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Welcome back.

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We'll now do another tension
problem and this one is just a

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slight increment harder than the
previous one just because

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we have to take out slightly
more sophisticated algebra

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tools than we did
in the last one.

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But it's not really
any harder.

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But you should actually see this
type of problem because

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you'll probably see
it on an exam.

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So let's figure out the
tension in the wire.

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So first of all, we know
that this point

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right here isn't moving.

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So the tension in this little
small wire right here is easy.

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It's trivial.

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The force of gravity is pulling
down at this point

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with 10 Newtons because you
have this weight here.

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And of course, since this
point is stationary, the

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tension in this wire has to
be 10 Newtons upward.

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That's an easy one.

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So let's just figure out the
tension in these two slightly

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more difficult wires to figure
out the tensions of.

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So once again, we know that this
point right here, this

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point is not accelerating
in any direction.

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It's not accelerating in the
x direction, nor is it

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accelerating in the vertical
direction or the y direction.

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So we know that the net forces
in the x direction need to be

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0 on it and we know the
net forces in the y

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direction need to be 0.

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So what are the net forces
in the x direction?

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Well they're going to be the x
components of these two-- of

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the tension vectors of
both of these wires.

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I guess let's draw the tension
vectors of the two wires.

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So this T1, it's pulling.

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The tension vector pulls in
the direction of the wire

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along the same line.

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So let's say that this is the
tension vector of T1.

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If that's the tension vector,
its x component will be this.

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Let me see how good
I can draw this.

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It's intended to be a straight
line, but that

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would be its x component.

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And its x component, let's
see, this is 30 degrees.

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This is 30 degrees right here.

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And hopefully this is a bit
second nature to you.

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If this value up here
is T1, what is the

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value of the x component?

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It's T1 cosine of 30 degrees.

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And you could do your
SOH-CAH-TOA.

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You know, cosine is adjacent
over hypotenuse.

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So the cosine of 30 degrees is
equal to-- This over T1 one is

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equal to the x component
over T1.

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And if you multiply both sides
by T1, you get this.

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This should be a little bit of
second nature right now.

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That the x component is going to
be the cosine of the angle

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between the hypotenuse and
the x component times the

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hypotenuse.

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And similarly, the x component
here-- Let me

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draw this force vector.

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So if this is T2, this would
be its x component.

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And very similarly, this is 60
degrees, so this would be T2

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cosine of 60.

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Now what do we know about
these two vectors?

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We know that their
net force is 0.

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Or that you also know that the
magnitude of these two vectors

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should cancel each other out
or that they're equal.

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I mean, they're pulling in
opposite directions.

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That's pretty obvious.

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And so you know that their
magnitudes need to be equal.

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So we know that T1 cosine
of 30 is going to equal

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T2 cosine of 60.

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So let's write that down.

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T1 cosine of 30 degrees is
equal to T2 cosine of 60.

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And then we could bring the
T2 on to this side.

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And actually, let's also-- I'm
trying to save as much space

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as possible because I'm guessing
this is going to take

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up a lot of room,
this problem.

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What's the cosine
of 30 degrees?

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If you haven't memorized
it already, it's square

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root of 3 over 2.

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So this becomes square root
of 3 over 2 times T1.

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That's the cosine
of 30 degrees.

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And then I'm going to bring
this on to this side.

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So the cosine of 60
is actually 1/2.

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You could use your calculator
if you forgot that.

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So this is 1/2 T2.

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Bring it on this side so
it becomes minus 1/2.

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I'm skipping more steps than
normal just because I don't

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want to waste too much space.

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And this equals 0.

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But if you seen the other
videos, hopefully I'm not

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creating too many gaps.

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And this is relatively
easy to follow.

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So we have the square root of
3 times T1 minus 1/2 T2 is

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equal to 0.

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So that gives us an equation.

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One equation with two unknowns,
so it doesn't help

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us much so far.

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But let's square that away
because I have a feeling this

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will be useful.

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Now what's going to be happening
on the y components?

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So let's say that this is the y
component of T1 and this is

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the y component of T2.

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What do we know?

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What what do we know about
the two y components?

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I could've drawn them here too
and then just shift them over

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to the left and the right.

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We know that their combined
pull upwards, the combined

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pull of the two vertical tension
components has to

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offset the force of gravity
pulling down because this

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point is stationary.

