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Algebra (all content)
Course: Algebra (all content) > Unit 19
Lesson 10: Applications of vectorsVectors word problem: tug of war
Sal solves a word problem where three vector forces act upon a ring in different magnitudes & directions. Created by Sal Khan.
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It seems to me that while Sal's answer is correct in that Team B won, his logic is not. The winning team was not supposed to be just who pulled the hardest (the magnitude of their vector), but whose area the ring ended up in. If Team B had pulled with the same magnitude, but in a direction closer to the X axis, they might have lost to team C. Am I missing something?(71 votes)
You are right, just from the magnitudes it's not possible to state which team would have won. After Sal sums the 3 vectors, we can see that the resulting vector-i + jis the resulting force on the ring, and from that we can say for certain that the ring will move into the area of team B.
In fact, calculating the individual magnitudes of each force is completely useless for this problem.(68 votes)
At8:20when Sal does the final calculation, why doesn't he switch to radian mode?(7 votes)
bellmanwells's comment is correct. The choice between degree mode and radian mode only determines how the trigonometric functions interpret their argument, or, in then case of the inverse trigonometric functions, how they return their result. Sal is just performing ordinary arithmetic operations, not trigonometric functions. The fact that the operations are on something whose units are in radians doesn't matter.(7 votes)
When deciding which team would win wouldn't it be simpler to just add the i and j components and find what quadrant the point ends up in? I know it works in the example video but I can't think of why it wouldn't apply in other examples.(7 votes)
I find the second question confusing. I interpreted it as the sum of all the vectors' magnitudes, so my math would've looked like: 4 + √20 + √18. If you are adding magnitudes, there should be no negative numbers, so why does he add it like that?(6 votes)
Sal took the sum of all three vectors. The sum of vectors is not the sum of their magnitudes! This is easy to see because first and foremost, the sum of vectors must be another vector, but if you are just adding magnitudes (scalar quantities) together you end up with a scalar!
So Sal basically took the vector components of each vector and add them up. In particular, he had:
𝐚 = 4ℹ + 0𝐣
𝐛 = -2ℹ + 4𝐣
𝐜 = -3ℹ – 3𝐣
So to add all three vectors, he simply added all three of these equations:
𝐚 + 𝐛 + 𝐜 = 4ℹ – 2ℹ – 3ℹ + 0𝐣 + 4𝐣 – 3𝐣
The key here is to recognize ℹ and 𝐣 as vectors, namely the standard basis unit vectors in ℝ². So on the LHS, we are adding vectors, and on the RHS, we are adding vectors that make up the summands on the LHS so they are both equal! This is like saying:
2 + 5 + 7 = (1 + 1) + (2 + 3) + (6 + 1)
We just split each number on the LHS into parts all of which we add on the RHS. With vectors, this is the easiest way we can add them because we can only add horizontal components with themselves and vertical components with themselves (you cannot add ℹ terms to 𝐣 terms).
Comment if you have questions!(4 votes)
When would the magnitude of a vector be negative?(3 votes)- The magnitude of a vector is always positive. Remember, magnitude of the vector a (a1,a2,a3...an) is calculated using sqrt((a1)^2+(a2)^2+(a3)^2+...+(an)^2). Since there is a square root and a bunch of squares, the magnitude will always be positive.(11 votes)
- How can the magnitude of combined forces be smaller than the magnitude of each force?(2 votes)
Think of it this way. (This example does not correlate to the video, but it will help explain vector addition and subtraction) Let's say each vector is equal to the distance and direction that you walked in a field. One was 3 feet north, the second was 3 feet south, and the other was 2 feet east. The idea behind adding vectors is that you get the displacement, not the total distance. So if you were to walk those directions and distances, your end point would be 2 feet east because 3 feet north and 3 feet south cancel each other out. What Sal is doing is basically the same thing I just did but just more complicated. The different magnitudes are cancelling each other out because they are pulling in opposite directions. The magnitude at the end is smaller because that is what is not cancelled out. I hope my explanation helps!(7 votes)
- In what direction is the ring getting pulled?Isn't the ring getting pulled in the direction of vector b?Because the force exerted by vector b is the strongest,the ring should be getting pulled in it's direction which can be obtained by taking the tan of y and x components of vector b.
