What is the magnitude of the resultant between these two vectors?

  • Thread starter IB
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  • #1
IB
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Vector A has a magnitude of 10 units and makes an angle of 30° with the horizontal
x-axis. Vector B has a magnitude of 25 units and makes an angle of 50° with the
negative x-axis. What is the magnitude of the resultant between these two vectors?

I'm not sure how to do this one. Are you supposed to add 10 and 25 to get the resultant vector or something?
 

Answers and Replies

  • #2
Diane_
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There are at least three ways to do this one. Try this: Draw a picture. Put the tail of vector A at the origin and sketch it out, then put the tail of vector B at the head of vector A and sketch it. The resultant will be the vector from the tail of A to the head of B. They will make a triangle. A bit of geometry and a bit of trigonometry should get you the solution.
 
  • #3
IB
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What are the other ways I can do for this kind of problem?
 
  • #4
IB
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Diane_ said:
There are at least three ways to do this one. Try this: Draw a picture. Put the tail of vector A at the origin and sketch it out, then put the tail of vector B at the head of vector A and sketch it. The resultant will be the vector from the tail of A to the head of B. They will make a triangle. A bit of geometry and a bit of trigonometry should get you the solution.
I got that A is a leg B is hypotenuse and Vector C, our resultant vector, is a leg. So is it [tex] {C^2 = B^2 - A^2} [/tex]? That way i'll get x^x = 30^2 - 10^2 so x = 900 - 100 = 800 => x = [tex] {\sqrt 800} [/tex] = 28.
 
  • #5
Diane_
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I don't see any right angles in there, luv. There is an 80 degree angle between A and B - perhaps that's what fooled you. I'm afraid Pythagoras won't be much use here - it's back to precalc and the Law of Cosines.
 
  • #6
IB
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so if it isn't a right triangle how do i know which angle is 30, which angle is 50 after having moved vector B?
 
  • #7
Tide
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Why don't you just resolve the vectors into components and add the components to find the resultant?
 
  • #8
Diane_
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There are so many times I miss having a blackboard on these forums.

I presume there are no problems sketching A, starting at the origin? OK, once you have that, lightly sketch in a new set of coordinate axes with the origin at the head of A. Use those to sketch in B. You'll have both sets of x-axes parallel, so all of the stuff you learned in geometry about "alternate interior angles" and so on will apply. I think you'll be able to see what the angles are fairly quickly.

Edit: Tide, I thought about having him do that, but I'm not sure he'd learn as much that way as through a purely geometric approach. It's the teacher in me, I guess.
 
  • #9
IB
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Diane_ said:
There are so many times I miss having a blackboard on these forums.

I presume there are no problems sketching A, starting at the origin? OK, once you have that, lightly sketch in a new set of coordinate axes with the origin at the head of A. Use those to sketch in B. You'll have both sets of x-axes parallel, so all of the stuff you learned in geometry about "alternate interior angles" and so on will apply. I think you'll be able to see what the angles are fairly quickly.

Edit: Tide, I thought about having him do that, but I'm not sure he'd learn as much that way as through a purely geometric approach. It's the teacher in me, I guess.
Hmm interesting! You're really a good teacher. Now I understand this vector problem. :)

You should teach Higher Level Maths, Further Maths, and Multivariable Calculus at secondary school instead of C.Sc. :wink:
 
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  • #10
Diane_
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I did. Math and physical science in American high schools for thirteen years. I've been thinking about going back - I miss the teenagers.

Anyway, glad to be of help.
 
  • #11
vector

Diane_ said:
There are at least three ways to do this one. Try this: Draw a picture. Put the tail ctor A at the origin and sketch it out, then put the tail of vector B at the head of vector A and sketch it.
 
  • #12
Tide
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Diane,

Without additional information it's often difficult to gauge what background students have when attempting to answer their questions particularly in an international forum such as this. In any case, IB seems happy with the results! :)
 
  • #13
IB said:
What are the other ways I can do for this kind of problem?

div(f grad g)=?
WE kow grad g= i dg/dx+j dg/dy+k dg/dy
 
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  • #15
IB
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Euh...you may need to explain more...
 
  • #16
Diane_
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Tide said:
Diane,

Without additional information it's often difficult to gauge what background students have when attempting to answer their questions particularly in an international forum such as this. In any case, IB seems happy with the results! :)
Oh, you're absolutely right, luv. If it seemed like I was criticizing, I apologize. Given the problem he was having, I did make an estimate as to his background, but the suggestion you offered was certainly reasonable.
 
  • #17
IB
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By the way, Diane, i somehow got sin20 = x/25 so x = 8.6 but i don't think that's true because the given choices are:
a. 20
b. 35
c. 15
d. 45
e. 25

so further help is appreciated! :)
 
  • #18
Tide
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Diane,

Oh, heavens no! I didn't take anything you said as criticism. It was merely an "editorial comment" on how difficult it is to teach with inadequate input from the student.
 
  • #19
Diane_
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IB - Can you tell me what you did to get to that point? I don't see any way there right off the bat.

Tide - True that. I think we understand each other.
 
  • #20
IB
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I have 20 degrees because I got 90 for the right angle at the tail of vector B at origin, then I got 50 - 30 = 20; the 30 is from the other origin, i used geometry to deduce that.

I think I'm doing it all wrong though...
 
  • #21
Diane_
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I'm afraid so. It's fairly natural - you try to get a right angle where you can. Unfortunately, there aren't any in this triangle.

Look at your diagram again. You know there is a 30 degree angle between A and the positive x-axis. If you look at the head of A - where B is drawn - and in particular at the duplicate x-axis you drew there, you'll see that there is an alternate interior angle between A and the duplicate. That's part of the angle between A and B. The rest of it is the 50 degree angle between the duplicate x-axis and B, so the angle in there is 80 degrees.

You know the length of A is 10 units and the length of B is 25 units, so you have enough information for the Law of Cosines to find the length of the resultant. Are you familiar with the Law of Cosines?
 
  • #22
IB
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yeah; cos = adj/hyp
 
  • #23
Diane_
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Ah. No, that's the right-triangle definition of the cosine. The Law of Cosines is basically the Pythagorean Theorem adapted to work with non-right triangles. Consider a triangle with sides a, b and c and angles A, B and C. Angle A is opposite side a, angle B is opposite side b, and (you probably see where this is going) angle C is opposite side C. The Law of Cosines says that

c^2 = a^2 + b^2 - 2ab cos C

This works with any triangle. In this case, you know that a = 10, b = 25 and C = 80 degrees. It's just a plug and chug from there.

Have you never seen that before?
 
  • #24
IB
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:blushing: dang, i've seen this in GCSE in britain, advanced alg/trig 2 and pre-calc in America, but somehow forgot all because I was trying to solve this vector problem at 3 o'clock in the morning. the problem is how to remember the formulae of these laws at such a time...

ok, got it. look at my diagram:

http://mail.google.com/mail/?view=att&disp=attd&attid=0.1&th=106a61dd3b4f0613

no student should try to do vector at 3 o'clock in the morning otherwise he will be like this: :grumpy:

i've finally solved it, yay! o:)
 

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