Introduction to Tension (Part 2) A slightly more difficult tension problem.
Introduction to Tension (Part 2)
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- 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
- 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
- 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.
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