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Diagram the vector combinations on graph paper

  1. Oct 23, 2004 #1
    I'm currently taking Physics in High School and having problems with Vectors.

    I apologize in advance if this is the wrong forum to post this on but here it is...

    I have to Diagram the vector combinations on graph paper and find the resultant and the equilibant.


    3 N at 120(degrees)
    4 N at 240(degrees)

    If someone could possibly teach me how to draw this and find the resultant and equilibrant it would be very helpful.
     
  2. jcsd
  3. Oct 23, 2004 #2
    Are these tensions pulling or pushing on an object (force vectors)? Or are they individual vectors, such as velocity vectors?
    --
    When you state "at 120" and "at 240", i assume it means from the horizontal. Is this an exact quote from the book?
     
  4. Oct 23, 2004 #3

    HallsofIvy

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    dekoi: they are forc vectors because they are given in Newtons. Also, it really doesn't matter if the "reference direction" is horizontal because you can give the answer relative to the reference direction, whatever it is.

    There are two ways to do problems like this:

    1) direct trigonometry:
    Draw a picture you have a line of length "3" making an angle of 120 degrees with your reference direction (which you might as well take to be horizontal) (that's actually 60 degrees above the horizontal to the left and makes a 30 degree angle with the vertical) and another line connected to that of length "4" at 240 degrees. 240= 180+ 60 so that will go off to the lower left. 120-90= 30 so that makes a 30 degree angle with the vertical and so a 60 degree angle with the first vector. Connect the tip of that vector to the base of the first vector to get a triangle.
    We now have a triangle with two sides of length "3" and "4" and angle between them 60 degrees. Use the "cosine law" to determine the length of the third side:
    c2= 32+ 42- 2*3*4*cos(60)= 9+ 16- 24*1/2= 25- 12= 13 and c= √(13) N. Now that we know that, we can use the sine law to find the angle inside the triangle at base of the first vector:
    sin(A)/a= sin(B)/b where A is the angle opposite side of length a, B is the angle opposite side of length b. The side opposite the angle we want is 4 and the side opposite the 60 degree angle is √(13). We have sin(A)/4= sin(60)/√(13), so sin(A)= 4 sin(60)/√(13)= 0.9608 and A= 74 degrees. Looking at your picture, you'll see that that is measured counter-clockwise from the first vector. The vector is 120+ 74= 194 degrees relative to the original reference direction.
    The resultant is &sqrt(13)= 3.6 N at 194 degrees.
    The "equilibrant", the force that will exactly off set that is the vector of the same strength in the opposite direction: 3.6 N at 14 degrees.

    2) Using components, more sophisticated but easier in my opinion.
    The first vector has length 3 N and is at 120 degrees. Its components, relative to the original reference direction, are <3 cos(120), 3 sin(120)>= <-1.5, 2.6>
    The second vector has length 4N and is at 240 degrees. Its components, relative to the original reference direction, are <4 cos(240), 4 sin(240)>= <-2, -3.5>.
    The resultant vector is the sum of those: <-3.5, -0.9>.
    The "length" of that is √((-3.5)2+ (-.9)2)= &radic;(13) just as before. The angle is given by arctan(-0.9/-3.5)= arctan(.2571)= 14 degrees.
    We have to be careful with that: tangent is a periodic function and we don't want the fundamental solution. Since the components are both negative, we are in the third quadrant and clearly want 180+ 14 = 194 degrees.
    Exactly as before, we find that the resultant is √(13) N at 194 degrees and the "equlibrant" is √(13) N at 14 degrees.
     
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