Determine Resultant Vector: Non-90deg Triangle Problem

[Chocolaty]

Ok I know how to mathematically find the resultant vector in any problem where they form a triangle with an angle of 90 degrees. But what do you do when you don't have a 90deg triangle?

The problem: Determine mathematically the resultant of this system of forces
F1 (100 N, 30deg)
F2 (150 N, 330deg)

I've searched through my geometry math book to see if there was a theorem I could use to do this, but to no avail.

Can anyone help me on this, please thx

 

[Diane_]

There are several ways to approach problems like that, depending on your level of "mathematical sophistication" (which is not an oxymoron).

1) You can use the Law of Sines and Law of Cosines to analyze the resulting triangles. Both of those work for any triangle. Given a triangle with sides a, b, and c and angles A (opposite side a), B (opposite side b) and C (opposite side c), the Law of Sines says

a/sin(A) = b/sin(B) = c/sin(C)

And the Law of Cosines says

c^2 = a^2 + b^2 - 2ab cos(C)

You'll note that's just the Pythagorean Theorem with a Finagle Factor to take into account the fact that the triangle doesn't have to be right.

2) You can convert the vectors from the polar coordinates in which you probably have them to Cartesian coordinates, add them, then convert back. This is equivalent to decomposing the vectors into the x- and y-directions. Since those components will be orthogonal, they can be added without worrying about interactions.

Variations on these themes also exist. Since I'm not sure where you are mathematically, I'm not sure if this will be any help or not, so: does this help? Does anything look worth pursuing further, or do you need another approach?

 

[Zee Student]

You see, if you plot your vector on a cartesian coordinate system, with the appropriate angle (30 degrees), you can turn it into a right triangle. Simply use the x-axis as the bottom of your triangle, and the y-axis as the side of it.

In terms of trig, your x-axis is your adjacent side, and the y-axis is the opposite side.

Once that is done, you can figure out the x and y components using algebraic manipulation of the trig functions.

(h is hypotenuse)

h • sin(theta) = opposite side or y component.

You can figure out the x component via the same route using the cos function.

Another route would be relating the x and y components by the relations between the sides of a 30-60-90 triangle.

You can do the same for the second angle if you realize that it can also be seen as -30 degrees.

Once you have the respective components, you can add them together and get your vector sum.

-Edited to say my bad, I hadn't seen Diane's response. I must say I like it better since it can work faster once you have it down and just seems cool.

Also wanted to say since this is all brand new to me, if someone can check what I just posted to make sure I didn't post a bunch of crap.

 

[Chocolaty]

There are several ways to approach problems like that, depending on your level of "mathematical sophistication" (which is not an oxymoron).
1) You can use the Law of Sines and Law of Cosines to analyze the resulting triangles. Both of those work for any triangle. Given a triangle with sides a, b, and c and angles A (opposite side a), B (opposite side b) and C (opposite side c), the Law of Sines says
a/sin(A) = b/sin(B) = c/sin(C)
And the Law of Cosines says
c^2 = a^2 + b^2 - 2ab cos(C)
You'll note that's just the Pythagorean Theorem with a Finagle Factor to take into account the fact that the triangle doesn't have to be right.
2) You can convert the vectors from the polar coordinates in which you probably have them to Cartesian coordinates, add them, then convert back. This is equivalent to decomposing the vectors into the x- and y-directions. Since those components will be orthogonal, they can be added without worrying about interactions.
Variations on these themes also exist. Since I'm not sure where you are mathematically, I'm not sure if this will be any help or not, so: does this help? Does anything look worth pursuing further, or do you need another approach?

Okay, I got:

Fr = cos(30) * (150+100) = 216,5
But i don't understand what you said about how to find out the direction of the vector. Btw, there is no cartesian coordinates, the problem I wrote is verbatim.

ps: I'm doing Calculus 1 math.

 

[HallsofIvy]

Okay, I got:
Fr = cos(30) * (150+100) = 216,5
But i don't understand what you said about how to find out the direction of the vector. Btw, there is no cartesian coordinates, the problem I wrote is verbatim.
ps: I'm doing Calculus 1 math.

The way you wrote it can't be "verbatim". Saying "30 deg" doesn't make sense without saying 30 deg to what!
Assuming the 30 deg and 330 deg are to some given line, the best way to go is to use the sine and cosine laws.
Given
F1 (100 N, 30deg)
F2 (150 N, 330deg)
Draw a picture and deduce by geometry that you have two sides of a triangle with an angle of 120 degrees between them. Use the cosine law to find the length of the side opposite the 120 degree angle (the magnitude of the resultant vector) and then use the sine law to find the angle that side makes with one of the others. You can use that information to find the angle made with your reference line.

 

[Chi Meson]

You see, if you plot your vector on a cartesian coordinate system, with the appropriate angle (30 degrees), you can turn it into a right triangle. Simply use the x-axis as the bottom of your triangle, and the y-axis as the side of it.
In terms of trig, your x-axis is your adjacent side, and the y-axis is the opposite side.
Once that is done, you can figure out the x and y components using algebraic manipulation of the trig functions.
(h is hypotenuse)
h • sin(theta) = opposite side or y component.
You can figure out the x component via the same route using the cos function.
Another route would be relating the x and y components by the relations between the sides of a 30-60-90 triangle.
You can do the same for the second angle if you realize that it can also be seen as -30 degrees.
Once you have the respective components, you can add them together and get your vector sum.
-Edited to say my bad, I hadn't seen Diane's response. I must say I like it better since it can work faster once you have it down and just seems cool.
Also wanted to say since this is all brand new to me, if someone can check what I just posted to make sure I didn't post a bunch of crap.

What you said is correct, but it could be more simply said.

At the introductory level, the most common method is the "componant method of vector addition" which is what Zee describes. In short:

You find the x and y components of your vectors, add the x's together, add the y's together to get "net x" and "net y". Do the Pythagoras thing on these to find the net resultant. The final angle will be the inverse-tan of the (y/x) values.

 

[Chocolaty]

thanks Chi, got it

 

[Chi Meson]

My pleasure. I used to teach the "laws of sines and cosines" to my (high school) students, but I find that the "componant method" is more conceptually helpful. The law of sines and law of cosines are indespensible when solving the more advanced vector problems though, so it's worth knowing them. But if you're just adding two vectors, "componant method" is quicker.