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## Precalculus

### Course: Precalculus > Unit 6

Lesson 7: Vector components from magnitude and direction- Vector components from magnitude & direction
- Vector components from magnitude & direction
- Vector components from magnitude & direction: word problem
- Vector components from magnitude & direction (advanced)
- Converting between vector components and magnitude & direction review

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# Vector components from magnitude & direction: word problem

Given the magnitudes and directions of the forces two people are applying to a box, find the overall force that is applied in the direction of the target. Created by Sal Khan.

## Want to join the conversation?

- I am not convinced with sal's explanation and this is how i feel. Please check what is wrong with this reasoning.

If we plot same scenario on 2D graph assuming box is at origin 0, 0 and target is at some positive number on y axis then :

Vertical component of vector a becomes cos(35) * 330 = 270 (approximated) and horizontal coponent become sin(35) * 330 = 189.

Since horizontal component is acting away from origin in negative direction, we can represent vector a = -189i + 270j.

Similarly vertical component of vector b becomes cos(15) * 300 = 290 (approximated) and horizontal coponent become sin(15) * 300 = 78.

Since horizontal component is acting away from origin in positive direction, we can represent vector b= 78i + 290j.

Note that vertical components of both vectors are acting towards target which we have assumed is in positive direction, so they are positive.

Hence resultant vector of and b can be given as a + b = (-189+78)i + (270+290)j = -111i + 560j.

So resultant vector is not going in the direction of target, rather it is moving away from target in negative direction horizontally.

To calaculate magnitude of resultant vector call it c.

c = sqrt(-111*-111 + 560*560) = 570.89 Newton

What is wrong with it ?(14 votes)- The problem is that we are only concerned with the the vectors in the direction of the target (perhaps the box is on rails), so all we are concerned with are the 270 and 290 force vectors and we can consider the remaining force (the i component in your case) as wasted energy.(35 votes)

- I know that the result is the same, but if we have considered the dashed line as our horizontal axis, wouldn't vector b have an angle of -15?(5 votes)
- Yes, but cos(-15) = cos(15).

Whats more interesting to me is that there is a force that is perpendicular to the line between the box and the target which could result in the box never reaching the target. Person A exerts a perpendicular force which is 330sin(35)= 189N and person B exerts a perpendicular force 300sin(-15)= -78N. Notice these two are in opposite directions. Since the perpendicular force of A is higher than the perpendicular force of B, we know the box will never reach the target. It will veer off to the left of the target.

So how much does person A need to back off on pushing so they exactly cancel out the components perpendicular to the line of motion? If we change person A force from 330N to F and we want all the forces perpendicular to the desired line of motion to be zero, we get Fsin(35) + 300sin(-15) = 0 or F = 135.4N. So the moral of the story is, if Person A wants to push at a angle more off course from target then person B, dont push so hard. Let person B do most of the work :)(6 votes)

- What is the whole "sub x" thing for?(0 votes)
**a**"sub x" means the amount of force in the x direction from vector**a****b**"sub x" means the amount of force in the x direction from vector**b**

Although this video didn't dwell on it,**a**"sub y" would be the amount (magnitude) of force in the y direction from vector**a**and so on(14 votes)

- Would the box really move towards the target exactly? Or it might end up at some other spot?(3 votes)
- It will definitely end up at some
**other**spot! The problem Sal is interested in is not**where**the box will end up, or**in which direction**it moves, but rather: what forces are exerted on the box**in the precise direction of the box**. This comes down to calculating the**component**of each force in this specific direction. The final question is: who exerts more force**directly toward the box**? (This question is asked at around2:20.)(7 votes)

- How do I find the magnitudes of two vectors, a, b, if I'm only given just the angles for a and b, and also given the magnitude of a third vector, c, s.t. a+b+c=0 ?

I've been running in circles trying to figure this out..(5 votes) - Will the box move in the given direction ( if it is
**not**moving in that direction currently ) ?(4 votes) - If you apply forces A and B simultaneously to the box, would it end up in the same place as you would get when you apply A and then B?(4 votes)
- Yes, think of it this way, force A exerts 20 Newtons of force on the box, then force B exerts a negative 30 Newtons of force on the box. Lets say every 10 Newtons the box travels 1 meter. So if the box at first was at the position of 0 meters, the position now would be negative 1 meter (because 20 newtons [force A] plus negative 30 newtons [force B] is equal to negative 10 newtons as its total or net force). So converting the net force into the length the box traveled, that would be negative one meter. The same exact thing occurs when you apply the forces simultaneously. The twenty newtons in force A cancels with the negative 20 newtons in force B, leaving just the negative 10 newtons in force B. So the answers are both the same.(1 vote)

- Why does Sal use a "sub x" instead of ||a "sub x"|| at3:30? Is there a specific benefit to not using vector notation or is it simply a matter of preference?(2 votes)
- According to the parallelogram law of addition of two vectors:

vector R = vector A + vector B

=> ||vector R|| = sqrt(A^2 + B^2 +2AB cos**theta**) [ where theta = the angle bet. vec. A and vec. B]

But, as per this formula the answer comes 571.115 N, why this difference ?(2 votes)- The vector R is not pointing directly at the target.(3 votes)

- Essentially what you did, in other words to find the combined force (270 +290) is add the vector a plus the vector b and then find the magnitude of the new vector?(0 votes)
- I don't think that's quite right: adding the two vectors themselves wouldn't (necessarily) produce a vector pointing to the target.

