Current time:0:00Total duration:10:19

0 energy points

Studying for a test? Prepare with these 7 lessons on Forces and Newton's laws of motion.

See 7 lessons

# Introduction to tension (part 2)

Video transcript

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.