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So we know these two y
components, when you add them

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together, the combined tension
in the vertical direction has

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to be 10 Newtons.

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Because it's offsetting
this force of gravity.

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So what's this y component?

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Well, this was T1
of cosine of 30.

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This should start to become a
little second nature to you

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that this is T1 sine of 30, this
y component right here.

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So T1-- Let me write it here.

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T1 sine of 30 degrees plus this
vector, which is T2 sine

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of 60 degrees.

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You could review your
trigonometry and your

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SOH-CAH-TOA.

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Frankly, I think, just seeing
what people get confused on is

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the trigonometry.

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But you can review the trig
modules and maybe some of the

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earlier force vector modules
that we did.

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And hopefully, these
will make sense.

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I'm skipping a few steps.

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And these will equal
10 Newtons.

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And let's rewrite this up here
where I substitute the values.

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So what's the sine of 30?

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Actually, let me do
it right here.

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What's the sine of 30 degrees?

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The sine of 30 degrees is 1/2 so
we get 1/2 T1 plus the sine

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of 60 degrees, which is square
root of 3 over 2.

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Square root of 3 over
2 T2 is equal to 10.

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And then I don't like
this, all these 2's

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and this 1/2 here.

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So let's multiply this
whole equation by 2.

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So 2 times 1/2, that's 1.

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So you get T1 plus the square
root of 3 T2 is equal to, 2

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times 10 , is 20.

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Similarly, let's take this
equation up here and let's

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multiply this equation by 2
and bring it down here.

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So this is the original
one that we got.

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So if we multiply this whole
thing by 2-- I'll do it in

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this color so that
you know that

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it's a different equation.

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So if you multiply square root
of 3 over 2 times 2-- I'm just

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doing this to get rid of the
2's in the denominator.

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So you get square root of 3 T1
minus T2 is equal to 0 because

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0 times 2 is 0.

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And let's see what
we could do.

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What if we take this top
equation because we want to

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start canceling out some terms.
Let's take this top

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equation and let's multiply
it by-- oh, I don't know.

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Let's multiply it by the
square root of 3.

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So you get the square
root of 3 T1.

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I'm taking this top equation
multiplied by the

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square root of 3.

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This is just a system
of equations

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that I'm solving for.

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And the square root of 3
times this right here.

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Square root of 3 times square
root of 3 is 3.

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So plus 3 T2 is equal to
20 square root of 3.

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And now what I want to do is
let's-- I know I'm doing a lot

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of equation manipulation here.

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But this is just hopefully, a
review of algebra for you.

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Let's subtract this equation
from this equation.

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So you can also view it as
multiplying it by negative 1

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and then adding the 2.

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So when you subtract this from
this, these two terms cancel

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out because they're the same.

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And so then you're left with
minus T2 from here.

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Minus this, minus 3 T2 is equal
to 0 minus 20 square

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roots of 3.

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And so this becomes minus 4 T2
is equal to minus 20 square

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roots of 3.

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And then, divide both sides by
minus 4 and you get T2 is

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equal to 5 square roots
of 3 Newtons.

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So that's the tension
in this wire.

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And now we can substitute
and figure out T1.

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Let's use this formula
right here because it

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looks suitably simple.

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So we have the square root
of 3 times T1 minus T2.

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Well T2 is 5 square
roots of 3.

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5 square roots of
3 is equal to 0.

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So we have the square root of
3 T1 is equal to five square

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roots of 3.

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Divide both sides by square
root of 3 and you get the

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tension in the first wire
is equal to 5 Newtons.

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So this is pulling with a force
or tension of 5 Newtons.

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Or a force.

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And this is pulling-- the second
wire --with a tension

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of 5 square roots
of 3 Newtons.

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So this wire right
here is actually

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doing more of the pulling.

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It's actually more of the force
of gravity is ending up

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on this wire.

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That makes sense because
it's steeper.

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So since it's steeper,
it's contributing

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more to the y component.

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It's good whenever you do these
problems to kind of do a

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reality check just to make sure
your numbers make sense.

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And if you think about it,
their combined tension is

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something more than
10 Newtons.

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And that makes sense because
some of the force that they're

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pulling with is wasted against
pulling each other in the

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horizontal direction.

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Anyway, I'll see you all
in the next video.

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