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Simple vector addition

  1. Sep 21, 2010 #1
    When adding the resultant of two or more forces i get confused when adding forces that are on 360 and 270 degrees. Like the question is 6 newtons on 360 degrees, i don't get how they get -6, when cos 360 is 1

    can anyone help me??
  2. jcsd
  3. Sep 21, 2010 #2
    Me neither, but it can depend on how you define your coordinate system and relative angles. Maybe you ought to just post the question. My advice for these angles is to draw them out, then you have to get the components right.
  4. Sep 21, 2010 #3

    Thats what i did and i completely understand the coordinate system but yea i hope someone knows how to answer this question
  5. Sep 21, 2010 #4
    This is unclear. Can you state the original problem?
  6. Sep 21, 2010 #5
    The original probem asks to calculate the resultant of the three forces

    F1 = 6kN at (0 or 360 degrees)
    F2 = 8 Kn at 240 degrees
    F3 = 10 Kn at 210 degrees

    The solution they show is:-
    Fx= 8cos240-6-10cos210

    I understand everything except how they get -6. And by the way all the forces are in tension(heading outwards)
  7. Sep 21, 2010 #6
    Must be an error. That isn't right. Bad solution.

    It should be Fx=8cos240+6+10cos210
  8. Sep 21, 2010 #7
    Yea i agree with you but they can't do the same error twice. This other questions has 4 forces:

    F1 = 60 at 40 degrees (tension)
    F2 = 50 at 110 degrees (compression)
    F3 = 30 at 160 degrees (tension)
    F4 = 40 at 270 degrees (compression)

    when adding the Fy components they have F4 as positive when it should be negative because sine 270 is negative. Do you think this question is all incorrect?
  9. Sep 22, 2010 #8
    From the information given, I agree with Mindscrape: the answer you have is a bad one.

    It is bad for 3 reasons:

    1) Sign. The vectors F1 and F3 should be added, not subtracted to/from F2.

    2) Components. The question, as asked, is looking for the resultant of three forces. The answer given is only the x-component of the resultant force.

    3) Units. Let's assume Mindscrape is correct in that all three vectors should be added together, rather than subtracting F1 and F3. Then then the x-component of the resultant force Fx would be: 6 Kn + 8 Kn * cos(240) + 10 Kn * cos(210)
    (at the very least, you need to include the "K" as this represents much larger magnitudes of "n")

    So the actual resultant should be:

    Fx = 6 Kn + 8 Kn * cos(240) + 10 Kn * cos(210) = 6 Kn + (-4) Kn + (-5)[itex]\sqrt{3}[/tex] Kn = (2 - 5[itex]\sqrt{3}[/tex]) Kn
    Fy = 8 Kn * sin(240) + 10 Kn * sin(210) = (-4)[itex]\sqrt{3}[/tex] Kn + (-5) Kn = (-5 - 4[itex]\sqrt{3}[/tex]) Kn
    (remember that sin(0) = sin(360) = 0)

    [itex]F_{resultant} \approx 13.66 Kn \angle (-119.18)^\circ[/tex]
  10. Sep 22, 2010 #9
    For this problem, I would have Fres = F1 - F2 + F3 - F4
    F2 and F4 are subtracted, since they are inward forces (compression vs. tension)

    What answer are you given?
  11. Sep 23, 2010 #10
    The answers are

    Rx = 34.87 Kn
    Ry = 41.84 Kn
    R = 54.5 Kn
    angle = 50.2

    Sorry for any inconvenience guys, but the answers given are correct because i asked my cousin and he said "hypythetically, if you have 3 Fy forces are going in one direction e.g. all facing down (which are negative) you can make them positive and make the other force negative. In other words you have to consistant with you the signs like make all forces in one direction positive or negative and vice verse to the other forces
  12. Sep 23, 2010 #11
    These values are correct, and they hold to my previous statement:

    [tex]F_{resultant} = F_1 - F_2 + F_3 - F_4[/tex]

    By your statement, when adding vectors, you can add their opposites instead? True, if you want to get the vector that is opposite of the resultant vector. But, if you do this, you have to take the opposite of all the vectors - you can't pick and choose.

    I suppose what you (and/or your cousin) were trying to say is that rather than subtracting a vector, you can add its opposite, which is true. There is nothing "hypothetical" about this - it is a fact.

