Triangle of forces

Question

Forces equal to $5P$, $12P$ and $13P$ acting on a particle are in equilibrium ;find ,by geometric construction and by calculation ,the angles between their directions?

I have an problem that, With three forces how can I use Lami's rule?

Answer 1

let $\vec x\parallel\vec F_1, \vec x \perp \vec y$

$\vec x:$

$\angle \vec F_2=\alpha$, $\angle \vec F_3=\beta$

$\vec y:$

$\angle \vec F_2=\frac {\pi}{2} - \alpha$, $\angle \vec F_3=\frac {\pi}{2} - \beta$

$\begin{equation*} \begin{cases} F_1 + F_2 \cdot \cos \alpha + F_3 \cdot \cos \beta = 0 \\ 0 + F_2 \cdot \sin \alpha + F_3 \cdot \sin \beta = 0 \end{cases} \end{equation*}$

$\begin{equation*} \begin{cases} F_1 + F_2 \cdot \cos \alpha + F_3 \cdot \cos \beta = 0 \\ 12 \cdot \sin \alpha = -13 \cdot \sin \beta \end{cases} \end{equation*}$

$\sin \alpha = - \frac {13}{12} \cdot \sin \beta$

$\cos^2 \alpha = 1 - \frac {169}{144} \cdot \sin^2 \beta=\frac {169}{144} \cdot \cos^2 \beta - \frac {25}{144}$

$ 5 + 12 \cdot \cos \alpha + 13 \cdot \cos \beta = 0$

$ 12 \cdot \cos \alpha =- 13 \cdot \cos \beta - 5$

$ 144 \cdot \cos^2 \alpha =(13 \cdot \cos \beta + 5)^2$

$169 \cdot \cos^2 \beta - 25 = 169 \cdot \cos^2 \beta + 130 \cdot \cos \beta + 25$

$130 \cdot \cos \beta = -50$

$\cos \beta = \frac {5}{13}, \beta < \pi$

$ \beta = 1.176$

$ \sin \alpha = -1$

$\alpha = \frac {3 \cdot \pi}{2}$

Answer:

$\angle \vec F_1 \vec F_2 = \frac {3 \cdot \pi}{2}, \angle \vec F_1 \vec F_3 = 1.176$

EDIT: Stupid mistake

Answer 2

Here is a more systematic approach. Suppose the $3$ forces acting on the particle are $\vec F_1$, $\vec F_2$, and $\vec F_3$. Then, we are given that the sum $\vec F$ is zero:

$$\vec F=\vec F_1+\vec F_2+\vec F_3=0$$

and are given the magnitudes of each of the forces. We are asked for the angles between the pairs. We need, therefore, three independent equations.

To do this, we form the inner product of $\vec F$ with each of the three forces. Proceeding we obtain,

$$F_1^2+F_1F_2\cos \phi_{12}+F_1F_3\cos \phi_{13} =0 \tag 1$$

$$F_1F_2\cos \phi_{12}+F_2^2+F_2F_3\cos \phi_{23} =0 \tag 2$$

$$F_1F_3\cos \phi_{13}+F_2F_3\cos \phi_{23}+F_3^2 =0 \tag 3$$

Equations $(1)-(3)$ constitute a linear system of $3$ equations in the $3$ unknow cosines. The solution to the system is given by

$$\begin{align} \cos \phi_{12}&=\frac{F_3^2-(F_1^2+F_2^2)}{2F_1F_2} \tag 4\\\\ \cos \phi_{13}&=\frac{F_2^2-(F_1^2+F_3^2)}{2F_1F_3} \tag 5\\\\ \cos \phi_{23}&=\frac{F_1^2-(F_2^2+F_3^2)}{2F_2F_3} \tag 6\\\\ \end{align}$$

For the specific problem herein, we are given that $F_1=5$, $F_2=12$, and $F_3=13$. Using these values in $(4)-(6)$, we obtain

$$\begin{align} \cos \phi_{12}&=0 \\\\ \cos \phi_{13}&=-\frac{5}{13} \\\\ \cos \phi_{23}&= -\frac{12}{13}\\\\ \end{align}$$

and therefore the angles between the force vectors are

$$\begin{align} \phi_{12}&=\frac{\pi}{2} \\\\ \phi_{13}&=\pi-\arccos\left(\frac{5}{13}\right) \\\\ \phi_{23}&= \pi-\arccos\left(\frac{12}{13}\right)\\\\ \end{align}$$