Instead why is Sal calculating the direction angle from the combined force?(3 votes)
We have to include all forces when calculating the movement of the metal ring. The ring will move in the direction of vector b only if neither team A nor team C are pulling in a different direction. Each team exerts its own force, and the ring moves according to the combined force.(3 votes)
Why didn't Sal have to change "degrees" to "radians" on the calculator when he was trying to find out the direction of the vector in radians?(3 votes)
That's a good question.
When you do a calculation involving π there is generally no reason to specify radians.
He is not using a trig function like sin, cos, tan etc in this case.
π is always just 3.14.... and so 2π is 6.28... and π/2 = 1.57...
It's when using the trig functions sin, cos, tan etc. that you really need to be careful because you will get completely different answers depending on your degree/radians setting.
For example
In radians:
sin π = 0
In degrees:
sin π = sin 3.14 = 0.055(2 votes)
Wouldn't the combined magnitude of the teams' forces be equal to 4+√20+√18?
:/(2 votes)
No, it will be correct only if all the forces have the same direction.
The combined magnitude of the teams' forces is as written, first you need to combine the forces, and then to find the magnitude of that vector.
As an example you can think of the case of two equal forces with opposite direction. It's clear that the combined magnitude is zero and not twice the force.(3 votes)
- if the forces remain the same would the ball accelerate in the direction of the combined force?i also think that the 3 teams should move with the same acceleration of ball in order to apply their forces on th ball and keep the ball moving in a constant acceleration.and i think we are also changing the direction of the ball.so we are both changing the magnitute and direction of the velocity.are we sure that the acceleration of the ball remains the same?(2 votes)
Yes, the ball would accelerate (Force = mass * acceleration) in the direction of the net force (the vector summation of all forces). Force and acceleration are the only vectors in that equation so they point in same direction, mass is a scalar. Acceleration is change in velocity per time so velocity vector is in the same direction as acceleration and force vector. Since force is constant and mass is constant, acceleration is constant. With constant acceleration, velocity is increasing.(1 vote)
8:20Video transcript
Voiceover: Teams A, B, and C are playing a game of strength. Each team has attached
a rope to a metal ring and is trying to pull the
ring into their own area. Team areas shown below. Let's see what that's ... We see that over here team A is pulling on the ring in that direction. Team B is pulling on the
ring in that direction. Team C is pulling on the
ring in that direction. They're each trying to get the ring into their respective areas, it's a three-way tug of war. They tell us team A pulls
with a force vector a of four times a unit vector i plus zero j. This is four times, there's
no vertical component here so it's going straight in the positive, I guess you can say x direction and we can see it right over there. A is going straight in
the positive x direction so just a multiple of the unit vector i. Team B pulls with force
vector negative two i plus four j, so it's to the left and up. We see that negative two i plus four j then we have team C
pulls with a force vector negative three i minus three j. Let's see, negative three i minus three j. Yup that seems consistent
with the diagram. Forces are given in
kilo newtons, all right. Which team will win the game? Think about which team will win the game if you think about which one is pulling with the most magnitude
in their directions. Let me get my little scratch pad out and let's think about the magnitudes of each of these vectors. Let's first think about
the magnitude of vector a, what is that going to be? Well, it's four times the unit vector i, so it's just going to
have a magnitude of four. The unit vector i has a magnitude of one so it's magnitude is just four times that, it's magnitude is equal to four. Let's put the units there,
it's four kilo newtons. What about the magnitude of vector b? We can use a Pythagorean theorem. It's negative two i plus four j so the magnitude squared is going to be negative two squared plus four squared which is really just coming
out of the Pythagorean theorem. The magnitude of vector b is going to be equal to the square root of
negative two squared plus four or one way to think about this
is the Pythagorean theorem or the distance formula between the origin and that point right over there. Plus four squared, that's four plus 16 is the square root of 20. Let's see, this is
