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## Physics library

# Introduction to tension (part 2)

A slightly more difficult tension problem. Created by Sal Khan.

## Want to join the conversation?

- But shouldn't the wire with the greater angle contain more pressure or force? Why are the two tension forces of T2cos60 and T1cos30 equal?(33 votes)
- Hi, again again, FirstLuminary... It appears that you have somewhat of a curious mind in pursuit of answers...I'm the same way. So anyway, if you are not already familiar with the great UNIT CIRCLE, let me introduce him. On the unit circle the x-coordinate represents cosine & the y-coordinate represents sine------ (x,y)=(cos,sin)------. So therefore anytime there is a physics problem dealing with angles, forces, or tension its safe to say that sine and cosine will get a word or two in. T2cos60 equals T1cos30 because the object is still...stationary...not moving...at rest. If they were not equal then the object would be swaying to one side (not at rest). The two horizontal forces pull in opposite directions with identical force causing the object to remain at rest and canceling eachother out. The reason it was brought up in this video was so he could have two equations, the T2sin60+T1sin30 and the cosine one that you asked about, with the two equations a substitution can be made and T2&T1 may be found.(61 votes)

- Cant we use Lami's rule here(13 votes)
- Lami's Theorem says that the ratio of the tension in the wire and the angle opposite for all three wires are equal. The angle opposite is the angle between the other two wires. See http://en.wikipedia.org/wiki/Lami's_theorem.

In this example the angle opposite T1 is 90 + 60, opposite T2 is 90 + 30 and opposite T0 (the tension in the wire attached to the weight) is 180 - 30 - 60 = 90. So:

T0/sin(90) =T1/sin(150) = T2/sin(120) or since we know T0:

T0/sin(90) =T1/sin(150) and

T0/sin(90) =T2/sin(120)

sin(90) is 1 and from the unit circle you may recall that sin(150) is .5 and sin(120) is sqrt(3)/2 so...

10/1 = T1/.5 (multiply both sides by .5)

5 = T1

and

10/1 = T2/(sqrt(3)/2) (multiply boith sides by sqrt(3)/2)

sqrt(3)/2 * 10 = T2 (10/2 is 5)

5*sqrt(3) = T2

So it works out the same.(18 votes)

- Seems like the easiest way to do this problem was just putting the value 10N up the middle between them, then taking 10sin(60*)=T2 and 10sin(30*) = T1.(12 votes)
- But they come out correctly, is it just a fluke?(4 votes)

- Couldn't you have just done,

T2 = 10Sin60° = 5√3N = 8.66N

and

T1 = 10Sin30° = 5N

Or is it just luck that this happens to work in this situation? Having to go through the way in the video can be a bit tedious.(9 votes) - What if I have more than 2 ropes, say 4. That would lead me to two equations with 4 unknowns. Or is it possible to derive two more equations with the increase of unknowns?(5 votes)
- Well, if you have 3 ropes, it could just be that 2 ropes are holding the weight, and the third is hanging slack, because it is too long. If you assume, that the ropes have the right length, that they are all under tension, or if you replace the ropes with bars (they support both tension and compression), it is solveable, but it gets complicated. btw this is called a "Statically Indeterminate Structure".

The way to do this is to calculate the deformation of the ropes/bars. Bars get a little longer if they are under tension and a little shorter under compression. I could make an example, but only if you care, it would be a bit of work.(2 votes)

- In the system of equations, how do you know which equation to subtract from the other? If I were doing this problem, I would have just subtracted the top equation from the bottom equation instead of the other way around, giving me 4T2 = 20√3, which basically gives me the same answer of T2 = 5√3.(3 votes)
- When solving a system of equations by elimination any of the two equations may be subtracted from another or added together. It does not matter if the top equation is subtracted from the bottom equation or vice versa and same for addition. The only thing that has to be seen is that a variable is eliminated.(6 votes)

- If i look at this problem i see that both y components must be equal because the vector has the same length. Using this you could solve the probelm much faster, couldn't you?(4 votes)
- Nothing would be wrong. The only thing you would have done is to change the question(1 vote)

- I was wondering on what contribution dose the rope on the bottom do to the overall tension supporting the block. I am talking about the rope that connects the mass and the point that attaches to t1 and t2. Dose the vertical wire contribute anything to the tension supporting the block or is t1 and t2 only responsible for pulling mass up against gravity. Is t1 and t2 divide the force of gravity that the bottom rope experinces?(3 votes)
- AT around3:56shouldnt the equation be sq root of 3 T1/T2=0 i.e. sq rooot of 3 T1 =T2(1 vote)
- Well..Isn't it more correct to do 5x / 3x = 0?(2 votes)

- At5:17, Why does the tension of the combined y components not equal 10N*9.8(gravity) ?(1 vote)
- Why would you multiply 10 N times 9.8? The 9.8 is 9.8 N/kg. It tells you how many newtons there are per kilogram, if you are on the surface of the earth. If you multiply 10 N * 9.8 N/kg, you have 98 N^2/kg, which doesn't make much sense.

If the object is just hanging, and it is not accelerating, the sum of the upward tension forces has to equal the downward force, which is the weight. All forces should be in newtons. To get the downward force if you only know mass, you would multiply the mass by 9.8 N/kg.(3 votes)

## 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.