Need help with this problem.

1. Create 3 Vectors, 1 with an angle degree of 0, the 2 other degrees unknown.
2. Use calculations to find the 2 other degrees needed for equilibrium.


I have these two equations so far.

A= mag. of first vector.
B =mag. of second vector.
C= mag. of third vector

Bcos(θ2)+Ccos(θ3)=-Acos(θ1)=-A
Bsin(θ2)+Csin(θ3)=-sin(θ1)=0

yes, looks good


there is several ways to solve it
however, i would just graph it in the calculator and find intercetions



This post was edited by saber_x3 on Sep 15 2013 09:05pm

We need to solve it, but I have no idea how too.

I tried the method of squaring both sides, but then I got stuck.

Ive been on this bad boy for like 8 hours now.


I must have made a mistake, cause I just kept working through the algebra after squaring and came up with b^2 * sqrt(1-(-csinθ/b)^2 = (a^2+2Ccosθ+c^2cos^2θ)/b^2

This post was edited by thestoryofisaac on Sep 15 2013 09:16pm

TRY looking at some trig id's, not necessarily the aa+bb=cc

try sum to product ones
not 100% sure, don't have time atm

This post was edited by saber_x3 on Sep 15 2013 09:39pm

View the cosines and sines as cos(2U) and sin(2U), while 2U=ϕ.

from wikipedia




Lastly, set a variable, t= tan(ϕ/2)

You will arrive at a quadratic, you will have to divide the roots by 2 and take the inverse tangent to get your final answer.


This is for an intro physics class?

This post was edited by saber_x3 on Sep 15 2013 11:01pm

Well a calculus based physics class.

Basic classical mechanics physics class.

I have no prior physics experience, and I guess this is expected.



-----

My school has 2 physics series, a intro physics for nonscience major, and this class/series.

Is this just homework?

Unfinished lab assignment.


I don't understand

b^2 = x^2 + c^2 + y^2 -2xc*cos(ϕ) -2yc*sin(ϕ)\


I ended up with B^2= x^2+c^2+y^2

I didn't foil. Oops

This post was edited by thestoryofisaac on Sep 15 2013 11:32pm

reread code 2
by squaring both sides, one side is simplified by Pythagorean theorem
while the other side will result in a quadratic with trig

b*cos(ψ)= x - c*cos(ϕ)
b*sin(ψ)= Y - c*sin(ϕ)

i picked the psi side to simplify
take in mind that one/some of the constants in your case wil be 0
mine was for a general case
Honestly, it's way too tedious of a problem for a regular physics lab

This post was edited by saber_x3 on Sep 15 2013 11:32pm

Yeah. I get it now.

I do think you did it the right way because the professor did say to use the Pythagorean.

I'm still working through the problem though.