1. The problem statement, all variables and given/known data I've got to determine the resultant(R) and the angle between the x-axis and the line of action of the resultant. My picture(don't know how to draw on here), is two vectors, with one pointing into quad 1, and the other into quad 2. For the right one, It is given that you have a 3y/4x/5hyp triangle for this vector. And, the magnitude is 240N. The vector comes directly out of the orgin, similar to a 45 deg angle. Vector 2 is given to be a magnitude of 180 N, and in a 1x/2y/3hyp triangle. 2. Relevant equations Using sin(90)/5 = sin(x)/3 I found the angle from the pos x axis to the vector to be 36 degrees.(vector 1) R^2 = F1^2 + F2^2 + 2F1F2cos(x) Law of sines/Law of cosines 3. The attempt at a solution If you can even understand this problem, I'd like some help as to where to start. I'm not sure if my geometry/trig is right. I've made a parallelogram around the vectors. Any help would be great.
Well, this is quite hard to explain without the use of a diagram! But you say you've made a quadrilateral out of the vectors. Now, the resultant will bisect the quadrilateral into two triangles, they will have sides R, 180 and 240. (Can you see this?) In order to find the angle, you need the angle from the x axis to the resultant. This will be the sum of the bottom left angle of the 3,4,5 triangle, and the bottom left angle of the triangle described above. The angle you have calculated above: is it the angle from the -x axis to the leftmost vector (i.e. the 180 one?) If so, I don't think it helps, as the question wants the angle from the x axis to the resultant (this angle is, strictly, from the -x axis and will not help you) Hope this makes sense! edit: But, if it doesn't make sense then it's probably because I'm imagining a different diagram to the one you've drawn- as radou says, it's hard to help with no diagram!
sorry guys!!!! Wow you are FAST!!! I drew one up, hope it helps a bit more.....if you can even see it.
the best way to attack this is to add components. Add the x component of one vector to the x of the other. Do the same with Y and you will get the components of your resultant vector (legs of the triangle). The magnitude is easy to find by use of right triangle rules. Also, use trig to find the angle. Now the only issue is knowing what the magnitudes of the components are. For this, see if you can use trig.
Can you give an example(using my pic) of what you are talking about. I can't seem to picture this(the x components and y components, making a triangle). Do I just add 4+1 for the x components(or is it 4+-1?)???
Ja4Coltrane's suggestion is good! You were right with 4+(-1). Do the same for the y components, then draw the diagram. You will be able to make a right angled triangle, and calculate the resultant and angle easily this way.
Ok guys, first thank you. Next.....I drew the triangle(and really like this approach). I came up with the resultant being 5.831 and I used law of sines to find an angle of 59.036. The thing that I'm wondering, is did I choose the correct angle? I don't know how this picture relates to my initial picture....so I'm having trouble figuring out which corner/angle would be "from the x-axis"......unless I have it right now that is.
The angle will be the correct one, since it will be the angle from the x to the resultant. Your resultant force is wrong. The easiest way to calculate it is to note that the magnitudes of the force form a right angled triangle with the resultant as the hypotenuse.
Oh ok Cristo....thank you for your help. I guess I wasn't ever taught how to do vectors very well. I'm learning though, slowy. So, I came up with 300N for the resultant.