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Force vectors acted on by two forces

  1. Jul 6, 2014 #1
    1. The problem statement, all variables and given/known data
    A body is acted on by two forces. Find the magnitude and direction of the resultant force by using the mathematical method. I'm not sure if I've done this correctly, any pointers would be appreciated :)

    Also what is the angle between the two vectors? Is it 90 or 76 degrees?

    Not sure how to describe this in words so I have the image for the diagram.
    http://s3.postimg.org/jkti9aqkj/P1130367.jpg



    2. Relevant equations



    3. The attempt at a solution

    Working out the first triangle.

    x=cos45 x 90 = 63.64
    y=sin^^^^^^^^^

    Working out the second triangle.
    x=cos45 x 10 = 7.07
    y= sin^^^^^^^^

    So for the system
    x=63.64 + 7.07= 70.71
    y= 63.64-7.07= 56.57

    √70.71^2+56.57^2 = 90.554 N
    ---------------------------------------------------------------------------------------------

    Direction

    Tanθ = opp/adj
    =70.71/56.57
    =1.25
    Tan-1 1.25 = 51.34° above the horizontal.
     
    Last edited: Jul 6, 2014
  2. jcsd
  3. Jul 6, 2014 #2

    ehild

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    Welcome to PF!

    What is the mathematical method of adding two vectors?


    ehild
     
  4. Jul 6, 2014 #3
    AC^2=AB + BC^2

    I think?
     
  5. Jul 6, 2014 #4

    Nathanael

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    Perhaps you meant AC^2=AB^2 + BC^2 ?

    This only applies if the vectors are perpendicular (it's pythagoreans theorem)
     
  6. Jul 6, 2014 #5
    Yes, thanks for the pointer.
     
  7. Jul 6, 2014 #6

    ehild

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    Look at the picture. You see the magnitudes of both vectors and the angles they enclose with the x axis. What are they?


    ehild
     
  8. Jul 6, 2014 #7
    The angle between them is 42 degrees?
     
  9. Jul 6, 2014 #8

    ehild

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    Yes, it is. But why do you need it?

    Write out the "relevant equations". How do you add vectors by components?

    ehild
     
  10. Jul 6, 2014 #9
    Is this the possible equation? sqrt((10cos 34°+ 90cos 76°)^2 + (10 sin 34° + 90sin 76°)^2)
     
  11. Jul 6, 2014 #10

    ehild

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    Yes, it is the magnitude of the sum vector.

    ehild
     
  12. Jul 6, 2014 #11
    So it = 97.66 N? The resultant force?

    Now I need to find the direction.
     
  13. Jul 6, 2014 #12

    ehild

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    The force is a vector, so the magnitude of the resultant force is 97.66 N.
    Yes, you have to find the angle it encloses with the positive x axis.


    ehild
     
  14. Jul 6, 2014 #13
    The direction is (10 sin 34° + 90sin 76°)/(10cos 34°+ 90cos 76°) = 3.09 degrees?

    Does this answer the 2nd part of the question of "A body is acted on by two forces. Find the magnitude and direction of the resultant force by calculation"?

    So the magnitude is ticked at 97.66 N.
    The direction is 3.09 degrees?
     
  15. Jul 6, 2014 #14

    ehild

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    No, it is the tangent of the angle. Calculate arctan of 3.09.

    If the x component of a vector a is ax, the y component is ay, then tan(θ)=ay/ax.
    Yes, you can give a vector by its magnitude and angle with respect to the horizontal, or you can define it by their x and y components.

    When vectors a and b add and the resultant is the vector c,
    cx=ax+bx; cy=ay+by.

    ehild
     
  16. Jul 6, 2014 #15
    So 3.09 degrees can be an answer with respect to the horizontal? By stating it is 3.09 degrees above the horizontal?

    I'm sorry but I don't quite understand what you stated. I'm not too knowledgeable with what I'm dealing here really.
     
    Last edited: Jul 6, 2014
  17. Jul 6, 2014 #16

    Nathanael

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    Analyzing the dimensions when you're unsure is often useful. You have units of degrees divided by degrees. This results in 3.09 being a pure (unitless) number.

    The reason it is unitless is because it's a ratio. Specifically (if you were to draw the triangle) it is the ratio of the vertical component to the horizontal component.

    Components are useful because they are perpendicular to eachother. This means we can use pythagoreans theorem and trigonemtric functions, because they create a right triangle (with the hypotenuse being the "resultant").

    If you recall from triginometry, the ratio of the sine to the cosine (y/x) is the definition of the tangent function.

    So you get [itex]tan(θ)=3.09[/itex] (so you find the angle by taking the arctangent)



    This is part of the idea that ehild was talking about (essentially about converting a vector from "component form" to "magnitude and direction (angle) form")

    As ehild said, it's just two ways of describing the same vector.
     
    Last edited: Jul 6, 2014
  18. Jul 6, 2014 #17
    So how do you go about answering the "direction of the resultant force"?
     
  19. Jul 6, 2014 #18

    Nathanael

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    How much triginometry have you learned?
     
  20. Jul 6, 2014 #19
    The basics I suppose? (Close to nothing) I'm tackling this question that is way out of my level. I find it confusing, because with this

    http://easycalculation.com/physics/classical-physics/resultant-vector.php

    I've worked out the resultant force, but can't figure out the 'direction' except that (10 sin 34° + 90sin 76°)/(10cos 34°+ 90cos 76°) = 3.09 has some part of it.
     
  21. Jul 6, 2014 #20

    Nathanael

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    Explain to me how you got this equation and why you think it works (sorry if this is an annoying request, it's just that a solid understanding of this is helpful)

    Have you tried drawing triangles to find the angle of the resultant?
     
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