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# Introduction to Tension : An introduction to tension. Solving for the tension(s) in a set of wires when a weight is hanging from them.

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- I will now introduce you to the concept of tension.
- So tension is really just the force that exists either
- within or applied by a string or wire.
- It's usually lifting something or pulling on something.
- So let's say I had a weight.
- Let's say I have a weight here.
- And let's say it's 100 Newtons.
- And it's suspended from this wire, which is right here.
- Let's say it's attached to the ceiling right there.
- Well we already know that the force-- if we're on this
- planet that this weight is being pull down by gravity.
- So we already know that there's a downward force on
- this weight, which is a force of gravity.
- And that equals 100 Newtons.
- But we also know that this weight isn't accelerating,
- it's actually stationary.
- It also has no velocity.
- But the important thing is it's not accelerating.
- But given that, we know that the net force on it must be 0
- by Newton's laws.
- So what is the counteracting force?
- You didn't have to know about tension to say well, the
- string's pulling on it.
- The string is what's keeping the weight from falling.
- So the force that the string or this wire applies on this
- weight you can view as the force of tension.
- Another way to think about it is that's also the force
- that's within the wire.
- And that is going to exactly offset the force of gravity on
- this weight.
- And that's what keeps this point right here stationery
- and keeps it from accelerating.
- That's pretty straightforward.
- Tension, it's just the force of a string.
- And just so you can conceptualize it, on a guitar,
- the more you pull on some of those higher-- what was it?
- The really thin strings that sound higher pitched.
- The more you pull on it, the higher the tension.
- It actually creates a higher pitched note.
- So you've dealt with tension a lot.
- I think actually when they sell wires or strings they'll
- probably tell you the tension that that wire or string can
- support, which is important if you're going to build a bridge
- or a swing or something.
- So tension is something that should be hopefully, a little
- bit intuitive to you.
- So let's, with that fairly simple example done, let's
- create a slightly more complicated example.
- So let's take the same weight.
- Instead of making the ceiling here, let's
- add two more strings.
- Let's add this green string.
- Green string there.
- And it's attached to the ceiling up here.
- That's the ceiling now.
- And let's see.
- This is the wall.
- And let's say there's another string right here
- attached to the wall.
- So my question to you is, what is the tension in these two
- strings So let's call this T1 and T2.
- Well like the first problem, this point right here, this
- red point, is stationary.
- It's not accelerating in either the left/right
- directions and it's not accelerating in the up/down
- directions.
- So we know that the net forces in both the x and y
- dimensions must be 0.
- My second question to you is, what is
- going to be the offset?
- Because we know already that at this point right here,
- there's going to be a downward force, which is the force of
- gravity again.
- The weight of this whole thing.
- We can assume that the wires have no weight for simplicity.
- So we know that there's going to be a downward force here,
- this is the force of gravity, right?
- The whole weight of this entire object of weight plus
- wire is pulling down.
- So what is going to be the upward force here?
- Well let's look at each of the wires.
- This second wire, T2, or we could call it w2, I guess.
- The second wire is just pulling to the left.
- It has no y components.
- It's not lifting up at all.
- So it's just pulling to the left.
- So all of the upward lifting, all of that's going to occur
- from this first wire, from T1.
- So we know that the y component of T1, so let's
- call-- so if we say that this vector here.
- Let me do it in a different color.
- Because I know when I draw these diagrams it starts to
- get confusing.
- Let me actually use the line tool.
- So I have this.
- Let me make a thicker line.
- So we have this vector here, which is T1.
- And we would need to figure out what that is.
- And then we have the other vector, which is its y
- component, and I'll draw that like here.
- This is its y component.
- We could call this T1 sub y.
- And then of course, it has an x component too, and I'll do
- that in-- let's see.
- I'll do that in red.
- Once again, this is just breaking up a force into its
- component vectors like we've-- a vector force into its x and
- y components like we've been doing in the last several
- problems. And these are just trigonometry problems, right?
- We could actually now, visually see that this is T
- sub 1 x and this is T sub 1 sub y.
- Oh, and I forgot to give you an important property of this
- problem that you needed to know before solving it.
- Is that the angle that the first wire forms with the
- ceiling, this is 30 degrees.
- So if that is 30 degrees, we also know that this is a
- parallel line to this.
- So if this is 30 degrees, this is also
- going to be 30 degrees.
- So this angle right here is also going to be 30 degrees.
- And that's from our-- you know, we know about parallel
- lines and alternate interior angles.
- We could have done it the other way.
- We could have said that if this angle is 30 degrees, this
- angle is 60 degrees.
- This is a right angle, so this is also 30.
- But that's just review of geometry
- that you already know.
- But anyway, we know that this angle is 30 degrees, so what's
- its y component?
- Well the y component, let's see.
- What involves the hypotenuse and the opposite side?
- Let me write soh cah toa at the top because this is really
- just trigonometry.
- soh cah toa in blood red.
- So what involves the opposite and the hypotenuse?
- So opposite over hypotenuse.
- So that we know the sine-- let me switch to the sine of 30
- degrees is equal to T1 sub y over the tension in the string
- going in this direction.
- So if we solve for T1 sub y we get T1 sine of 30 degrees is
- equal to T1 sub y.
- And what did we just say before we kind of
- dived into the math?
- We said all of the lifting on this point is being done by
- the y component of T1.
- Because T2 is not doing any lifting up or down, it's only
- pulling to the left.
- So the entire component that's keeping this object up,
- keeping it from falling is the y component of
- this tension vector.
- So that has to equal the force of gravity pulling down.
- This has to equal the force of gravity.
- That has to equal this or this point.
- So that's 100 Newtons.
- And I really want to hit this point home because it might be
- a little confusing to you.
- We just said, this point is stationery.
- It's not moving up or down.
- It's not accelerating up or down.
- And so we know that there's a downward force of 100 Newtons,
- so there must be an upward force that's being provided by
- these two wires.
- This wire is providing no upward force.
- So all of the upward force must be the y component or the
- upward component of this force vector on the first wire.
- So given that, we can now solve for the tension in this
- first wire because we have T1-- what's sine of 30?
- Sine of 30 degrees, in case you haven't memorized it, sine
- of 30 degrees is 1/2.
- So T1 times 1/2 is equal to 100 Newtons.
- Divide both sides by 1/2 and you get T1 is
- equal to 200 Newtons.
- So now we've got to figure out what the tension in this
- second wire is.
- And we also, there's another clue here.
- This point isn't moving left or right, it's stationary.
- So we know that whatever the tension in this wire must be,
- it must be being offset by a tension or some other force in
- the opposite direction.
- And that force in the opposite direction is the x component
- of the first wire's tension.
- So it's this.
- So T2 is equal to the x component of the
- first wire's tension.
- And what's the x component?
- Well, it's going to be the tension in the first wire, 200
- Newtons times the cosine of 30 degrees.
- It's adjacent over hypotenuse.
- And that's square root of 3 over 2.
- So it's 200 times the square root of 3 over 2, which equals
- 100 square root of 3.
- So the tension in this wire is 100 square root of 3, which
- completely offsets to the left and the x component of this
- wire is 100 square root of 3 Newtons to the right.
- Hopefully I didn't confuse you.
- See you in the next video.

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