What has been done in the video is to find the vector*components*that lie on a line going through the target and then adding those components. See what Sal says at around2:23.(6 votes)

## Video transcript

Voiceover: Let's say that
you have two folks that are trying to collectively
push a box across the snow towards a target, so
this is where the box is, right over here and this is
the target, right over here. Let me write that, that is the target. That's where they're
trying to get the box, and person A, because for
some reason, they can't push it from exactly behind the box, maybe there's not a good footing there, or I guess if they pressed
there, maybe the box squeezes a little bit,
so person A has to push in a direction that's not
exactly going in the direction of the target, so they push in a direction that looks like this, and so let me show you
this vector I'm drawing, essentially represents the
force that they're exerting. This is their force vector. This is person A's force vector. So this is person A's force vector, and we know the length of this vector, or another way to think
about it, the magnitude of vector A is 330 newtons, 330 newtons. And let's say person B, once again, because they can't push exactly in the direction of the target, maybe the box is really
soft right over here, person B is pushing at
this angle right over here, so that right over there, that vector is the force that person B is pushing onto it and the direction, and the
magnitude of that force, of that vector of person
B's pushing, is 300 newtons. And we know their angles that those make with the direction of the target. So if this is the direction
of the target right over here we know that this is a 35 degree angle, and we know that this
is a 15 degree angle. Now what I want you to
do, is pause this video and think about how much
of each of their force is going in the direction of getting the box towards the target, and then who is actually exerting more
force in that direction. We see that person A is exerting a total, a higher magnitude of
force in this direction than person B is doing in that direction. 330 newtons versus 300, but who's helping the box
go more in that direction? And by how much more? And also, what's the total force now pushing the box in that direction? So I'm assuming you had a go at it, and the key here is to find the component of each of these vectors, the magnitude of each of these vectors
in this direction, in the direction towards the target. And so, let's look first at vector A. So vector A looks like this,
and I'll just draw them separately just so we can
clean it up a little bit. Vector A looks like this. We know its magnitude -
the magnitude of vector A is equal to 330 newtons, and if we say that the target is in this
direction right over here, so that's the target
is some place out here, we've already been told
that this is 35 degrees. So what we really want to
do is find the component, the magnitude of the component going in that direction right over there. The way we can do that is just with our traditional trig functions. This right over here is a right triangle. We are looking for this
side right over here. We have - let me just call that A sub X. And we already know
that the magnitude of A is 330 newtons, so the magnitude - let me just say that the magnitude - let me just write it this
way so we don't confuse. So let's just say the
magnitude of the vector in the X direction, so this
vector right over here, we can write like this. The magnitude of that, we'll just write it without the vector notation. So how can we think about that? Well we know cosine is
adjacent over hypotenuse, so we could write cosine
of 35 degrees is equal to the length of the adjacent side. That would just be A sub X,
without the vector over it. We're just saying that
that's the magnitude of vector A sub X, over
the magnitude of vector A, over 330 newtons, or we
could say that A sub X is equal to 330 times
the cosine of 35 degrees. And we can make the exact
same argument for B. Vector B, so let me draw it like this. Vector B, you could, maybe
I'll draw it like this just to make it a little bit - let me do it a little bit different. If this is the direction to the target, so once again, I'll just draw
a horizontal line for that. Then relative to that, vector B looks something like this. Vector B looks something like that, so that is vector B. B sub X, in the direction of the target - so we would drop a
perpendicular like that. This would be - this
right over here would be the vector B sub X, and
so what is the magnitude of B sub X going to be equal to? Now we could say that the
magnitude of B sub X - we'll just call that B sub X
without the vector notation, same exact logic. This is 15 degrees. Cosine of 15 degrees is
going to be the length of the adjacent side over the length of the hypotenuse. So we already know that the length of the hypotenuse is 300 newtons. So we could write that
cosine of 15 degrees is equal to B sub X,
length of the adjacent side over the length of the hypotenuse. Or that B sub X is equal to 300 times cosine of 15 degrees. So let's get our calculator out and let's calculate what these things are. So, let's see, we have 330 times cosine of 35 degrees,
gets us to 270 newtons. So that's A sub X is 270 newtons, and B sub X is 300 times
cosine of 15 degrees. We get 289.777, so what we see is, even though B's magnitude is less than A's magnitude, the
component of vector B going the direction of the
target is actually larger than the component of
vector A going in it. So if we were rounding
to the nearest newton, this right over here, the
magnitude of this vector right over here, B sub X, that is, if we round to the nearest
newton, 290 newtons. So this is approximately
290 newtons length, or I guess you could say magnitude, while this one is a little bit shorter - it's a little bit shorter. We saw if we round to the nearest newton, it's about 270 newtons. The length of this one is
270 newtons, approximately. So if you were to say, how much more is person B pushing in the direction that we care about, it's about - well, if we want to be
a little more precise, we can subtract the two, so we can take 300 cosine of 15, minus 330 cosine of 35, and we get about 19.5 newtons difference. The blue, person B, is contributing 19.458 newtons more in that direction towards the target than person A is. But if we wanted to talk about, what is the total force
going in that direction, then we would take the
sum of these two things. So we would - the total
force in that direction is going to be 560 newtons if we round to the nearest newton. So if you add this blue component to this magenta component,
you get this one right over here, which is 560 newtons. So this whole vector right
over here, it's magnitude - so I could write that as
the magnitude of A sub X, plus B sub X, which is
the same thing as A sub X, plus B sub X - I already
said these are the equivalent of the magnitude of each of
these vectors is equal to - I could write approximately
equal 560 newtons.