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Calculate the resultant force

  1. Jul 21, 2013 #1

    Zondrina

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    1. The problem statement, all variables and given/known data

    Given ##\vec{F}_1 = 22N [N]##, ##\vec{F}_2 = 38N [E 35° S]##, and the angle between them is 125° calculate ##\vec{F}_R##.

    Picture : http://gyazo.com/a95dc91a95639edc7f1ef21f469976d1

    2. Relevant equations

    N for newtons.

    3. The attempt at a solution

    I used ##\vec{F}_R = \sqrt{\vec{F}_1^2 + \vec{F}_2^2 - 2\vec{F}_1\vec{F}_2 cos(125°)} = 53.7N##

    Now I want to find the direction. I think I should put my angle ##θ## between ##\vec{F}_1## and ##\vec{F}_R## ( I don't know how to explain why though which is my problem really ).

    Presuming that's the right way to go, I would then use the sin law to get ##θ = 30°##.

    ##∴\vec{F}_R = 53.7N [S 30° E]## <- I think, I mean it would make sense because the force in the SE direction is 16N greater than the one in the northern direction, right?
     
    Last edited: Jul 21, 2013
  2. jcsd
  3. Jul 21, 2013 #2
    Would it not be far easier to work with the components of the forces?
     
  4. Jul 21, 2013 #3

    Zondrina

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    I find modeling the problem and actually drawing what's happening to be effective for these types of problems. I don't see how you would do this with components though.
     
  5. Jul 22, 2013 #4

    collinsmark

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    There's nothing wrong with modeling the problem and drawing a free body diagram. I welcome and encourage that.

    But trust voko on this one. Use components. Treat East as the positive x-direction and North as the positive y-direction. Break up the vectors into their x- and y-components. That way you can add the vectors' respective components together directly.

    Once you know the resultant force's components, you can find the magnitude using the Pythagorean theorem. The direction can be found using the arc-sine, arc-cosine, or arc-tangent.

    In general, using the law of cosines (the approach you took) might help you find the magnitude of the resultant force, but it's cumbersome to use to find the direction.

    You're not using the correct angle in your application of the law of cosines. When adding vectors together you must place the vectors from head to tail (the first vector's head connected to the second vector's tail, etc). Then measure the distance from the first tail to the last head.

    You'll see when you do this properly, the angle between the constituent vectors is not 125o.

    Alternately, break up each vector into its components, add the respective components of all the vectors, then use the Pythagorean theorem.

    That's where using the vector's components really helps. Once you know the resultant vector's x- and y- components, it only takes an arc-tangent to find the angle, or an arc-sine or arc-cosine if you also know the magnitude. :wink:

    [Edit: by the way, the angle is not 30o.]
     
    Last edited: Jul 22, 2013
  6. Jul 22, 2013 #5
    Before Monsieur Descartes that was the only way, and it worked for centuries, or perhaps even for millenia, so obviously it works and is effective in a way.

    Yet we all stand on the shoulders of the giants, and there is no reason to neglect what they figured was the easier way to attack this sort of problem.

    You have two vectors and you need to find out their sum. So add one to the other component-wise. You will get the components of the result. Which you can then use to find out the magnitude and the direction of the result.
     
  7. Jul 22, 2013 #6

    Zondrina

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    Okay, I've never tried this in three directions, but I'll see what I can do. Hmm separating the components from the vectors I get :

    ##\vec{Δd}_N = 22N##
    ##\vec{Δd}_S = 38N sin(35°)##
    ##\vec{Δd}_E = 38N cos(35°)##

    The north and south directions should be added together here I think ( So I only have two directions to work with ). I should switch the direction of ##\vec{Δd}_S## to north before I do ( So I can actually add them together ).

    ##\vec{Δd}_V = \vec{Δd}_N - \vec{Δd}_S = 22N - 38N sin(35°) = 0.2N [N]##

    Is this on the right track? Then I would simply draw my components down and find the hypotenuse?
     
  8. Jul 22, 2013 #7
    Yes, this is correct in the principle. Your notation, however, is somewhat unusual. The arrow over a variable denotes a vector; but when you take a component of a vector in some direction, you typically drop the arrow and append the direction as a subscript. For example, the north component of ## \vec{F}_1 ## would typically be denoted as ##F_{1N}##, not ## \vec{\Delta d}_N## as you have it.
     
  9. Jul 22, 2013 #8

    Zondrina

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    Yes, I was used to doing this for displacements so I was recycling old notation ( though I agree it makes zero sense using it ).

    So I should use :

    ##F_N = 22N##
    ##F_S = 38N sin(35°)##
    ##F_E = 38N cos(35°)##

    Then ##\vec{F_V} = 0.2N [N]##.

    Now I draw my triangle with ##F_V## in the positive y direction and ##F_E## in the positive x direction.

    Thus ##F_R = \sqrt{(0.2N)^2 + (38N cos(35°))^2} = 31.13 N##

    Now I've got a nice right angle triangle with 3 sides labeled. I can find the direction of ##F_R## by using :

    ##sin(θ) = \frac{0.2}{31.13}##
    ##θ = 0.3681°##

    ##∴ \vec{F_R} = 31.13N [E 0.3681° N]##

    That should be it I believe?
     
    Last edited: Jul 22, 2013
  10. Jul 22, 2013 #9
    I think this is all correct.
     
  11. Jul 22, 2013 #10

    Zondrina

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    Okay, sounds good. Now I know exactly when to use arrows and subscripts in my notation.

