3 forces, 5N,6N and 7N ac on a particle which remains in a state of equilibrium.

[JessiMen]

3 forces, 5N,6N and 7N act on a particle which remains in a state of equilibrium. What is the angle between the 5N and 6N force?

 

[MarkFL]

I'm going to assume the 3 forces all act within the same plane. And so, I would choose one of the forces to act along the $x$-axis in the positive direction (I'll choose the one with the largest magnitude so our unknowns involve the vectors in question), and I would make the following definitions:

\(\displaystyle F_1=\langle x_1,y_1 \rangle\) where \(\displaystyle \left|F_1\right|=5\)

\(\displaystyle F_2=\langle x_2,y_2 \rangle\) where \(\displaystyle \left|F_2\right|=6\)

\(\displaystyle F_3=\langle 7,0 \rangle\)

Next, let the resultant force be:

\(\displaystyle F_R=F_1+F_2+F_3=\langle 0,0 \rangle\)

And so we obtain:

\(\displaystyle x_1+x_2+7=0\)

\(\displaystyle y_1+y_2=0\)

\(\displaystyle x_1^2+y_1^2=5^2\)

\(\displaystyle x_2^2+y_2^2=6^2\)

We have 4 equations in 4 unknowns...can you proceed?

 

[Evgeny.Makarov]

Here is an idea using dot product. I call the vectors $a$, $b$ and $c$ with $|a|=7$.
\[
b\cdot c=\frac12((b+c)\cdot(b+c)-b\cdot b-c\cdot c)
\]
Now use the facts that $b+c=-a$ and that $a\cdot a$, $b\cdot b$ and $c\cdot c$ are known to find $b\cdot c$ and therefore the angle between $b$ and $c$.