axis.

So it's hard to see the direction,
here, but you can

actually see it's up here.

You can see it in here, because
here is the direction.

And this is on a right triangle
that we know everything about.

So we know this side of this
right triangle is 0.6.

And we know this side is 0.15.

So since we know those two things,
we do know that angle.

So you've go to think back
to your sohcahtoa.

I don't know if they still....

Sohcahtoa is a word we learned to help
us do trig, at least when I was young.

So sine is opposite over hypotenuse.

Cosine is adjacent over hypotenuse.

And tangent is opposite over adjacent.

So we know that the tangent
of theta is 0.15/0.6.

So if you throw that in your calculator,
theta is about 14 degrees.

So that theta is the
same as that theta.

This is still the x-axis.

So now we can break it
into components.

And you can see that it's going to
have a negative y component and a

negative x component.

Let's see, so we could then
write this F 3-1.

So starting with what we had before,
F 3-1 would be the magnitude, 2.35,

times the sine and cosine components.

But it's also negative.

Just by inspection, just by looking at
it, we can tell, this axis has a

negative component.

This axis has a negative component.

So we're going to put in the
negatives, minus 2.35--

and remember it's Newtons, so
I'm not writing the units--

times the cosine of 14 degrees.

That is the i-hat.

That's the x-axis, because the cosine
is the adjacent component.

And then also minus 2.35 time the sine,
in this case, of 14 degrees.

And that's the y component.

Also negative, because this is pointing
towards negative x and

negative y.

So if you put those together, you get
minus 2.28 Newtons i-hat minus 0.57--

oops, I wasn't writing that--

Newtons j-hat.

So there we've broken the F 3-1 into the
two Cartesian components, on the

two Cartesian axes.

So now you do the entire sum.

So you say F1.

And you just go through.

And you add up the i-hat part
of each of those vectors.

And I'll give you the numbers.

The answer is minus 4.78 i-hat.

And then you add up all of the
j hat numbers, the minus 0.57

and the minus 40.

And you get minus 40.57 j hat.

And the whole thing is,
of course, Newtons.

So that's the final force on charge 1.

And I think the question asked for
the magnitude and the direction.

So the direction is 14 degrees
below the x-axis.

And if you wanted the magnitude, it
would be the square root of this one

squared plus that one squared.

If you wanted the total force.