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#1 2006-01-12 10:23:37

zara
Guest

vectors

sad:(

Please help me!!!

ive just started doing vectors and i cant get around to doing this question:

Forces of f1=3N at 90°  and  f2=170°  acting at a point.

Find F1 + F2 and F1 - F2

By calculation

#2 2006-01-12 13:04:28

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: vectors

f2=170°

You gave a direction, but not a magnitude.  You need both.


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#3 2006-01-14 09:18:56

zara
Guest

Re: vectors

I'll write the directions and magnitudes again.......


F1= 3N at 90° 

F2= 5N at 170°

#4 2006-01-14 09:51:00

Ricky
Moderator
Registered: 2005-12-04
Posts: 3,791

Re: vectors

First, we need to make these vectors.

The vectors x direction is cos(θ).  The y direction is sin(θ).

<x, y> = <cos(90°), sin(90°)>

However, this gives us a magntidue of 1.  We don't want 1, we want 3:

3 * <cos(90°), sin(90°)> = <3*cos(90°), 3*sin(90°)>

And that gives us it in vector form.

Now do the same for the other vector, and you'll have to vectors:

v1 = <x1, y1> and v2 = <x2, y2>

v1 + v2 = <x1+x2, y1+y2>
v1 - v2 = <x1-x2, y1-y2>


"In the real world, this would be a problem.  But in mathematics, we can just define a place where this problem doesn't exist.  So we'll go ahead and do that now..."

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#5 2006-01-14 10:02:57

mathsyperson
Moderator
Registered: 2005-06-22
Posts: 4,900

Re: vectors

Ricky's method is perfectly correct. A quicker, but harder, alternative is to make the two known forces into two sides of a triangle and then use trigonometric formulae to find the length and angle of the third side.

However, this only works if there are only two forces involved. Ricky's way will work for any number of forces.


Why did the vector cross the road?
It wanted to be normal.

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#6 2006-01-14 10:58:30

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: vectors

You can break up as many vectors as you want to add or subtract and use the Pythagorean theorem.

  ∑F = √[(∑x)² + (∑y)²]

  Specifically the resultant force is;

  F = √[(∑x)² + (∑y)²], arctan(∑y/∑x)

  I actually think that this method is easier because you can simply break all the forces into two component columns and add them up.  Once you do that you can just plug them in to the above equation.

  For the original equations posted this is what would happen.

  F1 = 3N, 90°  and F2 = 5N, 170°

  F = √[(3cos90° + 5cos170°)² + (3sin90° + 5sin170°)²]

  F = √[24.24615776 + 14.96328757], arctan(3.868240888/-4.924038765)

  F ≈ 6.26N, -38.15°

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#7 2006-01-14 11:13:05

irspow
Member
Registered: 2005-11-24
Posts: 455

Re: vectors

I'm sorry, I forgot about the F1 - F2 equation, but it really is no different than what I stated above.  My last post was F1 + F2.  The only difference in that calculation and the subtraction one is that you must use the negative components of F2.  All of the summation and equation still hold true.
 
  F = √[(3cos90° - 5cos170°)² + (3sin90° - 5sin170°)²]

  F = √[24.24615776 + 4.54439691], arctan(2.131759112/4.924038765)

  F ≈ 5.37N, 23.41°

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