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Find the angle between the forces given the magnitude

  1. Jun 22, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the angle between the forces given the magnitude of their resultant. (Hint: write the force one as as a vector in the direction of the positive x-axis and force two as a vector at an angle θ with the positive x-axis.)

    Force 1: 45 pounds
    Force 2: 60 pounds
    Resultant Force: 90 lbs

    2. Relevant equations

    3. The attempt at a solution

    Well, writing force one as a vector in the direction of the positive x-axis would make the first vector be <45,0>

    Then I know that when adding the first and second vectors the magnitude of the resulting vector must be 90 and the magnitude of the second vector must be 60.

    I don't know how to find the coordinates of the second vector such that it will have a magnitude of 60, and when added to the first will result in a vector with a magnitude of 90.

    Can someone point me in the right direction.
  2. jcsd
  3. Jun 22, 2010 #2


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    Okay, having drawn that you should see a triangle with sides of length 45, 60, and 90. And the angle you want, [itex]\theta[/itex], is the angle between the 45 and 60 length sides.

    Use the "cosine rule" from trigonometry: [itex]c^2= a^2+ b^2- 2ab cos(\theta)[/itex] where [itex]\theta[/itex] is the angle between sides of length a and b. Put in the values you are given for a, b, and c and solve for [itex]\theta[/itex].
  4. Jun 22, 2010 #3
    I solved for θ and got 117.28 degrees. The answer in the back of the book says the answer is 62.7 degrees. I don't think you use the law of cosines to solve this problem. a line with a distance of 90 drawn between two vectors will not find the angle θ of the vector c created by adding vector a and vector b with their own magnitudes but undefined directions that must make a vector with a magnitude of 90.

    The first vector has to have a direction of <0,45> with a magnitude of 45, that can't change

    The second vector has to have a magnitude of 60, but the direction is undetermined, the direction has to be the right direction so when you add vector one and vector two you get a new vector with a magnitude of 90. that vector has an angle θ that you have to find.
  5. Jun 22, 2010 #4
    You're thinking along the right lines, Mad Hatter, but you can solve the problem with the law of cosines. If you draw the vectors, using a parallelogram to represent vector addition, the resultant vector splits the paralellogram into two triangles. You know the lengths of all their sides. Find the angle between (45,0) and the resultant vector, then find the angle between the resultant vector and the one with magnitude 60. Then add those two angles.

    (117.28 degrees is the angle at the other corners of the parallelogram.)
  6. Jun 22, 2010 #5
    Let A, B be forces and C be the resultant. Summing components in the x-direction,
    we get Acos(a1)+Bcos(a2)=Ccos(a3), doing the same for y-direction we get
    Asin(a1)+Bsin(a2)=Csin(a3). We have two equations three unknowns, lets try to get the last one, a1+a2+a3=180. Now we have 3-equations, 3 unknowns. See if this helps.
  7. Jun 22, 2010 #6
    Ahhh... Ok, thanks. I understand now.
  8. Jun 22, 2010 #7
    In this case, if we set the x-axis parallel to the 45 pound vector, a1=0, a2=62.72 degrees, a3 = ArcCos((-60^2+45^2+90^2)/(2*45*90)) = 36.34 degrees.

    a1+a2+a3 = 0+62.72+36.34 = 99.06 degrees, not 180.
  9. Jun 22, 2010 #8
    Rasalhague, you are right. Looking at it more carefully that third equation is not correct. But perhaps is not needed since it is possible to assume one angle to be 0, leaving two unknowns. But the method you presented is better.
  10. Jun 22, 2010 #9


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    I don't know how you could have gotten that. With c= 90, a= 45, b= 60,
    [itex]c^2= a^2+ b^2- 2abcos(\theta)[/itex] becomes
    [itex]90^2= 45^2+ 60^2- 2(45)(60)cos(\theta)[/itex]
    [itex]cos(\theta)= (90^2- 45^2-60^2)/(2(45)(60))= .4533333[/itex]
    and that gives [itex]\theta= 62.7[/itex] degrees.

  11. Jun 23, 2010 #10
    ArcCos((90^2-45^2-60^2)/(-2*60*45)) = 117.3 degrees, the angle at the corner of the paralellogram where the sides of length 60 and 45 meet. This, I presume, is what Mad Hatter did as it seems the most natural interpretation of your hint: "Okay, having drawn that you should see a triangle with sides of length 45, 60, and 90. And the angle you want, [itex]\theta[/itex], is the angle between the 45 and 60 length sides."

    But we want the angle at the other vertex of the paralellogram, 360/2-117.3 = 62.7, the one at the vertex of a triangle made out of the 45 side, the 60 side, its third side being the other diagonal of the parallelogram, the one with length sqrt(45^2+60^2-2*45*60*cos(62.7 degrees)) = 56.1, rather than the diagonal of length 90.

    What's interesting, to me anyway, it's probably second nature to you ;-), is how you got this correct answer. You implicitly multiplied just one side side of the equation by -1 at the stage immediately before finding the angle:

    [tex]\cos \theta = \frac{90^2- 45^2-60^2}{-2 \cdot 45 \cdot 60}[/tex]

    multiplied by -1 becomes

    [tex]\cos \theta = \frac{90^2- 45^2-60^2}{2 \cdot 45 \cdot 60}[/tex]

    How does this work? Multiplying a negative cosine, as we have here, by -1 gives us [itex]\theta-2(\theta-90)=\theta(1-2)+180=180-\theta[/itex], in this case 180-117.3 = 62.7. (The interior angles of the rectangle sum to 360 degrees, and angles at opposite vertices are the same.)

    Or we could just, recognising this, begin by writing the equation we actually mean to solve:

    [tex]c^2= a^2+ b^2 + 2ab \cos \theta[/tex]

    which, if I've got all the signs right (fingers crossed!) equals

    [tex]a^2+ b^2 - 2ab \cos (\pi - \theta)[/itex]

    EDIT: Sorry I switched notation at the end there; to be consistent I should have written 180 degrees instead of pi (radians).
    Last edited: Jun 24, 2010
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