    I only have one question left to ask then. Imagine I didn't know how to use components in such a manner.

    Why was the angle between the two given forces not 125°? From the picture, It's heavily implied that the dotted line forms a right angle and then the lower portion is labeled 35°.

    How would I find this angle otherwise?
     
    Last edited: Jul 22, 2013
  12. Jul 22, 2013 #11
    Yes I agree with others , you should resolve the vectors in the form xi+yj ( where x & y are numbers and i is unit vector in x direction and j is unit vector in y direction . ) Then simply add respective components of the two vectors . If you get the result ai+bj , then the magnitude of the resultant vector is √a^2+b^2 . To get the direction simply use the fact that (y component of resultant vector / x component ) = tan θ.
     
  13. Jul 22, 2013 #12
    One could say that the angle between the forces is 125°.

    However, that is misleading because you do not just need some angle, but the angle in the force (vector) addition triangle. As mentioned by collinsmark, in vector addition you must have one vector's head at the other vector's tail. This is not so in the diagram. You have to parallel-transport one of the vectors so that the head-tail condition is satisfied, then form the triangle. What angle do you get this way?
     
  14. Jul 22, 2013 #13

    Zondrina

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    Okay, so lets say I drew the 22N vector to the north on its own. Now I add the 38N vector, which facing south east, to the head of the 22N vector.

    Then I draw my resultant force from the tail of the 22N vector to the head of the 38N vector forming my triangle.

    So the angle between them would be 90° - 35° = 55°?
     
  15. Jul 22, 2013 #14

    collinsmark

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    So far that looks right. :approve:

    That's also the right idea. You also could have used
    tan(θ) = 0.2/31.1. You could also have used the cosine equivalent, but that becomes a little tougher since it requires more precision is maintained, for this particular problem:
    cos(θ) = 31.1277777/31.1284468

    Ahh! :eek: Try that calculation again. Your angle is off by two orders of magnitude.
     
    Last edited: Jul 22, 2013
  16. Jul 22, 2013 #15

    Zondrina

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    I could have used arcsin, arccos or arctan in this case because all the sides of my triangle were labeled.

    arcsin( 0.2/31.1 ) = 0.3685°
    arccos( 38cos(35°)/31.1 ) = I typed this into wolfram and it gave me a weird answer due to the precision being off I guess.
    arctan( 0.2/38cos(35°) ) = 0.3681°

    arcsin and arctan are close enough, they're only off slightly due to my hypotenuse calculation.

    That's why I rounded up to 37° since it didn't matter too too much.
     
  17. Jul 22, 2013 #16
    Yes.
     
  18. Jul 22, 2013 #17
    But how does ~0.37 become 37?

    Note the north component of the resultant is just 0.2, clearly that means the angle (from the east) must be small.
     
  19. Jul 22, 2013 #18

    Zondrina

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    Sounds good. I'll solve without components and compare the answers.

    ##F_R = \sqrt{\vec{F}_1^2 + \vec{F}_2^2 - 2\vec{F}_1\vec{F}_2 cos(55°)} = 31.13N## same answer as the component hypotenuse calculation.

    Using the sine law I get :

    ##\frac{sin(55°)}{31.13N} = \frac{sin(θ)}{22N}##
    ???????
    ##θ = 35.37°##??

    It seems to be more than a whole degree off from my other answer.

    I was just rounding off to have easier numbers to deal with.

    EDIT : I double checked the calculations for the component method and the law of cosines method again and I'm still getting the same 1 degree off issue.

    I'm not sure why this is occuring?
     
    Last edited: Jul 22, 2013
  20. Jul 22, 2013 #19
    θ is the angle between what and what?

    You cannot round 0.37 off to 37. It makes no sense.
     
  21. Jul 22, 2013 #20

    collinsmark

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    Again, I highly recommend moving away from the law of cosines and law of sines, and instead simply use the vector components. More on that later. But for the moment, let me correct a mistake you made with the law of sines. Your method will work, but it's much easier to make mistakes compared to working with components.

    Take a look at the original diagram. In your mind (or better yet, on paper), move the 38.0 N vector up so that its tail meets the head of the 22.0 N vector. Then we can connect the tail of the 22.0 N vector with the head of the 38.0 N vector to find the resultant force. We know that the magnitude of that vector is 31.1284468 N. (I calculated with more precision.)

    We know the angle opposite the resultant vector, in this new triangle is 55o. This is the top angle in the new triangle. But what we don't know is the bottom left angle in this new triangle. But we do know the opposite of that angle is the 38.0 N vector, not the 22.0 N vector, and this angle is relative to North.

    [tex] \frac{\sin 55}{31.1284468} = \frac{\sin \theta}{38.0} [/tex]

    So [itex] \theta = 89.624326^o [/itex]. But that's East of North. If we want to express that as North of East, we have to subtract it from 90 degrees.

    90o - 89.624326o = ~0.376o, North of East.

    [Edit: And the only reason this is easy at all is because one of the original vectors was already pointing due North. If both of the original vectors were at arbitrary angles, the best you could do using the law of sines or law of cosines is find the angle relative so some other arbitrary angle. That adds even more steps in the process to the final solution.]

    But it's so much easier and saves time by just working from components from the beginning.

    You know that [itex] \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} [/itex]

    You know that [itex] \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} [/itex]

    You know that [itex] \tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} [/itex]

    And since you already broken everything down into its x- and y- components,

    y-component = opposite
    x=component = adjacent
    magnitude of resultant = hypotenuse
     
    Last edited: Jul 22, 2013
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