    If this statement is referring to your first question (the one with the 3 forces, all in tension), then the answer is still incorrect. And, by the way, just because "your cousin" said something, doesn't make it true.

    The fact is that in the first problem,

    [tex]F_x = F_{x1} + F_{x2} + F_{x3} = 6cos(0) + 8cos(240) + 10cos(210)[/tex] (leaving out the units)
  13. Sep 23, 2010 #12
    Wrong! F4 should not be negative "because sin(270) is negative." It is negative because it is in compression (i.e. it points inward, rather than outward).

    F2 should also be subtracted because it is in compression (even though sin(110) is positive).

    You should think of these forces as pulling or pushing on a single point. Think of that point as the origin. You then add all the forces that are pulling the point (tension) and subtract all forced that are pushing on it (compression).

    Consider a simpler scenario: Two forces pulling on a common point in opposite directions:
    F1 = 52n at 300 deg (or, more correctly -60 deg)
    F2 = 37n at 120 deg

    The resultant force, FR, is the vector sum of F1 and F2, that is F1 + F2

    To do this, we generally need to break down each vector into its x- and y- components, add the corresponding components together, and calculate the new force give the new components.

    So (leaving off units for simplicity - we can add them back in later),
    [itex]F1_x = 52 cos(300)[/tex]
    [itex]F2_x = 37 cos(120)[/tex]
    [itex]F1_y = 52 sin(300)[/tex]
    [itex]F2_y = 37 sin(120)[/tex]

    This gives:
    [itex]FR_x = F1_x + F2_x = 52 cos(300) + 37 cos(120)[/tex]
    [itex]FR_y = F1_y + F2_y = 52 sin(300) + 37 sin(120)[/tex]

    And finally, the magnitude of FR can be found by:

    [tex]\sqrt{(FR_x)^2 + (FR_y)^2}[/tex]

    with direction:

    [tex]tan^{-1} \left( \frac{FR_y}{FR_x} \right)[/tex]

    Which comes out to FR = 15n at 300 deg (-60 deg)

    This is what you should have already expected, since the forces are directly opposite each other (180 degrees apart).

    When 2 forces are pulling in opposite directions, there are 3 possible outcomes for the resultant force:
    1) the resultant force pulls in the same direction as F1 (with a magnitude less than that of F1)
    2) the resultant force pulls in the same direction as F2 (with a magnitude less than that of F2)
    3) there is no resultant force, because the 2 original forces are equal in magnitude, thus canceling each other out.

    Now, if instead we had 2 forces, one of which is pulling on a point and the other that is pushing on it from the opposite side, we essentially have to forces acting on that point in the same direction. Using the forces listed above, let's assume F1 is pulling on the point, while F2 is pushing. Then, the resultant force becomes FR = F1 - F2 because these forces are acting on a point in different ways. Whether the sine of one angle or another is negative has no bearing on the equation that you use to find the resultant.

    I think what your cousin may have been trying to point out is that you can look at a "negative force" as a positive force in the opposite direction. That is since we have F1 as a tension force and F2 as a compression force, we can consider F2 to be a "negative force" (in respect to the point being acted upon). Therefore, we can think of F2 as being either (-37)n at 120 deg or as (+37)n at 300 deg. If we do that, then we can add the forces. This really is the same as adding real numbers: A - B = A + (-B)
  14. Sep 24, 2010 #13

    Thanks for the reply i appreciate the example you gave. But i thought when it comes to vectors it doesn't matter if they're in tension or compression unless its a truss because we only want the magnitude of each force. F2 and F4 were both in compression however F2 was negative but F4 was positive
  15. Sep 25, 2010 #14
    Absolutely not true. The whole point of vectors is superposition, and the whole point of superposition is to be able to figure out the correct over all magnitude. The easiest way to figure out the correct vector projections is to simply break the vectors into the components, find the resultant components, then finally the resultant magnitude and direction.
  16. Sep 25, 2010 #15

    I saw a different question and yea you are right but then let me ask you this; If you have a 10 kN vector thats on 270 degrees and it in compression, will the positive right because -(-9)=9?? is that correct
  17. Sep 25, 2010 #16
    Request post to be deleted, put in a seperate topic please. Thanks.
  18. Sep 26, 2010 #17
    Why? its not even a question
  19. Sep 27, 2010 #18
    Not really sure what you are asking here. What do you mean by compression, the vector is acting towards the origin? If so, then you would have 10kN in the positive y-direction.
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