Welcome back.
We'll now do another tension problem and this one is just a
slight increment harder than the previous one just because
we have to take out slightly more sophisticated algebra
tools than we did in the last one.
But it's not really any harder.
But you should actually see this type of problem because
you'll probably see it on an exam.
So let's figure out the tension in the wire.
So first of all, we know that this point
right here isn't moving.
So the tension in this little small wire right here is easy.
It's trivial.
The force of gravity is pulling down at this point
with 10 Newtons because you have this weight here.
And of course, since this point is stationary, the
tension in this wire has to be 10 Newtons upward.
That's an easy one.
So let's just figure out the tension in these two slightly
more difficult wires to figure out the tensions of.
So once again, we know that this point right here, this
point is not accelerating in any direction.
It's not accelerating in the x direction, nor is it
accelerating in the vertical direction or the y direction.
So we know that the net forces in the x direction need to be
0 on it and we know the net forces in the y
direction need to be 0.
So what are the net forces in the x direction?
Well they're going to be the x components of these two-- of
the tension vectors of both of these wires.
I guess let's draw the tension vectors of the two wires.
So this T1, it's pulling.
The tension vector pulls in the direction of the wire
along the same line.
So let's say that this is the tension vector of T1.
If that's the tension vector, its x component will be this.
Let me see how good I can draw this.
It's intended to be a straight line, but that
would be its x component.
And its x component, let's see, this is 30 degrees.
This is 30 degrees right here.
And hopefully this is a bit second nature to you.
If this value up here is T1, what is the
value of the x component?
It's T1 cosine of 30 degrees.
And you could do your SOH-CAH-TOA.
You know, cosine is adjacent over hypotenuse.
So the cosine of 30 degrees is equal to-- This over T1 one is
equal to the x component over T1.
And if you multiply both sides by T1, you get this.
This should be a little bit of second nature right now.
That the x component is going to be the cosine of the angle
between the hypotenuse and the x component times the
hypotenuse.
And similarly, the x component here-- Let me
draw this force vector.
So if this is T2, this would be its x component.
And very similarly, this is 60 degrees, so this would be T2
cosine of 60.
Now what do we know about these two vectors?
We know that their net force is 0.
Or that you also know that the magnitude of these two vectors
should cancel each other out or that they're equal.
I mean, they're pulling in opposite directions.
That's pretty obvious.
And so you know that their magnitudes need to be equal.
So we know that T1 cosine of 30 is going to equal
T2 cosine of 60.
So let's write that down.
T1 cosine of 30 degrees is equal to T2 cosine of 60.
And then we could bring the T2 on to this side.
And actually, let's also-- I'm trying to save as much space
as possible because I'm guessing this is going to take
up a lot of room, this problem.
What's the cosine of 30 degrees?
If you haven't memorized it already, it's square
root of 3 over 2.
So this becomes square root of 3 over 2 times T1.
That's the cosine of 30 degrees.
And then I'm going to bring this on to this side.
So the cosine of 60 is actually 1/2.
You could use your calculator if you forgot that.
So this is 1/2 T2.
Bring it on this side so it becomes minus 1/2.
I'm skipping more steps than normal just because I don't
want to waste too much space.
And this equals 0.
But if you seen the other videos, hopefully I'm not
creating too many gaps.
And this is relatively easy to follow.
So we have the square root of 3 times T1 minus 1/2 T2 is
equal to 0.
So that gives us an equation.
One equation with two unknowns, so it doesn't help
us much so far.
But let's square that away because I have a feeling this
will be useful.
Now what's going to be happening on the y components?
So let's say that this is the y component of T1 and this is
the y component of T2.
What do we know?
What what do we know about the two y components?
I could've drawn them here too and then just shift them over
to the left and the right.
We know that their combined pull upwards, the combined
pull of the two vertical tension components has to
offset the force of gravity pulling down because this
point is stationary.
So we know these two y components, when you add them
together, the combined tension in the vertical direction has
to be 10 Newtons.
Because it's offsetting this force of gravity.
So what's this y component?
Well, this was T1 of cosine of 30.
This should start to become a little second nature to you
that this is T1 sine of 30, this y component right here.
So T1-- Let me write it here.
T1 sine of 30 degrees plus this vector, which is T2 sine
of 60 degrees.
You could review your trigonometry and your
SOH-CAH-TOA.
Frankly, I think, just seeing what people get confused on is
the trigonometry.
But you can review the trig modules and maybe some of the
earlier force vector modules that we did.
And hopefully, these will make sense.
I'm skipping a few steps.
And these will equal 10 Newtons.
And let's rewrite this up here where I substitute the values.
So what's the sine of 30?
Actually, let me do it right here.
What's the sine of 30 degrees?
The sine of 30 degrees is 1/2 so we get 1/2 T1 plus the sine
of 60 degrees, which is square root of 3 over 2.
Square root of 3 over 2 T2 is equal to 10.
And then I don't like this, all these 2's
and this 1/2 here.
So let's multiply this whole equation by 2.
So 2 times 1/2, that's 1.
So you get T1 plus the square root of 3 T2 is equal to, 2
times 10 , is 20.
Similarly, let's take this equation up here and let's
multiply this equation by 2 and bring it down here.
So this is the original one that we got.
So if we multiply this whole thing by 2-- I'll do it in
this color so that you know that
it's a different equation.
So if you multiply square root of 3 over 2 times 2-- I'm just
doing this to get rid of the 2's in the denominator.
So you get square root of 3 T1 minus T2 is equal to 0 because
0 times 2 is 0.
And let's see what we could do.
What if we take this top equation because we want to
start canceling out some terms. Let's take this top
equation and let's multiply it by-- oh, I don't know.
Let's multiply it by the square root of 3.
So you get the square root of 3 T1.
I'm taking this top equation multiplied by the
square root of 3.
This is just a system of equations
that I'm solving for.
And the square root of 3 times this right here.
Square root of 3 times square root of 3 is 3.
So plus 3 T2 is equal to 20 square root of 3.
And now what I want to do is let's-- I know I'm doing a lot
of equation manipulation here.
But this is just hopefully, a review of algebra for you.
Let's subtract this equation from this equation.
So you can also view it as multiplying it by negative 1
and then adding the 2.
So when you subtract this from this, these two terms cancel
out because they're the same.
And so then you're left with minus T2 from here.
Minus this, minus 3 T2 is equal to 0 minus 20 square
roots of 3.
And so this becomes minus 4 T2 is equal to minus 20 square
roots of 3.
And then, divide both sides by minus 4 and you get T2 is
equal to 5 square roots of 3 Newtons.
So that's the tension in this wire.
And now we can substitute and figure out T1.
Let's use this formula right here because it
looks suitably simple.
So we have the square root of 3 times T1 minus T2.
Well T2 is 5 square roots of 3.
5 square roots of 3 is equal to 0.
So we have the square root of 3 T1 is equal to five square
roots of 3.
Divide both sides by square root of 3 and you get the
tension in the first wire is equal to 5 Newtons.
So this is pulling with a force or tension of 5 Newtons.
Or a force.
And this is pulling-- the second wire --with a tension
of 5 square roots of 3 Newtons.
So this wire right here is actually
doing more of the pulling.
It's actually more of the force of gravity is ending up
on this wire.
That makes sense because it's steeper.
So since it's steeper, it's contributing
more to the y component.
It's good whenever you do these problems to kind of do a
reality check just to make sure your numbers make sense.
And if you think about it, their combined tension is
something more than 10 Newtons.
And that makes sense because some of the force that they're
pulling with is wasted against pulling each other in the
horizontal direction.
Anyway, I'll see you all in the next video.