Simple Vector Problem Help

[Moriarty]

Hi everyone,

First, let me say I can actually solve this problem without trouble using the prescribed method. I just don't like it, think it's kind of clunky and have been wondering about another way to go at it. Something more intuitive to me. As you can see from the pic, the problem to solve is pretty straightforward. What I was wondering is what would happen if the boat were to *not* compensate for the river current at all? Where would it end up? The triangle I drew up in the corner kind of illustrates what I think would happen. 15 represents the speed of the boat along the Y axis, and 4 is the perpindicular force acting upon it by the river current. Calculating the angle theta I think the boat would drift about 14.9 degrees to the right. So my questions are:

1. Is my assumption above correct, or am I misrepresenting what's happening? (I suspect so, but don't know why)
2. If my assumtion is correct, then why couldn't the boat steer 14.9 degrees to the left of the Y axis to compensate for the current? (The correct answer is 15.5)
3. I noted that the hypotenuse of the triangle I drew happened to measure 15.5. Is this a coincidence, or is there a connection between this and the correct answer to the problem, which is also 15.5?

Thanks!

Edited August 16, 2014 by Moriarty

[studiot]

 

Moriarty

and 4 is the perpindicular force acting upon it by the river current.

I really think it is far more important for you to concentrate on reading the question than on trying to pick holes in your instructor's exercises.

 

 

You seem to have a fundmental misconception about the addition of vectors.

 

Can you see what is wrong with your statement?

Edited August 16, 2014 by studiot

[Moriarty]

That answers question 1, thanks. As I mentioned, I suspected this was the case, but am trying to understand why. I can solve the problem and get the correct result, but that doesn't mean I truly understand everything that's happening. My response to this problem is purely robotic if I want to get the correct answer, yet I don't truly understand the nature of the problem. I'd like to. I don't see that as picking holes in it, personally.

Edited August 16, 2014 by Moriarty

[studiot]

 

You seem to have a fundmental misconception about the addition of vectors.

 

Let's clear this up first, as it will get in your way for the rest of vector analysis.

 

You can only add vectors that are of the same type.

 

You cannot add forces to velocities or accelerations to velocities or whatever.

 

Please also state whether you have been taught the triangle of vector addition or the parallelogram.

Edited August 16, 2014 by studiot

[Moriarty]

Maybe it would help to say I'm asking this question because I'm bothered by the problem. The class is complete and final grades have been posted. I got an A. Great. In most cases, I was able to kind of "dissect" problems to better understand them, which allowed me to approach them from new directions if I found myself stuck on something. In the case of vectors this never quite happened and I had to simply follow the steps. My particular frustration is in not understanding how these moving parts are connected and how they influence an outcome. I'd like to get this under my belt before I move on to the next course because I feel I couldn't properly formulate a real problem, as evidenced above.

I saw your reply above and thanks for that. Let me ask this question: if, hypothetically, the longitudinal axis of the boat were not going to turn left or right..as in the bow would remain perpindicular to the x axis..how might I find out where the boat would land on the opposite bank if no effort were made to correct the course? I think if I could to that correctly, I'd be on the path to understanding this. Thanks.

[studiot]

How can I help if you don't cooperate?

 

I don't want to confuse you by introducing stuff they didn't tell you.

Surely it's no big secret whether they told you to add vectors using triangles or parallelograms?

Edited August 16, 2014 by studiot

[Moriarty]

I'm not being uncooperative. I missed your question. Triangle, but that's a secret. Please keep it between us.

 

But by all means, confuse me. The more perspective, the more information, the greater the odds that I'll leave with what I came for.

[studiot]

OK

 

Let the boat be headed so as to make an angle (a) with the current, and with the dock assuming the dock is parallel to the current.

 

Then its velocity is the resultant of 15 km/hr in this direction and 4 km/hr in the direction of the current.

 

The component of velocity perpendicular to the current is 15 sin(a) and the component parallel to the current is 4 - 15 cos(a)

 

But to go straight across the velocity parallel to the current must be zero.

 

So 4 - 15cos(a) = 0

So cos(a) = 4/15

 

Having found (a) you can substitute into the this into 15sin(a) to obtain the crossing velocity and thus the time.

 

Now the triangle you have drawn is not the vector difference of 4 and cos sin(a), it is the vector difference of 4 and 15

Edited August 16, 2014 by studiot

[Moriarty]

Thanks, I'll play around with this tonight and let you know. Appreciate it!

[studiot]

Alternatively you can look at it like this

 

In the diagram the boat wants to travel straight across from A to B, but has to head towards D because of the current AC

 

Now it is important to place your arrows correctly and also indicate their direction because the boat at A has two components of velocity, its motor at 15 km/hr and the current at 4 km/hr.

These vectors are concurrent at A so we cannot draw the triangle ADC as this does not represent the sum or resultant.

 

We have to draw AD first then 'place' DE on the end of AD to and complete the triangle ADE, reversing the arrow AE so that we draw the correct triangle.

 

In the diagram the angle (a) for the other method is shown as is the angle in triangle ADE as (b). (a) + (b) is, of course, a right angle here.

 

Now we calcualte AE = AD²-DE² = 14.47 km/hr and thus cos(b) = AE/AE = 14.47/!5 and (b) = 15.47 degrees, which agress with the other (resolution) method whichs (a) as 74.53

 

 

Perhaps this shows more clearly why the boat should not steer your angle theta.

 

In the case of question above 15 is the resultant, and the hypotenuse of the vector triangle.

 

In the case you propose where the boat heads straight across the 15 is one of the components, and the resultant is sqrt(15²+4²), not the

sqrt(15² - 4²), we had in the question.

 

The angles are close because of the big difference between 15 and 4 so sqrt(15²+4²) is not very different from sqrt(15² - 4²) so the triangles are nearly the same.

Edited August 16, 2014 by studiot

[Moriarty]

There we go. I knew this was easy. What you've done is attack this problem the same way my text does, but with a much more straightforward approach that I like. For example, my text would create an equation that would lead to finding the direction of the boat relative to the water thus:

Vc = 4i (water current)
Vw = ai + bj, so ||Vw|| = sqrt(a^2 + b^2)
Vg = Vw + Vc (from the diagram above)
kj = ai + bj + 4i, where k is the magnitude of the vector Vg, so
kj = (a +4)i + bj, noting that for this equation to be true, the coefficients of the i and j components must be equal on both sides.
Solving, a = -4 and b = sqrt(209), so Vw = -4isqrt(209)j and Vg = kj = sqrt(209)j

We then find angle theta between Vw and Vg: COS theta = the dot product of Vw and Vg divided by ||VW|| ||VG|| = .9638, or 15.5 degrees.

It's functional, but I wouldn't consider it efficient when I can just draw a right triangle and be done with it. At least now I can, so thanks! I also see why my first attempt flubbed..