Three forces vec F_1,vec F_2,vec F_3 , each of magnitude a…
Three forces $\vec F_1,\vec F_2,\vec F_3$, each of magnitude $a$, act at one point, and every two of them make an angle of $60^\circ$. Find the magnitude of the resultant and the angle it makes with each force.
Explanation of Method
The resultant of three equal forces making 60° with each other.
Step 1: Products
\[ \vec F_i^{\,2}=a^2,\qquad \vec F_i\cdot\vec F_j=a^2\cos60^\circ=\frac{a^2}{2}\quad(i\neq j). \]
Step 2: Magnitude of the resultant
\[ |\vec R|^2=(\vec F_1+\vec F_2+\vec F_3)^2=3a^2+2\cdot3\cdot\frac{a^2}{2}=6a^2. \]
\[|\vec R|=a\sqrt6\]
Step 3: Angle with each force
\[ \vec R\cdot\vec F_1=a^2+\frac{a^2}{2}+\frac{a^2}{2}=2a^2, \]
\[ \cos\alpha=\frac{2a^2}{a\sqrt6\cdot a}=\frac{2}{\sqrt6}=\frac{\sqrt6}{3}\approx0.8165. \]
By symmetry the same holds for $\vec F_2$ and $\vec F_3$: $\alpha=\arccos\frac{\sqrt6}{3}\approx35.3^\circ$.
Model. The three face diagonals of a cube leaving one vertex are pairwise at $60^\circ$ (each pair spans an equilateral triangle), so the resultant lies along the space diagonal.
- $\vec R=\vec F_1+\vec F_2+\vec F_3,\ |\vec R|=a\sqrt6$
Result
\[\boxed{\,|\vec R|=a\sqrt6\,}\]
\[\boxed{\,\alpha=\arccos\frac{\sqrt6}{3}\approx35.3^\circ\,}\]
Remark
Correction. The cosine is $2/\sqrt6=\sqrt6/3$. A value such as $2/\sqrt3\approx1.15$ is impossible, since $|\cos\alpha|\le1$.