definitely greater than four. Four squared is just 16, this is some place between four and five. Let's see, team C, the
magnitude of vector c same logic is going to be the square
root of negative three squared plus negative three squared
which is equal to ... Let's see this is square root
of 18, right nine plus nine. Square root of 18. Square root of 20 is clearly larger than square root of 18. We already said that square
root of 20 is larger than four. Team B is going to win. Which team will win the game? Team B, and we can even, let's
just fill it out on our ... Team B is going to win the game. Now the next thing they ask us is what is the magnitude of
the teams combined forces acting on the ring? To figure that out we can just
add up all of these vectors. The combined force, so let's write this. The force, let's say force
combined, this is a vector it's going to be the sum of
vectors a plus vector b plus c which is going to be equal
to four i plus zero j, minus two i plus four j,
minus three i minus three j. Now we can just add up the
i's and add up the j's. We have four i minus two i minus three i, so four of something minus two you're going to have two left. Then you subtract another three you're going to have negative one of that, so that's going to be equal to negative, I could write negative
one i or just negative i and then we have zero j
plus four j minus three j so that's was zero plus four minus three is just going to be one j. Or you can say just plus j. That's the resulting vector
here is negative i plus j. It might look something like this. Negative i plus a j, gets us
a vector that looks like that. Let me actually draw it
a little bit bigger here. If I were to draw our coordinate axis that's my y-axis, this is my x-axis. It's negative one times i, so
let's say it looks like that that's negative i. That's negative i right over there. Negative i, let me write that that is negative i plus j. Plus the j unit vector which
might look something like this both of them have a magnitude of one, so that is the j unit vector. The sum of those two
is our combined force. Our combined force is going to
be going into this direction. Hence this is its magnitude. The length of it is this
magnitude, combined force and that goes in line with we're going to be going in that direction which takes us to the zone of B. They ask us what is the magnitude of the teams' combined force? Well the magnitude of this side is one, it's a unit vector where it's
going to the left direction. The magnitude of this side is one so the magnitude of this side we just use a Pythagorean theorem, it's a square root of one
squared plus one squared which is just the square root of two. The magnitude here, what is the magnitude of
the teams combined forces is square root of two. In what direction is
the ring getting pulled? Actually let me just fill out what they've already
asked us so we can ... What is the magnitude of
the teams' combined force acting on the ring? Round to the nearest tenth. Let's just write square root of two. Square root of two is equal to and round to the nearest
tenth 1.4 kilo newtons. In what direction is
the ring getting pulled and they want this is in radians. Round it to the nearest tenth, your answer should be between
zero and two pi radians. That, go back to our diagram, we need to figure out what
this angle is right over here. We need to figure out what
that angle is in radians and we know that this whole thing if we went from the positive x-axis all the way to the negative
x-axis that is pi radians. We also know what this
angle right over here is. How do we know that? Well, I've just constructed a ... Notice this is a right angle and this is an isosceles triangle, isosceles right triangle. It's going to be a 45, 45, 90 triangle but if you're thinking in terms of radians 45 degrees is pi over four. This is pi over four, pi over four, pi over two triangle. Same thing as 45, 45, 90. If this whole purple angle is pi and this is pi over four, you just subtract the
pi over four from the pi to get what theta is. Theta is going to be equal to ... Pi is the same thing as four pi over four minus pi over four. I just did that to get
a common denominator which is equal to three pi over four. Three pi over four radians. Let's go back over here. Now they want us to round
it to the nearest tenth, so let's take three
times pi divided by ... Just to make sure you can
see what I'm doing here so three times pi divided by four is equal to 2.356, round
it to the nearest tenth would be 2.4 radians. Approximately 2.4 radians
and we got it right.